blob_id string | repo_name string | path string | length_bytes int64 | score float64 | int_score int64 | text string | is_english bool |
|---|---|---|---|---|---|---|---|
01193520f4ca41afc35c5bd4e34b2f1ccaf42527 | VireshDoshi/pd_test | /PerspectumDiagnostics.py | 2,721 | 4.5 | 4 | #!/usr/bin/env python
import itertools
def strings_appear_in_multiple_lists(list_in):
""" This method will print out the list of Strings appearing in multiple
Lists.
Input: List of Lists ( 1..n)
Output: String
"""
# establish the size of the list
list_in_len = len(list_in)
# set the display_string
display_string = ''
final_list = []
for b in range(0,list_in_len -1):
for a in range(b,list_in_len-1):
new_list = [ x for x in list_in[b] if x in list_in[a + 1] ]
final_list.append(new_list)
# remove unwated empty lists
filtered_final_list = [ x for x in final_list if x]
for list in filtered_final_list:
# display_string = "'".join(map(str,list))[1:-1]
display_string = display_string + str(list)[1:-1] + ','
# strip out the last comma
display_string = display_string.rstrip(',')
# print "Strings appearing in multiple lists: {0}".format(display_string)
return display_string
def number_of_unique_strings(list_in):
"""
This method will return the number of unique strings in the list given
:param list_in:
:return: integer of unique strings
"""
# lets concatenate all the values into one large list for processing
merged_list = list(itertools.chain(*list_in))
# establish the length of items in the set
merged_set_len = len(set(merged_list))
# print "number of unique strings: {0}".format(merged_set_len)
return merged_set_len
def total_number_of_strings_processed(list_in):
"""
This method will return the number of strings processed
:param list_in:
:return: integer of strings processed
"""
# lets concatenate all the values into one large list for processing
merged_list = list(itertools.chain(*list_in))
merged_len = len(merged_list)
# print "Total number of strings processed {0}".format(merged_len)
return merged_len
def perspectum_diagnostics_test(list_in):
"""
This method displays the output to the screen based on the list
:param list_in:
:return: print output to the screen
"""
print "Strings appearing in multiple lists: {0}".format(strings_appear_in_multiple_lists(list_in))
print "number of unique strings: {0}".format(number_of_unique_strings(list_in))
print "Total number of strings processed {0}".format(total_number_of_strings_processed(list_in))
if __name__ == '__main__':
test_list_1 = [['a','b','c','dh'],['a','d','ha','e'],['f','g','h'],['c'],['dh'],['h','ha'],['e'],['d']]
test_list_2 = [['g', 'gh', 'ghj', 'g'], ['j', 'ju', 'gh', 'gk', 'gn']]
test_list_3 = [['a'],['f'],['f']]
perspectum_diagnostics_test(test_list_3)
| true |
9f1d61ffab774846d0c465ebe959ba3fe0e93275 | RiyazShaikAuxo/datascience_store | /PyhonProgramming/leapYear.py | 380 | 4.1875 | 4 | #If the year divisible by 4, 100 and 400 is a leap year elase not a leap year
year=int(input("Enter the year: "))
if (year%4)==0:
if(year%100)==0:
if(year%400)==0:
print(year, "is a leap year")
else:
print(year, "is not a leap year")
else:
print(year, "is not a leap year")
else:
print(year, "is not a leap year")
| false |
9402dc7956bd81a9f115c27f35dc1bb17e3d8363 | RiyazShaikAuxo/datascience_store | /1.python/Datastrutures_in_Python/tupple.py | 508 | 4.28125 | 4 | # tupples once craeted can not be modified at all
#Similar to lists
tuple=(30,'Riyaz',5.8)
print(tuple[0])
tuple1=("Banglore",29.3343,34)
print(tuple1)
#assigning new variable
tuple2=tuple1
print(tuple2)
#tupple not allow item assignment will throw error
#tuple2(1)="Riyaz"
retrieve1=tuple2[:-1]
print(retrieve1)
#Another way to create a tupple
new_tupple=1,2,3
print(new_tupple)
#Concatinate tupples
cocatinate_tupple=new_tupple,(1,2,3,4,5)
print(cocatinate_tupple)
print(cocatinate_tupple[0])
| true |
fd73446ae26bc74f87d35ce6c29b156da1130201 | mskyberg/Module7 | /fun_with_collections/sort_and_search_array.py | 2,712 | 4.59375 | 5 | """
Program: sort_and_search_array.py
Author: Michael Skyberg, mskyberg@dmacc.edu
Last date modified: June 2020
Purpose: Demonstrates the use of a basic array sorting and searching
"""
import array as arr
LIST_MAX = 3
def get_input():
"""
Description: Gets user input
:returns: returns a string of user input
:raises keyError: raises an exception
"""
# prompt user for input
return input('Enter an integer value: ')
def make_array(sort=False, reverse=False):
"""
Description: gets 3 user inputs and combines them into a list
:param sort: optional bool whether or not to sort the list
:param reverse: optional bool sort in reverse order or not
:returns: returns an array of 3 values
:raises ValueError: raises an exception if input is not numeric
"""
new_list = []
# asks for 3 user input in a loop by
for index in range(LIST_MAX):
user_input = get_input()
try:
# attempt to cast string to an integer
user_input = int(user_input)
if not 1 <= user_input <= 50:
raise ValueError(f'Input is out of range! {user_input}')
except ValueError:
raise
else:
# if successful, insert into list
new_list.insert(index, user_input)
new_array = arr.array('i', new_list)
if sort:
return sort_array(new_array, reverse)
else:
return new_array
def sort_array(a_arr, rev=False):
"""
Description: Sort a list in order with optional reverse
:param a_arr: an array to sort
:param rev: sort order, true if reverse sort
:returns: a sorted array
"""
# convert the array to a list since it is for sure same type
arr_as_list = a_arr.tolist()
# sort the list accordingly
arr_as_list.sort(reverse=rev)
# create new array with the integer list sorted
# I am creating a new array so going to return the new array
return arr.array('i', arr_as_list)
def search_array(a_arr, element):
"""
Description: Search for an element at provided index
:param a_arr: array to search
:param element: element to find
:returns: index of element
:raises ValueError: raises an exception if element does not exist
"""
try:
index = a_arr.index(element)
except ValueError:
return -1
else:
return index
if __name__ == '__main__':
try:
test_array = make_array(sort=False, reverse=False)
print(test_array)
element_index = int(input('Enter element to search:'))
print(f'Index: {search_array(test_array, element_index)}')
except ValueError as main_error:
print(f'failure in main: {main_error}')
| true |
95a6bdf3d1dfe5cdb2f78a857e494743df0aa0d3 | carepack/PycharmProjects | /exercise04.py | 566 | 4.28125 | 4 | # 13.11.2019
# identify divisors of number. ouput as list
in_num = int(input("Please enter number: "))
x_ra = range(2, in_num+1)
for element in x_ra:
res = in_num % element
if res == 0:
out_num = str(in_num)
out_element = str(element)
out_divisor = str(in_num / element)
print("Your number " + out_num + " is dividable by " + out_element + " result = " + out_divisor)
# output as list
num_div = []
for element in x_ra:
if in_num % element == 0:
num_div.append(element)
print("Divisors as list: " + str(num_div)) | false |
8421badfb54ed5238572c619a67f828dff41de39 | henry199101/6.00.1x_Files | /Midterm_Exam/Quiz/Problem_5/laceStringsFinished.py | 823 | 4.25 | 4 | def laceStrings(s1, s2):
"""
s1 and s2 are strings.
Returns a new str with elements of s1 and s2 interlaced,
beginning with s1. If strings are not of same length,
then the extra elements should appear at the end.
"""
# Your Code Here
if len(s1) > len(s2):
(s1, remainder) = (s1[0:len(s2)], s1[len(s2):])
elif len(s1) < len(s2):
(s2, remainder) = (s2[0:len(s1)], s2[len(s1):])
else:
remainder = ""
# s1 and s2 now start out as two equal length strings
chars = []
for i in range(len(s1)):
chars.append(s1[i])
chars.append(s2[i])
result = "".join(chars) + remainder
return result
# For example, if we lace 'abcd' and 'efghi', we would get the new string: 'aebfcgdhi'.
s1 = 'abcd'
s2 = 'efghi'
print laceStrings(s1, s2)
| true |
9bcdd2c0696cb4ed65f741fc02cf14f2472468d6 | Hari-97/task_py | /num square pattern.py | 245 | 4.1875 | 4 | row=3
for i in range(1,2*row):
for j in range(1,2*row):
if (i==1) or (i==2*row-1) or ((j==1 or j==2*row-1) and (i>1 and i<2*row-1)):
print(row,end=" ")
elif(i==3 and j==3):
print(1,end=" ")
else:
print(row-1,end=" ")
print()
| false |
229463b206f7594e56f94825645a236bcb48160b | PrabuddhaBanerjee/Python | /Chapter3/Ch3P9.py | 269 | 4.15625 | 4 | import math
def main():
print("This program calculates area of a triangle")
a, b, c = eval(input("Please enter 3 sides of triangle a, b, c:"))
s = (a + b + c)/2
area = math.sqrt(s * (s - a) * (s - b) * (s - c))
print("Area of the triangle is ", area)
main()
| true |
dca253cfa9ab3a66679770ec8879ca841c8cc616 | PrabuddhaBanerjee/Python | /Chapter3/Ch3P6.py | 289 | 4.125 | 4 | def main():
print("This program calculates slope between two points")
x1, y1 = eval(input("Please enter x, y coordinates for point 1:"))
x2, y2 = eval(input("Please enter x, y coordinates for point 2:"))
slope = (y2 - y1)/ (x2 - x1)
print("Slope of the line is ", slope)
main()
| true |
ef61b8338e2e3f0678e0b3174068bbeef8301723 | PrabuddhaBanerjee/Python | /Chapter3/Ch3P1.py | 320 | 4.28125 | 4 | import math
def main():
print("This program is to calculate the volume and surface area of a sphere")
radius = int(input("Please enter the radius for sphere:"))
vol = (4 / 3)* math.pi * (radius ** 3)
area = 4 * math.pi * (radius ** 2)
print("The area of the sphere is",area," and the volume is ",vol)
main()
| true |
3c0ca0d28398b31334d052ee7363271aef4efbf3 | devsave/docs | /Python实验室/list添加元素/list_append.py | 867 | 4.1875 | 4 | # a = [1, 2]
# b = [1, 2]
# print('The id of a: %d' % id(a) )
# a += [3]
# print('The id of a: %d' % id(a) )
# print('a = %s' % str(a))
# print('The id of b: %d' % id(b) )
# b = b + [3]
# print('The id of b: %d' % id(b) )
# print('b = %s' % str(b))
# b = (1, 2)
# c = b
# print('Before calculation')
# print('The id of b: %d' % id(b))
# print('The id of c: %d' % id(c))
# b += (3, )
# print('After calculation')
# print('The id of b: %d' % id(b) )
# print('The id of c: %d' % id(c))
# print('b = %s' % str(b))
# print('c = %s' % str(c))
b = (1, 2)
c = b
print('Before calculation')
print('The id of b: %d' % id(b))
print('The id of c: %d' % id(c))
print('b = %s' % str(b))
print('c = %s' % str(c))
b += (3, )
print()
print('After calculation')
print('The id of b: %d' % id(b) )
print('The id of c: %d' % id(c))
print('b = %s' % str(b))
print('c = %s' % str(c)) | false |
3d2bcf58c87f8995ddc8974f753a9d6ebc8c79ff | teago83/ExerciciosPython | /Calculadora.py | 1,175 | 4.125 | 4 | class calculadora():
def soma(self, x, y):
return x + y
def subtracao(self, x, y):
return x - y
def multiplicacao(self, x, y):
return x * y
def divisao(self, x, y):
return x / y
calc = calculadora()
def menu():
while True:
try:
op = int(input("1) Soma"
"\n2) Subtração"
"\n3) Multiplicação"
"\n4) Divisão"
"\n5) Vazar"
"\nDigite a operação desejada."))
except ValueError:
print("Valor inválido.")
break
xis = int(input("Digite o valor número 1:"))
ispu = int(input("Digite o valor número 2:"))
if op == 1:
print("Resultado da soma: %d" %(calc.soma(xis, ispu)))
elif op == 2:
print("Resultado da subtração: %d" % (calc.subtracao(xis, ispu)))
elif op == 3:
print("Resultado da multiplicação: %d" % (calc.multiplicacao(xis, ispu)))
elif op == 4:
print("Resultado da divisão: %d" % (calc.divisao(xis, ispu)))
elif op == 5:
exit()
while True:
menu()
| false |
fbdc4501fd424d15c0d478286655ce417924b01f | FaatimahM/Analyse-predict | /analysepredict/word_split.py | 505 | 4.28125 | 4 | def word_splitter(df):
"""
Splits the sentences in a dataframe's column into a list of the separate words.
The created lists should be placed in a column named 'Split Tweets' in the original dataframe.
parameters
----------
df: Dataframe
It should take a pandas dataframe as an input.
Returns
-------
df: DataFrame
The function should return the modified dataframe.
