code stringlengths 1 1.72M | language stringclasses 1 value |
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#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Mimic pyquick exercise -- optional extra exercise.
Google's Python Class
Read in the file specified on the command line.
Do a simple split() on whitespace to obtain all the words in the file.
Rather than read the file line by line, it's easier to read
it into one giant string and split it once.
Build a "mimic" dict that maps each word that appears in the file
to a list of all the words that immediately follow that word in the file.
The list of words can be be in any order and should include
duplicates. So for example the key "and" might have the list
["then", "best", "then", "after", ...] listing
all the words which came after "and" in the text.
We'll say that the empty string is what comes before
the first word in the file.
With the mimic dict, it's fairly easy to emit random
text that mimics the original. Print a word, then look
up what words might come next and pick one at random as
the next work.
Use the empty string as the first word to prime things.
If we ever get stuck with a word that is not in the dict,
go back to the empty string to keep things moving.
Note: the standard python module 'random' includes a
random.choice(list) method which picks a random element
from a non-empty list.
For fun, feed your program to itself as input.
Could work on getting it to put in linebreaks around 70
columns, so the output looks better.
"""
import random
import sys
import re
def mimic_dict(filename):
"""Returns mimic dict mapping each word to list of words which follow it."""
f = open(filename, 'r')
str = f.read()
if len(str) == 0:
return -1
str = re.sub("[!_?/\|<>{}():;,.@#$%^&*--+='~`]","",str)
str = str.lower().split()
ans = {"":[str[0]]}
index=0
for word in str[:-1]:
if ans.get(word):
ans[word].append(str[index+1])
else:
ans[word]=[str[index+1]]
index+=1
return ans
def print_mimic(mimic_dict, word):
"""Given mimic dict and start word, prints 200 random words."""
temp = ''
print word,
for i in range(200):
if mimic_dict.get(word) == None:
word = ''
temp = random.choice(mimic_dict[word])
print temp,
word = temp
return
# Provided main(), calls mimic_dict() and mimic()
def main():
if len(sys.argv) != 2:
print 'usage: ./mimic.py file-to-read'
sys.exit(1)
dict = mimic_dict(sys.argv[1])
print_mimic(dict, '')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python2.4 -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic string exercises
# D. verbing
# Given a string, if its length is at least 3,
# add 'ing' to its end.
# Unless it already ends in 'ing', in which case
# add 'ly' instead.
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
if len(s) < 3:
return s
elif s[-3:] == 'ing':
return s+'ly'
else:
return s+'ing'
# E. not_bad
# Given a string, find the first appearance of the
# substring 'not' and 'bad'. If the 'bad' follows
# the 'not', replace the whole 'not'...'bad' substring
# with 'good'.
# Return the resulting string.
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
if s.find('not') < s.find('bad') and s.find('not') != -1:
return s.replace(s[s.find('not'):s.find('bad')+3], 'good')
return s
# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back
def front_back(a, b):
return a[:len(a)/2+len(a)%2]+b[:len(b)/2+len(b)%2]+a[-len(a)/2+len(a)%2:]+b[-len(b)/2+len(b)%2:]
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# main() calls the above functions with interesting inputs,
# using the above test() to check if the result is correct or not.
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic list exercises
# D. Given a list of numbers, return a list where
# all adjacent == elements have been reduced to a single element,
# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
# modify the passed in list.
def remove_adjacent(nums):
ans=[]
i=0
while i < len(nums):
if nums[i] != nums[i-1]:
ans.append(nums[i])
i+=1
return ans
# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
ans=[]
i=0
j=0
while i+j < len(list1)+len(list2):
if i < len(list1) and j < len(list2):
if list1[i] < list2[j]:
ans.append(list1[i])
i+=1
else:
ans.append(list2[j])
j+=1
elif j >= len(list2):
ans.append(list1[i])
i+=1
else:
ans.append(list2[j])
j+=1
return ans
# Note: the solution above is kind of cute, but unforunately list.pop(0)
# is not constant time with the standard python list implementation, so
# the above is not strictly linear time.
# An alternate approach uses pop(-1) to remove the endmost elements
# from each list, building a solution list which is backwards.
# Then use reversed() to put the result back in the correct order. That
# solution works in linear time, but is more ugly.
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'remove_adjacent'
test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
test(remove_adjacent([]), [])
print
print 'linear_merge'
test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
['aa', 'aa', 'aa', 'bb', 'bb'])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Wordcount exercise
Google's Python class
The main() below is already defined and complete. It calls print_words()
and print_top() functions which you write.
1. For the --count flag, implement a print_words(filename) function that counts
how often each word appears in the text and prints:
word1 count1
word2 count2
...
Print the above list in order sorted by word (python will sort punctuation to
come before letters -- that's fine). Store all the words as lowercase,
so 'The' and 'the' count as the same word.
2. For the --topcount flag, implement a print_top(filename) which is similar
to print_words() but which prints just the top 20 most common words sorted
so the most common word is first, then the next most common, and so on.
Use str.split() (no arguments) to split on all whitespace.
Workflow: don't build the whole program at once. Get it to an intermediate
milestone and print your data structure and sys.exit(0).
When that's working, try for the next milestone.
Optional: define a helper function to avoid code duplication inside
print_words() and print_top().
"""
import sys
def read_file(filename):
f = open(filename, 'r')
ans=[]
buf = ''
temp= ''
for line in f:
for i in line:
if i == ' ' or i == '\n':
buf = buf.lower()
ans += buf.split()
buf=''
else:
buf += i
f.close
dict = {}
for i in ans:
if i in dict:
dict[i]+=1
else:
dict[i]=1
return dict
def print_words(filename):
f = read_file(filename)
for key in sorted(f.keys(), reverse = True, key=f.get):
print key,' ', f.get(key)
def print_top(filename):
dict = read_file(filename)
count=0
for i in sorted(dict.keys(), reverse = True, key=dict.get):
if count < len(dict) and count < 20:
print i, dict.get(i)
count+=1
else:
break
# +++your code here+++
# Define print_words(filename) and print_top(filename) functions.
# You could write a helper utility function that reads a file
# and builds and returns a word/count dict for it.
# Then print_words() and print_top() can just call the utility function.
###
# This basic command line argument parsing code is provided and
# calls the print_words() and print_top() functions which you must define.
def main():
if len(sys.argv) != 3:
print 'usage: ./wordcount.py {--count | --topcount} file'
sys.exit(1)
option = sys.argv[1]
filename = sys.argv[2]
if option == '--count':
print_words(filename)
elif option == '--topcount':
print_top(filename)
else:
print 'unknown option: ' + option
sys.exit(1)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python2.4 -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic string exercises
# D. verbing
# Given a string, if its length is at least 3,
# add 'ing' to its end.
# Unless it already ends in 'ing', in which case
# add 'ly' instead.
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
if len(s) < 3:
return s
elif s[-3:] == 'ing':
return s+'ly'
else:
return s+'ing'
# E. not_bad
# Given a string, find the first appearance of the
# substring 'not' and 'bad'. If the 'bad' follows
# the 'not', replace the whole 'not'...'bad' substring
# with 'good'.
# Return the resulting string.
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
if s.find('not') < s.find('bad') and s.find('not') != -1:
return s.replace(s[s.find('not'):s.find('bad')+3], 'good')
return s
# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back
def front_back(a, b):
return a[:len(a)/2+len(a)%2]+b[:len(b)/2+len(b)%2]+a[-len(a)/2+len(a)%2:]+b[-len(b)/2+len(b)%2:]
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# main() calls the above functions with interesting inputs,
# using the above test() to check if the result is correct or not.
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic list exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in list2.py.
# A. match_ends
# Given a list of strings, return the count of the number of
# strings where the string length is 2 or more and the first
# and last chars of the string are the same.
# Note: python does not have a ++ operator, but += works.
def match_ends(words):
count = 0
for word in words:
wordlen = len(word)
if wordlen>1 and word[0]==word[wordlen-1]:
count+=1
return count
# B. front_x
# Given a list of strings, return a list with the strings
# in sorted order, except group all the strings that begin with 'x' first.
# e.g. ['mix', 'xyz', 'apple', 'xanadu', 'aardvark'] yields
# ['xanadu', 'xyz', 'aardvark', 'apple', 'mix']
# Hint: this can be done by making 2 lists and sorting each of them
# before combining them.
def front_x(words):
x_list = []
other_list = []
for word in words:
if word[0]=='x':
x_list.append(word)
else:
other_list.append(word)
x_list.sort()
other_list.sort()
x_list.extend(other_list)
# +++your code here+++
return x_list
# C. sort_last
# Given a list of non-empty tuples, return a list sorted in increasing
# order by the last element in each tuple.
# e.g. [(1, 7), (1, 3), (3, 4, 5), (2, 2)] yields
# [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
# Hint: use a custom key= function to extract the last element form each tuple.
def sort_last(tuples):
return sorted(tuples,key=last)
def last(tuple):
last = tuple[len(tuple)-1]
return last
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'match_ends'
test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)
print
print 'front_x'
test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])
print
print 'sort_last'
test(sort_last([(1, 3), (3, 2), (2, 1)]),
[(2, 1), (3, 2), (1, 3)])
test(sort_last([(2, 3), (1, 2), (3, 1)]),
[(3, 1), (1, 2), (2, 3)])
test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
[(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic string exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in string2.py.
# A. donuts
# Given an int count of a number of donuts, return a string
# of the form 'Number of donuts: <count>', where <count> is the number
# passed in. However, if the count is 10 or more, then use the word 'many'
# instead of the actual count.
# So donuts(5) returns 'Number of donuts: 5'
# and donuts(23) returns 'Number of donuts: many'
def donuts(count):
if count < 10:
return 'Number of donuts: %d' %count
else:
return 'Number of donuts: many'
# B. both_ends
# Given a string s, return a string made of the first 2
# and the last 2 chars of the original string,
# so 'spring' yields 'spng'. However, if the string length
# is less than 2, return instead the empty string.
def both_ends(s):
if len(s) < 2:
return ''
else:
return s[0:2]+s[-2:]
return
# C. fix_start
# Given a string s, return a string
# where all occurences of its first char have
# been changed to '*', except do not change
# the first char itself.
# e.g. 'babble' yields 'ba**le'
# Assume that the string is length 1 or more.
# Hint: s.replace(stra, strb) returns a version of string s
# where all instances of stra have been replaced by strb.
def fix_start(s):
char = s[0]
word = char
for num in range(1,len(s)):
if s[num] == char:
word += '*'
else:
word += s[num]
return word
# D. MixUp
# Given strings a and b, return a single string with a and b separated
# by a space '<a> <b>', except swap the first 2 chars of each string.
# e.g.
# 'mix', pod' -> 'pox mid'
# 'dog', 'dinner' -> 'dig donner'
# Assume a and b are length 2 or more.
def mix_up(a, b):
return b[0:2]+a[2:]+ ' ' +a[0:2]+b[2:]
# Provided simple test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Provided main() calls the above functions with interesting inputs,
# using test() to check if each result is correct or not.
def main():
print 'donuts'
# Each line calls donuts, compares its result to the expected for that call.
test(donuts(4), 'Number of donuts: 4')
test(donuts(9), 'Number of donuts: 9')
test(donuts(10), 'Number of donuts: many')
test(donuts(99), 'Number of donuts: many')
print
print 'both_ends'
test(both_ends('spring'), 'spng')
test(both_ends('Hello'), 'Helo')
test(both_ends('a'), '')
test(both_ends('xyz'), 'xyyz')
print
print 'fix_start'
test(fix_start('babble'), 'ba**le')
test(fix_start('aardvark'), 'a*rdv*rk')
test(fix_start('google'), 'goo*le')
test(fix_start('donut'), 'donut')
print
print 'mix_up'
test(mix_up('mix', 'pod'), 'pox mid')
test(mix_up('dog', 'dinner'), 'dig donner')
test(mix_up('gnash', 'sport'), 'spash gnort')
test(mix_up('pezzy', 'firm'), 'fizzy perm')
# Standard boilerplate to call the main() function.
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic string exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in string2.py.
# A. donuts
# Given an int count of a number of donuts, return a string
# of the form 'Number of donuts: <count>', where <count> is the number
# passed in. However, if the count is 10 or more, then use the word 'many'
# instead of the actual count.
# So donuts(5) returns 'Number of donuts: 5'
# and donuts(23) returns 'Number of donuts: many'
def donuts(count):
if count < 10:
return 'Number of donuts: %d' %count
else:
return 'Number of donuts: many'
# B. both_ends
# Given a string s, return a string made of the first 2
# and the last 2 chars of the original string,
# so 'spring' yields 'spng'. However, if the string length
# is less than 2, return instead the empty string.
def both_ends(s):
if len(s) < 2:
return ''
else:
return s[0:2]+s[-2:]
return
# C. fix_start
# Given a string s, return a string
# where all occurences of its first char have
# been changed to '*', except do not change
# the first char itself.
# e.g. 'babble' yields 'ba**le'
# Assume that the string is length 1 or more.
# Hint: s.replace(stra, strb) returns a version of string s
# where all instances of stra have been replaced by strb.
def fix_start(s):
char = s[0]
word = char
for num in range(1,len(s)):
if s[num] == char:
word += '*'
else:
word += s[num]
return word
# D. MixUp
# Given strings a and b, return a single string with a and b separated
# by a space '<a> <b>', except swap the first 2 chars of each string.
# e.g.
# 'mix', pod' -> 'pox mid'
# 'dog', 'dinner' -> 'dig donner'
# Assume a and b are length 2 or more.
def mix_up(a, b):
return b[0:2]+a[2:]+ ' ' +a[0:2]+b[2:]
# Provided simple test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Provided main() calls the above functions with interesting inputs,
# using test() to check if each result is correct or not.
def main():
print 'donuts'
# Each line calls donuts, compares its result to the expected for that call.
test(donuts(4), 'Number of donuts: 4')
test(donuts(9), 'Number of donuts: 9')
test(donuts(10), 'Number of donuts: many')
test(donuts(99), 'Number of donuts: many')
print
print 'both_ends'
test(both_ends('spring'), 'spng')
test(both_ends('Hello'), 'Helo')
test(both_ends('a'), '')
test(both_ends('xyz'), 'xyyz')
print
print 'fix_start'
test(fix_start('babble'), 'ba**le')
test(fix_start('aardvark'), 'a*rdv*rk')
test(fix_start('google'), 'goo*le')
test(fix_start('donut'), 'donut')
print
print 'mix_up'
test(mix_up('mix', 'pod'), 'pox mid')
test(mix_up('dog', 'dinner'), 'dig donner')
test(mix_up('gnash', 'sport'), 'spash gnort')
test(mix_up('pezzy', 'firm'), 'fizzy perm')
# Standard boilerplate to call the main() function.
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic list exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in list2.py.
# A. match_ends
# Given a list of strings, return the count of the number of
# strings where the string length is 2 or more and the first
# and last chars of the string are the same.
# Note: python does not have a ++ operator, but += works.
def match_ends(words):
count = 0
for word in words:
wordlen = len(word)
if wordlen>1 and word[0]==word[wordlen-1]:
count+=1
return count
# B. front_x
# Given a list of strings, return a list with the strings
# in sorted order, except group all the strings that begin with 'x' first.
# e.g. ['mix', 'xyz', 'apple', 'xanadu', 'aardvark'] yields
# ['xanadu', 'xyz', 'aardvark', 'apple', 'mix']
# Hint: this can be done by making 2 lists and sorting each of them
# before combining them.
def front_x(words):
x_list = []
other_list = []
for word in words:
if word[0]=='x':
x_list.append(word)
else:
other_list.append(word)
x_list.sort()
other_list.sort()
x_list.extend(other_list)
# +++your code here+++
return x_list
# C. sort_last
# Given a list of non-empty tuples, return a list sorted in increasing
# order by the last element in each tuple.
# e.g. [(1, 7), (1, 3), (3, 4, 5), (2, 2)] yields
# [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
# Hint: use a custom key= function to extract the last element form each tuple.
def sort_last(tuples):
return sorted(tuples,key=last)
def last(tuple):
last = tuple[len(tuple)-1]
return last
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'match_ends'
test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)
print
print 'front_x'
test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])
print
print 'sort_last'
test(sort_last([(1, 3), (3, 2), (2, 1)]),
[(2, 1), (3, 2), (1, 3)])
test(sort_last([(2, 3), (1, 2), (3, 1)]),
[(3, 1), (1, 2), (2, 3)])
test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
[(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic list exercises
# D. Given a list of numbers, return a list where
# all adjacent == elements have been reduced to a single element,
# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
# modify the passed in list.
def remove_adjacent(nums):
if nums == []:
return nums
last = nums[0]
mlist = []
mlist.append(last)
for i in nums[1:]:
if i!= last:
mlist.append(i)
last = i
return mlist
# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
list1.extend(list2)
list1.sort()
return list1
# Note: the solution above is kind of cute, but unforunately list.pop(0)
# is not constant time with the standard python list implementation, so
# the above is not strictly linear time.
# An alternate approach uses pop(-1) to remove the endmost elements
# from each list, building a solution list which is backwards.
# Then use reversed() to put the result back in the correct order. That
# solution works in linear time, but is more ugly.
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'remove_adjacent'
test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
test(remove_adjacent([]), [])
print
print 'linear_merge'
test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
['aa', 'aa', 'aa', 'bb', 'bb'])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Wordcount exercise
Google's Python class
The main() below is already defined and complete. It calls print_words()
and print_top() functions which you write.
1. For the --count flag, implement a print_words(filename) function that counts
how often each word appears in the text and prints:
word1 count1
word2 count2
...
Print the above list in order sorted by word (python will sort punctuation to
come before letters -- that's fine). Store all the words as lowercase,
so 'The' and 'the' count as the same word.
2. For the --topcount flag, implement a print_top(filename) which is similar
to print_words() but which prints just the top 20 most common words sorted
so the most common word is first, then the next most common, and so on.
Use str.split() (no arguments) to split on all whitespace.
Workflow: don't build the whole program at once. Get it to an intermediate
milestone and print your data structure and sys.exit(0).
When that's working, try for the next milestone.
Optional: define a helper function to avoid code duplication inside
print_words() and print_top().
"""
import sys
import re
# +++your code here+++
# Define print_words(filename) and print_top(filename) functions.
# You could write a helper utility function that reads a file
# and builds and returns a word/count dict for it.
# Then print_words() and print_top() can just call the utility function.
def read_file(filename):
f=open(filename,'r')
text = f.read()
text = re.sub("[/.,\';:?!-]","",text)
text = text.lower()
words = text.split()
wordsMap = {}
for word in words:
if word in wordsMap:
wordsMap[word] +=1
else:
wordsMap[word] = 1
return wordsMap
def print_words(filename):
wordsMap = read_file(filename)
keys=sorted(wordsMap.keys())
for key in keys:
print key, wordsMap[key]
return
def print_top(filename):
wordsMap = read_file(filename)
items=sorted(wordsMap.items(),key=lambda (k, v): v,reverse=True)
for item in items[:20]:
print item[0], item[1]
###
# This basic command line argument parsing code is provided and
# calls the print_words() and print_top() functions which you must define.
def main():
if len(sys.argv) != 3:
print 'usage: ./wordcount.py {--count | --topcount} file'
sys.exit(1)
option = sys.argv[1]
filename = sys.argv[2]
if option == '--count':
print_words(filename)
elif option == '--topcount':
print_top(filename)
else:
print 'unknown option: ' + option
sys.exit(1)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python2.4 -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic string exercises
# D. verbing
# Given a string, if its length is at least 3,
# add 'ing' to its end.
# Unless it already ends in 'ing', in which case
# add 'ly' instead.
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
if len(s)<3:
return s
elif s[-3:] == 'ing':
return s + 'ly'
else:
return s + 'ing'
# E. not_bad
# Given a string, find the first appearance of the
# substring 'not' and 'bad'. If the 'bad' follows
# the 'not', replace the whole 'not'...'bad' substring
# with 'good'.
# Return the resulting string.
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
if s.find('bad') > s.find('not'):
return s.replace(s[s.find('not'):s.find('bad')+3], 'good')
else:
return s
# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back
def front_back(a, b):
af = (len(a)+1)/2
bf = (len(b)+1)/2
return a[0:af] + b[0:bf] + a[af:] + b[bf:]
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# main() calls the above functions with interesting inputs,
# using the above test() to check if the result is correct or not.
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic list exercises
# D. Given a list of numbers, return a list where
# all adjacent == elements have been reduced to a single element,
# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
# modify the passed in list.
def remove_adjacent(nums):
if nums == []:
return nums
last = nums[0]
mlist = []
mlist.append(last)
for i in nums[1:]:
if i!= last:
mlist.append(i)
last = i
return mlist
# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
list1.extend(list2)
list1.sort()
return list1
# Note: the solution above is kind of cute, but unforunately list.pop(0)
# is not constant time with the standard python list implementation, so
# the above is not strictly linear time.
# An alternate approach uses pop(-1) to remove the endmost elements
# from each list, building a solution list which is backwards.
# Then use reversed() to put the result back in the correct order. That
# solution works in linear time, but is more ugly.
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'remove_adjacent'
test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
test(remove_adjacent([]), [])
print
print 'linear_merge'
test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
['aa', 'aa', 'aa', 'bb', 'bb'])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Wordcount exercise
Google's Python class
The main() below is already defined and complete. It calls print_words()
and print_top() functions which you write.
