blob_id string | repo_name string | path string | length_bytes int64 | score float64 | int_score int64 | text string | is_english bool |
|---|---|---|---|---|---|---|---|
b942302e280c289fceac406b562b7fb05b7cab36 | 36rahu/codewar_works | /22-4-2015/product_of_diagonal.py | 341 | 4.3125 | 4 | '''
Description:
Given a list of rows of a square matrix, find the product of the main diagonal.
Examples:
main_diagonal_product([[1,0],[0,1]]) => 1
main_diagonal_product([[1,2,3],[4,5,6],[7,8,9]]) => 45
'''
def main_diagonal_product(mat):
result = 1
for i in range(len(mat[0])):
result *= mat[i][i]
return result
| true |
153705a0ea3ff05d7e59fcd5bee4f28ac60ce413 | muhammeta7/CodeAcademyPython | /strings_cons_output.py | 1,542 | 4.375 | 4 | # Create a string by surrounding it with quotes
string = "Hello there "
# For words with apostrophe's us a backslash \
print "There isn\'t flying, this is falling with style!"
# Accesing by index
print fifth_letter = "MONTY"[4] # Will return Y
# String methods
parrot = "Parrot"
print len(parrot) # will return number of letters in a string which in this case is 6
print parrot.lower() # will return parrot with all lowercase letters
print parrot.upper() # will return PARROT with all captialized letters
pi = 3.14
print str(pi) # will turn non-strings into strings
# String Concatenation
print "Green Eggs " + "and " + "ham!" # Must take into account spaces
# String formatting with %
name = "Moe"
state = "New Jersey"
print "Hello my name is %s! I am from %s" %(name, state)
# String concatination with raw_inputs
name = raw_input("What is your name?")
quest = raw_input("What is your quest?")
color = raw_input("What is your favorite color?")
# \ is simply a continuation marker
print "Ah, so your name is %s, your quest is %s, " \
"and your favorite color is %s." % (name, quest, color)
# Getting current data and time
from datetime import datetime
now = datetime.now
print now # will print date in following format year-month-day 23:43:37.127290
print now.year # will return current year
print now.month # will return current month
print now.day # will return current day
# To properly format everything simply use %s
print "%s/%s/%s" % (now.month, now.day, now.year)
# Same can be done for time using hour, minute, second
| true |
598768ac97d73068511606617b003142fcbcc7c1 | anbarasanv/PythonWB | /Merge Sort Recursive.py | 890 | 4.21875 | 4 | #Time complexity O(n log n)
def merge(left,right):
'''This Function will merge 2 array or list'''
result = []
i,j = 0,0
#Comparing two list for smallest element
while i < len(left) and j < len(right):
if left[i] <= right[j]:
result.append(left[i])
i +=1
else:
result.append(right[j])
j +=1
#left elements appends to the result list
result += left[i:]
result += right[j:]
return result
def mergeSort(lst):
'''This is the recursive sorting array'''
if(len(lst) <= 1):
return lst
#Finding the mid
mid = int(len(lst)/2)
#Recursive call for left and right array or list
left = mergeSort(lst[:mid])
right = mergeSort(lst[mid:])
#Merginging splited lists or array
return merge(left,right)
A = [1,-2,0,-3,10,999999,20]
print(mergeSort(A))
| true |
3000f9b65ad6926b0274a5d94ea41a7e2608781f | bimri/programming_python | /chapter_1/person_start.py | 1,107 | 4.125 | 4 | "Step 3: Stepping Up to OOP"
'Using Classes'
class Person:
def __init__(self, name, age, pay=0, job=None):
self.name = name
self.age = age
self.pay = pay
self.job = job
if __name__ == '__main__':
bob = Person('Bob Smith', 42, 30000, 'software')
sue = Person('Sue Jones', 45, 40000, 'hardware')
print(bob.name, sue.pay)
print(bob.name.split()[-1])
sue.pay *= 1.10
print(sue.pay)
'''
This isn’t a database yet, but we could stuff these objects into a list or dictionary as
before in order to collect them as a unit:
>>> from person_start import Person
>>> bob = Person('Bob Smith', 42)
>>> sue = Person('Sue Jones', 45, 40000)
>>> people = [bob, sue] # a "database" list
>>> for person in people:
print(person.name, person.pay)
>>> x = [(person.name, person.pay) for person in people]
>>> x
[('Bob Smith', 0), ('Sue Jones', 40000)]
>>> [rec.name for rec in people if rec.age >= 45] # SQL-ish query
['Sue Jones']
>>> [(rec.age ** 2 if rec.age >= 45 else rec.age) for rec in people]
[42, 2025]
'''
| true |
9ae6c269b41521b7ededa35ccae246557185174f | bimri/programming_python | /chapter_7/gui2.py | 492 | 4.28125 | 4 | "Adding Buttons and Callbacks"
import sys
from tkinter import *
widget = Button(None, text='Hello GUI world!', command=sys.exit)
widget.pack()
widget.mainloop()
'''
Here, instead of making a label, we create an instance of the tkinter Button class.
For buttons, the command option is the place where we specify a callback handler function
to be run when the button is later pressed. In effect, we use command to register an
action for tkinter to call when a widget’s event occurs.
'''
| true |
f46676e4e06c7657a92b10f6ac48f8a13f199d02 | bimri/programming_python | /chapter_1/tkinter102.py | 964 | 4.15625 | 4 | "Step 5: Adding a GUI"
'Using OOP for GUIs'
'''
In larger programs, it is often more useful to code a GUI as a subclass
of the tkinter Frame widget—a container for other widgets.
'''
from tkinter import *
from tkinter.messagebox import showinfo
class MyGui(Frame):
def __init__(self, parent=None):
Frame.__init__(self, parent)
button = Button(self, text='press', command=self.reply)
button.pack()
def reply(self):
showinfo(title='popup', message='Button pressed!')
if __name__ == '__main__':
window = MyGui()
window.pack()
window.mainloop()
'''
The button’s event handler is a bound method—self.reply, an object that remembers
both self and reply when later called.
but because it is now a subclass of
Frame, it automatically becomes an attachable component—i.e., we can add all of the
widgets this class creates, as a package, to any other GUI, just by attaching this Frame
to the GUI.
'''
| true |
73e6e2712fc736f21f7818bef2c63815de570750 | hymhanraj/lockdown_coding | /range.py | 266 | 4.40625 | 4 | ''' Given start and end of a range, write a Python program to print all negative numbers in given range.
Example: Input: start = -4, end = 5 Output: -4, -3, -2, -1 Input: start = -3, end = 4 Output: -3, -2, -1
'''
for num in range(-3,4):
if num<0:
print(num)
| true |
040cdf085336c553e8b0951a59a25028577999f7 | helper-uttam/Hacktober2020-5 | /Python/calc.py | 1,011 | 4.1875 | 4 | def add(x, y):
return x + y
def subtract(x, y):
return x - y
def multiply(x, y):
return x * y
def divide(x, y):
return x / y
print("Select operation.")
print("1.Add")
print("2.Subtract")
print("3.Multiply")
print("4.Divide")
while True:
num1 = int(input("Enter first number: "))
num2 = int(input("Enter second number: "))
print("Select operation.")
print("1. Add")
print("2. Subtract")
print("3. Multiply")
print("4. Divide")
choice = input("Enter choice(1/2/3/4):")
result = 0
if choice == '1':
result = add(num1, num2)
elif choice == '2':
result = subtract(num1, num2)
elif choice == '3':
result = multiply(num1, num2)
elif choice == '4':
if(num2 !=0):
result = divide(num1, num2)
else:
result="Divide by 0"
else:
print("Invalid Input")
print("Result = ", result)
quit = input("Do you want to continue (y/n) ?")
if quit == 'n':
break
| true |
20e5713218979589bc452c464c4e9d52216d2b2b | mishra28soumya/Hackerrank | /Python/Basic_Data_Types/second_largest.py | 673 | 4.125 | 4 | #Find the Runner-Up Score!
# Given the participants' score sheet for your University Sports Day, you are required to find the runner-up score.
# You are given n scores. Store them in a list and find the score of the runner-up.
# Input Format
# The first line contains N. The second line contains an array A[] of N integers each separated by a space.
# Constraints
# 2 <= N <= 10
# -100 <= A[i] <= 100
# Output Format
# Output the value of the second largest number.
if __name__ == '__main__':
n = int(raw_input())
l = list(map(int, raw_input().split()))
s = set(l) #to get all the unique numbers
result = sorted(list(s))
print(result[-2])
| true |
91e6153e75db60860bd2f486effad5dd468c456a | nath-btx/hi-python | /ex03_rps/rps.py | 1,377 | 4.125 | 4 | import random
moves=['rock','paper','scissors']
def convert_input(move):
if (move == "r"): return 'rock'
elif(move == "p"): return 'paper'
elif(move == "s"): return 'scissors'
else: return move
def game():
tie = True
while tie:
tie = False
AI_move = random.choice(moves)
move = convert_input(input('Do you want to play Rock, Paper, or Scissors (r or rock / p or paper / s or Scissors)'))
convert_input(move)
if (move == AI_move):
tie = True
print(f"You both played {move}, it's a tie !")
elif(move == "r" or move == "rock"):
if(AI_move == "scissors"): print("Rock > Scissors, you win !")
else: print("Rocks < Paper, you lose !")
elif(move == "p" or move == "paper"):
if(AI_move == "rock"): print("Paper > Rock, you win !")
else: print("Paper < Scissors, you lose !")
elif(move == "s" or move == "scissors"):
if(AI_move == "paper"): print("Scissors > Paper, you win !")
else: print("Scissors < Rock, you lose !")