"""
df['Split Tweets'] = df['Tweets'].str.lower().str.split()
return df | true |
48f9a0e5b3af6cc743bb3541af4ac6d67bf5fc47 | venkatreddymallidi/python | /leapyear.py | 350 | 4.125 | 4 | #program to find leap year or not
year=int(input("enter a year"))
if year%400==0 or (year%4==0 and year%100!=0):#1996 and 2004 is divisible by 4 and divisible by 100 so leap years 2100 not a leap year but divisible by 4 divisible by 100 so leap year divisible by 4 not divisible by 100
print("leap year")
else:
print("non leap year ")
| false |
0fefbee1a8189445eb6acd86da4b4bbdec8f612d | gapigo/CEV_Aulas_Python | /Aulas/Aula 17/Aula 17a.py | 279 | 4.1875 | 4 | num = [2, 5, 9, 1]
print(num)
num[2] = 3
print(num)
num.append(7)
print(num)
num.sort()
print(num)
num.sort(reverse=True)
num.insert(2, 0)
print(num)
num.pop(2)
print(num)
print(f'Essa lista tem {len(num)} elementos.')
num.insert(2, 2)
print(num)
num.remove(2)
print(num)
| false |
47e6d9d3a584db96db9eec5e8541b6d4043a56a5 | atena-data/Python-Bootcamp-Codes | /Day 4 - Rock, Paper, Scissors Game/main.py | 1,308 | 4.21875 | 4 | #import random module
import random
#ASCII arts for game choices
rock = '''
_______
---' ____)
(_____)
(_____)
(____)
---.__(___)
'''
paper = '''
_______
---' ____)____
______)
_______)
_______)
---.__________)
'''
scissors = '''
_______
---' ____)____
______)
__________)
(____)
---.__(___)
'''
print ("Welcome to Rock, Paper, Scissors!\nLets Play!\n")
#Ask user for their choice
user_choice = int (input ("What do you choose? Type 0 for Rock, 1 for Paper and 2 for Scissors.\n"))
#Generate random value as computer's choice
computer_choice = random.randint(0,2)
#List all possible choices
choices = [rock, paper, scissors]
#Check user and computer choices against each other
if user_choice >= 3 or user_choice < 0:
print ("You typed an invalid number. Please try again!")
else:
print(f"You chose:\n{choices[user_choice]}")
print(f"Computer chose:\n{choices[computer_choice]}")
if user_choice == 0 and computer_choice == 2:
print ("CONGRATULATIONS!\nYou won!")
elif user_choice > computer_choice:
print ("CONGRATULATIONS!\nYou won!")
elif computer_choice > user_choice:
print ("SORRY!\nYou lost!\nPlease try again.")
elif computer_choice == user_choice:
print ("It's a draw!")
| false |
89a0dad041bb636a7119e958f881d7c95d017346 | atena-data/Python-Bootcamp-Codes | /Day 1 - Band Name Generator/main.py | 435 | 4.375 | 4 | #1. Create a greeting for the program.
print("Hello there! Welcome to the band name generator :) \n")
#2. Ask the user for their favorite color and pet name.
favorite_color = input("What is your favorite color?\n")
pet_name = input("What is the name of your pet?\n")
#4. Combine the names to suggest a band name.
print("\nYour suggested band name is " + favorite_color + " " + pet_name + "!")
print("\nThanks for using this program!")
| true |
28085fc4229c92fcaf5f8ba852a8db66389ca5f5 | atena-data/Python-Bootcamp-Codes | /Day 12 - Guess a Number Game/main.py | 1,131 | 4.15625 | 4 | from art import logo
import random
#Function to compare user's guess to the number
def compare(user_guess, number):
"""Compares user's guess to the number and will return result"""
if user_guess > number:
global attempts
attempts -= 1
return "Too high."
elif user_guess < number:
attempts -= 1
return "Too low."
else:
return f"Congratulations! The answer was {number}."
print(logo)
print("Welcome to the number guessing game!")
print("I'm thinking of a number between 1 and 100")
#Generate a random number between 1 and 100
number = random.randrange(1, 101)
#Ask user for their desired difficulty level
level = input("Choose a dificulty level. Type 'easy' or 'hard': ")
if level == "easy":
attempts = 10
else:
attempts = 5
#Run the game
print(f"You have {attempts} attempts to guess the number")
while attempts !=0:
for attempt in range(attempts):
user_guess = int(input("Make a guess: "))
print(compare(user_guess, number))
if user_guess == number:
break
if attempts == 0 and user_guess != number:
print (f"Sorry! The answer was {number}. You lose.")
break | true |
1c4a20def81fddff2c3df963a8cf84cc2487a615 | kwhit2/holbertonschool-higher_level_programming | /0x07-python-test_driven_development/4-print_square.py | 740 | 4.53125 | 5 | #!/usr/bin/python3
""" This module contains a function that prints a square with #s """
def print_square(size):
""" print_square method
Args:
size - int (length of a side of the square)
Raises:
TypeError: if size is not an int, if size is a float and less than 0
ValueError: if size is less than 0
Returns:
None
"""
if type(size) is not int:
raise TypeError('size must be an integer')
elif size < 0:
raise ValueError('size must be >= 0')
elif type(size) is float and size < 0:
raise TypeError('size must be an integer')
else:
for x in range(size):
for y in range(size):
print("#", end="")
print()
| true |
42a554b961de6ca003d11b0a2e7cbb00837486ae | kwhit2/holbertonschool-higher_level_programming | /0x01-python-if_else_loops_functions/9-print_last_digit.py | 229 | 4.375 | 4 | #!/usr/bin/python3
def print_last_digit(number):
number = abs(number) # abs value needed for negative numbers
print((number % 10), end="") # print the last digit with no \n
return (number % 10) # return last digit
| true |
e19a22d3050519cb0cf86d4abfb867cecf945cb7 | 13522568615/zhouzx_python | /pythonBase/面向对象实例/po_game.py | 1,035 | 4.1875 | 4 | #面向过程编程
"""
虫子的初始位置
蚂蚁的初始位置
进入循环，条件为蚂蚁和虫子不再同一个位置
依照规则，蚂蚁和虫子移动位置
直到蚂蚁和虫子走到同一位置，程序结束
"""
import random
#蚂蚁
ant_point = random.randint(0,20)
#虫子
worm_point = random.randint(0,20)
#输出虫子与蚂蚁的初始位置
print("蚂蚁：", ant_point, "虫子：", worm_point)
#指定可以走的步数
step = [-2, +2, -3 ,+3]
#进入循环
while ant_point != worm_point:
#选择步数
astep = random.choice(step)
#判断蚂蚁走的步数是否超出了范围
if 0 <= ant_point + astep <=20:
#不超出范围从新修改位置
ant_point += astep
# 选择步数
astep = random.choice(step)
#判断虫子走的步数是否超出了范围
if 0 <= worm_point +astep <= 20:
# 不超出范围从新修改位置
worm_point += astep
#输出虫子与蚂蚁现有位置
print("蚂蚁：", ant_point, "虫子：", worm_point)
| false |
ad0e1c52f13e80b96eeea36d41f4c3d2d3aaab52 | ArthurZheng/python_hard_way | /month_list.py | 573 | 4.1875 | 4 |
def choose_month():
months = ['January', 'February', 'March', 'April', 'May', 'June', 'July', 'August', 'September', 'October','November', 'December']
month_input = int(raw_input("Enter a number for the month (1-12)"))
print "The month you pick is ", month_input, " month name: ", months[month_input-1]
def main():
print "Enter a month number to get your month name."
print "How many times do you want to play this game?"
times = int(raw_input(">>>"))
for i in range(times):
print "\nRound number ", i +1
choose_month()
print "\nEnd of the Game."
main()
| true |
ce558dd7444d0bf453e7d055db8d869bf912fd27 | gasamoma/cracking-code | /Strings/1.3.py | 2,226 | 4.5 | 4 | # URLify: Write a method to replace all spaces in a string with '%20'. You may assume that the string has sufficient space at the end to hold the additional characters, and that you are given the "true" length of the string. (Note: If implementing in Java, please use a character array so that you can perform this operation in place.)
# so Id ask if i have to make the swap from space to url space in place? YES
# double space has to be compressed? No, it should be two %20
#Ok so first I was thinking of each space I encounter move the rest of the string to the back so im going to do a quick implementations of that, but that wont be efficient
#IDK what means the true length of the str i would suppose that is excluding the extra space at the end so an example would be like
# 'asd asd ', 7 so two extra spaces for the extra %20 i need to add
def urilify(str1, str_len): # O(n^2)
max_len = len(str1)
if max_len == str_len:
return str1
for ith in range(max_len):
if ith == str_len:
return str1[:max_len]
if str1[ith]== " ":
str1=str1[:ith]+"%20"+str1[ith+1:] #if all the string is made of spaces**
ith+=2
Return str1[:max_len+1]
##if all the string is spaces then assuming the string is N i have to move n chars N times
#means that n^2
# im ignoring that they are givin me the extra space at the end because of some reason
def urilify_2(str1,str_len):#O(n)
str1 = invert(str1)#this is O(n)
ith = 0
jth = len(str1) - str_len
while jth < len(str1):#this is O(n)
if str1[jth]==" ":
str1[ith]="0"
str1[ith+1]="2"
str1[ith+2]="%"
ith+=2
else:
str1[ith]=str1[jth]
jth+=1
ith+=1
return ''.join(invert(str1))#this is O(2n)
def invert(str1):#inverting a list is O(n)
str_len = len(str1)
str1 = list(str1)#this is O(n)
for ith in range(int(str_len/2)):#this is O(n/2)
tempo=str1[ith]#space O(n)
str1[ith]= str1[str_len-ith-1]
str1[str_len-ith-1]=tempo
return str1
print(urilify_2("asdasd", 6))
print(urilify_2("asd asd ", 8))
print(urilify_2("asd a sd ", 9))
#so I could improve this solution and as soon as I start inverting the string start checking if its a space and swap with %20 and as soon as i finish i just revert it back
| true |
ee2339b680fadc5f869e3371dfb98513807fcb8f | gasamoma/cracking-code | /Strings/1.4.py | 2,415 | 4.1875 | 4 | #Palindrome Permutation: Given a string, write a function to check if it is a permutation of a palindrome. A palindrome is a word or phrase that is the same forwards and backwards. A permutation is a rearrangement of letters. The palindrome does not need to be limited to just dictionary words
#EXAMPLE
#Input: Tact Coa
#Output: True (permutations: "taco cat", "atco cta", etc.)
#The spaces do not matter at the palindrom A: No they don't
#Do caps matter? am I receiving caps on the imput A:they dont matter in the palindrome but
# you will be receiving them
# special caracters matter? like hyphens or else? A: No palindrome its just for letters but
# you could assume you will be getting them
#Can i use any other data structure like hash or classes? A:yes
# I would say that a palindrome ignoring spaces has the same amout of chars for the left
# and the right side, if is odd then there is a "center" that its char count must be odd
# asd ddas for EXAMPLE
# count of chars: a=2 d=3 s=2 this means that its permutations of multiple palindromes
# another example "counting continue" c=2 o=2 u=2 n=4 t=2 i=2 g=1 e=1 this was so close but it has 2 odd counts so id assume that there is palindrome here
def palindrome(str1):#Solution O(n)
ith=0
caps_to_lower = ord('a') - ord('A')
record=[0]*255
while ith < len(str1):# this is O(n)
ith_c = ord(str1[ith])
if not (ith_c < ord('A') or (ith_c > ord('Z') and ith_c < ord('a')) or ith > ord('z')):
#this means im interested in this char
if ith_c >= ord('A') and ith_c <= ord('Z'):
ith_c+=caps_to_lower
record[ith_c]+=1
ith+=1
odd_count=0
for char in record:# this is O(n)
if char%2 == 1:
odd_count+=1
if odd_count>1:
return False
return True
print (palindrome("asd dedas"))
print (palindrome("Tact Coa"))
# Another solution would be sorting and check letter by letter
def palindrome2(str1):
new_str=sorted(str1.lower())#this is o(nlogn)
ith = 0
odds=0
while ith < len(new_str):#this is O(n)
current=new_str[ith]
if ord(current) < ord('a') or ord(current) > ord('z'):
ith+=1
continue
count_current=0
while ith < len(new_str):
ith+=1
if new_str[ith]==current:
count_current+=1
else:
if count_current%2==1:
odds+=1
if odds > 1:
return False
break
ith+=1
print (palindrome2("asd dedas"))
print (palindrome2("Tact Coa"))
| true |
9454cb7e1942bb081cb24c502bb71ff0d7007edd | PatrickBrennan92/hangman | /word_file_setup.py | 516 | 4.21875 | 4 | # This module was used to write only the basic words to a new file.
# It removed all words that contained numbers or other characters.
# Also removed any words starting with a capital letter, such as names and
# placed etc.
words = []
with open("words.txt", "r") as all_words:
for word in all_words:
if word.strip().isalpha() and not word.strip()[0].isupper():
words.append(word.strip())
with open("only_words.txt", "w") as only_words:
for i in words:
print(i, file=only_words)
| true |
b539fd8ada7794d0233a4ce7d09ef20b827dc180 | ShahanKrakirian/Python_Stack | /Python_Fundamentals/Fun_With_Functions.py | 988 | 4.34375 | 4 | #---------------------------