1. For the --count flag, implement a print_words(filename) function that counts
how often each word appears in the text and prints:
word1 count1
word2 count2
...
Print the above list in order sorted by word (python will sort punctuation to
come before letters -- that's fine). Store all the words as lowercase,
so 'The' and 'the' count as the same word.
2. For the --topcount flag, implement a print_top(filename) which is similar
to print_words() but which prints just the top 20 most common words sorted
so the most common word is first, then the next most common, and so on.
Use str.split() (no arguments) to split on all whitespace.
Workflow: don't build the whole program at once. Get it to an intermediate
milestone and print your data structure and sys.exit(0).
When that's working, try for the next milestone.
Optional: define a helper function to avoid code duplication inside
print_words() and print_top().
"""
import sys
import re
# +++your code here+++
# Define print_words(filename) and print_top(filename) functions.
# You could write a helper utility function that reads a file
# and builds and returns a word/count dict for it.
# Then print_words() and print_top() can just call the utility function.
def read_file(filename):
f=open(filename,'r')
text = f.read()
text = re.sub("[/.,\';:?!-]","",text)
text = text.lower()
words = text.split()
wordsMap = {}
for word in words:
if word in wordsMap:
wordsMap[word] +=1
else:
wordsMap[word] = 1
return wordsMap
def print_words(filename):
wordsMap = read_file(filename)
keys=sorted(wordsMap.keys())
for key in keys:
print key, wordsMap[key]
return
def print_top(filename):
wordsMap = read_file(filename)
items=sorted(wordsMap.items(),key=lambda (k, v): v,reverse=True)
for item in items[:20]:
print item[0], item[1]
###
# This basic command line argument parsing code is provided and
# calls the print_words() and print_top() functions which you must define.
def main():
if len(sys.argv) != 3:
print 'usage: ./wordcount.py {--count | --topcount} file'
sys.exit(1)
option = sys.argv[1]
filename = sys.argv[2]
if option == '--count':
print_words(filename)
elif option == '--topcount':
print_top(filename)
else:
print 'unknown option: ' + option
sys.exit(1)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python2.4 -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic string exercises
# D. verbing
# Given a string, if its length is at least 3,
# add 'ing' to its end.
# Unless it already ends in 'ing', in which case
# add 'ly' instead.
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
if len(s)<3:
return s
elif s[-3:] == 'ing':
return s + 'ly'
else:
return s + 'ing'
# E. not_bad
# Given a string, find the first appearance of the
# substring 'not' and 'bad'. If the 'bad' follows
# the 'not', replace the whole 'not'...'bad' substring
# with 'good'.
# Return the resulting string.
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
if s.find('bad') > s.find('not'):
return s.replace(s[s.find('not'):s.find('bad')+3], 'good')
else:
return s
# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back
def front_back(a, b):
af = (len(a)+1)/2
bf = (len(b)+1)/2
return a[0:af] + b[0:bf] + a[af:] + b[bf:]
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# main() calls the above functions with interesting inputs,
# using the above test() to check if the result is correct or not.
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic list exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in list2.py.
# A. match_ends
# Given a list of strings, return the count of the number of
# strings where the string length is 2 or more and the first
# and last chars of the string are the same.
# Note: python does not have a ++ operator, but += works.
def match_ends(words):
# +++your code here+++
temp=0
for x in words:
if(len(x) > 1):
if(x[:1]==x[-1:]):
temp+=1
return temp
# B. front_x
# Given a list of strings, return a list with the strings
# in sorted order, except group all the strings that begin with 'x' first.
# e.g. ['mix', 'xyz', 'apple', 'xanadu', 'aardvark'] yields
# ['xanadu', 'xyz', 'aardvark', 'apple', 'mix']
# Hint: this can be done by making 2 lists and sorting each of them
# before combining them.
def front_x(words):
# +++your code here+++
temp1=[];
temp2=[];
for x in words:
if(x[:1]=='x'):
temp1.append(x)
else:
temp2.append(x)
return sorted(temp1)+ sorted(temp2)
# C. sort_last
# Given a list of non-empty tuples, return a list sorted in increasing
# order by the last element in each tuple.
# e.g. [(1, 7), (1, 3), (3, 4, 5), (2, 2)] yields
# [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
# Hint: use a custom key= function to extract the last element form each tuple.
def sort_last(tuples):
# +++your code here+++
return sorted(tuples, key=lambda tuples: tuples[-1:])
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'match_ends'
test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)
print
print 'front_x'
test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])
print
print 'sort_last'
test(sort_last([(1, 3), (3, 2), (2, 1)]),
[(2, 1), (3, 2), (1, 3)])
test(sort_last([(2, 3), (1, 2), (3, 1)]),
[(3, 1), (1, 2), (2, 3)])
test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
[(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Mimic pyquick exercise -- optional extra exercise.
Google's Python Class
Read in the file specified on the command line.
Do a simple split() on whitespace to obtain all the words in the file.
Rather than read the file line by line, it's easier to read
it into one giant string and split it once.
Build a "mimic" dict that maps each word that appears in the file
to a list of all the words that immediately follow that word in the file.
The list of words can be be in any order and should include
duplicates. So for example the key "and" might have the list
["then", "best", "then", "after", ...] listing
all the words which came after "and" in the text.
We'll say that the empty string is what comes before
the first word in the file.
With the mimic dict, it's fairly easy to emit random
text that mimics the original. Print a word, then look
up what words might come next and pick one at random as
the next work.
Use the empty string as the first word to prime things.
If we ever get stuck with a word that is not in the dict,
go back to the empty string to keep things moving.
Note: the standard python module 'random' includes a
random.choice(list) method which picks a random element
from a non-empty list.
For fun, feed your program to itself as input.
Could work on getting it to put in linebreaks around 70
columns, so the output looks better.
"""
import random
import sys
import re
def mimic_dict(filename):
"""Returns mimic dict mapping each word to list of words which follow it."""
f=open(filename,'r')
text=f.read()
text=re.sub("[(/.,!?:-;)`]","",text)
str1=text.lower().split()
mimic={"":str1[0]}
suf=[]
for i in range(len(str1)):
if mimic.get(str1[i])==None:
j=i
for j in range(len(str1)-1):
if str1[i]==str1[j]:
suf.append((str1[j+1]))
mimic.update({str1[i]:suf})
suf=[]
return mimic
def print_mimic(mimic_dict, word):
"""Given mimic dict and start word, prints 200 random words."""
# +++your code here+++
for i in range(200):
if word=='' or mimic_dict.get(word)==[]or mimic_dict.get(word)==None:
word=mimic_dict.get('')
sys.stdout.write("%s " %(word))
else:
word=random.choice(mimic_dict.get(word))
sys.stdout.write("%s " %(word))
return
# Provided main(), calls mimic_dict() and mimic()
def main():
if len(sys.argv) != 2:
print 'usage: ./mimic.py file-to-read'
sys.exit(1)
dict = mimic_dict(sys.argv[1])
print_mimic(dict, '')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic string exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in string2.py.
# A. donuts
# Given an int count of a number of donuts, return a string
# of the form 'Number of donuts: <count>', where <count> is the number
# passed in. However, if the count is 10 or more, then use the word 'many'
# instead of the actual count.
# So donuts(5) returns 'Number of donuts: 5'
# and donuts(23) returns 'Number of donuts: many'
def donuts(count):
# +++your code here+++
if (count < 10):
return 'Number of donuts: '+str(count)
else:
return 'Number of donuts: many'
# B. both_ends
# Given a string s, return a string made of the first 2
# and the last 2 chars of the original string,
# so 'spring' yields 'spng'. However, if the string length
# is less than 2, return instead the empty string.
def both_ends(s):
# +++your code here+++
if (len(s)>2):
return s[:2]+s[-2:]
else:
return ''
# C. fix_start
# Given a string s, return a string
# where all occurences of its first char have
# been changed to '*', except do not change
# the first char itself.
# e.g. 'babble' yields 'ba**le'
# Assume that the string is length 1 or more.
# Hint: s.replace(stra, strb) returns a version of string s
# where all instances of stra have been replaced by strb.
def fix_start(s):
# +++your code here+++
temp = s[1:].replace(s[:1],'*')
return s[:1]+temp
# D. MixUp
# Given strings a and b, return a single string with a and b separated
# by a space '<a> <b>', except swap the first 2 chars of each string.
# e.g.
# 'mix', pod' -> 'pox mid'
# 'dog', 'dinner' -> 'dig donner'
# Assume a and b are length 2 or more.
def mix_up(a, b):
# +++your code here+++
return a.replace(a[:2],b[:2])+' '+ b.replace(b[:2],a[:2])
# Provided simple test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Provided main() calls the above functions with interesting inputs,
# using test() to check if each result is correct or not.
def main():
print 'donuts'
# Each line calls donuts, compares its result to the expected for that call.
test(donuts(4), 'Number of donuts: 4')
test(donuts(9), 'Number of donuts: 9')
test(donuts(10), 'Number of donuts: many')
test(donuts(99), 'Number of donuts: many')
print
print 'both_ends'
test(both_ends('spring'), 'spng')
test(both_ends('Hello'), 'Helo')
test(both_ends('a'), '')
test(both_ends('xyz'), 'xyyz')
print
print 'fix_start'
test(fix_start('babble'), 'ba**le')
test(fix_start('aardvark'), 'a*rdv*rk')
test(fix_start('google'), 'goo*le')
test(fix_start('donut'), 'donut')
print
print 'mix_up'
test(mix_up('mix', 'pod'), 'pox mid')
test(mix_up('dog', 'dinner'), 'dig donner')
test(mix_up('gnash', 'sport'), 'spash gnort')
test(mix_up('pezzy', 'firm'), 'fizzy perm')
# Standard boilerplate to call the main() function.
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic string exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in string2.py.
# A. donuts
# Given an int count of a number of donuts, return a string
# of the form 'Number of donuts: <count>', where <count> is the number
# passed in. However, if the count is 10 or more, then use the word 'many'
# instead of the actual count.
# So donuts(5) returns 'Number of donuts: 5'
# and donuts(23) returns 'Number of donuts: many'
def donuts(count):
# +++your code here+++
if (count < 10):
return 'Number of donuts: '+str(count)
else:
return 'Number of donuts: many'
# B. both_ends
# Given a string s, return a string made of the first 2
# and the last 2 chars of the original string,
# so 'spring' yields 'spng'. However, if the string length
# is less than 2, return instead the empty string.
def both_ends(s):
# +++your code here+++
if (len(s)>2):
return s[:2]+s[-2:]
else:
return ''
# C. fix_start
# Given a string s, return a string
# where all occurences of its first char have
# been changed to '*', except do not change
# the first char itself.
# e.g. 'babble' yields 'ba**le'
# Assume that the string is length 1 or more.
# Hint: s.replace(stra, strb) returns a version of string s
# where all instances of stra have been replaced by strb.
def fix_start(s):
# +++your code here+++
temp = s[1:].replace(s[:1],'*')
return s[:1]+temp
# D. MixUp
# Given strings a and b, return a single string with a and b separated
# by a space '<a> <b>', except swap the first 2 chars of each string.
# e.g.
# 'mix', pod' -> 'pox mid'
# 'dog', 'dinner' -> 'dig donner'
# Assume a and b are length 2 or more.
def mix_up(a, b):
# +++your code here+++
return a.replace(a[:2],b[:2])+' '+ b.replace(b[:2],a[:2])
# Provided simple test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Provided main() calls the above functions with interesting inputs,
# using test() to check if each result is correct or not.
def main():
print 'donuts'
# Each line calls donuts, compares its result to the expected for that call.
test(donuts(4), 'Number of donuts: 4')
test(donuts(9), 'Number of donuts: 9')
test(donuts(10), 'Number of donuts: many')
test(donuts(99), 'Number of donuts: many')
print
print 'both_ends'
test(both_ends('spring'), 'spng')
test(both_ends('Hello'), 'Helo')
test(both_ends('a'), '')
test(both_ends('xyz'), 'xyyz')
print
print 'fix_start'
test(fix_start('babble'), 'ba**le')
test(fix_start('aardvark'), 'a*rdv*rk')
test(fix_start('google'), 'goo*le')
test(fix_start('donut'), 'donut')
print
print 'mix_up'
test(mix_up('mix', 'pod'), 'pox mid')
test(mix_up('dog', 'dinner'), 'dig donner')
test(mix_up('gnash', 'sport'), 'spash gnort')
test(mix_up('pezzy', 'firm'), 'fizzy perm')
# Standard boilerplate to call the main() function.
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic list exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in list2.py.
# A. match_ends
# Given a list of strings, return the count of the number of
# strings where the string length is 2 or more and the first
# and last chars of the string are the same.
# Note: python does not have a ++ operator, but += works.
def match_ends(words):
# +++your code here+++
temp=0
for x in words:
if(len(x) > 1):
if(x[:1]==x[-1:]):
temp+=1
return temp
# B. front_x
# Given a list of strings, return a list with the strings
# in sorted order, except group all the strings that begin with 'x' first.
# e.g. ['mix', 'xyz', 'apple', 'xanadu', 'aardvark'] yields
# ['xanadu', 'xyz', 'aardvark', 'apple', 'mix']
# Hint: this can be done by making 2 lists and sorting each of them
# before combining them.
def front_x(words):
# +++your code here+++
temp1=[];
temp2=[];
for x in words:
if(x[:1]=='x'):
temp1.append(x)
else:
temp2.append(x)
return sorted(temp1)+ sorted(temp2)
# C. sort_last
# Given a list of non-empty tuples, return a list sorted in increasing
# order by the last element in each tuple.
# e.g. [(1, 7), (1, 3), (3, 4, 5), (2, 2)] yields
# [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
# Hint: use a custom key= function to extract the last element form each tuple.
def sort_last(tuples):
# +++your code here+++
return sorted(tuples, key=lambda tuples: tuples[-1:])
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'match_ends'
test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)
print
print 'front_x'
test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])
print
print 'sort_last'
test(sort_last([(1, 3), (3, 2), (2, 1)]),
[(2, 1), (3, 2), (1, 3)])
test(sort_last([(2, 3), (1, 2), (3, 1)]),
[(3, 1), (1, 2), (2, 3)])
test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
[(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic list exercises
# D. Given a list of numbers, return a list where
# all adjacent == elements have been reduced to a single element,
# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
# modify the passed in list.
def remove_adjacent(nums):
# +++your code here+++
for x in nums:
if(nums.count(x) > 1):
nums.remove(x)
return nums
# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
# +++your code here+++
return sorted(list1+list2)
# Note: the solution above is kind of cute, but unforunately list.pop(0)
# is not constant time with the standard python list implementation, so
# the above is not strictly linear time.
# An alternate approach uses pop(-1) to remove the endmost elements
# from each list, building a solution list which is backwards.
# Then use reversed() to put the result back in the correct order. That
# solution works in linear time, but is more ugly.
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'remove_adjacent'
test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
test(remove_adjacent([]), [])
print
print 'linear_merge'
test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
['aa', 'aa', 'aa', 'bb', 'bb'])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Wordcount exercise
Google's Python class
The main() below is already defined and complete. It calls print_words()
and print_top() functions which you write.
1. For the --count flag, implement a print_words(filename) function that counts
how often each word appears in the text and prints:
word1 count1
word2 count2
...
Print the above list in order sorted by word (python will sort punctuation to
come before letters -- that's fine). Store all the words as lowercase,
so 'The' and 'the' count as the same word.
2. For the --topcount flag, implement a print_top(filename) which is similar
to print_words() but which prints just the top 20 most common words sorted
so the most common word is first, then the next most common, and so on.
Use str.split() (no arguments) to split on all whitespace.
Workflow: don't build the whole program at once. Get it to an intermediate
milestone and print your data structure and sys.exit(0).
When that's working, try for the next milestone.
Optional: define a helper function to avoid code duplication inside
print_words() and print_top().
"""
import sys
import re
# +++your code here+++
def print_words(filename):
fo = open(filename, "r+")
str1 = fo.read()
str1=re.sub("[(/.,!?:-;)`]", "",str1)
c={};
keyGen(str1,c)
fo.close()
printing(c)
def keyGen(str1,c):
i=0;
k=0
temp=str1.lower().split()
for x in temp:
c.update({temp[i]:temp.count(x)})
i+=1
def printing(c):
for k in c:
print k,'\t',c[k]
def print_top(filename):
fo = open(filename, "r+")
str1 = fo.read()
str1=re.sub("[/.,!?;]", "",str1)
c={};
keyGen(str1,c)
b = list(c.items())
b.sort(key=lambda item: item[1],reverse=True)
for item in b[:20]:
print(item[0] +' - '+ str(item[1]))
fo.close()
# Define print_words(filename) and print_top(filename) functions.
# You could write a helper utility function that reads a file
# and builds and returns a word/count dict for it.
# Then print_words() and print_top() can just call the utility function.
###
# This basic command line argument parsing code is provided and
# calls the print_words() and print_top() functions which you must define.
def main():
if len(sys.argv) != 3:
print 'usage: ./wordcount.py {--count | --topcount} file'
sys.exit(1)
option = sys.argv[1]
filename = sys.argv[2]
if option == '--count':
print_words(filename)
elif option == '--topcount':
print_top(filename)
else:
print 'unknown option: ' + option
sys.exit(1)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Mimic pyquick exercise -- optional extra exercise.
Google's Python Class
Read in the file specified on the command line.
Do a simple split() on whitespace to obtain all the words in the file.
Rather than read the file line by line, it's easier to read
it into one giant string and split it once.
Build a "mimic" dict that maps each word that appears in the file
to a list of all the words that immediately follow that word in the file.
The list of words can be be in any order and should include
duplicates. So for example the key "and" might have the list
["then", "best", "then", "after", ...] listing
all the words which came after "and" in the text.
We'll say that the empty string is what comes before
the first word in the file.
With the mimic dict, it's fairly easy to emit random
text that mimics the original. Print a word, then look
up what words might come next and pick one at random as
the next work.
Use the empty string as the first word to prime things.
If we ever get stuck with a word that is not in the dict,
go back to the empty string to keep things moving.
Note: the standard python module 'random' includes a
random.choice(list) method which picks a random element
from a non-empty list.
For fun, feed your program to itself as input.
Could work on getting it to put in linebreaks around 70
columns, so the output looks better.
"""
import random
import sys
import re
def mimic_dict(filename):
"""Returns mimic dict mapping each word to list of words which follow it."""
f=open(filename,'r')
text=f.read()
text=re.sub("[(/.,!?:-;)`]","",text)
str1=text.lower().split()
mimic={"":str1[0]}
suf=[]
for i in range(len(str1)):
if mimic.get(str1[i])==None:
j=i
for j in range(len(str1)-1):
if str1[i]==str1[j]:
suf.append((str1[j+1]))
mimic.update({str1[i]:suf})
suf=[]
return mimic
def print_mimic(mimic_dict, word):
"""Given mimic dict and start word, prints 200 random words."""
# +++your code here+++
for i in range(200):
if word=='' or mimic_dict.get(word)==[]or mimic_dict.get(word)==None:
word=mimic_dict.get('')
sys.stdout.write("%s " %(word))
else:
word=random.choice(mimic_dict.get(word))
sys.stdout.write("%s " %(word))
return
# Provided main(), calls mimic_dict() and mimic()
def main():
if len(sys.argv) != 2:
print 'usage: ./mimic.py file-to-read'
sys.exit(1)
dict = mimic_dict(sys.argv[1])
print_mimic(dict, '')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python2.4 -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic string exercises
# D. verbing
# Given a string, if its length is at least 3,
# add 'ing' to its end.
# Unless it already ends in 'ing', in which case
# add 'ly' instead.
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
# +++your code here+++
if (len(s) > 3):
if (s[-3:] == 'ing'):
return s+'ly'
else:
return s+'ing'
else:
return s
# E. not_bad
# Given a string, find the first appearance of the
# substring 'not' and 'bad'. If the 'bad' follows
# the 'not', replace the whole 'not'...'bad' substring
# with 'good'.
# Return the resulting string.
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
# +++your code here+++
no = s.find('not')
bad = s.find('bad')+3
if(no < bad):
return s.replace (s[no:bad],'good')
else:
return s
# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back
def front_back(a, b):
# +++your code here+++
if(len(a)%2 ==0 and len(b)%2 ==0):
return a[:len(a)/2]+b[:len(b)/2]+a[-len(a)/2:]+b[-len(b)/2:]
if(len(a)%2 ==0 and len(b)%2 !=0):
return a[:len(a)/2]+b[:len(b)/2+1]+a[-len(a)/2:]+b[-len(b)/2+1:]
if(len(a)%2 !=0 and len(b)%2 ==0):
return a[:len(a)/2+1]+b[:len(b)/2]+a[-len(a)/2+1:]+b[-len(b)/2:]
else:
return a[:len(a)/2+1]+b[:len(b)/2+1]+a[-len(a)/2+1:]+b[-len(b)/2+1:]
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# main() calls the above functions with interesting inputs,
# using the above test() to check if the result is correct or not.
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic list exercises
# D. Given a list of numbers, return a list where
# all adjacent == elements have been reduced to a single element,
# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
# modify the passed in list.
def remove_adjacent(nums):
# +++your code here+++
for x in nums:
if(nums.count(x) > 1):
nums.remove(x)
return nums
# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
# +++your code here+++
return sorted(list1+list2)
# Note: the solution above is kind of cute, but unforunately list.pop(0)
# is not constant time with the standard python list implementation, so
# the above is not strictly linear time.
# An alternate approach uses pop(-1) to remove the endmost elements
# from each list, building a solution list which is backwards.
# Then use reversed() to put the result back in the correct order. That
# solution works in linear time, but is more ugly.
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'remove_adjacent'
test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
test(remove_adjacent([]), [])
print
print 'linear_merge'
test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
['aa', 'aa', 'aa', 'bb', 'bb'])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Wordcount exercise
Google's Python class
The main() below is already defined and complete. It calls print_words()
and print_top() functions which you write.