else:
tie = True
print("Please select a valid input")
if __name__ == "__main__":
while True:
game()
play_again = input("Do you want to play again ? (y/n)")
if(play_again == "y"):continue
else: exit()
| false |
8763f32cd9ae69553a4554c0223283a51c5522be | crypto4808/PythonTutorial_CoreySchafer-master1 | /intro.py | 1,731 | 4.3125 | 4 | #import as a smaller name so you don't have to type 'my_modules' each time
# import my_modules
# #from my_module import find_index
# courses = ['history', 'math' , 'physics' , 'compsci' ]
# index = my_modules.find_index(courses,'math')#
# print(index)
#Use shorthand for my_modules so you don't have to type it out all the time
# import my_modules as mm
# courses = ['history', 'math' , 'physics' , 'compsci' ]
# index = mm.find_index(courses,'math')
# print(index)
# #import the function itself
# from my_modules import find_index
# courses = ['history', 'math' , 'physics' , 'compsci' ]
# index = find_index(courses,'math')
# print(index)
# #import the function itself
# #access the test variable from my_modules
# from my_modules import find_index,test
# courses = ['history', 'math' , 'physics' , 'compsci' ]
# index = find_index(courses,'math')
# print(index)
# print(test)
# #import the function itself
# #access the test variable from my_modules
# from my_modules import find_index as fi, test
# courses = ['history', 'math' , 'physics' , 'compsci' ]
# index = fi(courses,'math')
# print(index)
# print(test)
#import everything though it's frowned upon
#Reason is if something goes wrong with find_index
#it's virtually impossible to find which module it was imported from
#access the test variable from my_modules
# from my_modules import *
# courses = ['history', 'math' , 'physics' , 'compsci' ]
# index = find_index(courses,'math')
# print(index)
#how does python know where is the module?
#py checks multiple locations for this modules
from my_modules import find_index, test
import sys
courses = ['history', 'math' , 'physics' , 'compsci' ]
index = find_index(courses,'math')
print(index)
print(sys.path)
| true |
ec022a502128acdcdc7b933a49ee215814f39c71 | statusyaza/hacktoberfest2021-5 | /Python/diamond_star.py | 481 | 4.125 | 4 | rows = int(input("Enter you number:"))
n = 0
for i in range(1, rows + 1):
n = 0
# loop to print spaces
n = (rows - i)
print(" " * n, end = "")
# loop to print star
n = (2 * i - 1)
print(str(i) * n, end = "")
print()
####
for i in range(rows-1, 0, -1):
# loop to print spaces
n = (rows - i)
print(" " * n, end = "")
# loop to print star
n = (2 * i - 1)
print(str(i) * n, end ="")
print() | false |
b9b1a64e80fe8d1a4e9038e36d98f84b56f75ba8 | Cassie-bot/Module-3 | /Area.py | 340 | 4.40625 | 4 | #Casandra Villagran
#1/30/2020
#This is a program that computes the area of a circle using the radius and it prints a message to the user with the answer.
radius = int(input ("What is the radius of the circle? "))
Area= 3.14 * radius ** 2
print ("I loved helping you find the area of this circle, here is your answer " + str(Area))
| true |
13091fa05d4d93a37a34f95520e6de8746a93763 | avijitkumar2011/My-Python-programs | /middle_letter.py | 213 | 4.1875 | 4 | #To find middle letter in a string
def middle(strg):
if len(strg) % 2 == 1:
n = len(strg)//2
print('Middle letter in string is:', strg[n])
else:
print('enter string of length odd number')
middle('kuma')
| true |
dcd1d7304fd1d7dbf55602f6f95150d98d92ea66 | Ma11ock/cs362-asgn4 | /avg_list.py | 226 | 4.125 | 4 | #!/usr/bin/env python
# Calculate the average of elements in a list.
def findAvg(lst):
if(len(lst) == 0):
raise ValueError("List is empty")
avg = 0
for i in lst:
avg += i
return avg / len(lst)
| true |
af211c4fb797589561c443a8593c3cc3bf122d47 | purnimagupta17/python-excel-func | /string_indexing_slicing_usingpython.py | 967 | 4.65625 | 5 | print('Hello World')
#to shift the 'World' to next line, we can use \n in the symtax as follows:
print('Hello \nWorld')
#to count the number of characters in a string, we use len function as follows:
str1="Hello"
print(len(str1))
#to assign the index number to a string, for example:
ss="Hello World"
print(ss[0])
print(ss[3:8])
#to jump on the string:
str2="abcdefghijk"
print(str2[::])
#to exclude the 2nd character from string
print(str2[::2])
#to reverse the string
print(str2[::-1])
#to edit the name in a string
name='sam'
#name[0]='P'
last_letters=name[1:]
last_letters
'P'+last_letters
#to combine two sentences/stings
x='Hello World'
x=x+ " It is beautiful outside"
x
#to concatenate the string
letter='z'
letter*10
#to concatenate the string
2+3
'2'+'3'
#to change the string to uppercase and lowercase
x='Hello World'
x
x.upper()
x.lower()
#to split the string
x.split()
#to split the string based on some specific characters
x.split('o')
| true |
7a71cc280ae6302fc83eebd003a4693c17f8bc2a | krallnyx/NATO_Alphabet_Translation | /main.py | 527 | 4.1875 | 4 | import pandas
data = pandas.read_csv("nato_phonetic_alphabet.csv")
alphabet = {row.letter: row.code for (index, row) in data.iterrows()}
def generate_phonetic():
to_translate = input("Please enter the word you want to translate into NATO Alphabet.\n").upper()
try:
output = [alphabet[letter] for letter in to_translate]
except KeyError:
print("Please enter a valid word, only letters in the alphabet are allowed")
generate_phonetic()
else:
print(output)
generate_phonetic()
| true |
eb7076d9c7d75d4b0ce59b5a1f74c6cdfe311b7e | jessiecantdoanything/Week10-24 | /CodesAndStrings.py | 1,531 | 4.125 | 4 | # strings
# concatenation
firstName = "Jesus"
lastName = "Monte"
print(firstName + " " + lastName)
name = firstName + " " + lastName
lastFirst = lastName + ", " + firstName
print(lastFirst)
# repition
print("uh "*2 + "AAAAAAAAA")
def imTiredofSchool():
print("me, "*3 + "young Jessie,")
print("is extremely exhausted")
print("Why must you ask?")
print("school "*3)
imTiredofSchool()
# indexing
name = "Roy G Biv"
firstChar = name[0]
print(firstChar)
middleCharIndex = len(name) // 2
print(middleCharIndex)
print(name[middleCharIndex])
print(name[-3])
for i in range(0, len(name)):
print(name[i])
#slicing and dicing
print(name[0:3])
for i in range(0, len(name)+1):
print(name[0:i])
# searching
print("biv" in name)
print("y" not in name)
# String Methods to investigate:
# Method Use Example Explanation
# center aStr.center(w)
# ljust aStr.ljust(w)
# rjust aStr.rjust(w)
# upper aStr.upper()
# lower aStr.lower()
# index aStr.index(item)
# rindex aStr.rindex(item)
# find aStr.find(item)
# rfind aStr.rfind(item)
# replace aStr.replace(old, new)
# Be sure to include multiple examples of all of them in use
# Character functions
from mapper import *
print(letterToIndex('M'))
print(indexToLetter(24))
from crypto import *
print(scramble2Encrypt("THE MEETING IS AT FIVE OCLOCK"))
print(scramble2Decrypt("H ETN SA IEOLCTEMEIGI TFV COK"))
| false |
efb19779c940b3f7d1f8999d089c2b2849764f1c | rckc/CoderDojo | /Python-TurtleGraphics/PythonIntro2-1Sequences.py | 1,435 | 4.5625 | 5 | # Author: Robert Cheung
# Date: 29 March 2014
# License: CC BY-SA
# For CoderDojo WA
# Python Intro Turtle Graphics
# Python has a rich "library" of functions.
# We are going to use "turtle" to do some simple graphics.
import turtle
# see https://docs.python.org/2/library/turtle.html#filling
if __name__ == "__main__":
# Setup our drawning surface
turtle.setup(800,600) # Sets up the size of the window
turtle.Screen() # Turns on the graphics window
# Set how fast we want the turtle to draw
turtle.speed(6); # 0 (fastest) .. 10 (slowest)
width = 20 # Create a variable called "width"
# and make it equal to 10
# In programming, computer follows instructions
# A "sequence" of instructions is the most basic construct
# It is simply a list of instructions
turtle.forward(width) # Step 1: Move forward 20 steps
turtle.right(90) # Step 2: Turn right 90 degrees
turtle.forward(width) # Step 3: Move forward 20 steps
turtle.right(90) # ... and so on
turtle.forward(width)
turtle.right(90)
turtle.forward(width)
turtle.right(90)
ret = input("Press enter to exit.")