#Odd/Even
# def odd_even():
# for i in range(1,2001):
# # Even
# if i % 2 == 0:
# print "Number is:", i, "This is an even number."
# else:
# print "Number is:", i, "This is an odd number."
# odd_even()
#---------------------------
#Multiply
def multiply(my_list, number):
return list(map((lambda x: x*number), my_list))
# print multiply([1,2,3,4,5],2)
# Gives [2,4,6,8,10]
#Hacker Challenge
def layered_multiples(my_list, number):
working_list = multiply(my_list, number)
final_list = []
for i in range(0, len(working_list)):
print working_list
print "Working with the value:", working_list[i]
current_list = []
for j in range(0, working_list[i]):
current_list.append(1)
print "Adding this list:", current_list, "to:", final_list
final_list.append(current_list)
return final_list
print(layered_multiples([1,2,3],2))
| false |
aeecd07aeae8eae5ba81a766d8b8b120b2bc2baa | Sajid305/Python-practice | /Source code/Multithreading/Multithreading.py | 2,041 | 4.28125 | 4 |
# [1] multitasking
# Executing several task simultaneously is the concept of multitasking
# There are 2 types of multitasking
# [1] process based multitasking : Executing several task simultaneously where each task is
# a seperate independent process
# [2] Thread based multitasking : Executing several task simultaneously
# where each task is a seperate independent part of same programm and each independent part is called thread
# with multitasking
# [1] we can reduce Execution time and incris performance
# * animation
# * multi media graphics
# video games
# web servers and aplication servrs use this thread base
# gmail is from jboss server
# the ways of creating Theread Example
# [1] : Creating a therad without any class
# [2] : Creating a therad by extending therad class
# [3] : Creating a therad without extending therad class
# [1] creating a therad without any class
import threading
from threading import *
def whoIsthis():
for i in range(3):
print('Executing Therad : child') # it will show that it is executing main therad
c = Thread(target=whoIsthis) # this will make whoIsthis as a child therad
c.start()# now whoIsthis will replace main therad
# print(whoIsthis())
print('Executing Therad : main_thread')
# [2] : Creating a therad by extending therad class
class MyThread(Thread): # MyThread is child class of Therad
def run(self):# overwriting
for i in range(5):
print('Child Thread')
t=MyThread()
t.start()
print('This is ',current_thread().getName())
# [3] Creating a therad without extending therad class
class Test:
def another_class(self):
for i in range(2):
print('child thread')
obj = Test()
thread2= Thread(target=obj.another_class)
thread2.start()
| true |
398300731ca0b213825c12f0227f5fee7f171962 | Sajid305/Python-practice | /Source code/so many kind of function , like map,Enumerate,filter etc/any and all function .py | 1,877 | 4.40625 | 4 |
# any and all function
# finding even number from a list if all of the number
# is even then we will print true if one of them are odd then we will return false
# i will do it with the help of all function
# first with normal way of using all() function
# number1 = [2,4,6,8,10]
# even_number = []
# for num in number1:
# even_number.append(num % 2 == 0)
# print(all([True,True,True,True,True])) # -----> True
# print(all([True,True,True,False,True])) # ----> if we change value then it will show us false
# now with list comprehension of all() function
# number1 = [2,4,6,8,10]
# print(all([num%2==0 for num in number1]))
# same work with any() function
# number1 = [2,4,6,8,10]
# number2 = [1,3,5,7,9]
# even_number = []
# for num in number2:
# even_number.append(num%2 == 0)
# print(any(even_number)) # -----------> it will return us flase becuse any one of them is not true as given by condition
# now with list comprehension of any() function
# number1 = [2,4,6,8,10]
# print(any([num%2==0 for num in number1]))
# in this practice we will be make a function that allow us to sum of int
# but is there any string present then it will through us a messsg
# def is_int_or_float(*args):
# if all([(type(arg) == int or type(arg) == float) for arg in args]):
# total = 0 #------------> this block of code will run when input type is right if not then it will show wrong input
# for num in args:
# total += num
# return total
# else:
# return "wrong input type"
# print(is_int_or_float(1,2,3,4,5))
| true |
52a69f0f6abdd0093187519415589e6279960048 | Sajid305/Python-practice | /Source code/so many kind of function , like map,Enumerate,filter etc/function_extra_usefull_stuff.py | 505 | 4.28125 | 4 |
# Doc string
#''' this is doc string ''''
def func(a,b):
''' this is a doc string this function use tow argument and return addition of them'''
return a+b
print(func(1,3))
print(func.__doc__) # ----> this is how we can check our doc string
# We can allso see doc string of built in function with the help of help() function
print(help(len))
print(help(max)) # -----> this is how we can use help function
print(help(type))
print(help(help)) | true |
39c210d91f06287ec8dd4220b143a3660bc7d7cb | Sajid305/Python-practice | /Source code/Lambda Expression/Lambda Expression Practice .py | 1,045 | 4.28125 | 4 |
# Lambda Expression Practice
# def is_even(i):
# if i%2 == 0:
# return True
# else:
# return False
# print(is_even(5))
# def is_even(s):
# return s%2 == 0 # s%2 == 0 ----> true , false
# print(is_even(4))
# def return_last_carecter(s):
# return s[-1]
# print(return_last_carecter("shajid"))
# same function with lambda expression
# is_even = lambda s : s%2 == 0
# print(is_even(9))
# return_last_carecter = lambda s : s[-1]
# print(return_last_carecter("shajid"))
# lambda with if else
# def function(s):
# if len(s) > 5:
# return True
# else:
# return False
# print(function("sha"))
# def function(s):
# return len(s) > 5
# print(function("shajid"))
# Same function with if else
# function = lambda s : True if len(s)>5 else False
# print(function("sha"))
# function = lambda s : len(s)>5
# print(function("shajid"))
| false |
e97c86a58301197e130c6f615853a9491a09c1c2 | Sajid305/Python-practice | /Source code/Decoretor/Decorators with arguments .py | 916 | 4.125 | 4 |
# Decoretor with argument and nested Decoretor this decoretor will only take string as argument
from functools import wraps
def only_string_alow(data_type): # ----> this decoretor made for only take str as argument
def nested_decoretor(any_function):# ----> this decoretor is for taking function
@wraps(any_function)
def wrapper(*args,**kwargs):
if all([type(arg)== data_type for arg in args]):
return any_function(*args,**kwargs)
print('only string alowed')
return wrapper
return nested_decoretor
@only_string_alow(str)
def string_join(*args): # --> this function one or more argument and joind them together
joind = ''
for i in args:
joind += i
return joind
var = string_join('shajid',' rayhan') # --> if we pass a int on this this will return out else message
print(var)
| true |
23b43e444e3c4bb55cf56a3bd4f43f16cb9ec2be | Sajid305/Python-practice | /Source code/random practice/dict_summary.py | 1,316 | 4.375 | 4 | # summary dictionary
# what is dictionary
# unordered collection of data
d = {'name' : 'Shajid', 'age' : 23}
# or
d1 = dict(name = 'Shajid', age = 23)
# or
d2 = {
'name' : 'Shajid',
'age' : 23,
'fav_movies' : []
}
# how to access data from dictionary
# you cannot do like
# d[0] , because there is no order inside dictionary
# syntax
# print(dictname[keyname])
# print(d['name'])
# add data inside empty dict
empty_dict = {}
empty_dict['key1'] = 'value1'
empty_dict['key2'] = 'value2'
# print(empty_dict)
# check existence of values inside dict
# use in keyword to check for keys
# how to iterate over dictionary
# most common method
# for key , value in d.items():
# print(f' key is {key} and value is {value}')
# to print all keys
# for i in d:
# print(i)
# to print all values
# for i in d.values():
# print(i)
# most common dict methods
# get method
# to access a key and check existance
# print(d.get('name'))
# Q - why we use get
# A - to get rid of error
# example
# print(d['names'])
# print(d.get('names'))
# to delete item
# pop ---> take one argument which is keyname
# popped = d.pop('name')
# print(popped)
# print(d)
# popitem
popped = d.popitem()
print(popped)
print(d) | false |
7be42beca9ceebe90fd285ce02a3c575fe8ca11f | rsairam34/sample | /range.py | 251 | 4.1875 | 4 | #range function
print (range(10))
print (range(5,10))
print (range(0,10,3))
print (range(-10,-100,-30)
list = ["mary","had","a","little","lamb"]
for i in range(len(list)):
print i,list[i]
print "third edit locally"
print "fifth change locally" | false |
57f29174db1efe288074c09198d9ece7d3bfc950 | bhumphris/Word-Jumble | /Word Jumble.py | 651 | 4.15625 | 4 | import random
birds = ["Macaw", "Toucan", "Pheasant", "Painted Bunting", "Cardinal", "Crane", "Flamingo", "Parakeet", "Love Bird", "Mallard", "Finch", "Robin", "Dove", "Hawk", "Eagle"]
selection = random.choice(birds)
answer = selection
jumble = list(selection)
for current_index in range(len(jumble)):
random_index = random.randrange(0, len(jumble))
temp = jumble[current_index]
jumble[current_index] = jumble[random_index]
jumble[random_index] = temp
for letter in jumble:
print letter,
guess = raw_input("\nWhat kind of bird is jumbled?")
guess = guess.upper()
if guess == answer:
print("correct")
else:
print(answer)
| true |
ba66d8b9d3be607f5a1d710aafe320fa236c1300 | purushottamkaushik/DataStructuresUsingPython | /DyammicProgramming/BricksFilling.py | 1,067 | 4.34375 | 4 |
def BrickFilling(n):
"""
This is Problem which fills N*4 wall with 1*4 bricks
this is the recursive solution of the problem
:param n is the one dimension of the wall may be width or height
:return the number of ways you can put bricks on the wall
"""
if n==0 or n == 1 or n == 2 or n == 3 :
return 1
else:
return BrickFilling(n-1) + BrickFilling(n-4)
if __name__=="__main__":
n =5
print("Using recurison" , BrickFilling(n))
def BrickFillingDP(n):
"""
this is a function to implement brick filling problem using dynamic programming (Using bottom up Approach)
:param n: is the one of the dimensions of the wall
:return: integer : number of ways to fill the wall with the bricks
"""
table = [0] *(n+1)
# print(table)
for i in range(0,4):
table[i] = 1
# print(table)
for j in range(4,n+1):
table[j] = table[j-1] + table[j-4]
return table[n]
if __name__=="__main__":
n =5
print("Using recursive DP approach " , BrickFillingDP(n))
| true |
c432c895df59851779f957d02f1c9deb77441caa | purushottamkaushik/DataStructuresUsingPython | /ImplementTrieLeetcode208.py | 1,809 | 4.125 | 4 |
class Node:
def __init__(self,val,isWord=False):
self.val = val
self.child = {}
self.isWord = isWord
class Trie:
def __init__(self):
"""
Initialize your data structure here.
"""
self.root = Node("")
def insert(self, word: str) -> None:
"""
Inserts a word into the trie.
"""
# Setting current to the root Node
current = self.root
# Iterating character by character and checking it is not present in the child nodes if present then point to the node else create a new node with the character and point it towards the character.
for ch in word:
if ch not in current.child:
current.child[ch] = Node(ch)
current = current.child.get(ch)
current.isWord = True
def search(self, word: str) -> bool:
"""
Returns if the word is in the trie.
"""
current = self.root
for ch in word:
if ch in current.child:
current = current.child.get(ch)
else:
return False
return current.isWord
def startsWith(self, prefix: str) -> bool:
"""
Returns if there is any word in the trie that starts with the given prefix.
"""
current = self.root
for ch in prefix:
if ch in current.child:
current = current.child.get(ch)
else:
return False
return True
# Your Trie object will be instantiated and called as such:
# obj = Trie()
# obj.insert(word)
# param_2 = obj.search(word)
# param_3 = obj.startsWith(prefix) | true |
2244144b6ac2656b94a86bc65919def6364dec81 | purushottamkaushik/DataStructuresUsingPython | /ArraysProblem/Python/SelectionSort.py | 469 | 4.125 | 4 |
def selectionSort(a):
for i in range(len(a)):
"""Traverse through the whole array"""
min_index = i # storing the index as the minimum index
for j in range(i+1,len(a)): # from index i + 1 to last
if a[j] < a[min_index]:
min_index = j
a[i] , a[min_index ] = a[min_index ] , a[i] # swaping the current element
a = [3,2,4,51,1,43]
print("Original array" , a)
selectionSort(a)
print("After sorting " , a)
| true |
f522b05378e8ff60d3d22de6cbe3d927dceafd3d | rpedraza01/interview-prep | /Python/fizzbuzz_practice_again.py | 722 | 4.40625 | 4 | # print "Fizz" for numbers that are multiples of 3
# print "Buzz" for numbers that are multiples of 5
# print "FizzBuzz" for numbers that are multiples of 3 AND 5
# for num in range(1, 1001):
# if num % 3 == 0 and num % 5 == 0:
# # if num % 15 == 0:
# print("FizzBuzz")
# elif num % 3 == 0:
# print("Fizz")
# elif num % 5 == 0:
# print("Buzz")
# else:
# print(f'{num} is not a multiple of 3 and/or 5')
num = int(input('Please pick a number > '))
if num % 15 == 0:
print("FizzBuzz, your number is a multiple of 3 and 5")
elif num % 3 == 0:
print("Fizz, your number is a multiple of 3")
elif num % 5 == 0:
print("Buzz, your number is a multiple of 5")
else:
print(f'{num} is not a multiple of 3 and/or 5') | false |
93055cb2e17a911765c8d5851f748d21c6772a98 | aclogreco/lpthw | /ex20.py | 820 | 4.25 | 4 | # ex20.py
"""
Exercise 20 -- Learn Python the Hard Way -- Zed A. Shaw
A.C. LoGreco
"""
from sys import argv
# unpack cmd line arguments
script, input_file = argv
# print out the contents of file f
def print_all(f):
print f.read()
# set file cursor position to the begining of file f
def rewind(f):
f.seek(0)
# print a line from file f
def print_a_line(line_count, f):
print line_count, f.readline()
# open the file for reading
current_file = open(input_file)
print "First, let's print the whole file:\n"
print_all(current_file)
print "Now, let's rewind, kind of like a tape."
rewind(current_file)
print "Let's print three lines:"
current_line = 1
print_a_line(current_line, current_file)
current_line += 1
print_a_line(current_line, current_file)
current_line += 1
print_a_line(current_line, current_file)
| true |
2d5710d2e041ed64cc1502c61585800d1b25752f | lunettakim/Project | /turtle_game.py | 1,849 | 4.3125 | 4 | # Turtle game using the package 'turtle'
# Import relevant modules
import turtle
import random
import time
# Setting up a nice screen for our game
screen = turtle.Screen()
screen.bgcolor('pink') # Background colour
# We want two players and that whoever gets to the other sinde wins.
# Player one set up
player_one = turtle.Turtle()
# Colour of player one
player_one.color('red')
# Make this player a turtle
player_one.shape('turtle')
# Player two set up
player_two = player_one.clone()
player_two.color('green')
# Let's position our players
player_one.penup() # line will be removed
player_one.goto(-300,200)
player_two.penup()
player_two.goto(-300,-200)
# Let's draw a finish line
player_one.goto(300,-250)
player_one.left(90)
player_one.pendown()
player_one.color('black')
player_one.forward(500)
player_one.write('Finish', font=24)
# go back to the first place
player_one.penup()
player_one.color('red')
player_one.goto(-300,200)
player_one.right(90)
# We need to make sure both players have their pens down
player_one.pendown()
player_two.pendown()
# Let's create values for the die
die = [1,2,3,4,5,6]
# Create the game
for i in range(30): # put number high enough so turtle reach the finish line
if player_one.pos() >= (300,250):
print("Player One wins the race!")
break
elif player_two.pos() >= (300, -250):
print("Player Two wins the race!")
break
else:
die_roll = random.choice(die)
player_one.forward(30*die_roll)
time.sleep(1) #player moves and sleep for one second
#to make it a bit slower
die_roll2 = random.choice(die)
player_two.forward(30 * die_roll2)
time.sleep(1)
# this keeps the turtle drawing on the screen
turtle.done() | true |
61c5751ed2815fbbf28de27d166e258c0b06d7ee | Svagtlys/PythonExercises | /PasswordGenerator.py | 2,542 | 4.21875 | 4 | import random
import string
# Write a password generator in Python. Be creative with how you
# generate passwords - strong passwords have a mix of lowercase
# letters, uppercase letters, numbers, and symbols. The passwords
# should be random, generating a new password every time the user
# asks for a new password. Include your run-time code in a main method.
# Extra:
# Ask the user how strong they want their password to be.
# For weak passwords, pick a word or two from a list.
def weak(): #pick from word list
return random.choice(['strange','disgusting','chivalrous','decide','loud','vivacious','love','toothpaste','steal','defeated','wood','claim'])
def passgen(strength): #generator
'''
Takes in strength as an int defining how many characters to generate for the password.