1. For the --count flag, implement a print_words(filename) function that counts
how often each word appears in the text and prints:
word1 count1
word2 count2
...
Print the above list in order sorted by word (python will sort punctuation to
come before letters -- that's fine). Store all the words as lowercase,
so 'The' and 'the' count as the same word.
2. For the --topcount flag, implement a print_top(filename) which is similar
to print_words() but which prints just the top 20 most common words sorted
so the most common word is first, then the next most common, and so on.
Use str.split() (no arguments) to split on all whitespace.
Workflow: don't build the whole program at once. Get it to an intermediate
milestone and print your data structure and sys.exit(0).
When that's working, try for the next milestone.
Optional: define a helper function to avoid code duplication inside
print_words() and print_top().
"""
import sys
import re
# +++your code here+++
def print_words(filename):
fo = open(filename, "r+")
str1 = fo.read()
str1=re.sub("[(/.,!?:-;)`]", "",str1)
c={};
keyGen(str1,c)
fo.close()
printing(c)
def keyGen(str1,c):
i=0;
k=0
temp=str1.lower().split()
for x in temp:
c.update({temp[i]:temp.count(x)})
i+=1
def printing(c):
for k in c:
print k,'\t',c[k]
def print_top(filename):
fo = open(filename, "r+")
str1 = fo.read()
str1=re.sub("[/.,!?;]", "",str1)
c={};
keyGen(str1,c)
b = list(c.items())
b.sort(key=lambda item: item[1],reverse=True)
for item in b[:20]:
print(item[0] +' - '+ str(item[1]))
fo.close()
# Define print_words(filename) and print_top(filename) functions.
# You could write a helper utility function that reads a file
# and builds and returns a word/count dict for it.
# Then print_words() and print_top() can just call the utility function.
###
# This basic command line argument parsing code is provided and
# calls the print_words() and print_top() functions which you must define.
def main():
if len(sys.argv) != 3:
print 'usage: ./wordcount.py {--count | --topcount} file'
sys.exit(1)
option = sys.argv[1]
filename = sys.argv[2]
if option == '--count':
print_words(filename)
elif option == '--topcount':
print_top(filename)
else:
print 'unknown option: ' + option
sys.exit(1)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python2.4 -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic string exercises
# D. verbing
# Given a string, if its length is at least 3,
# add 'ing' to its end.
# Unless it already ends in 'ing', in which case
# add 'ly' instead.
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
# +++your code here+++
if (len(s) > 3):
if (s[-3:] == 'ing'):
return s+'ly'
else:
return s+'ing'
else:
return s
# E. not_bad
# Given a string, find the first appearance of the
# substring 'not' and 'bad'. If the 'bad' follows
# the 'not', replace the whole 'not'...'bad' substring
# with 'good'.
# Return the resulting string.
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
# +++your code here+++
no = s.find('not')
bad = s.find('bad')+3
if(no < bad):
return s.replace (s[no:bad],'good')
else:
return s
# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back
def front_back(a, b):
# +++your code here+++
if(len(a)%2 ==0 and len(b)%2 ==0):
return a[:len(a)/2]+b[:len(b)/2]+a[-len(a)/2:]+b[-len(b)/2:]
if(len(a)%2 ==0 and len(b)%2 !=0):
return a[:len(a)/2]+b[:len(b)/2+1]+a[-len(a)/2:]+b[-len(b)/2+1:]
if(len(a)%2 !=0 and len(b)%2 ==0):
return a[:len(a)/2+1]+b[:len(b)/2]+a[-len(a)/2+1:]+b[-len(b)/2:]
else:
return a[:len(a)/2+1]+b[:len(b)/2+1]+a[-len(a)/2+1:]+b[-len(b)/2+1:]
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# main() calls the above functions with interesting inputs,
# using the above test() to check if the result is correct or not.
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic list exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in list2.py.
# A. match_ends
# Given a list of strings, return the count of the number of
# strings where the string length is 2 or more and the first
# and last chars of the string are the same.
# Note: python does not have a ++ operator, but += works.
def match_ends(words):
n=0
for i in range (len(words)):
if len(words[i])>=2 and words[i][0]==words[i][-1:]:
n+=1
return n
# B. front_x
# Given a list of strings, return a list with the strings
# in sorted order, except group all the strings that begin with 'x' first.
# e.g. ['mix', 'xyz', 'apple', 'xanadu', 'aardvark'] yields
# ['xanadu', 'xyz', 'aardvark', 'apple', 'mix']
# Hint: this can be done by making 2 lists and sorting each of them
# before combining them.
def front_x(words):
list1=[]
list2=[]
for i in range (len(words)):
if words[i].startswith('x'):
list1.append(words[i])
else:
list2.append(words[i])
return sorted(list1)+sorted(list2)
# C. sort_last
# Given a list of non-empty tuples, return a list sorted in increasing
# order by the last element in each tuple.
# e.g. [(1, 7), (1, 3), (3, 4, 5), (2, 2)] yields
# [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
# Hint: use a custom key= function to extract the last element form each tuple.
def sort_last(tuples):
return sorted(tuples, key=lambda tuples:tuples[-1:])
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'match_ends'
test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)
print
print 'front_x'
test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])
print
print 'sort_last'
test(sort_last([(1, 3), (3, 2), (2, 1)]),
[(2, 1), (3, 2), (1, 3)])
test(sort_last([(2, 3), (1, 2), (3, 1)]),
[(3, 1), (1, 2), (2, 3)])
test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
[(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
main()
input()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Mimic pyquick exercise -- optional extra exercise.
Google's Python Class
Read in the file specified on the command line.
Do a simple split() on whitespace to obtain all the words in the file.
Rather than read the file line by line, it's easier to read
it into one giant string and split it once.
Build a "mimic" dict that maps each word that appears in the file
to a list of all the words that immediately follow that word in the file.
The list of words can be be in any order and should include
duplicates. So for example the key "and" might have the list
["then", "best", "then", "after", ...] listing
all the words which came after "and" in the text.
We'll say that the empty string is what comes before
the first word in the file.
With the mimic dict, it's fairly easy to emit random
text that mimics the original. Print a word, then look
up what words might come next and pick one at random as
the next work.
Use the empty string as the first word to prime things.
If we ever get stuck with a word that is not in the dict,
go back to the empty string to keep things moving.
Note: the standard python module 'random' includes a
random.choice(list) method which picks a random element
from a non-empty list.
For fun, feed your program to itself as input.
Could work on getting it to put in linebreaks around 70
columns, so the output looks better.
"""
import random
import sys
import re
def mimic_dict(filename):
"""Returns mimic dict mapping each word to list of words which follow it."""
f=open(filename,'r')
text=f.read()
text=re.sub("[/.,'`;:?!-()]","",text)
lines=text.lower().split()
if len(lines)==0:
return 0
mimic={"":lines[0]}
suf=[]
for i in range (len(lines)):
if mimic.get(lines[i])==None:
j=i
for j in range (len(lines)-1):
if lines[i]==lines[j]:
suf.append(lines[j+1])
mimic.update({lines[i]:suf})
suf=[]
return mimic
def print_mimic(mimic_dict, word):
"""Given mimic dict and start word, prints 200 random words."""
if mimic_dict==0:
print "Sorry, your file is empty :("
return
for i in range(200):
if word=='' or mimic_dict.get(word)==[]:
word=mimic_dict.get('')
sys.stdout.write("%s " %(word))
else:
word=random.choice(mimic_dict.get(word))
sys.stdout.write("%s " %(word))
return
# Provided main(), calls mimic_dict() and mimic()
def main():
if len(sys.argv) != 2:
print 'usage: ./mimic.py file-to-read'
sys.exit(1)
dict = mimic_dict(sys.argv[1])
print_mimic(dict, '')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic string exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in string2.py.
# A. donuts
# Given an int count of a number of donuts, return a string
# of the form 'Number of donuts: <count>', where <count> is the number
# passed in. However, if the count is 10 or more, then use the word 'many'
# instead of the actual count.
# So donuts(5) returns 'Number of donuts: 5'
# and donuts(23) returns 'Number of donuts: many'
def donuts(count):
if count<0:
return 'Error!!!'
elif count<10:
return 'Number of donuts: %d' %(count)
else:
return 'Number of donuts: many'
#return
# B. both_ends
# Given a string s, return a string made of the first 2
# and the last 2 chars of the original string,
# so 'spring' yields 'spng'. However, if the string length
# is less than 2, return instead the empty string.
def both_ends(s):
if len(s)<2:
return ''
else:
return s[0:2]+s[-2:]
# C. fix_start
# Given a string s, return a string
# where all occurences of its first char have
# been changed to '*', except do not change
# the first char itself.
# e.g. 'babble' yields 'ba**le'
# Assume that the string is length 1 or more.
# Hint: s.replace(stra, strb) returns a version of string s
# where all instances of stra have been replaced by strb.
def fix_start(s):
if len(s)<1:
return s
else:
return s[0]+s[1:].replace(s[0],'*')
# D. MixUp
# Given strings a and b, return a single string with a and b separated
# by a space '<a> <b>', except swap the first 2 chars of each string.
# e.g.
# 'mix', pod' -> 'pox mid'
# 'dog', 'dinner' -> 'dig donner'
# Assume a and b are length 2 or more.
def mix_up(a, b):
if len(a)<2 or len(b)<2:
return 'Error!!!'
else:
return '%s %s' %(a.replace(a[0:2],b[0:2]),b.replace(b[0:2],a[0:2]))
# Provided simple test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Provided main() calls the above functions with interesting inputs,
# using test() to check if each result is correct or not.
def main():
print 'donuts'
# Each line calls donuts, compares its result to the expected for that call.
test(donuts(4), 'Number of donuts: 4')
test(donuts(9), 'Number of donuts: 9')
test(donuts(10), 'Number of donuts: many')
test(donuts(99), 'Number of donuts: many')
print
print 'both_ends'
test(both_ends('spring'), 'spng')
test(both_ends('Hello'), 'Helo')
test(both_ends('a'), '')
test(both_ends('xyz'), 'xyyz')
print
print 'fix_start'
test(fix_start('babble'), 'ba**le')
test(fix_start('aardvark'), 'a*rdv*rk')
test(fix_start('google'), 'goo*le')
test(fix_start('donut'), 'donut')
print
print 'mix_up'
test(mix_up('mix', 'pod'), 'pox mid')
test(mix_up('dog', 'dinner'), 'dig donner')
test(mix_up('gnash', 'sport'), 'spash gnort')
test(mix_up('pezzy', 'firm'), 'fizzy perm')
# Standard boilerplate to call the main() function.
if __name__ == '__main__':
main()
input()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic string exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in string2.py.
# A. donuts
# Given an int count of a number of donuts, return a string
# of the form 'Number of donuts: <count>', where <count> is the number
# passed in. However, if the count is 10 or more, then use the word 'many'
# instead of the actual count.
# So donuts(5) returns 'Number of donuts: 5'
# and donuts(23) returns 'Number of donuts: many'
def donuts(count):
if count<0:
return 'Error!!!'
elif count<10:
return 'Number of donuts: %d' %(count)
else:
return 'Number of donuts: many'
#return
# B. both_ends
# Given a string s, return a string made of the first 2
# and the last 2 chars of the original string,
# so 'spring' yields 'spng'. However, if the string length
# is less than 2, return instead the empty string.
def both_ends(s):
if len(s)<2:
return ''
else:
return s[0:2]+s[-2:]
# C. fix_start
# Given a string s, return a string
# where all occurences of its first char have
# been changed to '*', except do not change
# the first char itself.
# e.g. 'babble' yields 'ba**le'
# Assume that the string is length 1 or more.
# Hint: s.replace(stra, strb) returns a version of string s
# where all instances of stra have been replaced by strb.
def fix_start(s):
if len(s)<1:
return s
else:
return s[0]+s[1:].replace(s[0],'*')
# D. MixUp
# Given strings a and b, return a single string with a and b separated
# by a space '<a> <b>', except swap the first 2 chars of each string.
# e.g.
# 'mix', pod' -> 'pox mid'
# 'dog', 'dinner' -> 'dig donner'
# Assume a and b are length 2 or more.
def mix_up(a, b):
if len(a)<2 or len(b)<2:
return 'Error!!!'
else:
return '%s %s' %(a.replace(a[0:2],b[0:2]),b.replace(b[0:2],a[0:2]))
# Provided simple test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Provided main() calls the above functions with interesting inputs,
# using test() to check if each result is correct or not.
def main():
print 'donuts'
# Each line calls donuts, compares its result to the expected for that call.
test(donuts(4), 'Number of donuts: 4')
test(donuts(9), 'Number of donuts: 9')
test(donuts(10), 'Number of donuts: many')
test(donuts(99), 'Number of donuts: many')
print
print 'both_ends'
test(both_ends('spring'), 'spng')
test(both_ends('Hello'), 'Helo')
test(both_ends('a'), '')
test(both_ends('xyz'), 'xyyz')
print
print 'fix_start'
test(fix_start('babble'), 'ba**le')
test(fix_start('aardvark'), 'a*rdv*rk')
test(fix_start('google'), 'goo*le')
test(fix_start('donut'), 'donut')
print
print 'mix_up'
test(mix_up('mix', 'pod'), 'pox mid')
test(mix_up('dog', 'dinner'), 'dig donner')
test(mix_up('gnash', 'sport'), 'spash gnort')
test(mix_up('pezzy', 'firm'), 'fizzy perm')
# Standard boilerplate to call the main() function.
if __name__ == '__main__':
main()
input()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic list exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in list2.py.
# A. match_ends
# Given a list of strings, return the count of the number of
# strings where the string length is 2 or more and the first
# and last chars of the string are the same.
# Note: python does not have a ++ operator, but += works.
def match_ends(words):
n=0
for i in range (len(words)):
if len(words[i])>=2 and words[i][0]==words[i][-1:]:
n+=1
return n
# B. front_x
# Given a list of strings, return a list with the strings
# in sorted order, except group all the strings that begin with 'x' first.
# e.g. ['mix', 'xyz', 'apple', 'xanadu', 'aardvark'] yields
# ['xanadu', 'xyz', 'aardvark', 'apple', 'mix']
# Hint: this can be done by making 2 lists and sorting each of them
# before combining them.
def front_x(words):
list1=[]
list2=[]
for i in range (len(words)):
if words[i].startswith('x'):
list1.append(words[i])
else:
list2.append(words[i])
return sorted(list1)+sorted(list2)
# C. sort_last
# Given a list of non-empty tuples, return a list sorted in increasing
# order by the last element in each tuple.
# e.g. [(1, 7), (1, 3), (3, 4, 5), (2, 2)] yields
# [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
# Hint: use a custom key= function to extract the last element form each tuple.
def sort_last(tuples):
return sorted(tuples, key=lambda tuples:tuples[-1:])
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'match_ends'
test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)
print
print 'front_x'
test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])
print
print 'sort_last'
test(sort_last([(1, 3), (3, 2), (2, 1)]),
[(2, 1), (3, 2), (1, 3)])
test(sort_last([(2, 3), (1, 2), (3, 1)]),
[(3, 1), (1, 2), (2, 3)])
test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
[(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
main()
input()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic list exercises
# D. Given a list of numbers, return a list where
# all adjacent == elements have been reduced to a single element,
# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
# modify the passed in list.
def remove_adjacent(nums):
i=1
newlist=[]
for i in range (len(nums)):
if nums[i]!=nums[i-1]:
newlist.append(nums[i])
return newlist
# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
list1.extend(list2)
return sorted(list1)
# Note: the solution above is kind of cute, but unforunately list.pop(0)
# is not constant time with the standard python list implementation, so
# the above is not strictly linear time.
# An alternate approach uses pop(-1) to remove the endmost elements
# from each list, building a solution list which is backwards.
# Then use reversed() to put the result back in the correct order. That
# solution works in linear time, but is more ugly.
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'remove_adjacent'
test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
test(remove_adjacent([]), [])
print
print 'linear_merge'
test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
['aa', 'aa', 'aa', 'bb', 'bb'])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Wordcount exercise
Google's Python class
The main() below is already defined and complete. It calls print_words()
and print_top() functions which you write.
1. For the --count flag, implement a print_words(filename) function that counts
how often each word appears in the text and prints:
word1 count1
word2 count2
...
Print the above list in order sorted by word (python will sort punctuation to
come before letters -- that's fine). Store all the words as lowercase,
so 'The' and 'the' count as the same word.
2. For the --topcount flag, implement a print_top(filename) which is similar
to print_words() but which prints just the top 20 most common words sorted
so the most common word is first, then the next most common, and so on.
Use str.split() (no arguments) to split on all whitespace.
Workflow: don't build the whole program at once. Get it to an intermediate
milestone and print your data structure and sys.exit(0).
When that's working, try for the next milestone.
Optional: define a helper function to avoid code duplication inside
print_words() and print_top().
"""
import sys
import re
def readfile(filename):
f=open(filename,'r')
text=f.read()
text=re.sub("[/.,\';:?!-]","",text)
lines=sorted(text.lower().split())
newlines=[]
i=1
for i in range(len(lines)):
if lines[i]!=lines[i-1]:
newlines.append(lines[i])
newlines.append(lines.count(lines[i]))
arr=[]
for i in range(len(newlines)/2):
arr.append(tuple(newlines[i*2:i*2+2]))
return arr
def print_words(filename):
arr=readfile(filename)
for i in range (len(arr)):
print arr[i][0],'\t',arr[i][1]
return
def print_top(filename):
arr=readfile(filename)
newarr=sorted(arr, key=lambda arr:arr[-1:])
for i in range (len(newarr)):
print newarr[i][0],'\t',newarr[i][1]
return
# Define print_words(filename) and print_top(filename) functions.
# You could write a helper utility function that reads a file
# and builds and returns a word/count dict for it.
# Then print_words() and print_top() can just call the utility function.
###
# This basic command line argument parsing code is provided and
# calls the print_words() and print_top() functions which you must define.
def main():
if len(sys.argv) != 3:
print 'usage: ./wordcount.py {--count | --topcount} file'
sys.exit(1)
option = sys.argv[1]
filename = sys.argv[2]
if option == '--count':
print_words(filename)
elif option == '--topcount':
print_top(filename)
else:
print 'unknown option: ' + option
sys.exit(1)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Mimic pyquick exercise -- optional extra exercise.
Google's Python Class
Read in the file specified on the command line.
Do a simple split() on whitespace to obtain all the words in the file.
Rather than read the file line by line, it's easier to read
it into one giant string and split it once.
Build a "mimic" dict that maps each word that appears in the file
to a list of all the words that immediately follow that word in the file.
The list of words can be be in any order and should include
duplicates. So for example the key "and" might have the list
["then", "best", "then", "after", ...] listing
all the words which came after "and" in the text.
We'll say that the empty string is what comes before
the first word in the file.
With the mimic dict, it's fairly easy to emit random
text that mimics the original. Print a word, then look
up what words might come next and pick one at random as
the next work.
Use the empty string as the first word to prime things.
If we ever get stuck with a word that is not in the dict,
go back to the empty string to keep things moving.
Note: the standard python module 'random' includes a
random.choice(list) method which picks a random element
from a non-empty list.
For fun, feed your program to itself as input.
Could work on getting it to put in linebreaks around 70
columns, so the output looks better.