turtle.Screen().bye() # Turn off the graphics window
# Library reference: http://docs.python.org/2/library/turtle.html
| true |
786073c2e842f81fd71fc391628f6161546b5a9e | yyuuliang/codingchallenges | /hackerrank/Reverse-a-doubly-linked-list.py | 2,556 | 4.15625 | 4 | '''
File: Reverse-a-doubly-linked-list.py
Source: hackerrank
Author: yyuuliang@github
-----
File Created: Thursday, 12th March 2020 4:21:54 pm
Last Modified: Thursday, 12th March 2020 4:21:58 pm
-----
Copyright: MIT
'''
# Reverse a doubly linked list
# https://www.hackerrank.com/challenges/reverse-a-doubly-linked-list/problem?h_l=interview&playlist_slugs%5B%5D=interview-preparation-kit&playlist_slugs%5B%5D=linked-lists
# Sample Input
# 4
# 1
# 2
# 3
# 4
# Sample Output
# 4 3 2 1
#!/bin/python3
import math
import os
import random
import re
import sys
class DoublyLinkedListNode:
def __init__(self, node_data):
self.data = node_data
self.next = None
self.prev = None
class DoublyLinkedList:
def __init__(self):
self.head = None
self.tail = None
def insert_node(self, node_data):
node = DoublyLinkedListNode(node_data)
if not self.head:
self.head = node
else:
self.tail.next = node
node.prev = self.tail
self.tail = node
def print_doubly_linked_list(node, sep, fptr):
while node:
fptr.write(str(node.data))
node = node.next
if node:
fptr.write(sep)
# Complete the reverse function below.
#
# For your reference:
#
# DoublyLinkedListNode:
# int data
# DoublyLinkedListNode next
# DoublyLinkedListNode prev
#
#
# use recursion to solve the problem
class node:
data=0
nextnode=None
prevnode=None
def __init__(self, dvalue):
self.data=dvalue
def printvalue(self):
print(self.data)
def printlinkedlist(head):
while head is not None:
head.printvalue()
head=head.nextnode
# double linked list
def reversedoublelinkedlist(head):
if head is None:
return None
tmp=reversedoublelinkedlist(head.nextnode)
if tmp is not None:
tmp.nextnode=head
tmp.nextnode.prevnode=tmp
tmp.nextnode.nextnode=None
return tmp.nextnode
else:
tmp=head
tmp.prevnode=None
tmp.nextnode=None
return tmp
def reverse(head):
tail =reversedoublelinkedlist(head)
# con=None
while tail.prevnode is not None:
# con=tail.prevnode
tail=tail.prevnode
return tail
n=node(1)
n2=node(2)
n3=node(3)
n4=node(4)
n.nextnode=n2
n2.nextnode=n3
n3.nextnode=n4
n2.prevnode=n
n3.prevnode=n2
n4.prevnode=n3
printlinkedlist(n)
t=reverse(n)
printlinkedlist(t)
# now I want to practice to reverse linkedlist in recursion
# which means dont use prev
| true |
7b04023a27193cf35c7c4db27c72cbb265193ddf | manusoler/code-challenges | /projecteuler/pe_1_multiples_3_and_5.py | 364 | 4.3125 | 4 | """
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
"""
def multiples_of(*nums, max=1000):
return [i for i in range(max) if any([i % n == 0 for n in nums])]
if __name__ == "__main__":
print(sum(multiples_of(3,5)))
| true |
74b4f4975aff361161aaa469e959a1652b22a77d | manusoler/code-challenges | /projecteuler/pe_9_special_pythagorean_triplet.py | 723 | 4.3125 | 4 | import functools
from math import sqrt
from utils.decorators import timer
"""
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which, a**2+b**2=c**2
a2 + b2 = c2
For example, 32 + 42 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
"""
def pythagorean_triplet(summ):
for a in range(1,summ):
for b in range(a+1, summ):
c = sqrt(a**2+b**2)
if a + b + c == summ:
return (a,b,int(c))
@timer
def main():
triplet = pythagorean_triplet(1000)
print("Triplet: {}, Product: {}".format(triplet, functools.reduce(lambda x,y: x*y, triplet)))
if __name__ == "__main__":
main()
| true |
ab110fc6c142c9981d0e034eba28dbae82a5afa6 | dersenok/python3_Les2 | /venv/HomeworkLes2/Les_2_HW_8.py | 717 | 4.125 | 4 | # 8. Посчитать, сколько раз встречается определенная цифра в
# # # введенной последовательности чисел. Количество вводимых
# # # чисел и цифра, которую необходимо посчитать,
# # # задаются вводом с клавиатуры.
n = int(input("Сколько будет чисел? "))
d = int(input("Какую цифру считать? "))
count = 0
for i in range(1, n + 1):
m = int(input("Число " + str(i) + ": "))
while m > 0:
if m % 10 == d:
count += 1
m = m // 10
print("Было введено %d цифр %d" % (count, d)) | false |
d3c3a3cf50793588756306288b30433842460b60 | littlejoe1216/Chapter-6-Exercies | /Python Chapter 6 Exercises/Chp 6 Excercise 4.py | 596 | 4.21875 | 4 | #Exercise 4:
#There is a string method called count that is similar to the function in the previous exercise.
#Read the documentation of this method at https://docs.python.org/3.5/library/stdtypes.html#string-methods
#and write an invocation that counts the number of times the letter a occurs in "banana".
word = "banana"
let = input("Enter a letter in the word banana.")
def counter():
count = 0
for letter in word:
if letter == let:
count = count + 1
return count
print (counter())
print ("is the number of times this letter is in the word banana.")
| true |
e3ddbeb215a0d6419b2ae83f2a0933ffc6fcb459 | mmengote/CentralParkZoo | /sketch_181013b/sketch_181013b.pyde | 971 | 4.34375 | 4 | #this library converts the random into integers
from random import randint
#lists of animals included in the simulation
#divided into two types, the zoo animals will have a starting state position of zoolocation.
#local animals will have a starting position of something else.
zoo_animals = ["Penguins", "Zebras", "Monkeys", "Sloths"]
local_animals = ["Humans", "Squirrels", "Racoons", "Coyotes"]
#this chooses the type of zoo animal that will escape randomly. they're stored in variables.
escapedAnimalRandom = zoo_animals[randint(0,3)]
numberOfescaped = randint(0,100) #maybe change the max to the capacity of population for the zoo
#this is the variable of which animal escaped and how many of them escaped
wantedAnimals = {
"animal escaped": escapedAnimalRandom,
"number escaped": numberOfescaped
}
#random escaped
#will need to visualize this instead of words eventually.
print wantedAnimals
## where would the penguin go##
| true |
4f03b52ca0f1f596bfe78b065f44ee8eb0f3b10c | markcharyk/data-structures | /tests/insert_sort_tests.py | 1,085 | 4.125 | 4 | import unittest
from data_structures.insert_sort import insertion_sort
class TestInsertSort(unittest.TestCase):
def setUp(self):
"""Set up some lists to be sorted"""
self.empty = []
self.single = [5]
self.sorted = [1, 2, 3, 10]
self.unsorted = [8, 1, 94, 43, 33]
def testEmpty(self):
"""Test an empty list"""
expected = []
actual = insertion_sort(self.empty)
self.assertEqual(expected, actual)
def testSingle(self):
"""Test a list of one item"""
expected = [5]
actual = insertion_sort(self.single)
self.assertEqual(expected, actual)
def testSorted(self):
"""Test an already sorted list"""
expected = [1, 2, 3, 10]
actual = insertion_sort(self.sorted)
self.assertEqual(expected, actual)
def testUnsorted(self):
"""Test an unsorted list"""
expected = [1, 8, 33, 43, 94]
actual = insertion_sort(self.unsorted)
self.assertEqual(expected, actual)
if __name__ == '__main__':
unittest.main()
| true |
9689a252589118db6d6623abb196aff8c3efa9ab | vielma24/playground | /bank.py | 1,337 | 4.40625 | 4 | #!/usr/bin/python
'''Program will intake customer information and account information. Account
management will also be added.'''
from datetime import date
#*******************************************************************************
class person:
"""
Returns a ```Person``` object with the given name, DOB, and phone number.
"""
import datetime
def __init__(self, name, DOB, phone_num):
self.name = name
self.DOB = date(DOB[0], DOB[1], DOB[2])
self.phone_num = phone_num
print("A student object is now created.")
def print_details(self):
"""
Prints the details of the student.
"""
print("Name:", self.name)
print("DOB:", self.DOB.year, self.DOB.month, self.DOB.day)
print("phone number:", self.phone_num)
#*******************************************************************************
def main():
# Test
'''This is how you create a new Person object and print the details.
The constructor must include 3 arguments as listed below.'''
aPerson = person('John Doe', (1982, 12, 15), 1)
aPerson.print_details()
''' TODO: Obtain the information of 5 new people and store them in a list
when complete print their details one at a time.'''
if __name__ == "__main__":
main()
| true |
9d625b9f273e27a48db58a2694b0b8e1eb537063 | dolapobj/interviews | /Data Structures/Array/MaximumProductSubarray.py | 1,155 | 4.3125 | 4 | #Maximum Product Subarray
#LC 152
"""
Given an integer array nums, find the contiguous subarray within an array
(containing at least one number) which has the largest product.
Example 1:
Input: [2,3,-2,4]
Output: 6
Explanation: [2,3] has the largest product 6.
Example 2:
Input: [-2,0,-1]
Output: 0
Explanation: The result cannot be 2, because [-2,-1] is not a subarray.