Returns a result containing the defined number of psuedo-randomly generated characters.
'''
password = ''
max = 2
for i in range(strength):
chartype = random.randint(0,max) #0 = num, 1 = letter, 2 = specialchar
if chartype == 0:
password += random.choice(string.digits) #this was randint(0,9), but when I was looking for the letters and punctuation, I found this, so
elif chartype == 1:
password += random.choice(string.ascii_letters)
else:
password += random.choice(string.punctuation)
max = 1 #I got tired of all special chars
return password
# apparently this is a python main function
if __name__ == '__main__':
# I looked it up, apparently __name__ is a variable
# the file has set to its name, until it's executed,
# then its name is '__main__', so this here is asking
# if the file is being executed, essentially. Smart
stronglength = 16
mediumlength = 8
#first, ask for strength
#Two methods, weak and other:
# weak will pick from word list
# other has random gen, which picks length depending on med or strong
strength = input("What strength do you want your password to be? (S)trong, (M)edium, (W)eak\n").lower()
while(strength != "s" and strength != "m" and strength != "w"):
strength = input("Please type s for strong, m for medium, or w for weak.\n").lower()
if(strength == "w"):
print("Your weak password is: " + weak())
elif(strength == "m"):
print("Your medium password is: " + passgen(mediumlength))
else:
print("Your strong password is: " + passgen(stronglength)) | true |
eff41102ff734ac7ca4579b47549adee5ac6114a | Svagtlys/PythonExercises | /Fibonacci.py | 649 | 4.34375 | 4 | # Write a program that asks the user how many Fibonnaci
# numbers to generate and then generates them. Take this
# opportunity to think about how you can use functions.
# Make sure to ask the user to enter the number of numbers
# in the sequence to generate
def gen_fib(length, mylist = []):
if(len(mylist) < 2):
mylist.append(1)
else:
mylist.append(mylist[len(mylist)-1] + mylist[len(mylist)-2])
if(len(mylist) == length):
return mylist
else:
return gen_fib(length, mylist)
print(gen_fib(int(input("How many numbers of the Fibonacci sequence would you like to generate?\n")))) | true |
77844a8b6a18dd00e1f2dba0cc8f0b4800d0b927 | Svagtlys/PythonExercises | /ElementSearch.py | 1,148 | 4.1875 | 4 | # Write a function that takes an ordered list of numbers (a list where
# the elements are in order from smallest to largest) and another number.
# The function decides whether or not the given number is inside the list
# and returns (then prints) an appropriate boolean.
# Extras:
# Use binary search.
def search(myset, num):
'''
Takes in an ordered list or set, and a number to search for
Uses binary search to see if the number is in the list
Return True if it is or False if it isn't
'''
foundnum = None
while len(myset) >= 1: #if the halved list is over 1 and we haven't found the number we're searching for, keep searching
foundnum = myset[int((len(myset)-1)/2)]
if foundnum == num:
return True
elif(foundnum < num):
myset = myset[int((len(myset)-1)/2)+1:len(myset)]
else: #foundnum > num
myset = myset[0:int((len(myset)-1)/2)]
return False
if __name__ == "__main__":
myseta = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]
mysetb = [1, 3, 5, 30, 42, 43, 500, 700]
num = 13
print(search(mysetb, num)) | true |
6f3ecd911dec9a1c3b037b361a851a9f7a38a156 | Pravin-N/python-scripts | /pro_movefiletype.py | 1,619 | 4.34375 | 4 | #! Python3
# Write a program that walks through a folder tree and searches for files with a certain file extension
# (such as .pdf or .jpg). Copy these files from whatever location they are in to a new folder.
#import necessary modules to be used.
import os, shutil
from pathlib import Path
# create function that moves user specified files from the user specified folder to a user specified new folder.
def movefile(folder, filetype, newfolder):
if not os.path.exists(newfolder):
os.mkdir(newfolder)
if os.path.exists(folder):
for root, subfolder, files in os.walk(folder):
print(f"Searching {filetype} in {root}")
for file in files:
if file.endswith(filetype):
print(str(os.path.join(root, file)) + ' moved to ' + (newfolder))
shutil.copy(os.path.join(root, file), newfolder)
else:
continue
# Request input from user from the folder the folders need to be moved.
oldfolderpath = input(r'Please enter the full path of the folder to be moved: ')
print(oldfolderpath)
# Request input from user for the type of files need to moved.
typeoffile = '.' + str(input(r'Please enter the extention of the file to be moved (pdf, txt, doc, jpg, docx, py): '))
print(typeoffile)
# Request the folder name that would be saved on the desktop.
newfoldername = str(Path(r'C:\Users\ACE\Desktop', str(input(r'Enter the name of the folder to where the files will be moved. This will be on the desktop: '))))
print(newfoldername)
movefile(oldfolderpath, typeoffile, newfoldername)
| true |
cf86bea9b198d357c9d02c6d99ecbbbde2654a25 | susejzepol/Proyectos_python | /Find.divisors.of a.number.py | 637 | 4.25 | 4 | """
Task...
Find the number of divisors of a positive integer n.
Random tests go up to n = 500000.
Examples
divisors(4) = 3 # 1, 2, 4
divisors(5) = 2 # 1, 5
divisors(12) = 6 # 1, 2, 3, 4, 6, 12
divisors(30) = 8 # 1, 2, 3, 5, 6, 10, 15, 30
"""
def divisors(n):
count = 0
for numer in range(n):
var = divmod(n,numer+1)
if(var[1] == 0):
count += 1
return count
print("Calling the function....")
print("The divisor function returns : " + str(divisors(4)))
"""
Created by...
Jesus Lopez Mesia (https://www.linkedin.com/in/susejzepol/)
At date...
2019-10-14
""" | true |
666d9427145c035218c161ac7d5a0a31ae26db50 | kishanameerali/Coding_Dojo_Python1 | /Foo_and_bar.py | 753 | 4.21875 | 4 | #Foo and Bar
"""
Write a program that prints all the prime numbers and all the perfect squares for all
numbers between 100 and 100000.
For all numbers between 100 and 100000 test that number for whether it is prime or a
perfect square. If it is a prime number print "Foo". If it is a perfect square print "Bar".
If it is neither print "FooBar". Do not use the python math library for this exercise.
For example, if the number you are evaluating is 25, you will have to figure out if it is
a perfect square. It is, so print "Bar".
"""
for num in range (2,100):
for divider in range(2, num/2):
if (num % divider is 0):
print num
print "FooBar"
#if (num % i) != 0:
#print "foo"
#elif num**0.5
| true |
ceae7932304264bcb33d200039f44dd1c76daf0e | kishanameerali/Coding_Dojo_Python1 | /Checkerboard_2.py | 283 | 4.46875 | 4 | #Checkerboard Assignment using nested for loops
#Write a program that prints a 'checkerboard' pattern to the console.
for row in range(1,9):
for col in range(1,9):
if (row % 2 is odd and col % 2 is 0):
print " "
if (row % 2 is 0 and col % 2 is 0):
| true |
a4e50b4103bdb1bbb9decddaf3042c9f9ae6f82a | schebiwot/Python-class | /lesson_8/lesson_2.py | 1,194 | 4.25 | 4 | #inheritance
class Animal:#parent class
def __init__(self,name,country):
self.name=name
self.country=country
def printDetails(self):
print("The name is :{} and country of origin is:{}".format(self.name,self.country))
class Goat(Animal):#child class
#pass #used when you dont want to add any code
def __init__(self,name,country,owner): #the child will no longer inherit parent's
self.name=name
self.country=country
self.owner=owner
def printDetails(self):
print("The name is :{} and country of origin is:{} ,the owner is {}".format(self.name,self.country,self.owner))
#object
g1=Goat("gush","kenya" ,"jon")
g1.printDetails()
g2=Goat("brayo","sudan",'doe')
g2.printDetails()
class Dog(Animal):
def __init__(self,name,country,color):
# Animal.__init__(self,name,country) #for a child to inherit from the parent class 'animal'
# inhert from parent
super().__init__(name,country)
self.color=color
def printDetails(self):
print("name: {} and country : {} and the color is:{}".format(self.name,self.country,self.color))
d1=Dog('smooth',"chiwawa","red")
d1.printDetails() | false |
ee17f742cf9883dde7031e7caa3b2ac50a8969b7 | libra202ma/cc150Python | /chap04 Trees and Graphs/4_5.py | 1,015 | 4.15625 | 4 | """
Implement a function to check if a binary tree is binary search
tree.
- Search. Modified, in-order DFS and keep track of a global variable
last_visited. Check if left branch is binary search tree, if false,
return false. Check if current data is less or equal to global min, if
yes, retrun false. Update the last_visited. Check if the right branch
is balanced. (???)
NOTE: check only left.data <= current.data < right.data is not
sufficient. The requirement is that ALL left data <= current data <
ALL right data.
- DFS. Use a range (min, max). When branch left, the max gets updated
using current data. When branch right, the min updated using current
data. Then the (min, max) passed to the recursion of left or right
check. If the current data does not satisfy the passed range, return
false.
"""
def isBST(t, min=float('-inf'), max=float('inf')):
if not min <= t.data < max:
return False
if t.left:
isBST(t.left, min, t.data)
if t.right:
isBST(t.right, t.data, max)
| true |
12a0da6e4b394d7f570ebded0c4d733c735d32d0 | libra202ma/cc150Python | /chap05 Bit Manipulation/5_1.py | 984 | 4.15625 | 4 | """
You are given two 32-bit number, N and M, and two bit positions, i
and j. Write a method to insert M into N such that M starts at bit j
and ends at bit i. You can assume that the bits j through i have
enough space to fit all of M. That is, if M = 10011, you can assume
that there are at least 5 bits between j and i. You would not, for
example, have j = 3 and i = 2, because M could not fully fit between 3
and 2.
EXAMPLE
Input: N = 10000000000, M = 10011, i = 2, j = 6
Output: N = 10001001100
- update Bits. Firstly clear bits [i:j] in N, using AND masks
like 1111000111. Then shift M to [i, j], perform an OR operation on N.
"""
def insert(N, M, i, j):
# clear bits from i to j
clearmask = 0
for k in range(i, j+1):
clearmask += 1 << k
clearmask = ~clearmask
N &= clearmask
# set bits
N |= M << i
return N
def test_insert():
N = 0b10000000000
M = 0b10011
i = 2
j = 6
assert insert(N, M, i, j) == 0b10001001100
| true |
30dd034601cb0942414a7d6d33931bc114d71a82 | libra202ma/cc150Python | /chap03 Stacks and Queues/3_5.py | 1,134 | 4.21875 | 4 | """
Implement a MyQueue class which implements a queue using two
stacks.
- Naive. For every enqueue operation, just enqueue it onto s1. For
every dequeue operation, we first pop all nodes from s1 to s2, then
pop from s2 to get the oldest element, then pop all nodes from s2
back to s1.
- lasy execuation. Again for enqueue operation, we push it onto
s1. But exec dequeue from s2. If s2 is empty, poping all nodes from
s1 to s2. If s2 is not empty, the top of s2 is oldest one, just pop
it. (WOW)
"""
class MyQueue:
def __init__(self):
self.newstack = []
self.oldstack = []
def push(self, data):
self.newstack.append(data)
def pop(self):
if self.oldstack:
return self.oldstack.pop()
elif self.newstack:
# dump newstack to oldstack
while self.newstack:
self.oldstack.append(self.newstack.pop())
return self.oldstack.pop()
else:
return None
def test_myqueue():
q = MyQueue()
q.push(1)
q.push(2)
assert q.pop() == 1
assert q.pop() == 2
assert q.pop() == None
| true |
3d2a1ac662fb5fa45415c0d438a8d6b1c91fcda6 | libra202ma/cc150Python | /chap03 Stacks and Queues/3_3.py | 1,800 | 4.125 | 4 | """
Imagine a (literal) stack of plates. If the stack gets too high,
it might topple. Therefore, in real life, we would likely start a new
stack when the previous stack exceeds some threshold. Implement a data
structure SetOfStacks that mimics this. SetOfStacks should be composed
of several stacks and should create a new stack once the previous one
exceeds capacity. SetOfStacks.push() and SetOfStacks.pop() should
behave identically to a single stack (that is, pop() should return
the same value as it would if there where just a single stack).
- naive. use variables/lists to log all stack and current active stack.
FOLLOWUP
Implement a function popAt(int index) which performs a pop operation
on a specific sub-stack.
- same as above.
"""
class Stack():
def __init__(self):
self.arrlen = 10
self.data = []
self.narrs = 0
self.addarr()
self.top = -1
def addarr(self):
self.data.append([-1] * self.arrlen)
self.narrs += 1
def push(self, data):
if self.top == self.arrlen - 1:
self.addarr()
self.top = 0
else:
self.top += 1
self.data[self.narrs-1][self.top] = data
def pop(self):
res = self.data[self.narrs-1][self.top]
if self.top == 0:
self.data.pop() # remove the last empty stack
self.narrs -= 1
self.top = self.arrlen - 1
else:
self.top -= 1
return res
def peek(self):
return self.data[self.narrs-1][self.top]
def test_stack():
s = Stack()
for i in range(15):
s.push(i)
assert s.peek() == 14
for i in range(5):
s.pop()
assert s.peek() == 9
for i in range(10):
s.pop()
s.push(1)
assert s.pop() == 1
| true |
0b65056f2e3025cc5812aa3763c9e6d2fbb9ecd4 | libra202ma/cc150Python | /chap09 Recursion and Dynamic Programming/9_09.py | 1,138 | 4.125 | 4 | """
Write an algorithm to print all ways of arranging eight queens on
an 8x8 chess board so that none of them share the same row, column or
diagonal. In this case, "diagonal" means all diagonals, not just the
two that bisect the board.