"""
import random
import sys
import re
def mimic_dict(filename):
"""Returns mimic dict mapping each word to list of words which follow it."""
f=open(filename,'r')
text=f.read()
text=re.sub("[/.,'`;:?!-()]","",text)
lines=text.lower().split()
if len(lines)==0:
return 0
mimic={"":lines[0]}
suf=[]
for i in range (len(lines)):
if mimic.get(lines[i])==None:
j=i
for j in range (len(lines)-1):
if lines[i]==lines[j]:
suf.append(lines[j+1])
mimic.update({lines[i]:suf})
suf=[]
return mimic
def print_mimic(mimic_dict, word):
"""Given mimic dict and start word, prints 200 random words."""
if mimic_dict==0:
print "Sorry, your file is empty :("
return
for i in range(200):
if word=='' or mimic_dict.get(word)==[]:
word=mimic_dict.get('')
sys.stdout.write("%s " %(word))
else:
word=random.choice(mimic_dict.get(word))
sys.stdout.write("%s " %(word))
return
# Provided main(), calls mimic_dict() and mimic()
def main():
if len(sys.argv) != 2:
print 'usage: ./mimic.py file-to-read'
sys.exit(1)
dict = mimic_dict(sys.argv[1])
print_mimic(dict, '')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python2.4 -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic string exercises
# D. verbing
# Given a string, if its length is at least 3,
# add 'ing' to its end.
# Unless it already ends in 'ing', in which case
# add 'ly' instead.
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
if len(s)<3:
return s
elif s[-3:]=='ing':
return s+'ly'
else:
return s+'ing'
# E. not_bad
# Given a string, find the first appearance of the
# substring 'not' and 'bad'. If the 'bad' follows
# the 'not', replace the whole 'not'...'bad' substring
# with 'good'.
# Return the resulting string.
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
if s.find('bad')>s.find('not'):
return s.replace(s[s.find('not'):s.find('bad')+3],'good')
else:
return s
# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back
def front_back(a, b):
if len(a)%2==0 and len(b)%2==0:
return a[0:len(a)/2]+b[0:len(b)/2]+a[-len(a)/2:]+b[-len(b)/2:]
if len(a)%2!=0 and len(b)%2==0:
return a[0:len(a)/2+1]+b[0:len(b)/2]+a[-len(a)/2+1:]+b[-len(b)/2:]
if len(a)%2==0 and len(b)%2!=0:
return a[0:len(a)/2]+b[0:len(b)/2+1]+a[-len(a)/2:]+b[-len(b)/2+1:]
if len(a)%2!=0 and len(b)%2!=0:
return a[0:len(a)/2+1]+b[0:len(b)/2+1]+a[-len(a)/2+1:]+b[-len(b)/2+1:]
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# main() calls the above functions with interesting inputs,
# using the above test() to check if the result is correct or not.
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
input()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic list exercises
# D. Given a list of numbers, return a list where
# all adjacent == elements have been reduced to a single element,
# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
# modify the passed in list.
def remove_adjacent(nums):
i=1
newlist=[]
for i in range (len(nums)):
if nums[i]!=nums[i-1]:
newlist.append(nums[i])
return newlist
# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
list1.extend(list2)
return sorted(list1)
# Note: the solution above is kind of cute, but unforunately list.pop(0)
# is not constant time with the standard python list implementation, so
# the above is not strictly linear time.
# An alternate approach uses pop(-1) to remove the endmost elements
# from each list, building a solution list which is backwards.
# Then use reversed() to put the result back in the correct order. That
# solution works in linear time, but is more ugly.
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'remove_adjacent'
test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
test(remove_adjacent([]), [])
print
print 'linear_merge'
test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
['aa', 'aa', 'aa', 'bb', 'bb'])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Wordcount exercise
Google's Python class
The main() below is already defined and complete. It calls print_words()
and print_top() functions which you write.
1. For the --count flag, implement a print_words(filename) function that counts
how often each word appears in the text and prints:
word1 count1
word2 count2
...
Print the above list in order sorted by word (python will sort punctuation to
come before letters -- that's fine). Store all the words as lowercase,
so 'The' and 'the' count as the same word.
2. For the --topcount flag, implement a print_top(filename) which is similar
to print_words() but which prints just the top 20 most common words sorted
so the most common word is first, then the next most common, and so on.
Use str.split() (no arguments) to split on all whitespace.
Workflow: don't build the whole program at once. Get it to an intermediate
milestone and print your data structure and sys.exit(0).
When that's working, try for the next milestone.
Optional: define a helper function to avoid code duplication inside
print_words() and print_top().
"""
import sys
import re
def readfile(filename):
f=open(filename,'r')
text=f.read()
text=re.sub("[/.,\';:?!-]","",text)
lines=sorted(text.lower().split())
newlines=[]
i=1
for i in range(len(lines)):
if lines[i]!=lines[i-1]:
newlines.append(lines[i])
newlines.append(lines.count(lines[i]))
arr=[]
for i in range(len(newlines)/2):
arr.append(tuple(newlines[i*2:i*2+2]))
return arr
def print_words(filename):
arr=readfile(filename)
for i in range (len(arr)):
print arr[i][0],'\t',arr[i][1]
return
def print_top(filename):
arr=readfile(filename)
newarr=sorted(arr, key=lambda arr:arr[-1:])
for i in range (len(newarr)):
print newarr[i][0],'\t',newarr[i][1]
return
# Define print_words(filename) and print_top(filename) functions.
# You could write a helper utility function that reads a file
# and builds and returns a word/count dict for it.
# Then print_words() and print_top() can just call the utility function.
###
# This basic command line argument parsing code is provided and
# calls the print_words() and print_top() functions which you must define.
def main():
if len(sys.argv) != 3:
print 'usage: ./wordcount.py {--count | --topcount} file'
sys.exit(1)
option = sys.argv[1]
filename = sys.argv[2]
if option == '--count':
print_words(filename)
elif option == '--topcount':
print_top(filename)
else:
print 'unknown option: ' + option
sys.exit(1)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python2.4 -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic string exercises
# D. verbing
# Given a string, if its length is at least 3,
# add 'ing' to its end.
# Unless it already ends in 'ing', in which case
# add 'ly' instead.
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
if len(s)<3:
return s
elif s[-3:]=='ing':
return s+'ly'
else:
return s+'ing'
# E. not_bad
# Given a string, find the first appearance of the
# substring 'not' and 'bad'. If the 'bad' follows
# the 'not', replace the whole 'not'...'bad' substring
# with 'good'.
# Return the resulting string.
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
if s.find('bad')>s.find('not'):
return s.replace(s[s.find('not'):s.find('bad')+3],'good')
else:
return s
# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back
def front_back(a, b):
if len(a)%2==0 and len(b)%2==0:
return a[0:len(a)/2]+b[0:len(b)/2]+a[-len(a)/2:]+b[-len(b)/2:]
if len(a)%2!=0 and len(b)%2==0:
return a[0:len(a)/2+1]+b[0:len(b)/2]+a[-len(a)/2+1:]+b[-len(b)/2:]
if len(a)%2==0 and len(b)%2!=0:
return a[0:len(a)/2]+b[0:len(b)/2+1]+a[-len(a)/2:]+b[-len(b)/2+1:]
if len(a)%2!=0 and len(b)%2!=0:
return a[0:len(a)/2+1]+b[0:len(b)/2+1]+a[-len(a)/2+1:]+b[-len(b)/2+1:]
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# main() calls the above functions with interesting inputs,
# using the above test() to check if the result is correct or not.
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
input()
| Python |
#!/usr/bin/python
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
import sys
import re
"""Baby Names exercise
Define the extract_names() function below and change main()
to call it.
For writing regex, it's nice to include a copy of the target
text for inspiration.
Here's what the html looks like in the baby.html files:
...
<h3 align="center">Popularity in 1990</h3>
....
<tr align="right"><td>1</td><td>Michael</td><td>Jessica</td>
<tr align="right"><td>2</td><td>Christopher</td><td>Ashley</td>
<tr align="right"><td>3</td><td>Matthew</td><td>Brittany</td>
...
Suggested milestones for incremental development:
-Extract the year and print it
-Extract the names and rank numbers and just print them
-Get the names data into a dict and print it
-Build the [year, 'name rank', ... ] list and print it
-Fix main() to use the extract_names list
"""
def extract_names(filename):
"""
Given a file name for baby.html, returns a list starting with the year string
followed by the name-rank strings in alphabetical order.
['2006', 'Aaliyah 91', Aaron 57', 'Abagail 895', ' ...]
"""
# +++your code here+++
return
def main():
# This command-line parsing code is provided.
# Make a list of command line arguments, omitting the [0] element
# which is the script itself.
args = sys.argv[1:]
if not args:
print 'usage: [--summaryfile] file [file ...]'
sys.exit(1)
# Notice the summary flag and remove it from args if it is present.
summary = False
if args[0] == '--summaryfile':
summary = True
del args[0]
# +++your code here+++
# For each filename, get the names, then either print the text output
# or write it to a summary file
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
import sys
import re
"""Baby Names exercise
Define the extract_names() function below and change main()
to call it.
For writing regex, it's nice to include a copy of the target
text for inspiration.
Here's what the html looks like in the baby.html files:
...
<h3 align="center">Popularity in 1990</h3>
....
<tr align="right"><td>1</td><td>Michael</td><td>Jessica</td>
<tr align="right"><td>2</td><td>Christopher</td><td>Ashley</td>
<tr align="right"><td>3</td><td>Matthew</td><td>Brittany</td>
...
Suggested milestones for incremental development:
-Extract the year and print it
-Extract the names and rank numbers and just print them
-Get the names data into a dict and print it
-Build the [year, 'name rank', ... ] list and print it
-Fix main() to use the extract_names list
"""
def extract_names(filename):
"""
Given a file name for baby.html, returns a list starting with the year string
followed by the name-rank strings in alphabetical order.
['2006', 'Aaliyah 91', Aaron 57', 'Abagail 895', ' ...]
"""
# +++your code here+++
return
def main():
# This command-line parsing code is provided.
# Make a list of command line arguments, omitting the [0] element
# which is the script itself.
args = sys.argv[1:]
if not args:
print 'usage: [--summaryfile] file [file ...]'
sys.exit(1)
# Notice the summary flag and remove it from args if it is present.
summary = False
if args[0] == '--summaryfile':
summary = True
del args[0]
# +++your code here+++
# For each filename, get the names, then either print the text output
# or write it to a summary file
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""A tiny Python program to check that Python is working.
Try running this program from the command line like this:
python hello.py
python hello.py Alice
That should print:
Hello World -or- Hello Alice
Try changing the 'Hello' to 'Howdy' and run again.
Once you have that working, you're ready for class -- you can edit
and run Python code; now you just need to learn Python!
"""
import sys
# Define a main() function that prints a little greeting.
def main():
# Get the name from the command line, using 'World' as a fallback.
if len(sys.argv) >= 2:
name = sys.argv[1]
else:
name = 'World'
print 'Hello', name
# This is the standard boilerplate that calls the main() function.
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""A tiny Python program to check that Python is working.
Try running this program from the command line like this:
python hello.py
python hello.py Alice
That should print:
Hello World -or- Hello Alice
Try changing the 'Hello' to 'Howdy' and run again.
Once you have that working, you're ready for class -- you can edit
and run Python code; now you just need to learn Python!
"""
import sys
# Define a main() function that prints a little greeting.
def main():
# Get the name from the command line, using 'World' as a fallback.
if len(sys.argv) >= 2:
name = sys.argv[1]
else:
name = 'World'
print 'Hello', name
# This is the standard boilerplate that calls the main() function.
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
import sys
import re
import os
import shutil
import commands
"""Copy Special exercise
"""
# +++your code here+++
# Write functions and modify main() to call them
def main():
# This basic command line argument parsing code is provided.
# Add code to call your functions below.
# Make a list of command line arguments, omitting the [0] element
# which is the script itself.
args = sys.argv[1:]
if not args:
print "usage: [--todir dir][--tozip zipfile] dir [dir ...]";
sys.exit(1)
# todir and tozip are either set from command line
# or left as the empty string.
# The args array is left just containing the dirs.
todir = ''
if args[0] == '--todir':
todir = args[1]
del args[0:2]
tozip = ''
if args[0] == '--tozip':
tozip = args[1]
del args[0:2]
if len(args) == 0:
print "error: must specify one or more dirs"
sys.exit(1)
# +++your code here+++
# Call your functions
if __name__ == "__main__":
main()
| Python |
#!/usr/bin/python
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
import sys
import re
import os
import shutil
import commands
"""Copy Special exercise
"""
# +++your code here+++
# Write functions and modify main() to call them
def main():
# This basic command line argument parsing code is provided.
# Add code to call your functions below.
# Make a list of command line arguments, omitting the [0] element
# which is the script itself.
args = sys.argv[1:]
if not args:
print "usage: [--todir dir][--tozip zipfile] dir [dir ...]";
sys.exit(1)
# todir and tozip are either set from command line
# or left as the empty string.
# The args array is left just containing the dirs.
todir = ''
if args[0] == '--todir':
todir = args[1]
del args[0:2]
tozip = ''
if args[0] == '--tozip':
tozip = args[1]
del args[0:2]
if len(args) == 0:
print "error: must specify one or more dirs"
sys.exit(1)
# +++your code here+++
# Call your functions
if __name__ == "__main__":
main()
| Python |
#!/usr/bin/python
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
import os
import re
import sys
import urllib
"""Logpuzzle exercise
Given an apache logfile, find the puzzle urls and download the images.
Here's what a puzzle url looks like:
10.254.254.28 - - [06/Aug/2007:00:13:48 -0700] "GET /~foo/puzzle-bar-aaab.jpg HTTP/1.0" 302 528 "-" "Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.1.6) Gecko/20070725 Firefox/2.0.0.6"
"""
def read_urls(filename):
"""Returns a list of the puzzle urls from the given log file,
extracting the hostname from the filename itself.
Screens out duplicate urls and returns the urls sorted into
increasing order."""
# +++your code here+++
def download_images(img_urls, dest_dir):
"""Given the urls already in the correct order, downloads
each image into the given directory.
Gives the images local filenames img0, img1, and so on.
Creates an index.html in the directory
with an img tag to show each local image file.
Creates the directory if necessary.
"""
# +++your code here+++
def main():
args = sys.argv[1:]
if not args:
print 'usage: [--todir dir] logfile '
sys.exit(1)
todir = ''
if args[0] == '--todir':
todir = args[1]
del args[0:2]
img_urls = read_urls(args[0])
if todir:
download_images(img_urls, todir)
else:
print '\n'.join(img_urls)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
import os
import re
import sys
import urllib
"""Logpuzzle exercise
Given an apache logfile, find the puzzle urls and download the images.
Here's what a puzzle url looks like:
10.254.254.28 - - [06/Aug/2007:00:13:48 -0700] "GET /~foo/puzzle-bar-aaab.jpg HTTP/1.0" 302 528 "-" "Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.1.6) Gecko/20070725 Firefox/2.0.0.6"
"""
def read_urls(filename):
"""Returns a list of the puzzle urls from the given log file,
extracting the hostname from the filename itself.
Screens out duplicate urls and returns the urls sorted into
increasing order."""
# +++your code here+++
def download_images(img_urls, dest_dir):
"""Given the urls already in the correct order, downloads
each image into the given directory.
Gives the images local filenames img0, img1, and so on.
Creates an index.html in the directory
with an img tag to show each local image file.
Creates the directory if necessary.
"""
# +++your code here+++
def main():
args = sys.argv[1:]
if not args:
print 'usage: [--todir dir] logfile '
sys.exit(1)
todir = ''
if args[0] == '--todir':
todir = args[1]
del args[0:2]
img_urls = read_urls(args[0])
if todir:
download_images(img_urls, todir)
else:
print '\n'.join(img_urls)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -
count = 0
for word in words:
if len(word) >= 2 and word[0] == word[-1]:
count = count + 1
return count
def front_x(words):
x_list = []
other_list = []
for w in words:
if w.startswith('x'):
x_list.append(w)
else:
other_list.append(w)
return sorted(x_list) + sorted(other_list)
def last(a):
return a[-1]
def sort_last(tuples):
return sorted(tuples, key=last)
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
def main():
print 'match_ends'
test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)
print
print 'front_x'
test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])
print
print 'sort_last'
test(sort_last([(1, 3), (3, 2), (2, 1)]),
[(2, 1), (3, 2), (1, 3)])
test(sort_last([(2, 3), (1, 2), (3, 1)]),
[(3, 1), (1, 2), (2, 3)])
test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
[(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python
def donuts(count):
if count < 10:
return 'Number of donuts: ' + str(count)
else:
return 'Number of donuts: many'
def both_ends(s):
if len(s) < 2:
return ''
first2 = s[0:2]
last2 = s[-2:]
return first2 + last2
def fix_start(s):
front = s[0]
back = s[1:]
fixed_back = back.replace(front, '*')
return front + fixed_back
def mix_up(a, b):
a_swapped = b[:2] + a[2:]
b_swapped = a[:2] + b[2:]
return a_swapped + ' ' + b_swapped
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
def main():
print 'donuts'
test(donuts(4), 'Number of donuts: 4')
test(donuts(9), 'Number of donuts: 9')
test(donuts(10), 'Number of donuts: many')
test(donuts(99), 'Number of donuts: many')
print
print 'both_ends'
test(both_ends('spring'), 'spng')
test(both_ends('Hello'), 'Helo')
test(both_ends('a'), '')
test(both_ends('xyz'), 'xyyz')
print
print 'fix_start'
test(fix_start('babble'), 'ba**le')
test(fix_start('aardvark'), 'a*rdv*rk')
test(fix_start('google'), 'goo*le')
test(fix_start('donut'), 'donut')
print
print 'mix_up'
test(mix_up('mix', 'pod'), 'pox mid')
test(mix_up('dog', 'dinner'), 'dig donner')
test(mix_up('gnash', 'sport'), 'spash gnort')
test(mix_up('pezzy', 'firm'), 'fizzy perm')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python
def donuts(count):
if count < 10:
return 'Number of donuts: ' + str(count)
else:
return 'Number of donuts: many'
def both_ends(s):
if len(s) < 2:
return ''
first2 = s[0:2]
last2 = s[-2:]
return first2 + last2
def fix_start(s):
front = s[0]
back = s[1:]
fixed_back = back.replace(front, '*')
return front + fixed_back
def mix_up(a, b):
a_swapped = b[:2] + a[2:]
b_swapped = a[:2] + b[2:]
return a_swapped + ' ' + b_swapped
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
def main():
print 'donuts'
test(donuts(4), 'Number of donuts: 4')
test(donuts(9), 'Number of donuts: 9')
test(donuts(10), 'Number of donuts: many')
test(donuts(99), 'Number of donuts: many')
print
print 'both_ends'
test(both_ends('spring'), 'spng')
test(both_ends('Hello'), 'Helo')
test(both_ends('a'), '')
test(both_ends('xyz'), 'xyyz')
print
print 'fix_start'
test(fix_start('babble'), 'ba**le')
test(fix_start('aardvark'), 'a*rdv*rk')
test(fix_start('google'), 'goo*le')
test(fix_start('donut'), 'donut')
print
print 'mix_up'
test(mix_up('mix', 'pod'), 'pox mid')
test(mix_up('dog', 'dinner'), 'dig donner')
test(mix_up('gnash', 'sport'), 'spash gnort')
test(mix_up('pezzy', 'firm'), 'fizzy perm')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -
count = 0
for word in words:
if len(word) >= 2 and word[0] == word[-1]:
count = count + 1
return count
def front_x(words):
x_list = []
other_list = []
for w in words:
if w.startswith('x'):
x_list.append(w)
else:
other_list.append(w)
return sorted(x_list) + sorted(other_list)
def last(a):
return a[-1]
def sort_last(tuples):
return sorted(tuples, key=last)
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
def main():
print 'match_ends'
test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)
print
print 'front_x'
test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])
print
print 'sort_last'
test(sort_last([(1, 3), (3, 2), (2, 1)]),
[(2, 1), (3, 2), (1, 3)])
test(sort_last([(2, 3), (1, 2), (3, 1)]),
[(3, 1), (1, 2), (2, 3)])
test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
[(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python
def remove_adjacent(nums):
result = []
for num in nums:
if len(result) == 0 or num != result[-1]:
result.append(num)
return result
def linear_merge(list1, list2):
result = []
while len(list1) and len(list2):
if list1[0] < list2[0]:
result.append(list1.pop(0))
else:
result.append(list2.pop(0))
result.extend(list1)
result.extend(list2)
return result
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
def main():
print 'remove_adjacent'
test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
test(remove_adjacent([]), [])
print
print 'linear_merge'
test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
['aa', 'aa', 'aa', 'bb', 'bb'])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python
import sys
def words(filename):
count = {}
input_file = open(filename, 'r')
for line in input_file:
words = line.split()
for word in words:
word = word.lower()
if not word in count:
count[word] = 1
else:
count[word] = count[word] + 1
input_file.close()
return count
def print_words(filename):
"""Prints one per line '<word> <count>' sorted by word for the given file."""
count = words(filename)
words = sorted(count.keys())
for word in words:
print word, count[word]
def get_count(count_tuple):
"""Returns the count from a dict word/count tuple -- used for custom sort."""
return count_tuple[1]
def print_top(filename):
"""Prints the top count listing for the given file."""
count = words(filename)
items = sorted(count.items(), key=get_count, reverse=True)
for item in items[:20]:
print item[0], item[1]
def main():
if len(sys.argv) != 3:
print 'usage: ./wordcount.py {--count | --topcount} file'
sys.exit(1)
option = sys.argv[1]
filename = sys.argv[2]
if option == '--count':
print_words(filename)
elif option == '--topcount':
print_top(filename)
else:
print 'unknown option: ' + option
sys.exit(1)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python
def verbing(s):
if len(s) >= 3:
if s[-3:] != 'ing': s = s + 'ing'
else: s = s + 'ly'
return s
def not_bad(s):
n = s.find('not')
b = s.find('bad')
if n != -1 and b != -1 and b > n:
s = s[:n] + 'good' + s[b+3:]
return s
def front_back(a, b):
a_middle = len(a) / 2
b_middle = len(b) / 2
if len(a) % 2 == 1:
a_middle = a_middle + 1
if len(b) % 2 == 1:
b_middle = b_middle + 1
return a[:a_middle] + b[:b_middle] + a[a_middle:] + b[b_middle:]
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python
def remove_adjacent(nums):
result = []
for num in nums:
if len(result) == 0 or num != result[-1]:
result.append(num)
return result
def linear_merge(list1, list2):
result = []
while len(list1) and len(list2):
if list1[0] < list2[0]:
result.append(list1.pop(0))
else:
result.append(list2.pop(0))
result.extend(list1)
result.extend(list2)
return result
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
def main():
print 'remove_adjacent'
test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
test(remove_adjacent([]), [])
print
print 'linear_merge'
test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
['aa', 'aa', 'aa', 'bb', 'bb'])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python
import sys
def words(filename):
count = {}
input_file = open(filename, 'r')
for line in input_file:
words = line.split()
for word in words:
word = word.lower()
if not word in count:
count[word] = 1
else:
count[word] = count[word] + 1
input_file.close()
return count
def print_words(filename):
"""Prints one per line '<word> <count>' sorted by word for the given file."""
count = words(filename)
words = sorted(count.keys())
for word in words:
print word, count[word]
def get_count(count_tuple):
"""Returns the count from a dict word/count tuple -- used for custom sort."""
return count_tuple[1]
def print_top(filename):
"""Prints the top count listing for the given file."""
count = words(filename)
items = sorted(count.items(), key=get_count, reverse=True)
for item in items[:20]:
print item[0], item[1]
def main():
if len(sys.argv) != 3:
print 'usage: ./wordcount.py {--count | --topcount} file'
sys.exit(1)
option = sys.argv[1]
filename = sys.argv[2]
if option == '--count':
print_words(filename)
elif option == '--topcount':
print_top(filename)
else:
print 'unknown option: ' + option
sys.exit(1)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python
def verbing(s):
if len(s) >= 3:
if s[-3:] != 'ing': s = s + 'ing'
else: s = s + 'ly'
return s
def not_bad(s):
n = s.find('not')
b = s.find('bad')
if n != -1 and b != -1 and b > n:
s = s[:n] + 'good' + s[b+3:]
return s
def front_back(a, b):
a_middle = len(a) / 2
b_middle = len(b) / 2
if len(a) % 2 == 1:
a_middle = a_middle + 1
if len(b) % 2 == 1:
b_middle = b_middle + 1
return a[:a_middle] + b[:b_middle] + a[a_middle:] + b[b_middle:]
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic list exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in list2.py.