"""
#idea--> at every step we have three options
# 1. we get a max product by multiplying by current max so far with current element
# 2. we get a max product by multiplying current min so far with current element (multiply negs)
# 3. the current element can be the start for the maximum product subarray
def maxProductSubarray(nums):
maxProduct,minProduct,ans = nums[0],nums[0],nums[0]
for i in range(1,len(nums)):
tempMax = max(nums[i], nums[i]*maxProduct,nums[i]*minProduct)
tempMin = min(nums[i], nums[i]*maxProduct,nums[i]*minProduct)
maxProduct,minProduct = tempMax,tempMin
ans = max(ans,maxProduct)
return ans
#RT: O(n) --> loop through once
#SC: O(1) --> store 4 variables that are updated in each iteration
| true |
4076f61a183fac23c05fbf162dde7289caff5a74 | sndp2510/PythonSelenium | /basic/_04Geometry.py | 432 | 4.28125 | 4 | sideCount=int(input("Enter Number of side: "))
if(sideCount==1):
print("Its square")
side=int(input("Enter length of side : "))
print("Area of the square is : " , (side*side))
elif(sideCount==2):
print("Its Rectangle")
length = int(input("Enter length of side : "))
breadth =int(input("Enter length of side : "))
print("Area of the square is : " , (length*breadth))
else:
print("Invalid Entry")
| true |
d60ccaf31fe2c293d566effaf936c5c17e6064e3 | ksu-is/Commute-Traffic-Twitter-Bot | /Practice/user_funct.py | 1,027 | 4.28125 | 4 |
## will create text file for new users with info on each line. (not able to do OOP yet reliably so this is the best I've got)
# do you have an account?
user_ask = input("Before we start, lets specify what user this is for\nHave you used this program before?\nType 'y' or 'n' only: ")
if user_ask.lower() == "n":
filename = input("Okay, new user! What is your name?\nType your first name here to create your username: ").lower() + ".txt"
user_file = open(filename, 'w+')
print("\nNow lets load up all the highways that you use for your commute into your user account.")
#at this point, i am able to make a custom text file for any new user
while True:
append = input("\nEnter one highway at a time\nType '75', '85, '285', '20', or '400'\nOr type 'end' to finish.\nEnter here: ")
if append.lower() == "end":
break
else:
user_file.write(append)
user_file.write("\n")
user_file.seek(0)
print(user_file.readlines())
#def make_user(): | true |
1e3cc57af263476e1346cccf480a6ed62e7c891d | jenshurley/python-practice | /python_early_examples.py | 1,836 | 4.125 | 4 | # Initial variable to track game play
user_play = "y"
# While we are still playing...
while user_play == "y":
# Ask the user how many numbers to loop through
user_number = input("How many numbers? ")
# Loop through the numbers. (Be sure to cast the string into an integer.)
for x in range(int(user_number) + 1):
# Print each number in the range
print(x)
# Once complete...
user_play = input("Continue: (y)es or (n)o? ")
# Ask for how many numbers
# Print them out
# Ask again or quit
# if you get more numbers, add that many to the highest original number
playing = True
current_num = 1
while playing:
number = input("How many numbers would you like or q to quit: ")
if number != 'q':
number = int(number)
for x in range(current_num, current_num + number):
print(x)
current_num += number
else:
playing = False
# In 2 lists, we have dogs and their meals
# eg, Chance eats steak, lucky eats chow, etc...
dogs = ['chance', 'scout', 'rover', 'lucky', 'boss']
meals = ['steak', 'chicken', 'bones', 'chow', 'vegan']
# If someone gives us a dog, how can we return their meal?
def find_meal(some_dog):
pass
#Return the meal the dog eats
# Hit, you'll have to use an index lookup and
# Use that to find the meals
#Rover --> Bones as an example
# Ask for how many numbers
# Print them out
# Ask again or quit
# if you get more numbers, add that many to the highest original number
playing = True
current_num = 1
while playing:
number = input("How many numbers would you like or q to quit: ")
if number != 'q':
number = int(number)
for x in range(current_num, current_num + number):
print(x)
current_num += number
else:
playing = False
3 CommentsCollapse
| true |
0be835d18fae63f8b63fbae296f65250f8b38ac4 | koaning/scikit-fairness | /skfair/common.py | 1,752 | 4.4375 | 4 | import collections
def as_list(val):
"""
Helper function, always returns a list of the input value.
:param val: the input value.
:returns: the input value as a list.
:Example:
>>> as_list('test')
['test']
>>> as_list(['test1', 'test2'])
['test1', 'test2']
"""
treat_single_value = str
if isinstance(val, treat_single_value):
return [val]
if hasattr(val, "__iter__"):
return list(val)
return [val]
def flatten(nested_iterable):
"""
Helper function, returns an iterator of flattened values from an arbitrarily nested iterable
>>> list(flatten([['test1', 'test2'], ['a', 'b', ['c', 'd']]]))
['test1', 'test2', 'a', 'b', 'c', 'd']
>>> list(flatten(['test1', ['test2']]))
['test1', 'test2']
"""
for el in nested_iterable:
if isinstance(el, collections.abc.Iterable) and not isinstance(
el, (str, bytes)
):
yield from flatten(el)
else:
yield el
def expanding_list(list_to_extent, return_type=list):
"""
Make a expanding list of lists by making tuples of the first element, the first 2 elements etc.
:param list_to_extent:
:param return_type: type of the elements of the list (tuple or list)
:Example:
>>> expanding_list('test')
[['test']]
>>> expanding_list(['test1', 'test2', 'test3'])
[['test1'], ['test1', 'test2'], ['test1', 'test2', 'test3']]
>>> expanding_list(['test1', 'test2', 'test3'], tuple)
[('test1',), ('test1', 'test2'), ('test1', 'test2', 'test3')]
"""
listed = as_list(list_to_extent)
if len(listed) <= 1:
return [listed]
return [return_type(listed[: n + 1]) for n in range(len(listed))]
| true |
7a5d6f9bdd12a2aa3d4d1ce7c50723742a23cd46 | Azoad/Coursera_py4e | /chapter7/ex_07_01.py | 220 | 4.375 | 4 | """Take a file as input and display it at uppercase"""
fname = input('Enter the file name: ')
try:
file = open(fname)
except:
print('Error in file!')
quit()
for line in file:
print(line.strip().upper())
| true |
7b6fd1b192bf2fdacb41ce57cef67bd6ca3db41c | Azoad/Coursera_py4e | /test.py | 1,944 | 4.1875 | 4 | print('this is the first one:')
print('let\'s do this!')
print('add 5 and -3')
f = 5
t = -3
print('addition of',str(f),'and',str(t),'is :',str(f+t))
print('-----------------------------')
print('this is the second one:')
x = 'It is not America but it is Bangladesh'
lens = len(x)
print('Length of \'' +x+ '\' is : '+str(lens))
print('------------------------------')
print('this is the third one:')
a = 5
if a<10:
print('Smaller!')
if a>20:
print('Larger!')
print('Finish')
print('------------------------------')
print('this is the fourth one:')
n = 5
while n>0:
print(n)
n = n-1
print('Blastoff!')
print('------------------------------')
print('this is the fifth one:')
hrs = input("Enter Hours:")
fhrs = float(hrs)
rate = input("Enter rate:")
frate = float(rate)
pay = fhrs*frate
spay = str(pay)
print("Pay:",spay)
print('------------------------------')
print('this is the sixth one:')
print('Multiplication table:')
num = input('Enter a number:')
for n in range (1,11):
print(str(num)+' X '+str(n)+' =',int(n)*int(num))
print('------------------------------')
print('this is the seventh one:')
x = 5
if x > 2:
print('Bigger than 2')
print('Still bigger')
print('Done with 2')
for i in range(5):
print(i)
if i > 2:
print('Bigger than 2')
print('Done with i',i)
print('All done')
print('------------------------------')
print('this is the eighth one:')
x = 15
if x < 2 :
print('Small')
elif x < 10 :
print('Medium')
elif x < 20 :
print('Big')
elif x < 40 :
print('Large')
elif x < 100 :
print('Huge')
else :
print('Ginormous')
print('------------------------------')
print('this is the ninth one:')
x = input('Enter a positive number : ')
try :
ix = int(x)
except :
ix = -1
print('The value is ',ix)
if ix > 0 :
print('Number')
else :
print('Not a number')
| false |
6dafc193f5e2f1731df5462295131e2cee3236b8 | victorrayoflight/Foudil | /Examen 2019-02-18/6.py | 582 | 4.15625 | 4 | x = True
y = False
z = False
# False OR True == True
# True OR False == True
# True OR True == True
# False OR False == False
# True AND True = True
# True AND False = False
# False AND True = False
# False AND False = False
# False XOR True == True
# True XOR False == True
# True XOR True == False
# False XOR False == False
if not x or y:
# False or False == False
print(1)
elif not x or not y and z:
# False or True and False == False
print(2)
elif not x or y or not y and x:
# False or False or True and True == True
print(3)
else:
print(4)
| true |
8b6e4666bf75fc6b7023e8c93717384f56183890 | hayuzi/py-first | /base/dealwithexp.py | 2,766 | 4.25 | 4 | # 异常处理
# 语法错误是错误
# 异常是 语法正确，但是执行时候出现错误。
# 异常处理使用 try except 格式
# 与其他一些语言一样，责任链的模式，如果不做 except获取，会反馈到上一层.
while True:
try:
x = int(input("Please enter a number: "))
break
except ValueError:
print("that was no valid number, try again ")
# 一个except子句可以同时处理多个异常，这些异常将被放在一个括号里成为一个元组，例如
# except (RuntimeError, TypeError, NameError):
# pass
# 最后一个except子句可以忽略异常的名称，它将被当作通配符使用。你可以使用这种方法打印一个错误信息，然后再次把异常抛出。
# try except 语句还有一个可选的else子句，如果使用这个子句，那么必须放在所有的except子句之后。
# 这个子句将在try子句没有发生任何异常的时候执行。
# try:
# ......
# except OSError:
# ......
# except:
# print("Unexpected error:", sys.exc_info()[0])
# raise
# else:
# ......
# 抛出异常使用 raise
# raise 的异常必须是一个异常的实例或者异常的类（ Exception的子类 ）
# raise NameError('HiThere')
# 用户自定义异常
# 你可以通过创建一个新的异常类来拥有自己的异常。
# 异常类继承自 Exception 类，可以直接继承，或者间接继承，例如:
class MyError(Exception):
def __init__(self, value):
self.value = value
def __str__(self):
return repr(self.value)
# 定义清理行为
# try 语句还有另外一个可选的子句，它定义了无论在任何情况下都会执行的清理行为。
# 如果一个异常在 try 子句里（或者在 except 和 else 子句里）被抛出，而又没有任何的 except 把它截住，
# 那么这个异常会在 finally 子句执行后被抛出
# 预定义的清理行为
# 一些对象定义了标准的清理行为，无论系统是否成功的使用了它，一旦不需要它了，那么这个标准的清理行为就会执行。
# 这面这个例子展示了尝试打开一个文件，然后把内容打印到屏幕上:
for line in open("myfile.txt"):
print(line, end="")
# 以上这段代码的问题是，当执行完毕后，文件会保持打开状态，并没有被关闭。
# 关键词 with 语句就可以保证诸如文件之类的对象在使用完之后一定会正确的执行他的清理方法:
with open("myfile.txt") as f:
for line in f:
print(line, end="")
# python3 assert 断言
# Python assert（断言）用于判断一个表达式，在表达式条件为 false 的时候触发异常。
# assert expression [, arguments]
# end
| false |
9d848849b6d688b2872bd9923a86319e3ad73807 | Anchals24/InterviewBit-Practice | /Topic - String/Longest Common Prefix.py | 1,216 | 4.25 | 4 | #Longest Common Prefix.