- classic. Firstly, the queens should be on different rows and
columns. That is, for every row or column,there is one queen
exactly. Thus from the point of row, the queen just have the freedom
of column. Then for-loop or recursion can be applied to find all
possible arrangement.
"""
def eightqueen(n):
if n == 1:
return [[i] for i in range(1, 9)]
arrangements = []
for subarr in eightqueen(n - 1):
for c in range(1, 9): # check all possible columns
if c in subarr: # column occupied
continue
for qr, qc in enumerate(subarr):
if abs(c - qc) == abs(n - qr - 1):
break
else: # not on same column and diagonal, Python's special 'for-else' clause
arrangements.append(subarr + [c])
return arrangements
def test_eightqueen():
assert len(eightqueen(8)) == 92
| true |
bc71c13906df3fbef02182d3962078664612d798 | HareshSankaliya/HareshPython | /HareshPy/HareshPractice/Datetimeformat.py | 539 | 4.21875 | 4 | from datetime import date
from datetime import time
from datetime import datetime
today=date.today() #today date
print(today)
print(today.strftime("%Y")) # Print only year with yyyy format
print(today.strftime("%y")) # Print only year with yy format
print(today.strftime("%a")) # Print today week name
print(today.strftime("%d")) # Print today date dd format
print(today.strftime("%B")) # Print month name
print(today.strftime("%a,%d %B,%y")) # Print week,date,month,year
print(today.strftime("%x")) # print today date in mm/dd/yy format
| true |
a94cce2fcc9f305be51aa53edd18b54745182aec | dsmilo/DATA602 | /hw2.py | 2,525 | 4.1875 | 4 | # Dan Smilowitz DATA 602 hw2
#1. fill in this class
# it will need to provide for what happens below in the
# main, so you will at least need a constructor that takes the values as (Brand, Price, Safety Rating),
# a function called showEvaluation, and an attribute carCount
class CarEvaluation:
'A simple class that represents a car evaluation'
#all your logic here
carCount = 0
def __init__(self, brand = '', price = '', safety = 0):
self.brand = brand
self.price = price
self.safety = safety
CarEvaluation.carCount += 1
def showEvaluation(self):
#The Ford has a High price and it's safety is rated a 2
print("The %s has a %s price and its safety is rated a %d" %(self.brand, self.price, self.safety))
#2. fill in this function
# it takes a list of CarEvaluation objects for input and either "asc" or "des"
# if it gets "asc" return a list of car names order by ascending price
# otherwise by descending price
def sortbyprice(car_list, order = ""):
sorted_cars = []
is_desc = True
if order.lower() == "asc": is_desc = False
price_num = {'High': 3, 'Med': 2, 'Low': 1}
car_list.sort(key= lambda x: price_num[x.price], reverse = is_desc)
for i in range(len(car_list)):
sorted_cars.append(car_list[i].brand)
return sorted_cars
#3. fill in this function
# it takes a list for input of CarEvaluation objects and a value to search for
# it returns true if the value is in the safety attribute of an entry on the list,
# otherwise false
def searchforsafety(car_list, car_rating):
found = False
for item in car_list:
if item.safety == car_rating:
found = True
return found
# This is the main of the program. Expected outputs are in comments after the function calls.
if __name__ == "__main__":
eval1 = CarEvaluation("Ford", "High", 2)
eval2 = CarEvaluation("GMC", "Med", 4)
eval3 = CarEvaluation("Toyota", "Low", 3)
print "Car Count = %d" % CarEvaluation.carCount # Car Count = 3
eval1.showEvaluation() #The Ford has a High price and its safety is rated a 2
eval2.showEvaluation() #The GMC has a Med price and its safety is rated a 4
eval3.showEvaluation() #The Toyota has a Low price and its safety is rated a 3
L = [eval1, eval2, eval3]
print sortbyprice(L, "asc"); #[Toyota, GMC, Ford]
print sortbyprice(L, "des"); #[Ford, GMC, Toyota]
print searchforsafety(L, 2); #true
print searchforsafety(L, 1); #false
| true |
98bb48d3f58796ed29294a615af14a314f4c8284 | carlgriffin57/python-exercises | /data_types_and_variables.py | 2,134 | 4.21875 | 4 | # You have rented some movies for your kids: The little mermaid (for 3 days), Brother Bear (for 5 days, they
# love it), and Hercules (1 day, you don't know yet if they're going to like it). If price for a movie per day
# is 3 dollars, how much will you have to pay?
num_of_little_mermaid_days = 3
num_of_brother_bear_days = 5
num_of_hercules_days = 1
price_per_day = 3
total_price = (num_of_little_mermaid_days + num_of_brother_bear_days + num_of_hercules_days) * price_per_day
print(total_price, "dollars")
# Suppose you're working as a contractor for 3 companies: Google, Amazon and Facebook, they pay you a different
# rate per hour. Google pays 400 dollars per hour, Amazon 380, and Facebook 350. How much will you receive in
# payment for this week? You worked 10 hours for Facebook, 6 hours for Google and 4 hours for Amazon.
google_pay_per_hour = 400
amazon_pay_per_hour = 380
facebook_pay_per_hour = 350
total_payment = (google_pay_per_hour * 6) + (amazon_pay_per_hour * 4) + (facebook_pay_per_hour * 10)
print(total_payment, "dollars")
# A student can be enrolled to a class only if the class is not full and the class schedule does not conflict
# with her current schedule.
class_full = True
conflicting_schedule = True
student_enrollment = not class_full or not conflicting_schedule
print(student_enrollment)
# A product offer can be applied only if people buys more than 2 items, and the offer has not expired. Premium
# members do not need to buy a specific amount of products.
item_count = 3
offer_expired = False
premium_member = True
product_offer = (item_count > 2 and not offer_expired) or (premium_member and not offer_expired)
print(product_offer)
username = 'codeup'
password = 'notastrongpassword'
# the password must be at least 5 characters
password_length = len(password) >= 5
print(password_length)
# the username must be no more than 20 characters
username_length = len(username) < 21
print(username_length)
# the password must not be the same as the username
different = password != username
print(different)
# bonus neither the username or password can start or end with whitespace
bonus_answer =
| true |
babec0f381668c48efa3dec3a386790cd4398241 | Geogy-fjq/Python | /Python/Basic/lab4-Array-2.py | 432 | 4.3125 | 4 | """
用程序实现两个字符串的比较、追加、拷贝。
具体流程如下：
Step1：输入两个字符串。
Step2：使用选择语句比较，使用简单符号计算追加、拷贝。
Step3：输出结果。
"""
str1=str(input("str1 is:"))
str2=str(input("str2 is:"))
if str1<str2:
print("str1<str2")
elif str1>str2:
print("str1>str2")
else:
print("str1=str2")
str1+=str2
print(str1)
str1=str2
print(str1)
| false |
bc6b770bb9a073c1794d688cff709c3dba067ea4 | Geogy-fjq/Python | /Python/Basic/lab1-Selection-1.py | 508 | 4.15625 | 4 | """
熟悉RAPTOR算法设计环境。设计一个程序，输出一个提示信息，请求用户输入一个数字，
如果大于0则输出“Positive”，如果小于0则输出“Negative”，如果等于0则输出“Zero”。
基本流程如下：
Step1：请求输入一个数字。
Step2：判断输入数字的正负性质，对应输出结果。
"""
num = eval(input("Please enter a number."))
if num > 0:
print("Positive")
elif num < 0:
print("Negative")
elif num == 0:
print("Zero")
| false |
2e4e4bf90864d841a3ff63b8e42d5ee8c319ff9f | arvindrocky/dataStructures | /trees/binary_search_tree.py | 1,228 | 4.21875 | 4 | class BST:
def __init__(self, value: int):
self.value: int = value
self.left_node: BST = None
self.right_node: BST = None
def insert_node(self, value: int) -> None:
if value < self.value:
if self.left_node:
self.left_node.insert_node(value)
else:
self.left_node = BST(value)
else:
if self.right_node:
self.right_node.insert_node(value)
else:
self.right_node = BST(value)
def print_in_order_tree(self) -> None:
if self.left_node:
self.left_node.print_in_order_tree()
print(self.value)
if self.right_node:
self.right_node.print_in_order_tree()
bst = BST(100)
print("Printing tree with only root:")
bst.print_in_order_tree()
bst.insert_node(50)
bst.insert_node(150)
bst.insert_node(20)
bst.insert_node(125)
bst.insert_node(130)
print("Printing In Order of a tree:")
bst.print_in_order_tree()
bst1 = BST(1)
bst1.insert_node(2)
bst1.insert_node(3)
bst1.insert_node(4)
bst1.insert_node(5)
bst1.insert_node(6)
bst1.insert_node(7)
print("Printing In Order of another tree which is BFT:")
bst1.print_in_order_tree()
| false |
a240a672532d266ab35092738f9a51726a4f5b41 | DVEC95/PY4E | /Chapter 5 - Iterations/PY4E_L5_Ex1.py | 812 | 4.1875 | 4 | # Exercise 1:
# Write a program which repeatedly reads numbers until the user enters “done”.
# Once “done” is entered, print out the total, count, and average of the numbers.
# If the user enters anything other than a number, detect their mistake using try and except and print an error message and skip to the next number.
numbers = []
while True:
user_input = input('Enter a number: \n')
try:
if user_input != 'done':
numbers.append(int(user_input))
except:
print('Please enter a numeric value or type "done" to exit.')
if user_input == 'done':
break
print('Numbers entered:', len(numbers))
print('Sum of numbers:', sum(numbers))
if len(numbers) > 0:
print('Average:', sum(numbers) / len(numbers))
else:
print('Average not available.') | true |
019feef40b84fc4261d3130d738fc2d44ad64019 | DVEC95/PY4E | /Chapter 7 - Files/PY4E_L7_Ex2.py | 943 | 4.3125 | 4 | # Exercise 2:
# Write a program to prompt for a file name, and then read through the file and look for lines of the form:
# X-DSPAM-Confidence: 0.8475
# When you encounter a line that starts with “X-DSPAM-Confidence:” pull apart the line to extract the floating-point number on the line.
# Count these lines and then compute the total of the spam confidence values from these lines.
# When you reach the end of the file, print out the average spam confidence.
file_name = input('Please enter a file name: \n')
try:
txt_file = open(file_name, 'r')
except:
if file_name == 'na na boo boo':
print('Grow up.')
print('Please enter a valid file name.')
exit()
num_array = []
for line in txt_file:
if line.startswith('X-DSPAM-Confidence:'):
num = line[line.find(':') + 2:line.find('\n')]
num_array.append(float(num))
print('Average spam confidence:', sum(num_array) / len(num_array)) | true |
4d12e59d208e7ddcaf9bf6df41c8f7436a4a3f23 | orvindemsy/python-practice | /19.08.16 beginner project/letterCapitalizeVer2.py | 769 | 4.21875 | 4 | '''
Written by: Orvin Demsy
Date: 17 August 2019
Source: https://coderbyte.com/solution/Letter%20Changes#Python
I'm rewriting it to get a better understanding
Challenge:
Have the function LetterCapitalize(str) take the str parameter being passed and
capitalize the first letter of each word. Words will be separated by only one space.
e.g.
Input:"hello world"
Output:"Hello World"
Input:"i ran there"
Output:"I Ran There"
'''
def LetterCapitalize(sentence):
#Split the sentence into array of words
word = sentence.split()
for i in range(len(word)):
#Create a word with first letter capitalized
word[i] = word[i][0].upper() + word[i][1:]
return " ".join(word)
sentence = "the man who can't be moved"
print(LetterCapitalize(sentence)) | true |
ad82ed794489e9e0326d8b6ffddde41c85f871d0 | orvindemsy/python-practice | /CodeWars 8kyu Challenges/isItPalindromeVer2.py | 353 | 4.125 | 4 | '''
Written by: Orvin Demsy
Date: 19 August 2019
Description:
Write function isPalindrome that checks if a given string (case insensitive) is a palindrome.
Only one-word input is accepted
e.g.
madam = True
Mom = True
walter = False
'''
def is_palindrome(word):
word = word.lower()
return word == word[::-1]
word = 'Mom'
print(is_palindrome(word)) | true |
59284127bf0405c4b2e4d14115a38d4e48bcb343 | orvindemsy/python-practice | /19.08.16 beginner project/letterChanges.py | 1,400 | 4.34375 | 4 | '''
Written by: Orvin Demsy
Date: 16 August 2019
Challenge :
Have the function LetterChanges(str) take the str parameter being passed and modify it using the following algorithm.
Replace every letter in the string with the letter following it in the alphabet (ie. c becomes d, z becomes a).
Then capitalize every vowel in this new string (a, e, i, o, u) and finally return this modified string.
e.g.
Input:"hello*3"
Output:"Ifmmp*3"
Input:"fun times!"
Output:"gvO Ujnft!"