# A. match_ends
# Given a list of strings, return the count of the number of
# strings where the string length is 2 or more and the first
# and last chars of the string are the same.
# Note: python does not have a ++ operator, but += works.
def match_ends(words):
count = 0
for word in words:
wordlen = len(word)
if wordlen>1 and word[0]==word[wordlen-1]:
count+=1
return count
# B. front_x
# Given a list of strings, return a list with the strings
# in sorted order, except group all the strings that begin with 'x' first.
# e.g. ['mix', 'xyz', 'apple', 'xanadu', 'aardvark'] yields
# ['xanadu', 'xyz', 'aardvark', 'apple', 'mix']
# Hint: this can be done by making 2 lists and sorting each of them
# before combining them.
def front_x(words):
x_list = []
other_list = []
for word in words:
if word[0]=='x':
x_list.append(word)
else:
other_list.append(word)
x_list.sort()
other_list.sort()
x_list.extend(other_list)
# +++your code here+++
return x_list
# C. sort_last
# Given a list of non-empty tuples, return a list sorted in increasing
# order by the last element in each tuple.
# e.g. [(1, 7), (1, 3), (3, 4, 5), (2, 2)] yields
# [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
# Hint: use a custom key= function to extract the last element form each tuple.
def sort_last(tuples):
return sorted(tuples,key=last)
def last(tuple):
last = tuple[len(tuple)-1]
return last
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'match_ends'
test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)
print
print 'front_x'
test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])
print
print 'sort_last'
test(sort_last([(1, 3), (3, 2), (2, 1)]),
[(2, 1), (3, 2), (1, 3)])
test(sort_last([(2, 3), (1, 2), (3, 1)]),
[(3, 1), (1, 2), (2, 3)])
test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
[(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic string exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in string2.py.
# A. donuts
# Given an int count of a number of donuts, return a string
# of the form 'Number of donuts: <count>', where <count> is the number
# passed in. However, if the count is 10 or more, then use the word 'many'
# instead of the actual count.
# So donuts(5) returns 'Number of donuts: 5'
# and donuts(23) returns 'Number of donuts: many'
def donuts(count):
if count < 10:
return 'Number of donuts: %d' %count
else:
return 'Number of donuts: many'
# B. both_ends
# Given a string s, return a string made of the first 2
# and the last 2 chars of the original string,
# so 'spring' yields 'spng'. However, if the string length
# is less than 2, return instead the empty string.
def both_ends(s):
if len(s) < 2:
return ''
else:
return s[0:2]+s[-2:]
return
# C. fix_start
# Given a string s, return a string
# where all occurences of its first char have
# been changed to '*', except do not change
# the first char itself.
# e.g. 'babble' yields 'ba**le'
# Assume that the string is length 1 or more.
# Hint: s.replace(stra, strb) returns a version of string s
# where all instances of stra have been replaced by strb.
def fix_start(s):
char = s[0]
word = char
for num in range(1,len(s)):
if s[num] == char:
word += '*'
else:
word += s[num]
return word
# D. MixUp
# Given strings a and b, return a single string with a and b separated
# by a space '<a> <b>', except swap the first 2 chars of each string.
# e.g.
# 'mix', pod' -> 'pox mid'
# 'dog', 'dinner' -> 'dig donner'
# Assume a and b are length 2 or more.
def mix_up(a, b):
return b[0:2]+a[2:]+ ' ' +a[0:2]+b[2:]
# Provided simple test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Provided main() calls the above functions with interesting inputs,
# using test() to check if each result is correct or not.
def main():
print 'donuts'
# Each line calls donuts, compares its result to the expected for that call.
test(donuts(4), 'Number of donuts: 4')
test(donuts(9), 'Number of donuts: 9')
test(donuts(10), 'Number of donuts: many')
test(donuts(99), 'Number of donuts: many')
print
print 'both_ends'
test(both_ends('spring'), 'spng')
test(both_ends('Hello'), 'Helo')
test(both_ends('a'), '')
test(both_ends('xyz'), 'xyyz')
print
print 'fix_start'
test(fix_start('babble'), 'ba**le')
test(fix_start('aardvark'), 'a*rdv*rk')
test(fix_start('google'), 'goo*le')
test(fix_start('donut'), 'donut')
print
print 'mix_up'
test(mix_up('mix', 'pod'), 'pox mid')
test(mix_up('dog', 'dinner'), 'dig donner')
test(mix_up('gnash', 'sport'), 'spash gnort')
test(mix_up('pezzy', 'firm'), 'fizzy perm')
# Standard boilerplate to call the main() function.
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic string exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in string2.py.
# A. donuts
# Given an int count of a number of donuts, return a string
# of the form 'Number of donuts: <count>', where <count> is the number
# passed in. However, if the count is 10 or more, then use the word 'many'
# instead of the actual count.
# So donuts(5) returns 'Number of donuts: 5'
# and donuts(23) returns 'Number of donuts: many'
def donuts(count):
if count < 10:
return 'Number of donuts: %d' %count
else:
return 'Number of donuts: many'
# B. both_ends
# Given a string s, return a string made of the first 2
# and the last 2 chars of the original string,
# so 'spring' yields 'spng'. However, if the string length
# is less than 2, return instead the empty string.
def both_ends(s):
if len(s) < 2:
return ''
else:
return s[0:2]+s[-2:]
return
# C. fix_start
# Given a string s, return a string
# where all occurences of its first char have
# been changed to '*', except do not change
# the first char itself.
# e.g. 'babble' yields 'ba**le'
# Assume that the string is length 1 or more.
# Hint: s.replace(stra, strb) returns a version of string s
# where all instances of stra have been replaced by strb.
def fix_start(s):
char = s[0]
word = char
for num in range(1,len(s)):
if s[num] == char:
word += '*'
else:
word += s[num]
return word
# D. MixUp
# Given strings a and b, return a single string with a and b separated
# by a space '<a> <b>', except swap the first 2 chars of each string.
# e.g.
# 'mix', pod' -> 'pox mid'
# 'dog', 'dinner' -> 'dig donner'
# Assume a and b are length 2 or more.
def mix_up(a, b):
return b[0:2]+a[2:]+ ' ' +a[0:2]+b[2:]
# Provided simple test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Provided main() calls the above functions with interesting inputs,
# using test() to check if each result is correct or not.
def main():
print 'donuts'
# Each line calls donuts, compares its result to the expected for that call.
test(donuts(4), 'Number of donuts: 4')
test(donuts(9), 'Number of donuts: 9')
test(donuts(10), 'Number of donuts: many')
test(donuts(99), 'Number of donuts: many')
print
print 'both_ends'
test(both_ends('spring'), 'spng')
test(both_ends('Hello'), 'Helo')
test(both_ends('a'), '')
test(both_ends('xyz'), 'xyyz')
print
print 'fix_start'
test(fix_start('babble'), 'ba**le')
test(fix_start('aardvark'), 'a*rdv*rk')
test(fix_start('google'), 'goo*le')
test(fix_start('donut'), 'donut')
print
print 'mix_up'
test(mix_up('mix', 'pod'), 'pox mid')
test(mix_up('dog', 'dinner'), 'dig donner')
test(mix_up('gnash', 'sport'), 'spash gnort')
test(mix_up('pezzy', 'firm'), 'fizzy perm')
# Standard boilerplate to call the main() function.
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic list exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in list2.py.
# A. match_ends
# Given a list of strings, return the count of the number of
# strings where the string length is 2 or more and the first
# and last chars of the string are the same.
# Note: python does not have a ++ operator, but += works.
def match_ends(words):
count = 0
for word in words:
wordlen = len(word)
if wordlen>1 and word[0]==word[wordlen-1]:
count+=1
return count
# B. front_x
# Given a list of strings, return a list with the strings
# in sorted order, except group all the strings that begin with 'x' first.
# e.g. ['mix', 'xyz', 'apple', 'xanadu', 'aardvark'] yields
# ['xanadu', 'xyz', 'aardvark', 'apple', 'mix']
# Hint: this can be done by making 2 lists and sorting each of them
# before combining them.
def front_x(words):
x_list = []
other_list = []
for word in words:
if word[0]=='x':
x_list.append(word)
else:
other_list.append(word)
x_list.sort()
other_list.sort()
x_list.extend(other_list)
# +++your code here+++
return x_list
# C. sort_last
# Given a list of non-empty tuples, return a list sorted in increasing
# order by the last element in each tuple.
# e.g. [(1, 7), (1, 3), (3, 4, 5), (2, 2)] yields
# [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
# Hint: use a custom key= function to extract the last element form each tuple.
def sort_last(tuples):
return sorted(tuples,key=last)
def last(tuple):
last = tuple[len(tuple)-1]
return last
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'match_ends'
test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)
print
print 'front_x'
test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])
print
print 'sort_last'
test(sort_last([(1, 3), (3, 2), (2, 1)]),
[(2, 1), (3, 2), (1, 3)])
test(sort_last([(2, 3), (1, 2), (3, 1)]),
[(3, 1), (1, 2), (2, 3)])
test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
[(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic list exercises
# D. Given a list of numbers, return a list where
# all adjacent == elements have been reduced to a single element,
# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
# modify the passed in list.
def remove_adjacent(nums):
if nums == []:
return nums
last = nums[0]
mlist = []
mlist.append(last)
for i in nums[1:]:
if i!= last:
mlist.append(i)
last = i
return mlist
# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
list1.extend(list2)
list1.sort()
return list1
# Note: the solution above is kind of cute, but unforunately list.pop(0)
# is not constant time with the standard python list implementation, so
# the above is not strictly linear time.
# An alternate approach uses pop(-1) to remove the endmost elements
# from each list, building a solution list which is backwards.
# Then use reversed() to put the result back in the correct order. That
# solution works in linear time, but is more ugly.
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'remove_adjacent'
test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
test(remove_adjacent([]), [])
print
print 'linear_merge'
test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
['aa', 'aa', 'aa', 'bb', 'bb'])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Wordcount exercise
Google's Python class
The main() below is already defined and complete. It calls print_words()
and print_top() functions which you write.
1. For the --count flag, implement a print_words(filename) function that counts
how often each word appears in the text and prints:
word1 count1
word2 count2
...
Print the above list in order sorted by word (python will sort punctuation to
come before letters -- that's fine). Store all the words as lowercase,
so 'The' and 'the' count as the same word.
2. For the --topcount flag, implement a print_top(filename) which is similar
to print_words() but which prints just the top 20 most common words sorted
so the most common word is first, then the next most common, and so on.
Use str.split() (no arguments) to split on all whitespace.
Workflow: don't build the whole program at once. Get it to an intermediate
milestone and print your data structure and sys.exit(0).
When that's working, try for the next milestone.
Optional: define a helper function to avoid code duplication inside
print_words() and print_top().
"""
import sys
import re
# +++your code here+++
# Define print_words(filename) and print_top(filename) functions.
# You could write a helper utility function that reads a file
# and builds and returns a word/count dict for it.
# Then print_words() and print_top() can just call the utility function.
def read_file(filename):
f=open(filename,'r')
text = f.read()
text = re.sub("[/.,\';:?!-]","",text)
text = text.lower()
words = text.split()
wordsMap = {}
for word in words:
if word in wordsMap:
wordsMap[word] +=1
else:
wordsMap[word] = 1
return wordsMap
def print_words(filename):
wordsMap = read_file(filename)
keys=sorted(wordsMap.keys())
for key in keys:
print key, wordsMap[key]
return
def print_top(filename):
wordsMap = read_file(filename)
items=sorted(wordsMap.items(),key=lambda (k, v): v,reverse=True)
for item in items[:20]:
print item[0], item[1]
###
# This basic command line argument parsing code is provided and
# calls the print_words() and print_top() functions which you must define.
def main():
if len(sys.argv) != 3:
print 'usage: ./wordcount.py {--count | --topcount} file'
sys.exit(1)
option = sys.argv[1]
filename = sys.argv[2]
if option == '--count':
print_words(filename)
elif option == '--topcount':
print_top(filename)
else:
print 'unknown option: ' + option
sys.exit(1)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python2.4 -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic string exercises
# D. verbing
# Given a string, if its length is at least 3,
# add 'ing' to its end.
# Unless it already ends in 'ing', in which case
# add 'ly' instead.
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
if len(s)<3:
return s
elif s[-3:] == 'ing':
return s + 'ly'
else:
return s + 'ing'
# E. not_bad
# Given a string, find the first appearance of the
# substring 'not' and 'bad'. If the 'bad' follows
# the 'not', replace the whole 'not'...'bad' substring
# with 'good'.
# Return the resulting string.
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
if s.find('bad') > s.find('not'):
return s.replace(s[s.find('not'):s.find('bad')+3], 'good')
else:
return s
# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back
def front_back(a, b):
af = (len(a)+1)/2
bf = (len(b)+1)/2
return a[0:af] + b[0:bf] + a[af:] + b[bf:]
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# main() calls the above functions with interesting inputs,
# using the above test() to check if the result is correct or not.
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic list exercises
# D. Given a list of numbers, return a list where
# all adjacent == elements have been reduced to a single element,
# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
# modify the passed in list.
def remove_adjacent(nums):
if nums == []:
return nums
last = nums[0]
mlist = []
mlist.append(last)
for i in nums[1:]:
if i!= last:
mlist.append(i)
last = i
return mlist
# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
list1.extend(list2)
list1.sort()
return list1
# Note: the solution above is kind of cute, but unforunately list.pop(0)
# is not constant time with the standard python list implementation, so
# the above is not strictly linear time.
# An alternate approach uses pop(-1) to remove the endmost elements
# from each list, building a solution list which is backwards.
# Then use reversed() to put the result back in the correct order. That
# solution works in linear time, but is more ugly.
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'remove_adjacent'
test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
test(remove_adjacent([]), [])
print
print 'linear_merge'
test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
['aa', 'aa', 'aa', 'bb', 'bb'])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Wordcount exercise
Google's Python class
The main() below is already defined and complete. It calls print_words()
and print_top() functions which you write.
1. For the --count flag, implement a print_words(filename) function that counts
how often each word appears in the text and prints:
word1 count1
word2 count2
...
Print the above list in order sorted by word (python will sort punctuation to
come before letters -- that's fine). Store all the words as lowercase,
so 'The' and 'the' count as the same word.
2. For the --topcount flag, implement a print_top(filename) which is similar
to print_words() but which prints just the top 20 most common words sorted
so the most common word is first, then the next most common, and so on.
Use str.split() (no arguments) to split on all whitespace.
Workflow: don't build the whole program at once. Get it to an intermediate
milestone and print your data structure and sys.exit(0).
When that's working, try for the next milestone.
Optional: define a helper function to avoid code duplication inside
print_words() and print_top().
"""
import sys
import re
# +++your code here+++
# Define print_words(filename) and print_top(filename) functions.
# You could write a helper utility function that reads a file
# and builds and returns a word/count dict for it.
# Then print_words() and print_top() can just call the utility function.
def read_file(filename):
f=open(filename,'r')
text = f.read()
text = re.sub("[/.,\';:?!-]","",text)
text = text.lower()
words = text.split()
wordsMap = {}
for word in words:
if word in wordsMap:
wordsMap[word] +=1
else:
wordsMap[word] = 1
return wordsMap
def print_words(filename):
wordsMap = read_file(filename)
keys=sorted(wordsMap.keys())
for key in keys:
print key, wordsMap[key]
return
def print_top(filename):
wordsMap = read_file(filename)
items=sorted(wordsMap.items(),key=lambda (k, v): v,reverse=True)
for item in items[:20]:
print item[0], item[1]
###
# This basic command line argument parsing code is provided and
# calls the print_words() and print_top() functions which you must define.
def main():
if len(sys.argv) != 3:
print 'usage: ./wordcount.py {--count | --topcount} file'
sys.exit(1)
option = sys.argv[1]
filename = sys.argv[2]
if option == '--count':
print_words(filename)
elif option == '--topcount':
print_top(filename)
else:
print 'unknown option: ' + option
sys.exit(1)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python2.4 -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic string exercises
# D. verbing
# Given a string, if its length is at least 3,
# add 'ing' to its end.
# Unless it already ends in 'ing', in which case
# add 'ly' instead.
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
if len(s)<3:
return s
elif s[-3:] == 'ing':
return s + 'ly'
else:
return s + 'ing'
# E. not_bad
# Given a string, find the first appearance of the
# substring 'not' and 'bad'. If the 'bad' follows
# the 'not', replace the whole 'not'...'bad' substring
# with 'good'.
# Return the resulting string.
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
if s.find('bad') > s.find('not'):
return s.replace(s[s.find('not'):s.find('bad')+3], 'good')
else:
return s
# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back
def front_back(a, b):
af = (len(a)+1)/2
bf = (len(b)+1)/2
return a[0:af] + b[0:bf] + a[af:] + b[bf:]
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# main() calls the above functions with interesting inputs,
# using the above test() to check if the result is correct or not.
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic list exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in list2.py.
# A. match_ends
# Given a list of strings, return the count of the number of
# strings where the string length is 2 or more and the first
# and last chars of the string are the same.
# Note: python does not have a ++ operator, but += works.
def match_ends(words):
# +++your code here+++
n=0
for i in range (len(words)):
if len(words[i])>=2 and words[i][0]==words[i][-1:]:
n+=1
return n
# B. front_x
# Given a list of strings, return a list with the strings
# in sorted order, except group all the strings that begin with 'x' first.
# e.g. ['mix', 'xyz', 'apple', 'xanadu', 'aardvark'] yields
# ['xanadu', 'xyz', 'aardvark', 'apple', 'mix']
# Hint: this can be done by making 2 lists and sorting each of them
# before combining them.
def front_x(words):
# +++your code here+++
list1=[]
list2=[]
for i in range (len(words)):
if words[i].startswith('x'):
list1.append(words[i])
else:
list2.append(words[i])
return sorted(list1)+sorted(list2)
return
# C. sort_last
# Given a list of non-empty tuples, return a list sorted in increasing
# order by the last element in each tuple.
# e.g. [(1, 7), (1, 3), (3, 4, 5), (2, 2)] yields
# [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
# Hint: use a custom key= function to extract the last element form each tuple.
def sort_last(tuples):
# +++your code here+++
return sorted(tuples, key=lambda tuples:tuples[-1:])
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'match_ends'
test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)
print
print 'front_x'
test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])
print
print 'sort_last'
test(sort_last([(1, 3), (3, 2), (2, 1)]),
[(2, 1), (3, 2), (1, 3)])
test(sort_last([(2, 3), (1, 2), (3, 1)]),
[(3, 1), (1, 2), (2, 3)])
test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
[(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
main()
input()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Mimic pyquick exercise -- optional extra exercise.
Google's Python Class
Read in the file specified on the command line.
Do a simple split() on whitespace to obtain all the words in the file.
Rather than read the file line by line, it's easier to read
it into one giant string and split it once.
Build a "mimic" dict that maps each word that appears in the file
to a list of all the words that immediately follow that word in the file.
The list of words can be be in any order and should include
duplicates. So for example the key "and" might have the list
["then", "best", "then", "after", ...] listing
all the words which came after "and" in the text.
We'll say that the empty string is what comes before
the first word in the file.