#Arr = ["aa" , "aab" , "aabvde"]
"""
Problem Description >>
Given the array of strings A, you need to find the longest string S which is the prefix of ALL the strings in the array.
Longest common prefix for a pair of strings S1 and S2 is the longest string S which is the prefix of both S1 and S2.
For Example: longest common prefix of "abcdefgh" and "abcefgh" is "abc".
"""
#Solution : 1
def longestcommonprefix(Arr):
smallest = min(Arr) #Find the smallest element of the array
#print(smallest)
lmini = len(smallest) #length of the smallest element of the array
i = 0
common = ""
while i < lmini:
for A in Arr:
if A[i] != smallest[i]:
return common
common += smallest[i]
i += 1
return common
print(longestcommonprefix(["aaaaaa" , "aab" , "aabvde" , "aabb"]))
#Solution : 2
def longestcommonprefix2(Arr):
minele = min(Arr)
maxele = max(Arr)
count = 0
for i in range(min(len(maxele) , len(minele))):
if minele[i] == maxele[i]:
count += 1
return minele[:count]
print(longestcommonprefix2(["aa" , "aabse" , "a" ]))
| true |
ac81f1e984700b6b3993e1b60e29812dc82e69c4 | emlemmon/emlemmon.github.io | /python/Week6/check06b.py | 1,121 | 4.4375 | 4 | class Phone:
"""Parent class that has 3 member variables and 2 methods"""
def __init__(self):
self.area_code = 0
self.prefix = 0
self.suffix = 0
def prompt_number(self):
self.area_code = input("Area Code: ")
self.prefix = input("Prefix: ")
self.suffix = input("Suffix: ")
def display(self):
print("Phone info:")
print("({}){}-{}".format(self.area_code, self.prefix, self.suffix))
class SmartPhone(Phone):
"""Child class of Phone that includes the member variables from Phone as well as an email address"""
def __init__(self):
super().__init__()
self.email = ""
def prompt(self):
Phone.prompt_number(self)
self.email = input("Email: ")
def display(self):
Phone.display(self)
print(self.email)
def main():
newPhone = Phone()
newCell = SmartPhone()
print("Phone:")
newPhone.prompt_number()
print()
newPhone.display()
print()
print("Smart phone:")
newCell.prompt()
print()
newCell.display()
if __name__ == "__main__":
main() | true |
0e821086fd74aebd7e245509f1507fc3ac6ccec2 | Jrufino/ListaRepeticaoPython | /ex22.py | 638 | 4.21875 | 4 | sair='N'
divisivel=[]
while sair !='S':
num=int(input('Digite um numero inteiro: '))
if num==2 or num==3 or num==5 or num==7:
print('Primo')
else:
if num%2==0 or num%3==0 or num%5==0 or num%7==0:
if num%2==0:
divisivel.append(2)
if num%3==0:
divisivel.append(3)
if num%5==0:
divisivel.append(5)
if num%7==0:
divisivel.append(7)
print('Nao eh primo. Divisivel por {}'.format(divisivel))
else:
print('Primo')
sair=str.upper(input('S- Sair, outra tecla continua')) | false |
41f61ba806ae6942a62b114ae624ab00ce0d6b7d | kobecow/algo | /NoAdjacentRepeatingCharacters/problem.py | 543 | 4.15625 | 4 | """
Given a string, rearrange the string so that no character next to each other are the same.
If no such arrangement is possible, then return None.
Example:
Input: abbccc
Output: cbcbca
"""
def rearrangeString(s: str) -> str or None:
pass
print(rearrangeString("gahoeaddddggrea"))
# no order required
# ahgogrgededadad
print(rearrangeString("cccciiiiiddddssskkdddcc"))
# no order required
# ckckcscscsdidididididcd
# improvement
# this is too complicated comarared to answer.py.
# use only slice operation, not priority queue. | true |
b6a1309a07f6203b3d183552e30e4e05c5582713 | Kishy-nivas/ds-in-python | /linked list.py | 522 | 4.15625 | 4 | # linked list implementation
class Node(object):
def__init__(self,value,pointer=None)
self.value =value
self.next =None
class LinkedList(object):
def __init__(self):
self.head =None
self.tail =None
def add(self,value ):
if self.head==None:
self.head =self.tail = Node(value )
self.head.next =self.tail.next =None
else:
self.head.next = Node(value )
self.head = self.head.next
def traverse(self):
r= self.tail
while r:
print(r.value )
r=r.next
if __name__ == '__main__': | false |
c47370697303df60b4e50e91092a5564b047a718 | stephenforsythe/crypt | /shift.py | 1,172 | 4.125 | 4 | """
This is essentially a substitution cypher designed to avoid fequency analysis.
The idea is to shift the char to the right by an increasing number, until n = 25,
at which time the loop will start again.
Spaces are treated as a character at the end of the alphabet.
"""
import string
FULL_ALPHABET = string.lowercase + ' '
MOD_NUM = len(FULL_ALPHABET)
def encrypt_decrypt(instr, direction):
shift = 0
encrypted_message = ''
for char in instr.lower():
pos = string.index(FULL_ALPHABET, char)
mod_pos = pos % MOD_NUM
if direction == 'en':
shift_pos = (mod_pos + shift) % MOD_NUM
else:
shift_pos = (mod_pos - shift) % MOD_NUM
encrypted_message += FULL_ALPHABET[shift_pos]
shift += 1
return encrypted_message
eord = raw_input('Encrypt or decrypt? (e/d):\n')
if eord == 'e':
user_input = raw_input('Message to be encrypted:\n')
print 'encrypted message:\n'
print encrypt_decrypt(user_input, 'en')
elif eord == 'd':
user_input = raw_input('Message to be decrypted:\n')
print 'plain text:\n'
print encrypt_decrypt(user_input, 'de')
else:
print 'error'
| true |
dfed06b4bbf51b8d57511753a08f09e26e3605d0 | MuhammetEser/Class4-PythonModule-Week4 | /4- Mis Calculator.py | 2,631 | 4.28125 | 4 | # As a user, I want to use a program which can calculate basic mathematical operations. So that I can add, subtract, multiply or divide my inputs.
# Acceptance Criteria:
# The calculator must support the Addition, Subtraction, Multiplication and Division operations.
# Define four functions in four files for each of them, with two float numbers as parameters.
# To calculate the answer, use math.ceil() and get the next integer value greater than the result
# Create a menu using the print command with the respective options and take an input choice from the user.
# Using if/elif statements for cases and call the appropriate functions.
# Use try/except blocks to verify input entries and warn the user for incorrect inputs.
# Ask user if calculate numbers again. To implement this, take the input from user Y or N.
import math
from addition import Addition
from subtraction import Subtraction
from multiplication import Multiplication
from division import Division
afronden = math.ceil
def num_input():
while True:
try:
number = float(input("Please enter a float number: "))
break
except ValueError:
print("""Please enter a valid float number using "." """)
return number
def operant():
while True:
try:
symbol = input("Please enter a mathematical operator symbol (+,-,*,/): ")
if symbol != '+' and symbol != '-' and symbol != '*' and symbol != '/':
raise ValueError
break
except ValueError:
print("Please enter a valid mathematical operator!")
return symbol
def yes_no():
while True:
try:
letter = input("Would you like to make another calculation (Y,N): ")
if letter != 'Y' and letter != 'y' and letter != 'N' and letter != 'n':
raise ValueError
break
except ValueError:
print("Oops! That was no valid symbol. Try again...")
return letter
while True:
symbol = operant()
if symbol == '+':
print("Result: " + str(afronden(Addition(num_input(), num_input()))))
elif symbol == '-':
print("Result: " + str(afronden(Subtraction(num_input(), num_input()))))
elif symbol == '*':
print("Result: " + str(afronden(Multiplication(num_input(), num_input()))))
elif symbol == '/':
print("Result: " + str(afronden(Division(num_input(), num_input()))))
antwoord = yes_no()
if antwoord == 'N' or antwoord == 'n':
break | true |
444256bdb2ae5b3aa29aec5acf93bfbb5875a8c7 | ivanakonatar/MidtermExample | /task4.py | 985 | 4.21875 | 4 |
"""
=================== TASK 4 ====================
* Name: Convert To Upper
*
* Write a function `convert_2_upper` that will take
* a string as an argument. The function should
* convert all lowercase letter to uppercase without
* usage of built-in function `upper()`.