All capitalized letter will be lower-cased
'''
import string
alphabet = string.ascii_lowercase
alphabet = list(alphabet)
word = input("Enter a sentence or words : ")
word = word.lower()
new_word = ""
for i in range(len(word)):
#check if the character is alphabet
if word[i].isalpha():
#check if the character is z
if word[i] == "z":
new_word = new_word + "A"
#check if the character is not z
elif word[i] != "z":
new_idx = alphabet.index(word[i]) + 1
#check if the new character is alphabet
if alphabet[new_idx].lower() in "aiueo":
new_word = new_word + alphabet[new_idx].upper()
else:
#check if the new character is not alphabet
new_word = new_word + alphabet[new_idx]
#check if the character is not alphabet
else:
new_word = new_word + word[i]
print(new_word)
| true |
4bc7c98ee2f36bc3bb74268cad48108be82b5f44 | orvindemsy/python-practice | /19.08.16 beginner project/reverseString.py | 737 | 4.375 | 4 | '''
Written by: Orvin Demsy
Date: 16 August 2019
Create a function that takes a string as an argument
and spit out that string in reverse
e.g:
hello -> olleh
'''
#The first method extended slice syntax
def reverse_word_1(word):
return word[::-1] #Return olleH
print("The reversed input is = " + reverse_word_1(input("Enter a sentences/ a word : ")))
#The second method
def reverse_word_2(word):
return ''.join(reversed(word))
print("The reversed input is = " + reverse_word_2(input("Enter a sentences/ a word : ")))
#Third method using loop
def reverse_word_3(word):
str = ""
for i in word:
str = i + str
return str
print("The reversed input is = " + reverse_word_3(input("Enter a sentences/ a word : ")))
| true |
3552eb59106c8bcf07113cd7d23c516e179b55b7 | muradsamadov/python_learning | /python_exercises_practice_solution/python_basic/part_25.py | 238 | 4.25 | 4 | # Write a Python program to concatenate all elements in a list into a string and return it.
def function(list):
result = ""
for i in list:
result = result + str(i)
return result
print(function([1, 5, 12, 2])) | true |
92907bac9ec611f8f5550bfa200d6c9b11d093d7 | muradsamadov/python_learning | /python_exercises_practice_solution/python_modules/module_random/part_4.py | 927 | 4.53125 | 5 | # Write a Python program to generate a random integer between 0 and 6 - excluding 6, random integer between 5 and 10 - excluding 10, random integer between 0 and 10, with a step of 3 and random date between two dates. Use random.randrange()
import random
import datetime
print("Generate a random integer between 0 and 6:")
print(random.randrange(5))
print("Generate random integer between 5 and 10, excluding 10:")
print(random.randrange(start=5, stop=10))
print("Generate random integer between 0 and 10, with a step of 3:")
print(random.randrange(start=0, stop=10, step=3))
print("\nRandom date between two dates:")
start_dt = datetime.date(2019, 2, 1)
end_dt = datetime.date(2019, 3, 1)
time_between_dates = end_dt - start_dt
days_between_dates = time_between_dates.days
random_number_of_days = random.randrange(days_between_dates)
random_date = start_dt + datetime.timedelta(days=random_number_of_days)
print(random_date)
| true |
0088dd0024d83b49b2ce07683de3a55b9bb53fd5 | muradsamadov/python_learning | /other_advanced_concepts/list_set_dict_compherension.py | 566 | 4.15625 | 4 | #List / Set / Dictionary comprehensions
#Instead of...
list1 = []
for i in range(10):
j = i ** 2
list1.append(j)
print(list1) #yuxarida qeyd olunmus misalda sade list uzerinde loop yaziriq
#...we can use a list comprehension
list2 = [x ** 2 for x in range(10)] #bu misalda ise comprehensions-dan istifade ederek daha qisa sekilde yaziriq
list3 = [x ** 2 for x in range(10) if x > 5] #with a conditional statament
set1 = {x ** 2 for x in range(10)} #set comprehension
dict1 = {x: x * 2 for x in range(10)} #dictionary comprehension | false |
0c6b8cb7548695d44b7ed9fecea11395a19ee010 | mellab/holbertonschool-higher_level_programming | /0x01-python-if_else_loops_functions/8-uppercase.py | 255 | 4.125 | 4 | #!/usr/bin/python3
def uppercase(str):
upper = 0
for up in str:
if ord(up) >= ord('a') and ord(up) <= ord('z'):
upper = 32
else:
upper = 0
print("{:c}".format(ord(up) - upper), end='')
print("")
| false |
fdc453807fbadc53426aa1ec020cd68e8253067f | Stefanos1312/Hello-World | /task 1 section 3.py | 384 | 4.25 | 4 | print ("This program will add 2 different numbers and will find the results")
#define numbers as integer
number1 = int(input("please enter the first number"))
number2 = int(input("please enter the second number"))
print ("thank you")
#now I will define the results by just adding '+' the numbers together
result = number1 + number2
print ("the result is")
print (result)
| true |
97bd113620df5f0f1b70d6be454811cfe743f0ce | SakshiBheda/Apni_Dictionary | /apni_dictionary.py | 457 | 4.125 | 4 | print("This is My own Dictionary : 'Apni Dictionary'")
print("key ,"" book "", comb "" ,mouse "" , joey ")
dic1={"key":"something which unlocks a lock",
"book":"a compiled set of pages which describes a certain topic",
"comb":"a plastic item to set hairs",
"mouse":"an animal which is afraid of cats",
"joey":"kid of a kangroo"}
print("Please enter the word whose meaning you want")
ax=input()
print("It means : " ,dic1[ax]) | true |
993fd1bcd8d453af376f78ad4357e32e3f38800a | DianaLuca/Algorithms | /Leetcode_WeeklyContests/contest_49/ImplementMagicDictionary676.py | 1,964 | 4.125 | 4 | """
Implement a magic directory with buildDict, and search methods.
For the method buildDict, you'll be given a list of non-repetitive words to build a dictionary.
For the method search, you'll be given a word, and judge whether if you modify exactly one character into another character
in this word, the modified word is in the dictionary you just built.
Example 1:
Input: buildDict(["hello", "leetcode"]), Output: Null
Input: search("hello"), Output: False
Input: search("hhllo"), Output: True
Input: search("hell"), Output: False
Input: search("leetcoded"), Output: False
Note:
You may assume that all the inputs are consist of lowercase letters a-z.
For contest purpose, the test data is rather small by now.
You could think about highly efficient algorithm after the contest.
Please remember to RESET your class variables declared in class MagicDictionary,
as static/class variables are persisted across multiple test cases.
"""
from collections import defaultdict
class MagicDictionary(object):
def __init__(self):
"""
Initialize your data structure here.
"""
self.dictionary = defaultdict(list)
def buildDict(self, dict):
"""
Build a dictionary through a list of words
:param dict: List[str]
:return: void
"""
for word in dict:
self.dictionary[len(word)].append(word)
def search(self, word):
"""
Returns if there is any word in the trie that equals to the given word after modifying exactly one character
:param word: str
:return: bool
"""
if len(self.dictionary[len(word)]) == 0:
return False # word is not in dictionary
for eachword in self.dictionary[len(word)]:
diff = 0
for i in range(len(word)):
if eachword[i] != word[i]:
diff += 1
if diff == 1:
return True
return False
| true |
5efa072bc226c13308ed74407f0e6c86fcd4c72e | DianaLuca/Algorithms | /Leetcode_WeeklyContests/contest_60/AsteroidCollision735.py | 1,306 | 4.1875 | 4 | """
We are given an array asteroids of integers representing asteroids in a row.
For each asteroid, the absolute value represents its size, and the sign represents its direction
(positive meaning right, negative meaning left). Each asteroid moves at the same speed.
Find out the state of the asteroids after all collisions. If two asteroids meet, the smaller one will explode.
If both are the same size, both will explode. Two asteroids moving in the same direction will never meet.
Example 1:
Input:
asteroids = [5, 10, -5]
Output: [5, 10]
Explanation:
The 10 and -5 collide resulting in 10. The 5 and 10 never collide.
Note:
The length of asteroids will be at most 10000.
Each asteroid will be a non-zero integer in the range [-1000, 1000]..
"""
class Solution(object):
res = []
def asteroidCollision(self, asteroids):
"""
:type asteroids: List[int]
:rtype: List[int]
"""
stk = []
for a in asteroids:
if a > 0:
stk.append(a)
else:
while stk and (stk[-1] > 0) and (abs(a) > stk[-1]):
stk.pop()
if not stk or stk[-1] < 0:
stk.append(a)
elif stk[-1] == abs(a):
stk.pop()
return stk
| true |
70b56e0356fa654b58a288e4de9c01e85d5e6d04 | narendra-manchala/GFG-DSA | /bitwise algorithms/power_of_2.py | 518 | 4.53125 | 5 | def power_of_2(n):
"""Find if the number is power of 2 or not.
Arguments:
n {int} -- The number to find if it is power of 2 or not.
Explanation:
1. In binary a power of 2 has only one 1. So doing a bitwise-and with n-1
will give us 0 if it is power of 2.
eg: n = 4 (100)
n-1 = 3(011)
4&3 => 000
2. For n is 0, explecitly need to return false.
"""
if n and not n&n-1:
print(True)
else: print(False) | true |
e9e9b78d7898e71dd2a62962bcde4a206bf13d74 | rwisecar/scrabble_cheat | /scrabble.py | 1,687 | 4.15625 | 4 | """ This is a simple python script to take a Scrabble rack of 7 letters and return all possible word iterations, with the associated scores."""
import re
# Dictionary of letters and values
scores = {
"a": 1,
"c": 3,
"b": 3,
"e": 1,
"d": 2,
"g": 2,
"f": 4,
"i": 1,
"h": 4,
"k": 5,
"j": 8,
"m": 3,
"l": 1,
"o": 1,
"n": 1,
"q": 10,
"p": 3,
"s": 1,
"r": 1,
"u": 1,
"t": 1,
"w": 4,
"v": 4,
"y": 4,
"x": 8,
"z": 10
}
# Script to open and read the sowpods.txt word file, and strip of extraneous characters
sowpods = []
score = 0
def compile_sowpods():
access_sowpods = open("sowpods.txt", "r+")
for i in access_sowpods:
i = i.strip(' \n\r')
sowpods.append(i)
access_sowpods.close()
# Get the scrabble rack from the user
def intro():
user_name = raw_input("Hello there! I see you want to win at Scrabble. I can help. First, tell me your name.")
print "Nice to meet you, %s." % user_name
def get_rack():
input = raw_input("Now tell me, what 7 letters are we working with here? ").upper()
rack = ""
if len(input) < 7 or len(input) > 7:
print("I'm sorry, you haven't entered a full rack of 7 letters. Please try again.")
exit()
else:
for i in input:
if re.match(r'^[A-Z]', i):
rack = rack + i
else:
print("I'm sorry, that is not a valid Scrabble rack. Please try again.")
exit()
print "Your rack is: %s" % rack
def print_options():
compile_sowpods()
intro()
get_rack()
for word in sowpods:
for r in rack:
| true |
3d1d244bef39869f8192536a369f6124ff594960 | ciaranmccormick/advent-of-code-2019 | /2/main.py | 2,303 | 4.15625 | 4 | #! /bin/env python3
from argparse import ArgumentParser, Namespace
from typing import List, Tuple
from computer import Computer
STOP = 99
ADD = 1
MULTIPLY = 2
def load_instructions(filename: str) -> List[int]:
"""Read in a list of comma separated integers"""
with open(filename, "r") as f:
codes = f.readline().split(",")
codes = [int(code) for code in codes]
return codes
def process_opcodes(opcodes: List[int]) -> List[int]:
"""
Processes the list of opcodes and applies instructions.
Returns the modified opcode list.
"""
i = 0
inputs = list(opcodes) # make a copy of the list
while i < len(inputs):
opcode = inputs[i]
if opcode == STOP:
break
pos1, pos2, dest = inputs[i + 1 : i + 4]
val1 = inputs[pos1]
val2 = inputs[pos2]
if opcode == ADD:
result = val1 + val2
elif opcode == MULTIPLY:
result = val1 * val2
inputs[dest] = result
i += 4
return inputs
def part_two(intcodes: List[int]) -> Tuple[int, int]:
"""Get solution to Part Two"""
for noun in range(1, 100):
for verb in range(1, 100):
intcodes[1] = noun
intcodes[2] = verb
comp = Computer(intcodes)
processed = comp.run()
if processed[0] == 19690720:
return noun, verb
def part_one(opcodes: List[int]) -> int:
"""
Get solution to Part One
"""
opcodes[1] = 12
opcodes[2] = 2
comp = Computer(opcodes)
codes = comp.run()
return codes[0]
def parse_args() -> Namespace:
"""Parse arguments."""
parser = ArgumentParser()
parser.add_argument("filename", help="File containing problem inputs.", type=str)
parser.add_argument(
"multiplier", help="The number multiply noun and verb by.", type=int
)
args = parser.parse_args()
return args
def main():
args = parse_args()
opcodes = load_instructions(args.filename)
zero_value = part_one(opcodes)
print(f"Part One: answer={zero_value}")
noun, verb = part_two(opcodes)
part_two_answer = args.multiplier * noun + verb
print(f"Part Two: noun={noun}, verb={verb}, answer={part_two_answer}")
if __name__ == "__main__":
main()
| true |
47f34d52f596047310cbf2fa23a934181b81ff12 | lucasbpaixao/lista_exercicios_python | /ex09.py | 323 | 4.28125 | 4 | #Faça um Programa que peça a temperatura em graus Fahrenheit, transforme e mostre a temperatura em graus Celsius.