With the mimic dict, it's fairly easy to emit random
text that mimics the original. Print a word, then look
up what words might come next and pick one at random as
the next work.
Use the empty string as the first word to prime things.
If we ever get stuck with a word that is not in the dict,
go back to the empty string to keep things moving.
Note: the standard python module 'random' includes a
random.choice(list) method which picks a random element
from a non-empty list.
For fun, feed your program to itself as input.
Could work on getting it to put in linebreaks around 70
columns, so the output looks better.
"""
import random
import sys
import re
def mimic_dict(filename):
"""Returns mimic dict mapping each word to list of words which follow it."""
# +++your code here+++
f=open(filename,'r')
text=f.read()
text=re.sub("[(/.,!?:-;`)]","",text)
lines=text.lower().split()
#if len(lines)==0:
#return 0
mimic={"":lines[0]}
suf=[]
for i in range(len(lines)):
if mimic.get(lines[i])==None:
j=i
for j in range(len(lines)-1):
if lines[i]==lines[j]:
suf.append(lines[j+1])
mimic.update({lines[i]:suf})
suf=[]
#print(mimic)
return mimic
def print_mimic(mimic_dict, word):
"""Given mimic dict and start word, prints 200 random words."""
# +++your code here+++
if mimic_dict==0:
print "File is empty"
return
for i in range(200):
if word=='' or mimic_dict.get(word)==[]:
word=mimic_dict.get('')
sys.stdout.write("%s " %(word))
else:
word=random.choice(mimic_dict.get(word))
sys.stdout.write("%s " %(word))
return
# Provided main(), calls mimic_dict() and mimic()
def main():
if len(sys.argv) != 2:
print 'usage: ./mimic.py file-to-read'
sys.exit(1)
dict = mimic_dict(sys.argv[1])
print_mimic(dict, '')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic string exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in string2.py.
# A. donuts
# Given an int count of a number of donuts, return a string
# of the form 'Number of donuts: <count>', where <count> is the number
# passed in. However, if the count is 10 or more, then use the word 'many'
# instead of the actual count.
# So donuts(5) returns 'Number of donuts: 5'
# and donuts(23) returns 'Number of donuts: many'
def donuts(count):
# +++your code here+++
if count<0:
return 'err'
elif count<10:
return 'Number of donuts: %d' %(count)
else:
return 'Number of donuts: many'
# B. both_ends
# Given a string s, return a string made of the first 2
# and the last 2 chars of the original string,
# so 'spring' yields 'spng'. However, if the string length
# is less than 2, return instead the empty string.
def both_ends(s):
# +++your code here+++
if len(s)<2:
return ''
else:
return s[0:2]+s[-2:]
# C. fix_start
# Given a string s, return a string
# where all occurences of its first char have
# been changed to '*', except do not change
# the first char itself.
# e.g. 'babble' yields 'ba**le'
# Assume that the string is length 1 or more.
# Hint: s.replace(stra, strb) returns a version of string s
# where all instances of stra have been replaced by strb.
def fix_start(s):
# +++your code here+++
if len(s)<1:
return s
else:
return s[0]+s[1:].replace(s[0],'*')
# D. MixUp
# Given strings a and b, return a single string with a and b separated
# by a space '<a> <b>', except swap the first 2 chars of each string.
# e.g.
# 'mix', pod' -> 'pox mid'
# 'dog', 'dinner' -> 'dig donner'
# Assume a and b are length 2 or more.
def mix_up(a, b):
if len(a)<2 or len(b)<2:
return 'Err'
else:
return '%s %s' %(a.replace(a[0:2],b[0:2]),b.replace(b[0:2],a[0:2]))
# Provided simple test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Provided main() calls the above functions with interesting inputs,
# using test() to check if each result is correct or not.
def main():
print 'donuts'
# Each line calls donuts, compares its result to the expected for that call.
test(donuts(4), 'Number of donuts: 4')
test(donuts(9), 'Number of donuts: 9')
test(donuts(10), 'Number of donuts: many')
test(donuts(99), 'Number of donuts: many')
print
print 'both_ends'
test(both_ends('spring'), 'spng')
test(both_ends('Hello'), 'Helo')
test(both_ends('a'), '')
test(both_ends('xyz'), 'xyyz')
print
print 'fix_start'
test(fix_start('babble'), 'ba**le')
test(fix_start('aardvark'), 'a*rdv*rk')
test(fix_start('google'), 'goo*le')
test(fix_start('donut'), 'donut')
print
print 'mix_up'
test(mix_up('mix', 'pod'), 'pox mid')
test(mix_up('dog', 'dinner'), 'dig donner')
test(mix_up('gnash', 'sport'), 'spash gnort')
test(mix_up('pezzy', 'firm'), 'fizzy perm')
# Standard boilerplate to call the main() function.
if __name__ == '__main__':
main()
input()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic string exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in string2.py.
# A. donuts
# Given an int count of a number of donuts, return a string
# of the form 'Number of donuts: <count>', where <count> is the number
# passed in. However, if the count is 10 or more, then use the word 'many'
# instead of the actual count.
# So donuts(5) returns 'Number of donuts: 5'
# and donuts(23) returns 'Number of donuts: many'
def donuts(count):
# +++your code here+++
if count<0:
return 'err'
elif count<10:
return 'Number of donuts: %d' %(count)
else:
return 'Number of donuts: many'
# B. both_ends
# Given a string s, return a string made of the first 2
# and the last 2 chars of the original string,
# so 'spring' yields 'spng'. However, if the string length
# is less than 2, return instead the empty string.
def both_ends(s):
# +++your code here+++
if len(s)<2:
return ''
else:
return s[0:2]+s[-2:]
# C. fix_start
# Given a string s, return a string
# where all occurences of its first char have
# been changed to '*', except do not change
# the first char itself.
# e.g. 'babble' yields 'ba**le'
# Assume that the string is length 1 or more.
# Hint: s.replace(stra, strb) returns a version of string s
# where all instances of stra have been replaced by strb.
def fix_start(s):
# +++your code here+++
if len(s)<1:
return s
else:
return s[0]+s[1:].replace(s[0],'*')
# D. MixUp
# Given strings a and b, return a single string with a and b separated
# by a space '<a> <b>', except swap the first 2 chars of each string.
# e.g.
# 'mix', pod' -> 'pox mid'
# 'dog', 'dinner' -> 'dig donner'
# Assume a and b are length 2 or more.
def mix_up(a, b):
if len(a)<2 or len(b)<2:
return 'Err'
else:
return '%s %s' %(a.replace(a[0:2],b[0:2]),b.replace(b[0:2],a[0:2]))
# Provided simple test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Provided main() calls the above functions with interesting inputs,
# using test() to check if each result is correct or not.
def main():
print 'donuts'
# Each line calls donuts, compares its result to the expected for that call.
test(donuts(4), 'Number of donuts: 4')
test(donuts(9), 'Number of donuts: 9')
test(donuts(10), 'Number of donuts: many')
test(donuts(99), 'Number of donuts: many')
print
print 'both_ends'
test(both_ends('spring'), 'spng')
test(both_ends('Hello'), 'Helo')
test(both_ends('a'), '')
test(both_ends('xyz'), 'xyyz')
print
print 'fix_start'
test(fix_start('babble'), 'ba**le')
test(fix_start('aardvark'), 'a*rdv*rk')
test(fix_start('google'), 'goo*le')
test(fix_start('donut'), 'donut')
print
print 'mix_up'
test(mix_up('mix', 'pod'), 'pox mid')
test(mix_up('dog', 'dinner'), 'dig donner')
test(mix_up('gnash', 'sport'), 'spash gnort')
test(mix_up('pezzy', 'firm'), 'fizzy perm')
# Standard boilerplate to call the main() function.
if __name__ == '__main__':
main()
input()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic list exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in list2.py.
# A. match_ends
# Given a list of strings, return the count of the number of
# strings where the string length is 2 or more and the first
# and last chars of the string are the same.
# Note: python does not have a ++ operator, but += works.
def match_ends(words):
# +++your code here+++
n=0
for i in range (len(words)):
if len(words[i])>=2 and words[i][0]==words[i][-1:]:
n+=1
return n
# B. front_x
# Given a list of strings, return a list with the strings
# in sorted order, except group all the strings that begin with 'x' first.
# e.g. ['mix', 'xyz', 'apple', 'xanadu', 'aardvark'] yields
# ['xanadu', 'xyz', 'aardvark', 'apple', 'mix']
# Hint: this can be done by making 2 lists and sorting each of them
# before combining them.
def front_x(words):
# +++your code here+++
list1=[]
list2=[]
for i in range (len(words)):
if words[i].startswith('x'):
list1.append(words[i])
else:
list2.append(words[i])
return sorted(list1)+sorted(list2)
return
# C. sort_last
# Given a list of non-empty tuples, return a list sorted in increasing
# order by the last element in each tuple.
# e.g. [(1, 7), (1, 3), (3, 4, 5), (2, 2)] yields
# [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
# Hint: use a custom key= function to extract the last element form each tuple.
def sort_last(tuples):
# +++your code here+++
return sorted(tuples, key=lambda tuples:tuples[-1:])
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'match_ends'
test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)
print
print 'front_x'
test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])
print
print 'sort_last'
test(sort_last([(1, 3), (3, 2), (2, 1)]),
[(2, 1), (3, 2), (1, 3)])
test(sort_last([(2, 3), (1, 2), (3, 1)]),
[(3, 1), (1, 2), (2, 3)])
test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
[(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
main()
input()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic list exercises
# D. Given a list of numbers, return a list where
# all adjacent == elements have been reduced to a single element,
# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
# modify the passed in list.
def remove_adjacent(nums):
# +++your code here+++
i=1
newlist=[]
for i in range (len(nums)):
if nums[i]!=nums[i-1]:
newlist.append(nums[i])
return newlist
# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
# +++your code here+++
list1.extend(list2)
return sorted(list1)
# Note: the solution above is kind of cute, but unforunately list.pop(0)
# is not constant time with the standard python list implementation, so
# the above is not strictly linear time.
# An alternate approach uses pop(-1) to remove the endmost elements
# from each list, building a solution list which is backwards.
# Then use reversed() to put the result back in the correct order. That
# solution works in linear time, but is more ugly.
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'remove_adjacent'
test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
test(remove_adjacent([]), [])
print
print 'linear_merge'
test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
['aa', 'aa', 'aa', 'bb', 'bb'])
if __name__ == '__main__':
main()
input()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Wordcount exercise
Google's Python class
The main() below is already defined and complete. It calls print_words()
and print_top() functions which you write.
1. For the --count flag, implement a print_words(filename) function that counts
how often each word appears in the text and prints:
word1 count1
word2 count2
...
Print the above list in order sorted by word (python will sort punctuation to
come before letters -- that's fine). Store all the words as lowercase,
so 'The' and 'the' count as the same word.
2. For the --topcount flag, implement a print_top(filename) which is similar
to print_words() but which prints just the top 20 most common words sorted
so the most common word is first, then the next most common, and so on.
Use str.split() (no arguments) to split on all whitespace.
Workflow: don't build the whole program at once. Get it to an intermediate
milestone and print your data structure and sys.exit(0).
When that's working, try for the next milestone.
Optional: define a helper function to avoid code duplication inside
print_words() and print_top().
"""
import sys
import re
# +++your code here+++
def readfile(filename):
f=open(filename,'r')
text=f.read()
text=re.sub("[/.,!?;]", "",text)
lines=sorted(text.lower().split())
newlines=[]
i=1
#newlines.append(lines[0])
#newlines.append(lines.count(lines[0]))
for i in range(len(lines)):
if lines[i]!=lines[i-1]:
newlines.append(lines[i])
newlines.append(lines.count(lines[i]))
arr=[]
for i in range(len(newlines)/2):
arr.append(tuple(newlines[i*2:i*2+2]))
return arr
def print_words(filename):
arr=readfile(filename)
for i in range (len(arr)):
print arr[i][0],'\t',arr[i][1]
return
def print_top(filename):
arr=sorted(readfile(filename), key=lambda tuples:tuples[-1:])
for i in range (len(arr)):
print arr[i][0],'\t',arr[i][1]
return
# Define print_words(filename) and print_top(filename) functions.
# You could write a helper utility function that reads a file
# and builds and returns a word/count dict for it.
# Then print_words() and print_top() can just call the utility function.
###
# This basic command line argument parsing code is provided and
# calls the print_words() and print_top() functions which you must define.
def main():
if len(sys.argv) != 3:
print 'usage: ./wordcount.py {--count | --topcount} file'
sys.exit(1)
option = sys.argv[1]
filename = sys.argv[2]
if option == '--count':
print_words(filename)
elif option == '--topcount':
print_top(filename)
else:
print 'unknown option: ' + option
sys.exit(1)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Mimic pyquick exercise -- optional extra exercise.
Google's Python Class
Read in the file specified on the command line.
Do a simple split() on whitespace to obtain all the words in the file.
Rather than read the file line by line, it's easier to read
it into one giant string and split it once.
Build a "mimic" dict that maps each word that appears in the file
to a list of all the words that immediately follow that word in the file.
The list of words can be be in any order and should include
duplicates. So for example the key "and" might have the list
["then", "best", "then", "after", ...] listing
all the words which came after "and" in the text.
We'll say that the empty string is what comes before
the first word in the file.
With the mimic dict, it's fairly easy to emit random
text that mimics the original. Print a word, then look
up what words might come next and pick one at random as
the next work.
Use the empty string as the first word to prime things.
If we ever get stuck with a word that is not in the dict,
go back to the empty string to keep things moving.
Note: the standard python module 'random' includes a
random.choice(list) method which picks a random element
from a non-empty list.
For fun, feed your program to itself as input.
Could work on getting it to put in linebreaks around 70
columns, so the output looks better.
"""
import random
import sys
import re
def mimic_dict(filename):
"""Returns mimic dict mapping each word to list of words which follow it."""
# +++your code here+++
f=open(filename,'r')
text=f.read()
text=re.sub("[(/.,!?:-;`)]","",text)
lines=text.lower().split()
#if len(lines)==0:
#return 0
mimic={"":lines[0]}
suf=[]
for i in range(len(lines)):
if mimic.get(lines[i])==None:
j=i
for j in range(len(lines)-1):
if lines[i]==lines[j]:
suf.append(lines[j+1])
mimic.update({lines[i]:suf})
suf=[]
#print(mimic)
return mimic
def print_mimic(mimic_dict, word):
"""Given mimic dict and start word, prints 200 random words."""
# +++your code here+++
if mimic_dict==0:
print "File is empty"
return
for i in range(200):
if word=='' or mimic_dict.get(word)==[]:
word=mimic_dict.get('')
sys.stdout.write("%s " %(word))
else:
word=random.choice(mimic_dict.get(word))
sys.stdout.write("%s " %(word))
return
# Provided main(), calls mimic_dict() and mimic()
def main():
if len(sys.argv) != 2:
print 'usage: ./mimic.py file-to-read'
sys.exit(1)
dict = mimic_dict(sys.argv[1])
print_mimic(dict, '')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python2.4 -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic string exercises
# D. verbing
# Given a string, if its length is at least 3,
# add 'ing' to its end.
# Unless it already ends in 'ing', in which case
# add 'ly' instead.
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
# +++your code here+++
if len(s)<3:
return s
elif s[-3:]=='ing':
return s+'ly'
else:
return s+'ing'
# E. not_bad
# Given a string, find the first appearance of the
# substring 'not' and 'bad'. If the 'bad' follows
# the 'not', replace the whole 'not'...'bad' substring
# with 'good'.
# Return the resulting string.
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
# +++your code here+++
if s.find('bad')>s.find('not'):
return s.replace(s[s.find('not'):s.find('bad')+3],'good')
else:
return s
return
# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back
def front_back(a, b):
# +++your code here+++
if len(a)%2==0 and len(b)%2==0:
return a[0:len(a)/2]+b[0:len(b)/2]+a[-len(a)/2:]+b[-len(b)/2:]
if len(a)%2!=0 and len(b)%2==0:
return a[0:len(a)/2+1]+b[0:len(b)/2]+a[-len(a)/2+1:]+b[-len(b)/2:]
if len(a)%2==0 and len(b)%2!=0:
return a[0:len(a)/2]+b[0:len(b)/2+1]+a[-len(a)/2:]+b[-len(b)/2+1:]
if len(a)%2!=0 and len(b)%2!=0:
return a[0:len(a)/2+1]+b[0:len(b)/2+1]+a[-len(a)/2+1:]+b[-len(b)/2+1:]
return
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# main() calls the above functions with interesting inputs,
# using the above test() to check if the result is correct or not.
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
input()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic list exercises
# D. Given a list of numbers, return a list where
# all adjacent == elements have been reduced to a single element,
# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
# modify the passed in list.
def remove_adjacent(nums):
# +++your code here+++
i=1
newlist=[]
for i in range (len(nums)):
if nums[i]!=nums[i-1]:
newlist.append(nums[i])
return newlist
# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
# +++your code here+++
list1.extend(list2)
return sorted(list1)
# Note: the solution above is kind of cute, but unforunately list.pop(0)
# is not constant time with the standard python list implementation, so
# the above is not strictly linear time.
# An alternate approach uses pop(-1) to remove the endmost elements
# from each list, building a solution list which is backwards.
# Then use reversed() to put the result back in the correct order. That
# solution works in linear time, but is more ugly.
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'remove_adjacent'
test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
test(remove_adjacent([]), [])
print
print 'linear_merge'
test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
['aa', 'aa', 'aa', 'bb', 'bb'])
if __name__ == '__main__':
main()
input()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Wordcount exercise
Google's Python class
The main() below is already defined and complete. It calls print_words()
and print_top() functions which you write.
1. For the --count flag, implement a print_words(filename) function that counts
how often each word appears in the text and prints:
word1 count1
word2 count2
...
Print the above list in order sorted by word (python will sort punctuation to
come before letters -- that's fine). Store all the words as lowercase,
so 'The' and 'the' count as the same word.
2. For the --topcount flag, implement a print_top(filename) which is similar
to print_words() but which prints just the top 20 most common words sorted
so the most common word is first, then the next most common, and so on.
Use str.split() (no arguments) to split on all whitespace.
Workflow: don't build the whole program at once. Get it to an intermediate
milestone and print your data structure and sys.exit(0).
When that's working, try for the next milestone.
Optional: define a helper function to avoid code duplication inside
print_words() and print_top().
"""
import sys
import re
# +++your code here+++
def readfile(filename):
f=open(filename,'r')
text=f.read()
text=re.sub("[/.,!?;]", "",text)
lines=sorted(text.lower().split())
newlines=[]
i=1
#newlines.append(lines[0])
#newlines.append(lines.count(lines[0]))
for i in range(len(lines)):
if lines[i]!=lines[i-1]:
newlines.append(lines[i])
newlines.append(lines.count(lines[i]))
arr=[]
for i in range(len(newlines)/2):
arr.append(tuple(newlines[i*2:i*2+2]))
return arr
def print_words(filename):
arr=readfile(filename)
for i in range (len(arr)):
print arr[i][0],'\t',arr[i][1]
return
def print_top(filename):
arr=sorted(readfile(filename), key=lambda tuples:tuples[-1:])
for i in range (len(arr)):
print arr[i][0],'\t',arr[i][1]
return
# Define print_words(filename) and print_top(filename) functions.
# You could write a helper utility function that reads a file
# and builds and returns a word/count dict for it.
# Then print_words() and print_top() can just call the utility function.
###
# This basic command line argument parsing code is provided and
# calls the print_words() and print_top() functions which you must define.
def main():
if len(sys.argv) != 3:
print 'usage: ./wordcount.py {--count | --topcount} file'
sys.exit(1)
option = sys.argv[1]
filename = sys.argv[2]
if option == '--count':
print_words(filename)
elif option == '--topcount':
print_top(filename)
else:
print 'unknown option: ' + option
sys.exit(1)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python2.4 -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic string exercises
# D. verbing
# Given a string, if its length is at least 3,
# add 'ing' to its end.
# Unless it already ends in 'ing', in which case
# add 'ly' instead.
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
# +++your code here+++
if len(s)<3:
return s
elif s[-3:]=='ing':
return s+'ly'
else:
return s+'ing'
# E. not_bad
# Given a string, find the first appearance of the
# substring 'not' and 'bad'. If the 'bad' follows
# the 'not', replace the whole 'not'...'bad' substring
# with 'good'.
# Return the resulting string.
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
# +++your code here+++
if s.find('bad')>s.find('not'):
return s.replace(s[s.find('not'):s.find('bad')+3],'good')
else:
return s
return
# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back
def front_back(a, b):
# +++your code here+++
if len(a)%2==0 and len(b)%2==0:
return a[0:len(a)/2]+b[0:len(b)/2]+a[-len(a)/2:]+b[-len(b)/2:]
if len(a)%2!=0 and len(b)%2==0:
return a[0:len(a)/2+1]+b[0:len(b)/2]+a[-len(a)/2+1:]+b[-len(b)/2:]
if len(a)%2==0 and len(b)%2!=0:
return a[0:len(a)/2]+b[0:len(b)/2+1]+a[-len(a)/2:]+b[-len(b)/2+1:]
if len(a)%2!=0 and len(b)%2!=0:
return a[0:len(a)/2+1]+b[0:len(b)/2+1]+a[-len(a)/2+1:]+b[-len(b)/2+1:]
return
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# main() calls the above functions with interesting inputs,
# using the above test() to check if the result is correct or not.
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
input()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic list exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in list2.py.