* Note: Please describe in details possible cases
* in which your solution might not work.
*
* Use main() function to test your solution.
===================================================
"""
def mala_u_velika(recenica):
if not isinstance(recenica,str):
print("Pogresan unos! ")
nova_recenica = ''
for karakter in recenica:
broj_chr = ord(karakter)
if broj_chr > 96 and broj_chr < 123:
broj_veliko_slova = broj_chr - 32
karakter = chr (broj_veliko_slova)
nova_recenica+= karakter
return nova_recenica
def main():
recenica="IvaNa KONatAr"
print("Velikim slovima napisana je: ", mala_u_velika(recenica))
pass
main() | false |
5cae2691b29e74917c55007405b67e7fcf1a60de | KennethNielsen/presentations | /python_and_beer_lesser_known_gems/ex6_namedtuple.py | 2,143 | 4.21875 | 4 | """namedtuple example"""
from __future__ import print_function
from collections import namedtuple
def arange(start, end, step):
"""Arange dummy function"""
print('arange', start, end, step)
def get_data_from_db(data_id):
"""Get data from db dummy function"""
print('get_data_from_db', data_id)
def plot(x, y, title):
"""plot dummy function"""
print('plot with title: "{}"'.format(title))
################################### Without named tuple
def without_named_tuple():
"""Without named tuple"""
print('Without named tuple')
# A dataset is: data_id, comment, step, start, end
datasets = (
(1234, 'this is the one', 0.1, 45, 55),
(1345, 'no wait this is', 0.2, 40, 60),
(1456, 'I know I got it now', 0.2, 40, 60),
(1567, 'OK maybe not .. let\'s try something else', 0.05, 40, 60),
(1678, 'here we go again', 0.05, 40, 62),
)
for dataset in datasets:
print('\n### Process dataset', dataset)
# Remember a dataset is: data_id, comment, step, start, end
x = arange(dataset[3], dataset[4], dataset[2])
y = get_data_from_db(data_id=dataset[0])
plot(x, y, title=dataset[1])
####################################### With named tuple
def with_named_tuple():
"""With named tuple"""
print('\n\nWith named tuple')
dataset = namedtuple('dataset', 'data_id comment step start end')
#dataset = namedtuple('dataset', ['data_id', 'comment', 'step', 'start', 'end'])
datasets = (
dataset(1234, 'this is the one', 0.1, 45, 55),
dataset(1345, 'no wait this is', 0.2, 40, 60),
dataset(1456, 'I know I got it now', 0.2, 40, 60),
dataset(1567, 'OK maybe not .. let\'s try something else', 0.05, 40, 60),
dataset(1678, 'here we go again', 0.05, 40, 62),
)
for dataset in datasets:
print('\n### Process', dataset)
x = arange(dataset.start, dataset.end, dataset.step)
y = get_data_from_db(data_id=dataset.data_id)
plot(x, y, title=dataset.comment)
if __name__ == '__main__':
#without_named_tuple()
with_named_tuple()
| true |
de31bc761ac04f9c15ac4a74328d74b907fbd822 | oliviamillard/CS141 | /coinAdder.py | 1,690 | 4.28125 | 4 | #Olivia Millard - Lab 3a
#This program will add up a user's coins based off of their input.
#Short welcome message.
print("Hello. This program will add up all of your change!\nCool! Let's get started.\n")
"""
This function, first, take input from the user asking how many coins they have. Next, it takes input for what type of coin they have.
And last, it adds up the total coins depending on their value
### Variables ###
value = how much the individual coin is worth.
coinNumber = which coin of theirs they are inputting the type for (penny, nickel, dime, quarter).
total = the total amount of their coins added together.
dollarAmount = the quotient from doing integer division on the total.
centsAmount = the remainder from using modulus on the total.
"""
def addUpCoins():
value = 0
coinNumber = 1
total = 0
numOfCoins = int(input("How many coins do you have? "))
while coinNumber <= numOfCoins:
coin = str(input("What is coin #" +str(coinNumber)+ "? (q for quarter, d for dime, n for nickel, p for penny): "))
coinNumber = coinNumber + 1
if coin == "p":
value = 0.01
total = total + value
elif coin == "n":
value = 0.05
total = total + value
elif coin == "d":
value = 0.10
total = total + value
elif coin == "q":
value = 0.25
total = total + value
dollarAmount = total // 100
centsAmount = total % 100
print("\n")
print("Whoa... you have a lot of monies!\n" +str(dollarAmount)+ " dollars and " +str(centsAmount)+ " cents if we're being exact.")
print("Now go buy yourself something nice.")
addUpCoins()
| true |
e287eb63af5781999f148ef1aecf4dc2752e2ed1 | a-ruzin/python-base2 | /lesson_11/homework/task5.py | 2,064 | 4.15625 | 4 | """
5. Продолжить работу над первым заданием. Разработайте методы, которые отвечают
за приём оргтехники на склад и передачу в определённое подразделение компании.
Для хранения данных о наименовании и количестве единиц оргтехники, а также
других данных, можно использовать любую подходящую структуру (например, словарь).
"""
class StoreError(Exception):
pass
class Store:
def __init__(self):
self.store = {}
def add(self, entity, department):
self.store.setdefault(department, [])
self.store[department].append(entity)
def remove(self, entity, department):
try:
self.store[department].remove(entity)
except:
raise StoreError(f'На складе в департаменте {department} нет техники {entity}')
def move(self, entity, department_from, department_to):
self.remove(entity, department_from)
self.add(entity, department_to)
class Entity:
def __init__(self, name, price):
self.name = name
self.price = price
def __str__(self):
return self.name
def __repr__(self):
return self.name
class Printer(Entity):
def __init__(self, name, price, print_size):
super().__init__(name, price)
self.print_size = print_size
class Scanner(Entity):
def __init__(self, name, price, is_color):
super().__init__(name, price)
self.is_color = is_color
class Copier(Entity):
def __init__(self, name, price, is_multipage):
super().__init__(name, price)
self.is_multipage = is_multipage
store = Store()
copier = Copier('Xerox', 20000, True)
store.add(copier, 'Бухгалтерия')
print(store.store)
store.move(copier, 'Бухгалтерия', 'Продажи')
print(store.store)
| false |
52c44bc5f786ab5c741fce402acf47ee901c5818 | a-ruzin/python-base2 | /lesson_3/homework/task3.py | 1,367 | 4.15625 | 4 | """
3. Написать функцию thesaurus(), принимающую в качестве аргументов имена сотрудников
и возвращающую словарь, в котором ключи — первые буквы имен, а значения — списки,
содержащие имена, начинающиеся с соответствующей буквы. Например:
>>> thesaurus("Иван", "Мария", "Петр", "Илья")
{
"И": ["Иван", "Илья"],
"М": ["Мария"], "П": ["Петр"]
}
Подумайте: полезен ли будет вам оператор распаковки?
Сможете ли вы вернуть отсортированный по ключам словарь?
"""
def thesaurus(*args):
dictionary = {}
for name in args:
first_letter = name[0]
dictionary.setdefault(first_letter, [])
dictionary[first_letter].append(name)
return dictionary
d = thesaurus("Иван", "Мария", "Петр", "Илья")
for first_letter in sorted(d.keys()):
print(f'{first_letter}: {d[first_letter]}')
# Подумайте: полезен ли будет вам оператор распаковки? нет
# Сможете ли вы вернуть отсортированный по ключам словарь? нет
| false |
4660fcbd2bc5df305214bffd4b4bc073abe6c1f5 | samwata/Python-Is-Easy | /Dictionaries-And-Sets/dictionary-assignment.py | 531 | 4.1875 | 4 | #Songs dictionary
Songs = {"Laugh Now, Cry Later":"Drake","Dark Lane Demo Tapes":"Drake",
"Sounds from the Other Side":"Wizkid","Joro":"Wizkid","jeje":"Diamond Platnumz","A Boy from Tandale":"Diamond Platnumz",
"Tambarare":"Eunice Njeri"}
#Function to check if the Song exists in Songs dictionary
while (True):
SongName = input("Please enter the song title and artist seperated by colon i.e song_name:artist :")
List=SongName.split(":")
if List[0] in Songs and List[1] in Songs.values():
print("True")
else:
print("False")
| false |
173ecefb3bd3070bff1b3cd83e91dab2dd2eb4bd | renan-suetsugu/WorkshopPythonOnAWS | /Labs/Loops/lab_6_step_1_Simple_Loops.py | 545 | 4.125 | 4 | #fruit = ['apples','oranges','bananas']
#for item in fruit:
# print(f'The best fruit now is {item}')
#numbers = [0,1,2,3,4,5,6,7,8,9,10]
#for number in numbers:
# print(f'The next number is {number}')
#for number in range(10):
# print(f'The next number is{number}')
#for number in range(10):
# print(f'The actual number is {number} and next number is {number+1}')
#for number in range(1,10):
# print(f'The next number is {number}')
for number in range(1,10,2):
print(f'The next number is {number}')
| true |
2a4dd3b8574142a991f73761bd43d6e2688c83a0 | blackicetee/Python | /python_games/simple_games/MyValidator.py | 451 | 4.125 | 4 | """This self written python module will handle different kinds of user input and validate it"""
# This is a module for regular expressions see python documentation "Regular expression operations"
import re
class MyValidator:
@staticmethod
def is_valid_from_zero_to_three(user_input):
regex_pattern = r'[0-3]'
if re.match(regex_pattern, user_input) is not None:
return True
else:
return False
| true |
8190dd1f19c1c918cd16997aae5a2c0b77c60607 | CoitThomas/Milestone_7 | /7.1/test_string.py | 480 | 4.3125 | 4 | """Verify the correct returned string value of a list of chars given
to the funtion string().