#C = 5 * ((F-32) / 9).
fahrenheit = float(input('Informe a temperatura em Fahrenheit: '))
celsius = 5 * ((fahrenheit - 32) / 9)
print('A temperatura em celsius é {celsius:.2f}'.format(celsius = celsius)) | false |
af45aee1ab6efe5194766784c94439a4c201852f | JaimePazLopes/dailyCodingProblem | /problem086.py | 1,561 | 4.21875 | 4 | # Problem #86
# Given a string of parentheses, write a function to compute the minimum number of parentheses to be removed to make
# the string valid (i.e. each open parenthesis is eventually closed).
#
# For example, given the string "()())()", you should return 1. Given the string ")(", you should return 2, since we
# must remove all of them.
def parentheses_to_remove(sequence):
# number of parentheses to remove
count = 0
# number of parentheses already open
opened = 0
# for each char in the sequence
for char in sequence:
# if it is opening, add it to the opened count
if char == "(":
opened += 1
# if it is closing
if char == ")":
# and there is opened parentheses, close it by removing it of the opened count
if opened > 0:
opened -= 1
# if there is no open parentheses, add the number of parentheses to remove
else:
count += 1
# return the number of parentheses to remove + the number of opened parentheses not closed
return count + opened
assert (parentheses_to_remove("")) == 0
assert (parentheses_to_remove("()")) == 0
assert (parentheses_to_remove("()())()")) == 1
assert (parentheses_to_remove("()()()()")) == 0
assert (parentheses_to_remove(")(")) == 2
assert (parentheses_to_remove("(((")) == 3
assert (parentheses_to_remove(")))")) == 3
assert (parentheses_to_remove(")))")) == 3
assert (parentheses_to_remove("stuff)in)the)middle")) == 3
| true |
8c12755934693f84fe28b32352fb780bd773d622 | JaimePazLopes/dailyCodingProblem | /problem034.py | 2,217 | 4.25 | 4 | # Problem #34 [Medium]
# Given a string, find the palindrome that can be made by inserting the fewest number of characters as possible
# anywhere in the word. If there is more than one palindrome of minimum length that can be made, return the
# lexicographically earliest one (the first one alphabetically).
#
# For example, given the string "race", you should return "ecarace", since we can add three letters to it
# (which is the smallest amount to make a palindrome). There are seven other palindromes that can be made from "race"
# by adding three letters, but "ecarace" comes first alphabetically.
#
# As another example, given the string "google", you should return "elgoogle".
def create_palindrome(word):
# if word is equal to its reverse, return it
if word == word[::-1]:
return word
# if the first and last char are the same, try to make a palindrome with the middle "word"
if word[0] == word[-1]:
return word[0] + create_palindrome(word[1:-1]) + word[-1]
# if they are not the same, make them the same
else:
# force the first char to be the last char too, try to make a palindrome with the middle "word"
by_first = word[0] + create_palindrome(word[1:]) + word[0]
# force the last char to be the first too, try to make a palindrome with the middle "word"
by_last = word[-1] + create_palindrome(word[:-1]) + word[-1]
# return the answer with less chars
if len(by_first) > len(by_last):
return by_last
elif len(by_first) < len(by_last):
return by_first
# return the answer in alphabetic order
if by_first < by_last:
return by_first
return by_last
assert create_palindrome("arara") == "arara"
assert create_palindrome("a") == "a"
assert create_palindrome("race") == "ecarace"
assert create_palindrome("google") == "elgoogle"
# it looked easy at first, but after finding my solution i read again the problem, it says "ANYWHERE in the word",
# i did for the beggining or ending only. couldnt think on a solution in time and looked for one online, it still a
# little weird will try to come back later
| true |
e8c505c66a0b152fed47d1c24164095d20d591a8 | JaimePazLopes/dailyCodingProblem | /problem083.py | 866 | 4.3125 | 4 | # Problem #83
# Invert a binary tree.
#
# For example, given the following tree:
#
# a
# / \
# b c
# / \ /
# d e f
#
# should become:
#
# a
# / \
# c b
# \ / \
# f e d
class Node:
def __init__(self, value):
self.value = value
self.left = None
self.right = None
def invert(node):
if not node:
return
# change left with right
node.left, node.right = node.right, node.left
# do the same with both sides
invert(node.left)
invert(node.right)
f = Node("f")
e = Node("e")
d = Node("d")
c = Node("c")
b = Node("b")
a = Node("a")
a.left = b
a.right = c
b.left = d
b.right = e
c.left = f
invert(a)
assert a.left == c
assert a.right == b
assert a.left.right == f
assert a.right.left == e
assert a.right.right == d
| false |
c43cc3f44200da29beaf525c92f11ade48fbe515 | JaimePazLopes/dailyCodingProblem | /problem046.py | 1,091 | 4.15625 | 4 | # Problem #46 [Hard]
# Given a string, find the longest palindromic contiguous substring.
# If there are more than one with the maximum length, return any one.
#
# For example, the longest palindromic substring of "aabcdcb" is "bcdcb".
# The longest palindromic substring of "bananas" is "anana".
def longest_palindromic(word):
if word == word[::-1]:
return word
without_first = longest_palindromic(word[1:])
without_last = longest_palindromic(word[:-1])
if len(without_first) > len(without_last):
return without_first
return without_last
print(longest_palindromic("aabcdcb"))
print(longest_palindromic("bananas"))
print(longest_palindromic(""))
print(longest_palindromic("a"))
print(longest_palindromic("abcde"))
print(longest_palindromic("abccccde"))
print(longest_palindromic("abcdeeee"))
# i think i understood something wrong here, they say that is a hard problem, but this is one of the easiest problem
# they sent. i finished in 5 minutes, but spent more 10 to be sure i was doing what i was supposed to do
| true |
3bd671741d9d5a5f079fcaae9b2a900885912df9 | JaimePazLopes/dailyCodingProblem | /problem109.py | 814 | 4.125 | 4 | # Problem #109
#
# Given an unsigned 8-bit integer, swap its even and odd bits. The 1st and 2nd bit should be swapped, the 3rd and 4th
# bit should be swapped, and so on.
#
# For example, 10101010 should be 01010101. 11100010 should be 11010001.
#
# Bonus: Can you do this in one line?
def swap_bits(x):
return (x & 0b10101010) >> 1 | (x & 0b01010101) << 1
assert swap_bits(0b10101010) == 0b01010101
assert swap_bits(0b11100010) == 0b11010001
# i didnt even understand how to represent a unsigned 8-bit int in python, have to google that. i had some multiple
# step ideas on how to do the problem, but no clue on how to do it in one line. after some google i found an
# explanation on the daily coding problem oficial site https://www.dailycodingproblem.com/blog/neat-bitwise-trick/
| true |
5915005ff5af0c7f85f45d9e80eac07f5d6d32cf | JaimePazLopes/dailyCodingProblem | /problem033.py | 2,805 | 4.3125 | 4 | # Problem #33 [Easy]
#
# Compute the running median of a sequence of numbers. That is, given a stream of numbers,
# print out the median of the list so far on each new element.
#
# Recall that the median of an even-numbered list is the average of the two middle numbers.
#
# For example, given the sequence [2, 1, 5, 7, 2, 0, 5], your algorithm should print out:
#
# 2
# 1.5
# 2
# 3.5
# 2
# 2
# 2
import statistics
import heapq
sequence = [2, 1, 5, 7, 2, 0, 5]
# my first solution, looked online for a python way to do median
def running_median(sequence):
for index in range(len(sequence)):
print(statistics.median(sequence[:index+1]))
print()
running_median(sequence)
# second solution, made my own median function
def median(sequence):
sort = sorted(sequence)
if len(sort) % 2 == 0:
return (sort[int(len(sort)/2-1)] + sort[int(len(sort)/2)]) / 2
else:
return sort[int((len(sort)-1)/2)]
def running_median2(sequence):
for index in range(len(sequence)):
print(median(sequence[:index+1]))
print()
running_median2(sequence)
# third solution, using min heap and max heap
def running_medians3(arr):
if not arr:
return None
smallest_values = list() # max heap of smallest values, multiply everything for -1 to get this effect
biggest_values = list() # min heap of biggest values
for x in arr:
# put it as a big value, so you can take the smallest value later
heapq.heappush(biggest_values, x)
# if they are no balanced, take one value of biggest to put on smallest
if len(biggest_values) > len(smallest_values) + 1:
# take the smallest element of the biggest values and add on the max heap of smallest values
heapq.heappush(smallest_values, heapq.heappop(biggest_values) * -1)
# if both have the same len
if len(biggest_values) == len(smallest_values):
# add the smallest element on biggest with the biggest element on smallest, divided by 2
print((biggest_values[0] + (smallest_values[0] * -1)) / 2)
else:
# the middle element is the smallest element on biggest
print(biggest_values[0])
running_medians3([2, 1, 5, 7, 2, 0, 5])
# the first 2 solution were easy and quick, but also quite bad. From the beginning i knew that i could do it in linear
# time, but to get there was hard, i got the idea of dividing on two parts but couldnt think in a way of keeping the
# lists ordered. Online I found heapq again, but it give only the smallest number, looking online for a max heap I
# found the * - 1 trick. It was not hard, but i didnt have to data structure know how. took one hour to finish
| true |
4919f1c45cb054ba3bab2a99ced47bed8105db1d | bryanyaggi/Coursera-Algorithms | /course1/pa1.py | 2,533 | 4.3125 | 4 | #!/usr/bin/env python3
'''
Programming Assignment 1
In this programming assignment you will implement one or more of the integer
multiplication algorithms described in lecture.
To get the most out of this assignment, your program should restrict itself to
multiplying only pairs of single-digit numbers. You can implement the
grade-school algorithm if you want, but to get the most out of the assignment
you'll want to implement recursive integer multiplication and/or Karatsuba's
algorithm.
So: what's the product of the following two 64-digit numbers?
3141592653589793238462643383279502884197169399375105820974944592
2718281828459045235360287471352662497757247093699959574966967627
[TIP: before submitting, first test the correctness of your program on some
small test cases of your own devising. Then post your best test cases to the
discussion forums to help your fellow students!]
[Food for thought: the number of digits in each input number is a power of 2.
Does this make your life easier? Does it depend on which algorithm you're
implementing?]
The numeric answer should be typed in the space below. So if your answer is
1198233847, then just type 1198233847 in the space provided without any space
/ commas / any other punctuation marks.
(We do not require you to submit your code, so feel free to use any programming
language you want --- just type the final numeric answer in the following
space.)
'''
'''
Implements Karatsuba's algorithm recursively.
See https://en.wikipedia.org/wiki/Karatsuba_algorithm for details.
intX string representation of first integer
intY string representation of second integer
returns integer result
'''
def karatsuba(intX, intY):
digsX = len(intX)
digsY = len(intY)
# Base case
if digsX < 2 or digsY < 2:
return int(intX) * int(intY)
m = 0
if digsX <= digsY:
m = digsX // 2
else:
m = digsY // 2
x1 = intX[:(len(intX) - m)]
x0 = intX[(len(intX) - m):]
y1 = intY[:(len(intY) - m)]
y0 = intY[(len(intY) - m):]
z2 = karatsuba(x1, y1)
z0 = karatsuba(x0, y0)
z1 = karatsuba(str(int(x1) + int(x0)), str(int(y0) + int(y1))) - z0 - z2
return z2 * 10 ** (2 * m) + z1 * 10 ** m + z0
def testResult(intX, intY):
return int(intX) * int(intY)
if __name__ == '__main__':
intA = '3141592653589793238462643383279502884197169399375105820974944592'
intB = '2718281828459045235360287471352662497757247093699959574966967627'
print(testResult(intA, intB))
print(karatsuba(intA, intB))
| true |
88057b2faecda38326d7a00ffa36b0ae4e5761b5 | sibimathews3010/S1-python-programs | /large3.py | 233 | 4.1875 | 4 | print("largest of three numbers")
X=int(input("enter first no="))
Y=int(input("enter second no="))
Z=int(input("enter third no="))
if(X>Y):
if(X>Z):
print("X largest;",X)
else:
print("Z larget;",Z)
else:
print("Y largest",Y)
| false |
9bcc693359ef87094813b860f056ebea87c035e4 | sibimathews3010/S1-python-programs | /sum.py | 306 | 4.15625 | 4 | num1=input("enter the 1st number")
num1=int(num1)
num2=input("enter the 2nd number")
num2=int(num2)
num3=input("enter the 3rd number")
num3=int(num3)
sum=int(num1)+int(num2)+int(num3)
average=sum/3
print("sum=",sum)
print("average=",average)
9
9
| false |
2944a8ea3698fff6e626881e634b69a3f3ca0fc4 | sibimathews3010/S1-python-programs | /28.py | 311 | 4.1875 | 4 | def greatest_three_number(a,b,c):
if a>b:
if a>c:
return a
else:
return c
else:
if b>c:
return b
else:
return c
a=int(input("enter the number1="))
b=int(input("enter the number2="))
c=int(input("enter the number3="))
G=greatest_three_number(a,b,c)
print("greatest of three no is",G)
| false |
c6ea854dc477ae6d5c0b5cbf226f8f71f1cfbb2a | ximuwang/Python_Crash_Course | /Chap9_Classes/Practice/Restaurant.py | 1,288 | 4.125 | 4 | # A module named restaurant
class Restaurant():
'''A class representing a restaurant'''
def __init__(self, name, cuisine_type):
'''Initialize the restaurant'''
self.name = name
self.cuisine_type = cuisine_type
self.number_served = 0
def describe_restaurant(self):
'''Display a summary of the restaurant'''
print(f"Welcome to {self.name}!")
print(f"We have the following cuisine type: {self.cuisine_type}")
def open_restaurant(self):
'''Display a message indicates that the restaurant is open.'''
print("\nWe are open now!")
def set_number_served(self, number_served):
'''allow user to set the number of customers that have been served'''
if self.number_served >= 0:
self.number_served = number_served
print('We served ' + str(self.number_served) + ' people today.')
def increment_number_served(self, add_customers):
'''allow user to increment the number of customers served.'''
if add_customers >= 0:
self.number_served += add_customers
print('We now serve ' + str(self.number_served) + ' person')
else:
print('Please add a positive number.')
| true |
daa35e217038a987d89248fa3de548c77ac61e5a | Someshwaran/Python-Basics | /employee_dataSaver.py | 2,069 | 4.5 | 4 | from employee import Employee
class employee_data_saver:
# to store the all employee objects
employee_data = []
def __init__(self):
self.employee_d = Employee()
# this method for read and store the data for a employee
def getting_values(self):
self.employee_d.set__name(input('Enter the name of the employee '))
self.employee_d.set__id(input('Enter the id '))
self.employee_d.set__salary(input('Enter the salary '))
self.employee_d.set__mailid(input('Enter the mailid '))
# entering the data of the employee to data list
employee_data_saver.employee_data.append(self.employee_d)
#this method is used for printing the data of the employee from the list
def printing_details_of_employee(self):
# iterate through the for loop
for employee_object in self.employee_data:
print()
print('Name : ',employee_object.get__name())
print('Id : ',employee_object.get__id())
print('Salary : ',employee_object.get__salary())
print('Mail_ID : ',employee_object.get__mailid())
print()
# to remove the details of the employee
# def remove_employee_details(self,id):
"""
we can add the remove and update method up_coming
"""
#definition of main method
def main():
# here we creating the object for the employee_data_saver
while True:
choice = input('enter the choice for the \n 1: Enter the data for Emplpoyee \n 2: for printing ')
employee_data_save_object = employee_data_saver()
if choice == '1':
#getting values
employee_data_save_object.getting_values()
elif choice == '2':
#printing all values
employee_data_save_object.printing_details_of_employee()
else:
break
#def main
if __name__ == '__main__':
main()
| true |
2b2854a5ccee8aebd3b674ea50233464aa77b9ea | MayLSantos/Refor-o-na-programa-o-para-ED | /Q4-Carnes.py | 1,290 | 4.15625 | 4 |
print("---- Tabela de carnes ----")
print("""File Duplo = [1]
Alcatra = [2]
Picanha = [3]""")
print(" ")
print("---- Tipo de pagamentos ----")
print("""Á vista = [1]
Cartão de crédito = [2]
Cartão Tabajara = [3] (5% de Desconto)""")
print(" ")
tipo= int(input("Tipo da carne: "))
quantidade = float(input("Quantidade de carne:[kg] "))
valor = 0
print(" ")
if tipo == 1:
print("{}Kg de File Duplo".format(quantidade))
if quantidade <= 5:
valor= 4.9 * quantidade
if quantidade > 5:
valor = 5.8 * quantidade
elif tipo == 2:
print("{}Kg de Alcantra".format(quantidade))
if quantidade <= 5:
valor= 5.9 * quantidade
if quantidade > 5:
valor = 6.8 * quantidade
elif tipo == 3:
print("{}Kg de Picanha".format(quantidade))
if quantidade <= 5:
valor= 6.9 * quantidade
if quantidade > 5:
valor = 7.8 * quantidade
print ("Preço total: R${:.2f}".format(valor))
pagamento = int(input("Tipo de pagamento: "))
if pagamento == 3:
desconto= (valor -(valor * 5/100))
print("5% de Desconto")
print("Valor a pagar: R${:.2f}".format(desconto))
print(" ")
print("Obrigado(a), volte sempre!")