# A. match_ends
# Given a list of strings, return the count of the number of
# strings where the string length is 2 or more and the first
# and last chars of the string are the same.
# Note: python does not have a ++ operator, but += works.
def match_ends(words):
# +++your code here+++
size=len(words);
i=0;
count=0;
while i < size:
if len(words[i]) >= 2:
if words[i][0] == words[i][-1]:
count=count+1;
i=i+1;
return count;
# B. front_x
# Given a list of strings, return a list with the strings
# in sorted order, except group all the strings that begin with 'x' first.
# e.g. ['mix', 'xyz', 'apple', 'xanadu', 'aardvark'] yields
# ['xanadu', 'xyz', 'aardvark', 'apple', 'mix']
# Hint: this can be done by making 2 lists and sorting each of them
# before combining them.
def front_x(words):
# +++your code here+++
words.sort();
size=len(words);
i=0;
list_x=[ ];
count=0;
while i < size:
if words[i][0] == 'x':
list_x.append(words[i]);
i=i+1;
index=words.index(list_x[0]);
newlist=list_x+words[0:index];
return newlist;
# C. sort_last
# Given a list of non-empty tuples, return a list sorted in increasing
# order by the last element in each tuple.
# e.g. [(1, 7), (1, 3), (3, 4, 5), (2, 2)] yields
# [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
# Hint: use a custom key= function to extract the last element form each tuple.
def sortfunc(tuples):
return tuples[-1];
def sort_last(tuples):
# +++your code here+++
tuples.sort(key=sortfunc);
return tuples;
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'match_ends'
test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)
print
print 'front_x'
test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])
print
print 'sort_last'
test(sort_last([(1, 3), (3, 2), (2, 1)]),
[(2, 1), (3, 2), (1, 3)])
test(sort_last([(2, 3), (1, 2), (3, 1)]),
[(3, 1), (1, 2), (2, 3)])
test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
[(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic string exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in string2.py.
# A. donuts
# Given an int count of a number of donuts, return a string
# of the form 'Number of donuts: <count>', where <count> is the number
# passed in. However, if the count is 10 or more, then use the word 'many'
# instead of the actual count.
# So donuts(5) returns 'Number of donuts: 5'
# and donuts(23) returns 'Number of donuts: many'
def donuts(count):
# +++your code here+++
string='Number of donuts: ';
if count < 10:
string=string+str(count);
return string;
else:
string=string+'many';
return string;
# B. both_ends
# Given a string s, return a string made of the first 2
# and the last 2 chars of the original string,
# so 'spring' yields 'spng'. However, if the string length
# is less than 2, return instead the empty string.
def both_ends(s):
# +++your code here+++
if len(s) <= 2:
return '';
else:
string=s[0:2]+s[-2: ];
return string;
# C. fix_start
# Given a string s, return a string
# where all occurences of its first char have
# been changed to '*', except do not change
# the first char itself.
# e.g. 'babble' yields 'ba**le'
# Assume that the string is length 1 or more.
# Hint: s.replace(stra, strb) returns a version of string s
# where all instances of stra have been replaced by strb.
def fix_start(s):
# +++your code here+++
letter=s[0];
newstring=s[1:].replace(letter,'*');
newstring=letter+newstring;
return newstring;
# D. MixUp
# Given strings a and b, return a single string with a and b separated
# by a space '<a> <b>', except swap the first 2 chars of each string.
# e.g.
# 'mix', pod' -> 'pox mid'
# 'dog', 'dinner' -> 'dig donner'
# Assume a and b are length 2 or more.
def mix_up(a, b):
# +++your code here+++
newstring=b[0:2]+a[2:]+' '+a[0:2]+b[2:];
return newstring;
# Provided simple test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Provided main() calls the above functions with interesting inputs,
# using test() to check if each result is correct or not.
def main():
print 'donuts'
# Each line calls donuts, compares its result to the expected for that call.
test(donuts(4), 'Number of donuts: 4')
test(donuts(9), 'Number of donuts: 9')
test(donuts(10), 'Number of donuts: many')
test(donuts(99), 'Number of donuts: many')
print
print 'both_ends'
test(both_ends('spring'), 'spng')
test(both_ends('Hello'), 'Helo')
test(both_ends('a'), '')
test(both_ends('xyz'), 'xyyz')
print
print 'fix_start'
test(fix_start('babble'), 'ba**le')
test(fix_start('aardvark'), 'a*rdv*rk')
test(fix_start('google'), 'goo*le')
test(fix_start('donut'), 'donut')
print
print 'mix_up'
test(mix_up('mix', 'pod'), 'pox mid')
test(mix_up('dog', 'dinner'), 'dig donner')
test(mix_up('gnash', 'sport'), 'spash gnort')
test(mix_up('pezzy', 'firm'), 'fizzy perm')
# Standard boilerplate to call the main() function.
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic string exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in string2.py.
# A. donuts
# Given an int count of a number of donuts, return a string
# of the form 'Number of donuts: <count>', where <count> is the number
# passed in. However, if the count is 10 or more, then use the word 'many'
# instead of the actual count.
# So donuts(5) returns 'Number of donuts: 5'
# and donuts(23) returns 'Number of donuts: many'
def donuts(count):
# +++your code here+++
string='Number of donuts: ';
if count < 10:
string=string+str(count);
return string;
else:
string=string+'many';
return string;
# B. both_ends
# Given a string s, return a string made of the first 2
# and the last 2 chars of the original string,
# so 'spring' yields 'spng'. However, if the string length
# is less than 2, return instead the empty string.
def both_ends(s):
# +++your code here+++
if len(s) <= 2:
return '';
else:
string=s[0:2]+s[-2: ];
return string;
# C. fix_start
# Given a string s, return a string
# where all occurences of its first char have
# been changed to '*', except do not change
# the first char itself.
# e.g. 'babble' yields 'ba**le'
# Assume that the string is length 1 or more.
# Hint: s.replace(stra, strb) returns a version of string s
# where all instances of stra have been replaced by strb.
def fix_start(s):
# +++your code here+++
letter=s[0];
newstring=s[1:].replace(letter,'*');
newstring=letter+newstring;
return newstring;
# D. MixUp
# Given strings a and b, return a single string with a and b separated
# by a space '<a> <b>', except swap the first 2 chars of each string.
# e.g.
# 'mix', pod' -> 'pox mid'
# 'dog', 'dinner' -> 'dig donner'
# Assume a and b are length 2 or more.
def mix_up(a, b):
# +++your code here+++
newstring=b[0:2]+a[2:]+' '+a[0:2]+b[2:];
return newstring;
# Provided simple test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Provided main() calls the above functions with interesting inputs,
# using test() to check if each result is correct or not.
def main():
print 'donuts'
# Each line calls donuts, compares its result to the expected for that call.
test(donuts(4), 'Number of donuts: 4')
test(donuts(9), 'Number of donuts: 9')
test(donuts(10), 'Number of donuts: many')
test(donuts(99), 'Number of donuts: many')
print
print 'both_ends'
test(both_ends('spring'), 'spng')
test(both_ends('Hello'), 'Helo')
test(both_ends('a'), '')
test(both_ends('xyz'), 'xyyz')
print
print 'fix_start'
test(fix_start('babble'), 'ba**le')
test(fix_start('aardvark'), 'a*rdv*rk')
test(fix_start('google'), 'goo*le')
test(fix_start('donut'), 'donut')
print
print 'mix_up'
test(mix_up('mix', 'pod'), 'pox mid')
test(mix_up('dog', 'dinner'), 'dig donner')
test(mix_up('gnash', 'sport'), 'spash gnort')
test(mix_up('pezzy', 'firm'), 'fizzy perm')
# Standard boilerplate to call the main() function.
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic list exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in list2.py.
# A. match_ends
# Given a list of strings, return the count of the number of
# strings where the string length is 2 or more and the first
# and last chars of the string are the same.
# Note: python does not have a ++ operator, but += works.
def match_ends(words):
# +++your code here+++
size=len(words);
i=0;
count=0;
while i < size:
if len(words[i]) >= 2:
if words[i][0] == words[i][-1]:
count=count+1;
i=i+1;
return count;
# B. front_x
# Given a list of strings, return a list with the strings
# in sorted order, except group all the strings that begin with 'x' first.
# e.g. ['mix', 'xyz', 'apple', 'xanadu', 'aardvark'] yields
# ['xanadu', 'xyz', 'aardvark', 'apple', 'mix']
# Hint: this can be done by making 2 lists and sorting each of them
# before combining them.
def front_x(words):
# +++your code here+++
words.sort();
size=len(words);
i=0;
list_x=[ ];
count=0;
while i < size:
if words[i][0] == 'x':
list_x.append(words[i]);
i=i+1;
index=words.index(list_x[0]);
newlist=list_x+words[0:index];
return newlist;
# C. sort_last
# Given a list of non-empty tuples, return a list sorted in increasing
# order by the last element in each tuple.
# e.g. [(1, 7), (1, 3), (3, 4, 5), (2, 2)] yields
# [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
# Hint: use a custom key= function to extract the last element form each tuple.
def sortfunc(tuples):
return tuples[-1];
def sort_last(tuples):
# +++your code here+++
tuples.sort(key=sortfunc);
return tuples;
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'match_ends'
test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)
print
print 'front_x'
test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])
print
print 'sort_last'
test(sort_last([(1, 3), (3, 2), (2, 1)]),
[(2, 1), (3, 2), (1, 3)])
test(sort_last([(2, 3), (1, 2), (3, 1)]),
[(3, 1), (1, 2), (2, 3)])
test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
[(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic list exercises
# D. Given a list of numbers, return a list where
# all adjacent == elements have been reduced to a single element,
# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
# modify the passed in list.
def remove_adjacent(nums):
# +++your code here+++
size=len(nums);
if size == 0:
return nums;
i=0;
j=0;
count=0;
size_lst=0;
lst=[];
lst.append(nums[0]);
while i< size:
size_lst=len(lst);
while j < size_lst:
if nums[i] != lst[j]:
count=count+1;
j=j+1;
if count == size_lst:
lst.append(nums[i]);
count=0;
j=0;
size_lst=0;
i=i+1;
return lst;
# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
# +++your code here+++
list1=list1+list2;
list1.sort();
return list1;
# Note: the solution above is kind of cute, but unforunately list.pop(0)
# is not constant time with the standard python list implementation, so
# the above is not strictly linear time.
# An alternate approach uses pop(-1) to remove the endmost elements
# from each list, building a solution list which is backwards.
# Then use reversed() to put the result back in the correct order. That
# solution works in linear time, but is more ugly.
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'remove_adjacent'
test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
test(remove_adjacent([]), [])
print
print 'linear_merge'
test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
['aa', 'aa', 'aa', 'bb', 'bb'])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Wordcount exercise
Google's Python class
The main() below is already defined and complete. It calls print_words()
and print_top() functions which you write.
1. For the --count flag, implement a print_words(filename) function that counts
how often each word appears in the text and prints:
word1 count1
word2 count2
...
Print the above list in order sorted by word (python will sort punctuation to
come before letters -- that's fine). Store all the words as lowercase,
so 'The' and 'the' count as the same word.
2. For the --topcount flag, implement a print_top(filename) which is similar
to print_words() but which prints just the top 20 most common words sorted
so the most common word is first, then the next most common, and so on.
Use str.split() (no arguments) to split on all whitespace.
Workflow: don't build the whole program at once. Get it to an intermediate
milestone and print your data structure and sys.exit(0).
When that's working, try for the next milestone.
Optional: define a helper function to avoid code duplication inside
print_words() and print_top().
"""
import sys
# +++your code here+++
# Define print_words(filename) and print_top(filename) functions.
# You could write a helper utility function that reads a file
# and builds and returns a word/count dict for it.
# Then print_words() and print_top() can just call the utility function.
def ReadFile(filename):
line="a";
array=[];
words=[[0,""]];
i=0;
j=0;
count=0;
fl=open(filename);
while line != "":
line=fl.readline();
line=line.lower();
array=line.split();
size_array=len(array);
size_words=len(words);
while i<size_array:
while j<size_words:
if array[i] != words[j][1]:
count=count+1;
else:
words[j][0]=words[j][0]+1;
count=0;
break;
j=j+1;
if count == size_words:
words.append([1,array[i]]);
size_words=len(words);
count=0;
j=0;
i=i+1;
i=0;
j=0;
fl.close();
i=0;
words.remove([0,""]);
words.sort(reverse=True);
return words;
def print_words(filename):
i=0;
words=ReadFile(filename);
size_words=len(words);
while i< size_words:
print '{0} {1}'.format(words[i][1],words[i][0]);
i=i+1;
def print_top(filename):
i=0;
words=ReadFile(filename);
size_words=len(words);
if size_words < 20:
while i< size_words:
print '{0} {1}'.format(words[i][1],words[i][0]);
i=i+1;
else:
while i< 20:
print '{0} {1}'.format(words[i][1],words[i][0]);
i=i+1;
###
# This basic command line argument parsing code is provided and
# calls the print_words() and print_top() functions which you must define.
def main():
if len(sys.argv) != 3:
print 'usage: ./wordcount.py {--count | --topcount} file'
sys.exit(1)
option = sys.argv[1]
filename = sys.argv[2]
if option == '--count':
print_words(filename)
elif option == '--topcount':
print_top(filename)
else:
print 'unknown option: ' + option
sys.exit(1)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic string exercises
# D. verbing
# Given a string, if its length is at least 3,
# add 'ing' to its end.
# Unless it already ends in 'ing', in which case
# add 'ly' instead.
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
# +++your code here+++
if len(s) >= 3:
if s[-3:] == 'ing':
newstring=s+'ly';
return newstring;
else:
newstring=s+'ing';
return newstring;
else:
return s;
# E. not_bad
# Given a string, find the first appearance of the
# substring 'not' and 'bad'. If the 'bad' follows
# the 'not', replace the whole 'not'...'bad' substring
# with 'good'.
# Return the resulting string.
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
# +++your code here+++
if s.find('not') != -1:
first=s.find('not');
if s.find('bad') != -1:
second=s.find('bad');
if first <= second:
newstring=s.replace(s[first:second+3],'good');
return newstring;
else:
return s;
else:
return s;
else:
return s;
# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back
def front_back(a, b):
# +++your code here+++
if len(a)%2 == 0:
a_front=a[0:len(a)/2];
a_back=a[len(a)/2:];
else:
a_front=a[0:len(a)/2+1];
a_back=a[len(a)/2+1:];
if len(b)%2 == 0:
b_front=b[0:len(b)/2];
b_back=b[len(b)/2:];
else:
b_front=b[0:len(b)/2+1];
b_back=b[len(b)/2+1:];
return a_front+b_front+a_back+b_back;
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# main() calls the above functions with interesting inputs,
# using the above test() to check if the result is correct or not.
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic list exercises
# D. Given a list of numbers, return a list where
# all adjacent == elements have been reduced to a single element,
# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
# modify the passed in list.
def remove_adjacent(nums):
# +++your code here+++
size=len(nums);
if size == 0:
return nums;
i=0;
j=0;
count=0;
size_lst=0;
lst=[];
lst.append(nums[0]);
while i< size:
size_lst=len(lst);
while j < size_lst:
if nums[i] != lst[j]:
count=count+1;
j=j+1;
if count == size_lst:
lst.append(nums[i]);
count=0;
j=0;
size_lst=0;
i=i+1;
return lst;
# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
# +++your code here+++
list1=list1+list2;
list1.sort();
return list1;
# Note: the solution above is kind of cute, but unforunately list.pop(0)
# is not constant time with the standard python list implementation, so
# the above is not strictly linear time.
# An alternate approach uses pop(-1) to remove the endmost elements
# from each list, building a solution list which is backwards.
# Then use reversed() to put the result back in the correct order. That
# solution works in linear time, but is more ugly.
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'remove_adjacent'
test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
test(remove_adjacent([]), [])
print
print 'linear_merge'
test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
['aa', 'aa', 'aa', 'bb', 'bb'])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Wordcount exercise
Google's Python class
The main() below is already defined and complete. It calls print_words()
and print_top() functions which you write.
1. For the --count flag, implement a print_words(filename) function that counts
how often each word appears in the text and prints:
word1 count1
word2 count2
...
Print the above list in order sorted by word (python will sort punctuation to
come before letters -- that's fine). Store all the words as lowercase,
so 'The' and 'the' count as the same word.
2. For the --topcount flag, implement a print_top(filename) which is similar
to print_words() but which prints just the top 20 most common words sorted
so the most common word is first, then the next most common, and so on.
Use str.split() (no arguments) to split on all whitespace.
Workflow: don't build the whole program at once. Get it to an intermediate
milestone and print your data structure and sys.exit(0).
When that's working, try for the next milestone.
Optional: define a helper function to avoid code duplication inside
print_words() and print_top().
"""
import sys
# +++your code here+++
# Define print_words(filename) and print_top(filename) functions.
# You could write a helper utility function that reads a file
# and builds and returns a word/count dict for it.
# Then print_words() and print_top() can just call the utility function.
def ReadFile(filename):
line="a";
array=[];
words=[[0,""]];
i=0;
j=0;
count=0;
fl=open(filename);
while line != "":
line=fl.readline();
line=line.lower();
array=line.split();
size_array=len(array);
size_words=len(words);
while i<size_array:
while j<size_words:
if array[i] != words[j][1]:
count=count+1;
else:
words[j][0]=words[j][0]+1;
count=0;
break;
j=j+1;
if count == size_words:
words.append([1,array[i]]);
size_words=len(words);
count=0;
j=0;
i=i+1;
i=0;
j=0;
fl.close();
i=0;
words.remove([0,""]);
words.sort(reverse=True);
return words;
def print_words(filename):
i=0;
words=ReadFile(filename);
size_words=len(words);
while i< size_words:
print '{0} {1}'.format(words[i][1],words[i][0]);
i=i+1;
def print_top(filename):
i=0;
words=ReadFile(filename);
size_words=len(words);
if size_words < 20:
while i< size_words:
print '{0} {1}'.format(words[i][1],words[i][0]);
i=i+1;
else:
while i< 20:
print '{0} {1}'.format(words[i][1],words[i][0]);
i=i+1;
###
# This basic command line argument parsing code is provided and
# calls the print_words() and print_top() functions which you must define.
def main():
if len(sys.argv) != 3:
print 'usage: ./wordcount.py {--count | --topcount} file'
sys.exit(1)
option = sys.argv[1]
filename = sys.argv[2]
if option == '--count':
print_words(filename)
elif option == '--topcount':
print_top(filename)
else:
print 'unknown option: ' + option
sys.exit(1)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic string exercises
# D. verbing
# Given a string, if its length is at least 3,
# add 'ing' to its end.
# Unless it already ends in 'ing', in which case
# add 'ly' instead.
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
# +++your code here+++
if len(s) >= 3:
if s[-3:] == 'ing':
newstring=s+'ly';
return newstring;
else:
newstring=s+'ing';
return newstring;
else:
return s;
# E. not_bad
# Given a string, find the first appearance of the
# substring 'not' and 'bad'. If the 'bad' follows
# the 'not', replace the whole 'not'...'bad' substring
# with 'good'.
# Return the resulting string.
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
# +++your code here+++
if s.find('not') != -1:
first=s.find('not');
if s.find('bad') != -1:
second=s.find('bad');
if first <= second:
newstring=s.replace(s[first:second+3],'good');
return newstring;
else:
return s;
else:
return s;
else:
return s;
# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back
def front_back(a, b):
# +++your code here+++
if len(a)%2 == 0:
a_front=a[0:len(a)/2];
a_back=a[len(a)/2:];
else:
a_front=a[0:len(a)/2+1];
a_back=a[len(a)/2+1:];
if len(b)%2 == 0:
b_front=b[0:len(b)/2];
b_back=b[len(b)/2:];
else:
b_front=b[0:len(b)/2+1];
b_back=b[len(b)/2+1:];
return a_front+b_front+a_back+b_back;
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# main() calls the above functions with interesting inputs,
# using the above test() to check if the result is correct or not.
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic list exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in list2.py.
# A. match_ends
# Given a list of strings, return the count of the number of
# strings where the string length is 2 or more and the first
# and last chars of the string are the same.
# Note: python does not have a ++ operator, but += works.
def match_ends(words):
number = 0
for word in words:
if len(word) >= 2 and word[0] == word[-1]:
number += 1
return number
# B. front_x
# Given a list of strings, return a list with the strings
# in sorted order, except group all the strings that begin with 'x' first.
# e.g. ['mix', 'xyz', 'apple', 'xanadu', 'aardvark'] yields
# ['xanadu', 'xyz', 'aardvark', 'apple', 'mix']
# Hint: this can be done by making 2 lists and sorting each of them
# before combining them.
def front_x(words):
others = [];
start_x = [];
for word in words:
if word.startswith('x'):
start_x.append(word)
else:
others.append(word);
start_x.sort();
others.sort();
return start_x + others
# C. sort_last
# Given a list of non-empty tuples, return a list sorted in increasing
# order by the last element in each tuple.
# e.g. [(1, 7), (1, 3), (3, 4, 5), (2, 2)] yields
# [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
# Hint: use a custom key= function to extract the last element form each tuple.
def last_el(list_el):
return list_el[-1]
def sort_last(tuples):
return sorted(tuples, key=last_el)
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'match_ends'
test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)
print
print 'front_x'
test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])
print
print 'sort_last'
test(sort_last([(1, 3), (3, 2), (2, 1)]),
[(2, 1), (3, 2), (1, 3)])
test(sort_last([(2, 3), (1, 2), (3, 1)]),
[(3, 1), (1, 2), (2, 3)])
test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
[(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic string exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in string2.py.