"""
from reverse_string import string
def test_string():
"""Assert the correct returned string values for various lists of
chars given to the string() function.
"""
# Anticipated input.
assert string(['H', 'e', 'l', 'l', 'o', ' ', 'w', 'o', 'r', 'l', 'd']) == 'Hello world'
# 1 input.
assert string(['A']) == 'A'
# No input.
assert string([]) == ''
| true |
f4ad480c6b9e7d8754506f060649b77b99049268 | taisei-s/nlp100 | /swings/swing09.py | 1,051 | 4.15625 | 4 | # coding:utf-8
"""
09. Typoglycemia
スペースで区切られた単語列に対して，各単語の先頭と末尾の文字は残し，それ以外の文字の順序をランダムに並び替えるプログラムを作成せよ．ただし，長さが４以下の単語は並び替えないこととする．適当な英語の文（例えば"I couldn't believe that I could actually understand what I was reading : the phenomenal power of the human mind ."）を与え，その実行結果を確認せよ．
"""
import random
def shuffle_inside(word):
if len(word) <= 4:
return word
inside = list(word[1:-1])
random.shuffle(inside)
new_word = word[0] + ''.join(inside) + word[-1]
return new_word
def swing09(str):
word_list = str.split()
shuffled = ' '.join([shuffle_inside(word) for word in word_list])
return shuffled
if __name__ == "__main__":
str = "I couldn't believe that I could actually understand what I was reading : the phenomenal power of the human mind ."
print(swing09(str)) | false |
313947b60813bec8b6956a808f7c5ee1efed7618 | scottwedge/code_hilfen | /ONLINE-KURSE/Become a Python Developer/5 - Learning the Python 3 Standard Library/code-training/3 - Python Input_Output/input_output.py | 2,875 | 4.1875 | 4 | #1 - Command line arguments
#python3 input_output.py 0 1 2
#in CLI ist alles immer ein String
import sys #um auf die CLI-Param zugreifen zu können
sys.argv
print("Number of arguments: " + str(len(sys.argv)))
print("Arguments: ", sys.argv)
#<- erstes Argument ist Name (bzw. Pfad der Datei)
sys.argv.remove(sys.argv[0]) #erstes Argument löschen. Funktioniet mit remove, da argv ist eine Liste
print("Arguments: ", sys.argv)
arguments = sys.argv
sum = 0
for i in arguments:
try: #hier wird veruch in Zahl zu konvertieren
number = int(i)
sum = sum + number
except e:
print("Keine Zahl")
print(sum)
print()
#2 - Input and Output
print("Hello !")
color = input("What is your favourite color? ") #Eingaben aus CLI einlesen
print("Your color: " + color)
print()
#3 - Files and file writing
myFile = open("scores.txt", "w")
#w -> write; r -> read; r+ -> read and write; a -> append
#Eigenschften der Datei
print("Name " + myFile.name)
print(("Mode " + myFile.mode))
myFile.write("GBJ : 100 \nKHD : 99 \nBBB : 89\n")
myFile.close()
myFile = open("scores.txt", "r")
print("Reading ...\n" + myFile.read())
myFile.seek(0, 0)
print("Reading... " + myFile.read(10)) #liest nur 10 Char.
myFile.close()
print()
#4 - File seeking in Python
#Seek-Pointer = zeigt auf Char in Datei, wenn man aus/in Datei liest/schreibt => wird dieser Seek-Pointer verschoben
#Beim öffnen der Datei wird Seek-Pointer auf 0 gesetzt.
myFile = open("scores.txt", "r")
print("Reading... " + myFile.read(10)) #liest nur 10 Char.
myFile.seek(0)
print("Reading... " + myFile.read(10)) #liest nur 10 Char.
myFile.close()
print()
#5 - Iterative files
# = Datei zeilenweise lesen
myFile = open("scores.txt", "r")
print("One line : " + myFile.readline())
myFile.seek(0)
print("One line : " + myFile.readline())
for line in myFile:
newHeighScorer = line.replace("BBB", "PDJ") # in der Zeile: BBB durch PDJ ersetzen <- !! nicht in der Datei erstetz sondern im String line
print(newHeighScorer)
myFile.close()
print()
#6 - Tempfile
import tempfile # um Temporäre Dateien zu benutzen
tempFile = tempfile.TemporaryFile()
tempFile.write(b"Save this special number for me: 567896") #in Tempfile schreiben, braucht aber Byte-Object => b davor
tempFile.seek(0)
print(tempFile.read()) #alles aus dem Tempfile lesen
print()
#7 - Manipulate zip files in Python
import zipfile
zip = zipfile.ZipFile("archiv.zip", "r") #zip-Datei öffnen
print(zip.namelist())
#Metadaten auslesen
for meta in zip.infolist():
print(meta)
info = zip.getinfo("purchased.txt") #Metadaten bestimmter Datei
print(info)
print(zip.read("scores.txt"))
with zip.open("scores.txt") as f: #öffnet wishlist.txt in zip-Datei und konvertiert es zu file (as f)
print(f.read())
#zip.extract("purchase.txt") # Datei aus Zip auspacken und speichern
#zip.extractall()
zip.close() | false |
52e700eccf6b8b453f6680c9a991b150774033e4 | scottwedge/code_hilfen | /ONLINE-KURSE/Become a Python Developer/4 - Python Essential Training/code-training/11 - String Objects/string_objects.py | 1,442 | 4.28125 | 4 | # 1 - Overview of string objects
#String sind Objekte
# Strin-Methoden,
print("TestString".upper())
print("TestString".swapcase())
print("TestString {}".format(4))
print('''Test
Str ing {}'''.format(4))
s = "Test String {}"
print(s.format(4))
class myString(str):
def __str__(self):
return self[::-1]
s = myString("TestString")
# 2 - Common string methods
print("\n # - 2\n")
print("Test string".title())
print("TestString".casefold()) # auch für UNICODE
#Stirng ist imuteble = nicht veränderbar
s1 = "Hello World1"
s2 = "Hello World2".upper()
print(id(s1))
print(id(s2))
#Zwei Strings verbinden
print(s1 +' '+ s2)
print("TestString")
# 3 - Formatting strings
print("Test {xx} String{bb}".format(xx = 10, bb = 20))
print("Test {1} String{0}".format(10, 20))
print("Test {0:<5} String{1:+05}".format(10, 20))
print("Test {} String{:,}".format(10, 20000000))
print("Test {} String{:,}".format(10, 20000000).replace(",", "."))
# mit ...f kann man Nachkommastellen spezifizieren
print("Test {:f} String{}".format(10, 20))
#hexadezimal
print("Test {:x} String{}".format(10, 20))
#oktal
print("Test {:o} String{}".format(10, 20))
#binär
print("Test {:b} String{}".format(10, 20))
#print(f'Test String {10}')
x = 10
#print(F'Test String{x}')
# 4 - Splitting and Joining
s = "grosser String mit Leerzeilen"
print(s.split())
print(s.split('e'))
#Teilt String in Liste auf
l = s.split()
#Verbinden + statt _ :
s2 = ":".join(l)
| false |
bc58fb8b971f9a907786caac451b8d603a7ec1c0 | llduyll10/backup | /Backup/BigOGreen/classPython.py | 1,960 | 4.34375 | 4 | #Create Class
class MyClass:
x=5
print(MyClass)
#Create object
p1 = MyClass()
print(p1.x)
#__innit__() function
class Person:
def __init__(self,name,age):
self.name = name
self.age = age
p1 = Person("DuyNND",22)
print(p1.name)
print(p1.age)
#Object METHOD
#Object can also cotain method
class People:
def __init__(self,name,age):
self.name = name
self.age = age
def myFunction(self):
print("Hello my name is: " + self.name)
p1 = People("DuyNguyenVJPPRO",20)
p1.myFunction()
#Self parameter
#parameter is a reference to the current instance of the class,
# and is used to access variables that belong to the class.
#Use the words mysillyobject and abc instead of self
class UpdatePerson:
def __init__(duyobject,name,age):
duyobject.name = name
duyobject.age = age
def myFunction(abc):
print("Hello my name is: " + abc.name)
p1 = UpdatePerson("John", 30)
p1.myFunction()
#Delete Objects
# p2 = UpdatePerson("Test", 20)
# del p2.age
# print(p2.age)
#Python Inheritance
class ParentClass:
def __init__(self,fname,lname):
self.firstName = fname
self.lastName = lname
def printName(self):
print(self.firstName, self.lastName)
P1 = ParentClass("Nguyen","Duy")
P1.printName()
#Create child class
class ChildClass(ParentClass):
pass
P2 = ChildClass("Duy","Pro")
P2.printName()
#Add properties in Child class
#And add method in Child class
class People:
def __init__(self,fname,lname):
self.firstName = fname
self.lastName = lname
def printName(self):
print(self.firstName, self.lastName)
class Student(People):
def __init__(self,fname,lname,year):
super().__init__(fname,lname)
self.graduationyear = year
def welcomeNewbie(self):
print(self.firstName, self.lastName, self.graduationyear)
x = Student("DuyNND","Vippro",2020)
x.welcomeNewbie()
print(x.graduationyear) | true |
56dac0836a399dfb331f7dad49cde574aea01446 | Clobbster/hackerrank | /Python/If_Else/test_sample_code.py | 529 | 4.3125 | 4 | # For a input variable, write if/then ladder for the following:
# If is odd, print Weird
# If is even and in the inclusive range of to , print Not Weird
# If is even and in the inclusive range of to , print Weird
# If is even and greater than , print Not Weird
N = int(input("Please enter a number: "))
if N % 2 != 0:
print("Weird")
elif N % 2 == 0:
if (N >= 2) & (N <= 5):
print("Not Weird")
elif (N >= 6) & (N <= 20):
print("Here")
print("Weird")
else:
print("Not Weird") | false |
b210549d4b32f01909c03f2a85f0127a3c4c3c1f | Clobbster/hackerrank | /Python/Write_A_Function/test_sample_code.py | 406 | 4.25 | 4 | # Write a function that determines if a given input year is a leap year.
def is_leap(year):
leap = ""
# Write your logic here
if year % 4 == 0:
if year % 100 == 0:
if year % 400 == 0:
leap = True
else:
leap = False
else:
leap = True
else:
leap = False
print(leap)
is_leap(1900)
| false |
950af6a1c4cd4ffe12ff99061bb44cfd0104fec6 | regismagnus/challenge_python | /is_list_palindrome.py | 1,390 | 4.15625 | 4 | '''
Note: Try to solve this task in O(n) time using O(1) additional space,
where n is the number of elements in l, since this is what you'll be asked to do during an interview.