| false |
82a0e559c5056ad3f4c8f04819c110f92638fc64 | AkshaySingh566/task-1 | /reversed string.py | 201 | 4.15625 | 4 | def rev():
word = input("Enter a string: ")
length = len(word)
s = " "
for i in reversed(range(length)):
s = s + word[i]
print("Reversed String is:", s)
rev() | false |
65c95e58fb3cdaa31db7ab4ec6626378a376c8b0 | K-Ellis/airline_seating_assignment | /Rough work folder/kron_sql_test_file.py | 2,225 | 4.15625 | 4 | import sqlite3
#Connect to the database using sqlite2's .connect() method which returns a
#connection object
conn = sqlite3.connect("airline_seating_test.db")
#From the connection we get a cursor object
cur = conn.cursor()
#create a database with different name, dimensin, col letters
def create_db(rows, col_letters):
cur.execute("CREATE TABLE metrics(passengers_refused INT, "
"passengers_separated INT)")
cur.execute("INSERT INTO metrics VALUES(0,0)")
cur.execute("CREATE TABLE rows_cols(nrows INT,seats TEXT)")
cur.execute("INSERT INTO rows_cols VALUES(?,?)", (rows, col_letters))
cur.execute("CREATE TABLE seating(row INT, seat TEXT, name TEXT)")
cur.execute("INSERT INTO seating VALUES(?,?)", (rows, col_letters))
def print_entries(name):
# Prints all of the tables
if name == "tables":
with conn:
#Tables names are stored in sqlite_master
cur.execute("SELECT name FROM sqlite_master WHERE type='table'")
list_of_tables = cur.fetchall()
for table1 in list_of_tables:
print(table1[0])
#The tables stored in airline_seating.db are: metrics, row_cols and seating
#print out the info in metrics
elif name == "metrics":
with conn:
cur.execute("SELECT * FROM metrics")
rows = cur.fetchall()
for row in rows:
print(row)
# print out the info in rows_cols
elif name == "rows_cols":
with conn:
cur.execute("SELECT * FROM rows_cols")
rows = cur.fetchall()
for row in rows:
print(row)
else:
# print out the info in seating
with conn:
cur.execute("SELECT * FROM seating")
rows = cur.fetchall()
print(type(rows[0][0]))
print(type(rows[0][1]))
print(type(rows[0][2]))
# for row in rows:
# print(row)
print(len(rows))
print(rows)
print_entries("seating")
#60 entries, two seats taken
# 1 A Donald Trump and 1 C Hilary Clinton
print_entries("rows_cols")
# 15*4
# A C D F
# 0 1 2 4
print_entries("metrics")
print_entries("tables")
conn.close() | true |
660abd61196b475b705ea4c63e4f5fef0a5b23df | golkedj/Complete_Python_Masterclass | /ProgramFlowChallenge/challenge.py | 2,311 | 4.46875 | 4 | # Create a program that takes an IP address entered at the keyboard
# and prints out the number of segments it contains, and the length of each segment.
#
# An IP address consists of 4 numbers, separated from each other with a full stop. But
# your program should just count however many are entered
# Examples of the input you may get are:
# 127.0.0.1
# .192.168.0.1
# 10.0.123456.255
# 172.16
# 255
#
# So your program should work even with invalid IP Addresses. We're just interested in the
# number of segments and how long each one is.
#
# Once you have a working program, here are some more suggestions for invalid input to test:
# .123.45.678.91
# 123.4567.8.9.
# 123.156.289.10123456
# 10.10t.10.10
# 12.9.34.6.12.90
# '' - that is, press enter without typing anything
#
# This challenge is intended to practise for loops and if/else statements, so although
# you could use other techniques (such as splitting the string up), that's not the
# approach we're looking for here.
# My solution
# ip_address = input("Please enter an IP address: ")
#
# segments = 0
# length = 0
# for char in ip_address:
# if char == '.':
# segments += 1
# print("The length of the segment was {} digits".format(length))
# length = 0
# else:
# length += 1
#
# if segments == 0:
# segments = 1
#
# print("The length of the segment was {} digits".format(length))
# print("There were {} segments in the IP address".format(segments))
# Instructor solution
input_prompt = ("Please enter an IP address. An IP address consists of 4 numbers, "
"separated from each other with a full stop: ")
ip_address = input(input_prompt)
if ip_address[-1] != '.':
ip_address += '.'
segment = 1
segment_length = 0
# character = ''
for character in ip_address:
if character == '.':
print("segment {} contains {} characters".format(segment, segment_length))
segment += 1
segment_length = 0
else:
segment_length += 1
# else:
# print("segment {} contains {} characters".format(segment, segment_length))
# unless the final character in the string was a . then we haven't printed the last segment
# if character != '.':
# print("segment {} contains {} characters".format(segment, segment_length))
| true |
0dd7705fb79d54f3343b7fe77c4c267e49e58ca5 | DKCisco/Starting_Out_W_Python | /2_7.py | 592 | 4.34375 | 4 | """
7. Miles-per-Gallon
A car's miles-per-gallon (MPG) can be calculated with the following formula:
MPG 5 Miles driven 4 Gallons of gas used
Write a program that asks the user for the number of miles driven and the gallons of gas
used. It should calculate the car's MPG and display the result.
"""
# Receive input
miles_driven = float(input('\nEnter number of miles driven: '))
gallons_of_gas_used = float(input('Enter gallons of gas used: '))
# Calculate MPG
mpg = miles_driven / gallons_of_gas_used
# Output info
print('Miles per gallon = ', format(mpg, ',.2f'))
| true |
264997cbfa001842a434411ed7743781afdc7a93 | DKCisco/Starting_Out_W_Python | /3_1PE.py | 502 | 4.3125 | 4 | """
Write a program that asks the user to enter an integer. The program should display
“Positive” if the number is greater than 0, “Negative” if the number is less than 0, and
“Zero” if the number is equal to 0. The program should then display “Even” if the number
is even, and “Odd” if the number is odd.
"""
# Input variable x from user
x = int(input('Enter a number positive or negative: '))
# Process
if x > 0:
print('Positive')
else:
print('Negative') | true |
d0e5ab73ec31f809d5db4c820fb91835abe86424 | DKCisco/Starting_Out_W_Python | /average_rainfall.py | 1,324 | 4.4375 | 4 | """
5. Average Rainfall
Write a program that uses nested loops to collect data and calculate the average rainfall over
a period of years. The program should first ask for the number of years. The outer loop will
iterate once for each year. The inner loop will iterate twelve times, once for each month.
Each iteration of the inner loop will ask the user for the inches of rainfall for that month.
After all iterations, the program should display the number of months, the total inches of
rainfall, and the average rainfall per month for the entire period.
"""
# Get number of years from the user
years_total = int(input('Enter the number of years to track: '))
# Assign the number of months in a year
months = 12
# Accumulator
total = 0.0
for years_rain in range(years_total):
print('\nYear Number', years_rain + 1 )
print('------------------------------')
for month in range(months):
print('How many inches for month ', month + 1, end='')
rain = int(input(' Enter rain amount: '))
total += rain
number_months = years_total * months
average = total / number_months
print('The total inches of rain was ', format(total,'.2f'))
print('The number of months measured was', number_months)
print('The average rainfall was', format(average, '.2f'), 'inches') | true |
e0f5e232bfc86a8b66c68aa464ecf980c02c6740 | DKCisco/Starting_Out_W_Python | /test_average.py | 728 | 4.25 | 4 | """
This program calculates the average of 3 test scores using decision structure.
"""
# Assign varibale to score above 95% average.
HIGH_AVERAGE = 95
# Get the test scores from user.
test1 = int(input('Enter the score for test 1: ' ))
test2 = int(input('Enter the score for test 2: ' ))
test3 = int(input('Enter the score for test 3: ' ))
# Calculate the average test score
average = (test1 + test2 + test3) / 3
# Print the average.
print("The average score is: %s %%\n" % average)
# Congratulate the user based on if the score is above or equal to 95.
if average >= HIGH_AVERAGE:
print('Congratulations on scoring 95% or better, Einstein would be proud')
print('That is a great average!')
| true |
f659140d86797009dce119565357affb2c41a62f | zhaobf1990/MyPhpDemo | /first/com/zhaobf/test6.py | 1,105 | 4.3125 | 4 | # Python3 函数
# def add(a, b):
# return a + b
#
#
# def aera(a, b):
# return a * b
#
#
# print(add(1, 2))
#
# print(add("1", "3"))
#
# print( aera("a",8))
# 标题:按值传递参数和按引用传递参数
# 在 Python 中，所有参数（变量）都是按引用传递。如果你在函数里修改了参数，那么在调用这个函数的函数里，原始的参数也被改变了。例如：
def changeme(data):
"修改传入的列表"
data.append([1, 2, 3, 4]);
print("函数内取值: ", data)
return data
# 调用changeme函数
mylist = [10, 20, 30];
c = changeme(mylist);
print (c)
print ("函数外取值: ", mylist)
# # 可写函数说明
# def printinfo(arg1, *vartuple):
# "打印任何传入的参数"
# print("输出arg1: ", arg1)
#
# for var in vartuple:
# print("vartuple", var)
# return;
#
#
# # 调用printinfo 函数
# # printinfo(10);
# printinfo(70, 60, 50);
# sum1=lambda arg1,arg2:arg1+arg2;
#
# print("相加后的值为:",sum1(10,20))
# sum()
def add(a):
a = 2
return a
b = 1
c = add(b)
print(c)
print(b)
| false |
1b304e5f6ae0a2c208377a15b31189ae990c09e1 | fizzywonda/CodingInterview | /Recursion&Dynamic Program/RecursiveMultiply.py | 1,549 | 4.5 | 4 | """
Recursive Multiply: Write a recursive function to multiply two positive integers without using the
*operator.You can use addition, subtraction, and bit shifting, but you should minimize the number
of those operations.
"""
"""Brute force Approach"""
def multiply(x, y):
if y == 0:
return 0
result = x + multiply(x, y - 1)
return result
"""improved approach"""
def multiply2(x, y):
# get the bigger number
small = x if x < y else y
big = y if y > x else x
return multiply_helper(small, big)
def multiply_helper(small, big):
if small == 0:
return 0
if small == 1:
return big
# divide small by 2
s = small >> 1
side1 = multiply2(s, big)
side2 = side1
if small % 2 == 1:
side2 = (side1 << 1) + big
return side1 + side2
return side1 + side2
""" optimized Approach (memoized)"""
def multiply3(x, y):
small = x if x < y else y
big = y if y > x else x
memo = [None for i in range(small + 1)]
return multiply_helper2(small, big, memo)
def multiply_helper2(small, big, memo):
if small == 0:
return 0
if small == 1:
return big
if memo[small] is not None:
return memo[small]
s = small >> 1
side1 = multiply3(s, big)
side2 = side1
if small % 2 == 1:
side2 = (side1 << 1) + big
memo[small] = side1 + side2
memo[small] = side1 + side2
return memo[small]
if __name__ == '__main__':
x = 2
y = 3
result = multiply3(x, y)
print(result)
| true |
26e3c6573df720de85538db9c86d0f24d469b13a | fizzywonda/CodingInterview | /Recursion&Dynamic Program/RobotInGrid.py | 1,727 | 4.46875 | 4 | """
Robot in a Grid: Imagine a robot sitting on the upper left corner of grid with r rows and c columns.
The robot can only move in two directions, right and down, but certain cells are "off limits" such that
the robot cannot step on them. Design an algorithm to find a path for the robot from the top left to
the bottom right.
"""
def find_path1(maze):
path = []
if find_path(maze, 0, 0, path):
return path
return None
def find_path(maze, row, col, path):
if row > len(maze) - 1 or col > len(maze[0]) - 1 or not maze[row][col]:
return False
end = row == len(maze) - 1 and col == len(maze[0]) -1
if end or find_path(maze, row + 1, col, path) or find_path(maze, row, col + 1, path):
p = [row, col]
path.append(p)
return True
return False
"""
Dynamic programming approach
"""
def find_path_2(maze):
path = []
visited_point = set()
if find_path2(maze, 0, 0, path, visited_point):
return path
return None
def find_path2(maze, row, col, path, visited_point):
if row > len(maze) - 1 or col > len(maze[0]) - 1 or not maze[row][col]:
return False
p = (row, col)
if p in visited_point:
return False
end = row == len(maze) - 1 and col == len(maze[0]) -1
if end or find_path2(maze, row + 1, col, path, visited_point) or find_path2(maze, row, col + 1, path, visited_point):
path.append(p)
return True
visited_point.add(p)
return False
if __name__ == '__main__':
maze = []
maze.append(["s", False, False])
maze.append([True, False, False])
maze.append([True, True, False])
maze.append([False, True, "e"])
path = find_path1(maze)
print(path)
| true |
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