# A. donuts
# Given an int count of a number of donuts, return a string
# of the form 'Number of donuts: <count>', where <count> is the number
# passed in. However, if the count is 10 or more, then use the word 'many'
# instead of the actual count.
# So donuts(5) returns 'Number of donuts: 5'
# and donuts(23) returns 'Number of donuts: many'
def donuts(count):
if count <10:
return "Number of donuts: %d" % count
else:
return "Number of donuts: many"
# B. both_ends
# Given a string s, return a string made of the first 2
# and the last 2 chars of the original string,
# so 'spring' yields 'spng'. However, if the string length
# is less than 2, return instead the empty string.
def both_ends(s):
if len(s) < 2:
return ""
else:
return s[:2] + s[-2:];
# +++your code here+++
# C. fix_start
# Given a string s, return a string
# where all occurences of its first char have
# been changed to '*', except do not change
# the first char itself.
# e.g. 'babble' yields 'ba**le'
# Assume that the string is length 1 or more.
# Hint: s.replace(stra, strb) returns a version of string s
# where all instances of stra have been replaced by strb.
def fix_start(s):
# +++your code here+++
if len(s) > 2:
return s[0] + s[1:].replace(s[0], '*');
else:
return s;
# D. MixUp
# Given strings a and b, return a single string with a and b separated
# by a space '<a> <b>', except swap the first 2 chars of each string.
# e.g.
# 'mix', pod' -> 'pox mid'
# 'dog', 'dinner' -> 'dig donner'
# Assume a and b are length 2 or more.
def mix_up(a, b):
# +++your code here+++
return a.replace(a[:2], b[:2]) + ' ' + b.replace(b[:2], a[:2]);
# Provided simple test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Provided main() calls the above functions with interesting inputs,
# using test() to check if each result is correct or not.
def main():
print 'donuts'
# Each line calls donuts, compares its result to the expected for that call.
test(donuts(4), 'Number of donuts: 4')
test(donuts(9), 'Number of donuts: 9')
test(donuts(10), 'Number of donuts: many')
test(donuts(99), 'Number of donuts: many')
print
print 'both_ends'
test(both_ends('spring'), 'spng')
test(both_ends('Hello'), 'Helo')
test(both_ends('a'), '')
test(both_ends('xyz'), 'xyyz')
print
print 'fix_start'
test(fix_start('babble'), 'ba**le')
test(fix_start('aardvark'), 'a*rdv*rk')
test(fix_start('google'), 'goo*le')
test(fix_start('donut'), 'donut')
print
print 'mix_up'
test(mix_up('mix', 'pod'), 'pox mid')
test(mix_up('dog', 'dinner'), 'dig donner')
test(mix_up('gnash', 'sport'), 'spash gnort')
test(mix_up('pezzy', 'firm'), 'fizzy perm')
# Standard boilerplate to call the main() function.
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic string exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in string2.py.
# A. donuts
# Given an int count of a number of donuts, return a string
# of the form 'Number of donuts: <count>', where <count> is the number
# passed in. However, if the count is 10 or more, then use the word 'many'
# instead of the actual count.
# So donuts(5) returns 'Number of donuts: 5'
# and donuts(23) returns 'Number of donuts: many'
def donuts(count):
if count <10:
return "Number of donuts: %d" % count
else:
return "Number of donuts: many"
# B. both_ends
# Given a string s, return a string made of the first 2
# and the last 2 chars of the original string,
# so 'spring' yields 'spng'. However, if the string length
# is less than 2, return instead the empty string.
def both_ends(s):
if len(s) < 2:
return ""
else:
return s[:2] + s[-2:];
# +++your code here+++
# C. fix_start
# Given a string s, return a string
# where all occurences of its first char have
# been changed to '*', except do not change
# the first char itself.
# e.g. 'babble' yields 'ba**le'
# Assume that the string is length 1 or more.
# Hint: s.replace(stra, strb) returns a version of string s
# where all instances of stra have been replaced by strb.
def fix_start(s):
# +++your code here+++
if len(s) > 2:
return s[0] + s[1:].replace(s[0], '*');
else:
return s;
# D. MixUp
# Given strings a and b, return a single string with a and b separated
# by a space '<a> <b>', except swap the first 2 chars of each string.
# e.g.
# 'mix', pod' -> 'pox mid'
# 'dog', 'dinner' -> 'dig donner'
# Assume a and b are length 2 or more.
def mix_up(a, b):
# +++your code here+++
return a.replace(a[:2], b[:2]) + ' ' + b.replace(b[:2], a[:2]);
# Provided simple test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Provided main() calls the above functions with interesting inputs,
# using test() to check if each result is correct or not.
def main():
print 'donuts'
# Each line calls donuts, compares its result to the expected for that call.
test(donuts(4), 'Number of donuts: 4')
test(donuts(9), 'Number of donuts: 9')
test(donuts(10), 'Number of donuts: many')
test(donuts(99), 'Number of donuts: many')
print
print 'both_ends'
test(both_ends('spring'), 'spng')
test(both_ends('Hello'), 'Helo')
test(both_ends('a'), '')
test(both_ends('xyz'), 'xyyz')
print
print 'fix_start'
test(fix_start('babble'), 'ba**le')
test(fix_start('aardvark'), 'a*rdv*rk')
test(fix_start('google'), 'goo*le')
test(fix_start('donut'), 'donut')
print
print 'mix_up'
test(mix_up('mix', 'pod'), 'pox mid')
test(mix_up('dog', 'dinner'), 'dig donner')
test(mix_up('gnash', 'sport'), 'spash gnort')
test(mix_up('pezzy', 'firm'), 'fizzy perm')
# Standard boilerplate to call the main() function.
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic list exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in list2.py.
# A. match_ends
# Given a list of strings, return the count of the number of
# strings where the string length is 2 or more and the first
# and last chars of the string are the same.
# Note: python does not have a ++ operator, but += works.
def match_ends(words):
number = 0
for word in words:
if len(word) >= 2 and word[0] == word[-1]:
number += 1
return number
# B. front_x
# Given a list of strings, return a list with the strings
# in sorted order, except group all the strings that begin with 'x' first.
# e.g. ['mix', 'xyz', 'apple', 'xanadu', 'aardvark'] yields
# ['xanadu', 'xyz', 'aardvark', 'apple', 'mix']
# Hint: this can be done by making 2 lists and sorting each of them
# before combining them.
def front_x(words):
others = [];
start_x = [];
for word in words:
if word.startswith('x'):
start_x.append(word)
else:
others.append(word);
start_x.sort();
others.sort();
return start_x + others
# C. sort_last
# Given a list of non-empty tuples, return a list sorted in increasing
# order by the last element in each tuple.
# e.g. [(1, 7), (1, 3), (3, 4, 5), (2, 2)] yields
# [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
# Hint: use a custom key= function to extract the last element form each tuple.
def last_el(list_el):
return list_el[-1]
def sort_last(tuples):
return sorted(tuples, key=last_el)
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'match_ends'
test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)
print
print 'front_x'
test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])
print
print 'sort_last'
test(sort_last([(1, 3), (3, 2), (2, 1)]),
[(2, 1), (3, 2), (1, 3)])
test(sort_last([(2, 3), (1, 2), (3, 1)]),
[(3, 1), (1, 2), (2, 3)])
test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
[(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic list exercises
# D. Given a list of numbers, return a list where
# all adjacent == elements have been reduced to a single element,
# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
# modify the passed in list.
def remove_adjacent(nums):
new_nums = []
nums.sort();
num=1
for num in range(len(nums)):
if nums[num] != nums[num-1]:
new_nums.append(nums[num])
return new_nums
# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
# +++your code here+++
merged_list = list1 + list2
merged_list.sort()
return merged_list
# Note: the solution above is kind of cute, but unforunately list.pop(0)
# is not constant time with the standard python list implementation, so
# the above is not strictly linear time.
# An alternate approach uses pop(-1) to remove the endmost elements
# from each list, building a solution list which is backwards.
# Then use reversed() to put the result back in the correct order. That
# solution works in linear time, but is more ugly.
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'remove_adjacent'
test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
test(remove_adjacent([]), [])
print
print 'linear_merge'
test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
['aa', 'aa', 'aa', 'bb', 'bb'])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Wordcount exercise
Google's Python class
The main() below is already defined and complete. It calls print_words()
and print_top() functions which you write.
1. For the --count flag, implement a print_words(filename) function that counts
how often each word appears in the text and prints:
word1 count1
word2 count2
...
Print the above list in order sorted by word (python will sort punctuation to
come before letters -- that's fine). Store all the words as lowercase,
so 'The' and 'the' count as the same word.
2. For the --topcount flag, implement a print_top(filename) which is similar
to print_words() but which prints just the top 20 most common words sorted
so the most common word is first, then the next most common, and so on.
Use str.split() (no arguments) to split on all whitespace.
Workflow: don't build the whole program at once. Get it to an intermediate
milestone and print your data structure and sys.exit(0).
When that's working, try for the next milestone.
Optional: define a helper function to avoid code duplication inside
print_words() and print_top().
"""
import sys
# +++your code here+++
# Define print_words(filename) and print_top(filename) functions.
# You could write a helper utility function that reads a file
# and builds and returns a word/count dict for it.
# Then print_words() and print_top() can just call the utility function.
###
# This basic command line argument parsing code is provided and
# calls the print_words() and print_top() functions which you must define.
def words_n_counts(filename):
_file = open(filename, 'r')
map_words = {}
for line in _file:
words = line.split()
for word in words:
word = word.lower()
if word in map_words:
map_words[word] += 1
else:
map_words[word] = 1
return map_words
def print_words(filename):
map_words = words_n_counts(filename)
words = sorted(map_words.keys())
for word in words:
print word, map_words[word]
def word_key(word):
return word[1]
def print_top(filename):
word_dict = words_n_counts(filename)
words = sorted(word_dict.items(), key=word_key, reverse=True)
for word in words[:20]:
print word[0], word[1]
def main():
if len(sys.argv) != 3:
print 'usage: ./wordcount.py {--count | --topcount} file'
sys.exit(1)
option = sys.argv[1]
filename = sys.argv[2]
if option == '--count':
print_words(filename)
elif option == '--topcount':
print_top(filename)
else:
print 'unknown option: ' + option
sys.exit(1)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python2.4 -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic string exercises
# D. verbing
# Given a string, if its length is at least 3,
# add 'ing' to its end.
# Unless it already ends in 'ing', in which case
# add 'ly' instead.
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
if len(s) >= 3:
if s.endswith('ing'):
return s + 'ly'
else:
return s + 'ing'
else: return s
# E. not_bad
# Given a string, find the first appearance of the
# substring 'not' and 'bad'. If the 'bad' follows
# the 'not', replace the whole 'not'...'bad' substring
# with 'good'.
# Return the resulting string.
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
ans = s
bad = s.find('bad')
no = s.find('not')
if bad != -1 and no != -1 and bad > no:
ans = s.replace(s[no:(bad + 3)], 'good')
return ans
# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back
def front_back(a, b):
a_m = len(a)/2
b_m = len(b)/2
if len(a)%2 == 1:
a_m = a_m + 1
if len(b)%2 == 1:
b_m = b_m + 1
return a[:a_m] + b[:b_m] + a[a_m:] + b[b_m:]
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# main() calls the above functions with interesting inputs,
# using the above test() to check if the result is correct or not.
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic list exercises
# D. Given a list of numbers, return a list where
# all adjacent == elements have been reduced to a single element,
# so [1, 2, 2, 3] returns [1, 2, 3]. You may create a new list or
# modify the passed in list.
def remove_adjacent(nums):
new_nums = []
nums.sort();
num=1
for num in range(len(nums)):
if nums[num] != nums[num-1]:
new_nums.append(nums[num])
return new_nums
# E. Given two lists sorted in increasing order, create and return a merged
# list of all the elements in sorted order. You may modify the passed in lists.
# Ideally, the solution should work in "linear" time, making a single
# pass of both lists.
def linear_merge(list1, list2):
# +++your code here+++
merged_list = list1 + list2
merged_list.sort()
return merged_list
# Note: the solution above is kind of cute, but unforunately list.pop(0)
# is not constant time with the standard python list implementation, so
# the above is not strictly linear time.
# An alternate approach uses pop(-1) to remove the endmost elements
# from each list, building a solution list which is backwards.
# Then use reversed() to put the result back in the correct order. That
# solution works in linear time, but is more ugly.
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'remove_adjacent'
test(remove_adjacent([1, 2, 2, 3]), [1, 2, 3])
test(remove_adjacent([2, 2, 3, 3, 3]), [2, 3])
test(remove_adjacent([]), [])
print
print 'linear_merge'
test(linear_merge(['aa', 'xx', 'zz'], ['bb', 'cc']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'xx'], ['bb', 'cc', 'zz']),
['aa', 'bb', 'cc', 'xx', 'zz'])
test(linear_merge(['aa', 'aa'], ['aa', 'bb', 'bb']),
['aa', 'aa', 'aa', 'bb', 'bb'])
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
"""Wordcount exercise
Google's Python class
The main() below is already defined and complete. It calls print_words()
and print_top() functions which you write.
1. For the --count flag, implement a print_words(filename) function that counts
how often each word appears in the text and prints:
word1 count1
word2 count2
...
Print the above list in order sorted by word (python will sort punctuation to
come before letters -- that's fine). Store all the words as lowercase,
so 'The' and 'the' count as the same word.
2. For the --topcount flag, implement a print_top(filename) which is similar
to print_words() but which prints just the top 20 most common words sorted
so the most common word is first, then the next most common, and so on.
Use str.split() (no arguments) to split on all whitespace.
Workflow: don't build the whole program at once. Get it to an intermediate
milestone and print your data structure and sys.exit(0).
When that's working, try for the next milestone.
Optional: define a helper function to avoid code duplication inside
print_words() and print_top().
"""
import sys
# +++your code here+++
# Define print_words(filename) and print_top(filename) functions.
# You could write a helper utility function that reads a file
# and builds and returns a word/count dict for it.
# Then print_words() and print_top() can just call the utility function.
###
# This basic command line argument parsing code is provided and
# calls the print_words() and print_top() functions which you must define.
def words_n_counts(filename):
_file = open(filename, 'r')
map_words = {}
for line in _file:
words = line.split()
for word in words:
word = word.lower()
if word in map_words:
map_words[word] += 1
else:
map_words[word] = 1
return map_words
def print_words(filename):
map_words = words_n_counts(filename)
words = sorted(map_words.keys())
for word in words:
print word, map_words[word]
def word_key(word):
return word[1]
def print_top(filename):
word_dict = words_n_counts(filename)
words = sorted(word_dict.items(), key=word_key, reverse=True)
for word in words[:20]:
print word[0], word[1]
def main():
if len(sys.argv) != 3:
print 'usage: ./wordcount.py {--count | --topcount} file'
sys.exit(1)
option = sys.argv[1]
filename = sys.argv[2]
if option == '--count':
print_words(filename)
elif option == '--topcount':
print_top(filename)
else:
print 'unknown option: ' + option
sys.exit(1)
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python2.4 -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Additional basic string exercises
# D. verbing
# Given a string, if its length is at least 3,
# add 'ing' to its end.
# Unless it already ends in 'ing', in which case
# add 'ly' instead.
# If the string length is less than 3, leave it unchanged.
# Return the resulting string.
def verbing(s):
if len(s) >= 3:
if s.endswith('ing'):
return s + 'ly'
else:
return s + 'ing'
else: return s
# E. not_bad
# Given a string, find the first appearance of the
# substring 'not' and 'bad'. If the 'bad' follows
# the 'not', replace the whole 'not'...'bad' substring
# with 'good'.
# Return the resulting string.
# So 'This dinner is not that bad!' yields:
# This dinner is good!
def not_bad(s):
ans = s
bad = s.find('bad')
no = s.find('not')
if bad != -1 and no != -1 and bad > no:
ans = s.replace(s[no:(bad + 3)], 'good')
return ans
# F. front_back
# Consider dividing a string into two halves.
# If the length is even, the front and back halves are the same length.
# If the length is odd, we'll say that the extra char goes in the front half.
# e.g. 'abcde', the front half is 'abc', the back half 'de'.
# Given 2 strings, a and b, return a string of the form
# a-front + b-front + a-back + b-back
def front_back(a, b):
a_m = len(a)/2
b_m = len(b)/2
if len(a)%2 == 1:
a_m = a_m + 1
if len(b)%2 == 1:
b_m = b_m + 1
return a[:a_m] + b[:b_m] + a[a_m:] + b[b_m:]
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# main() calls the above functions with interesting inputs,
# using the above test() to check if the result is correct or not.
def main():
print 'verbing'
test(verbing('hail'), 'hailing')
test(verbing('swiming'), 'swimingly')
test(verbing('do'), 'do')
print
print 'not_bad'
test(not_bad('This movie is not so bad'), 'This movie is good')
test(not_bad('This dinner is not that bad!'), 'This dinner is good!')
test(not_bad('This tea is not hot'), 'This tea is not hot')
test(not_bad("It's bad yet not"), "It's bad yet not")
print
print 'front_back'
test(front_back('abcd', 'xy'), 'abxcdy')
test(front_back('abcde', 'xyz'), 'abcxydez')
test(front_back('Kitten', 'Donut'), 'KitDontenut')
if __name__ == '__main__':
main()
| Python |
#!/usr/bin/python -tt
# Copyright 2010 Google Inc.
# Licensed under the Apache License, Version 2.0
# http://www.apache.org/licenses/LICENSE-2.0
# Google's Python Class
# http://code.google.com/edu/languages/google-python-class/
# Basic list exercises
# Fill in the code for the functions below. main() is already set up
# to call the functions with a few different inputs,
# printing 'OK' when each function is correct.
# The starter code for each function includes a 'return'
# which is just a placeholder for your code.
# It's ok if you do not complete all the functions, and there
# are some additional functions to try in list2.py.
# A. match_ends
# Given a list of strings, return the count of the number of
# strings where the string length is 2 or more and the first
# and last chars of the string are the same.
# Note: python does not have a ++ operator, but += works.
def match_ends(words):
count = 0
for word in words:
if len(word) > 1:
if word[0] == word[-1] :
count +=1
return count
# B. front_x
# Given a list of strings, return a list with the strings
# in sorted order, except group all the strings that begin with 'x' first.
# e.g. ['mix', 'xyz', 'apple', 'xanadu', 'aardvark'] yields
# ['xanadu', 'xyz', 'aardvark', 'apple', 'mix']
# Hint: this can be done by making 2 lists and sorting each of them
# before combining them.
def front_x(words):
array_x = []
other_words = []
for word in words:
if word[0] == "x":
array_x.append(word)
else:
other_words.append(word)
return sorted(array_x) + sorted(other_words)
# C. sort_last
# Given a list of non-empty tuples, return a list sorted in increasing
# order by the last element in each tuple.
# e.g. [(1, 7), (1, 3), (3, 4, 5), (2, 2)] yields
# [(2, 2), (1, 3), (3, 4, 5), (1, 7)]
# Hint: use a custom key= function to extract the last element form each tuple.
def sort_last(tuples):
def last(s):
return s[-1]
return sorted(tuples, key=last)
# Simple provided test() function used in main() to print
# what each function returns vs. what it's supposed to return.
def test(got, expected):
if got == expected:
prefix = ' OK '
else:
prefix = ' X '
print '%s got: %s expected: %s' % (prefix, repr(got), repr(expected))
# Calls the above functions with interesting inputs.
def main():
print 'match_ends'
test(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']), 3)
test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 2)
test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)
print
print 'front_x'
test(front_x(['bbb', 'ccc', 'axx', 'xzz', 'xaa']),
['xaa', 'xzz', 'axx', 'bbb', 'ccc'])
test(front_x(['ccc', 'bbb', 'aaa', 'xcc', 'xaa']),
['xaa', 'xcc', 'aaa', 'bbb', 'ccc'])
test(front_x(['mix', 'xyz', 'apple', 'xanadu', 'aardvark']),
['xanadu', 'xyz', 'aardvark', 'apple', 'mix'])
print
print 'sort_last'
test(sort_last([(1, 3), (3, 2), (2, 1)]),
[(2, 1), (3, 2), (1, 3)])
test(sort_last([(2, 3), (1, 2), (3, 1)]),
[(3, 1), (1, 2), (2, 3)])
test(sort_last([(1, 7), (1, 3), (3, 4, 5), (2, 2)]),
[(2, 2), (1, 3), (3, 4, 5), (1, 7)])
if __name__ == '__main__':
main()
| Python |
Subsets and Splits
SQL Console for ajibawa-2023/Python-Code-Large
Provides a useful breakdown of language distribution in the training data, showing which languages have the most samples and helping identify potential imbalances across different language groups.