Given a singly linked list of integers, determine whether or not it's a palindrome.
Note: in examples below and tests preview linked lists are presented as arrays just for simplicity of visualization:
in real data you will be given a head node l of the linked list
Example
For l = [0, 1, 0], the output should be
isListPalindrome(l) = true;
For l = [1, 2, 2, 3], the output should be
isListPalindrome(l) = false.
'''
# Singly-linked lists are already defined with this interface:
class ListNode(object):
def __init__(self, x):
self.value = x
self.next = None
def isListPalindrome(l):
a=convertToArray(l)
for i in range(int(len(a)/2)):
if a[i]!=a[len(a)-i-1]:
return False
return True
def convertToArray(l):
a=[]
if(l!=None):
a.append(l.value)
c=l.next
while(c!=None):
a.append(c.value)
c=c.next
return a
''''l=ListNode(1)
l.next=ListNode(1000000000)
l.next.next=ListNode(-1000000000)
l.next.next.next=ListNode(-1000000000)
l.next.next.next.next=ListNode(1000000000)
l.next.next.next.next.next=ListNode(1)'''
l=ListNode(0)
l.next=ListNode(1)
l.next.next=ListNode(0)
print(isListPalindrome(l)) | true |
0f9ba273804db8997bd5ec73e7c233c11d58eac5 | regismagnus/challenge_python | /longest_word.py | 611 | 4.34375 | 4 | '''
Define a word as a sequence of consecutive English letters. Find the longest word from the given string.
Example
For text = "Ready, steady, go!", the output should be
longestWord(text) = "steady".
best solution by dnl-blkv:
return max(re.split('[^a-zA-Z]', text), key=len)
'''
def longestWord(text):
rex='[\[\],!@#$%&{}+-=~^`_*/:()<>|?*]|(\\\)'
m=None
w=None
text=re.sub(rex, ' ',text)
for word in text.split():
word=re.sub(rex, '',word)
if m==None or m<len(word):
m=len(word)
w=word
return w
print(longestWord('Ready, steady, go!')) | true |
bee95c39ec36fa669a574732be32e6a85d024117 | bvishny/basic-algos | /mergesort_inversions.py | 2,008 | 4.25 | 4 | # Basic Merge Sort Implementation
# Components
# 1. split_array(array) array array - takes an array and splits it at the midpoint into two other arrays
# 2. merge(array, array) array - takes two split arrays and merges them together sorted
# 3. mergesort(array) array - takes an array and sorts it using mergesort
def split_array(array):
split_pt = (len(array) / 2)
return (array[:split_pt], array[split_pt:])
# Tests:
# Empty Array:
arr1, arr2 = split_array([])
if not (len(arr1) == 0 and len(arr2) == 0):
print("Empty array failed")
# 1 Elem Array
arr1, arr2 = split_array([1])
if not (len(arr1) == 0 and len(arr2) == 1 and arr2[0] == 1):
print("1 Elem array failed")
# Odd Elem Array
arr1, arr2 = split_array([1,2,4])
if not (len(arr1) == 1 and len(arr2) == 2 and arr2[-1] == 4):
print("Odd Elem Array Failed")
# Even Elem Array
arr1, arr2 = split_array([1,2,3, 4])
if not (len(arr1) == 2 and len(arr2) == 2 and arr1[-1] == 2):
print("Even Elem Array Failed")
# MERGE:
class Counter(object):
def __init__(self):
self.total = 0
def incr(self, amt):
self.total += amt
def count(self):
return self.total
def merge(array1, array2, counter):
idx1, idx2, result = 0, 0, []
while idx1 < len(array1) or idx2 < len(array2):
if idx2 >= len(array2) or (idx1 < len(array1) and array1[idx1] <= array2[idx2]):
result.append(array1[idx1])
idx1 += 1
else:
result.append(array2[idx2])
idx2 += 1
counter.incr(len(array1) - idx1)
return result
# MERGESORT:
def mergesort(array, counter = Counter()):
if len(array) > 1:
arr1, arr2 = split_array(array)
else:
return array
return merge(mergesort(arr1, counter), mergesort(arr2, counter), counter)
c = Counter()
f = [int(x) for x in open("/Users/10gen/Downloads/IntegerArray.txt", 'r').read().split("\r\n")[:-1]]
result3 = mergesort(f, c)
print(c.count())
| true |
b98c45e58694e05b7fadee7721ce57a4e625c603 | shubhamPrakashJha/selfLearnPython | /6.FunctionalProgramming/6.Recursion.py | 328 | 4.1875 | 4 | def factorial(x):
if x == 1:
return 1
else:
return x*factorial(x-1)
print(factorial(int(input("enter the number: "))))
def is_even(x):
if x == 0:
return True
else:
return is_odd(x-1)
def is_odd(x):
return not is_even(x)
print(is_even(23))
print(is_odd(1))
print(is_odd(2)) | true |
f9a34ec2cd69deb1bc020afd803e3abd5f3ac877 | juechen-zzz/LeetCode | /python/0207.Course Schedule.py | 2,706 | 4.125 | 4 | """
There are a total of numCourses courses you have to take, labeled from 0 to numCourses-1.
Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]
Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?
Example 1:
Input: numCourses = 2, prerequisites = [[1,0]]
Output: true
Explanation: There are a total of 2 courses to take.
To take course 1 you should have finished course 0. So it is possible.
Example 2:
Input: numCourses = 2, prerequisites = [[1,0],[0,1]]
Output: false
Explanation: There are a total of 2 courses to take.
To take course 1 you should have finished course 0, and to take course 0 you should
also have finished course 1. So it is impossible.
"""
# BFS
class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
# 存储有向图
edges = collections.defaultdict(list)
# 存储每个节点的入度
indeg = [0] * numCourses
# 存储答案
visited = 0
for info in prerequisites:
edges[info[1]].append(info[0])
indeg[info[0]] += 1
# 将所有入度为 0 的节点放入队列中
q = collections.deque([u for u in range(numCourses) if indeg[u] == 0])
while q:
visited += 1
u = q.popleft()
for v in edges[u]:
indeg[v] -= 1
if indeg[v] == 0:
q.append(v)
return visited == numCourses
# DFS
class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
# 存储有向图
edges = collections.defaultdict(list)
# 标记每个节点的状态：0=未搜索，1=搜索中，2=已完成
visited = [0] * numCourses
# 用数组来模拟栈，下标 0 为栈底，n-1 为栈顶
result = list()
# 判断有向图中是否有环
valid = True
for info in prerequisites:
edges[info[1]].append(info[0])
def dfs(u: int):
nonlocal valid
visited[u] = 1
for v in edges[u]:
if visited[v] == 0:
dfs(v)
if not valid:
return
elif visited[v] == 1:
valid = False
return
visited[u] = 2
result.append(u)
for i in range(numCourses):
if valid and not visited[i]:
dfs(i)
return valid
| true |
837c95fb0a7b12b26dd16d0c3c693ac16ed1bfd1 | juechen-zzz/LeetCode | /python/0075.Sort Colors.py | 1,075 | 4.21875 | 4 | '''
Given an array with n objects colored red, white or blue, sort them in-place so that objects of the same color are adjacent, with the colors in the order red, white and blue.
Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.
Note: You are not suppose to use the library's sort function for this problem.
Example:
Input: [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]
'''
class Solution:
def sortColors(self, nums: List[int]) -> None:
'''
荷兰三色旗问题解
'''
# 对于所有 idx < p0 : nums[idx < p0] = 0
# curr是当前考虑元素的下标
p0 = curr = 0
# 对于所有 idx > p2 : nums[idx > p2] = 2
p2 = len(nums) - 1
while curr <= p2:
if nums[curr] == 0:
nums[p0], nums[curr] = nums[curr], nums[p0]
p0 += 1
curr += 1
elif nums[curr] == 2:
nums[curr], nums[p2] = nums[p2], nums[curr]
p2 -= 1
else:
curr += 1
| true |
de7e65a78aeb655fb91579b349f8f2c2c697e0aa | Phoenix795/lesson1 | /ld.py | 986 | 4.1875 | 4 | # step 5(Комплексные типы данных: списки)
my_list = [3,5,7,9,10.5]
print('My list: ', my_list)
my_list.append('Python')
print('Length of my_list: ', len(my_list))
print('First element: ', my_list[0])
print('Last element: ', my_list[-1])
print('From second to fourth element: ',my_list[1:4])
print('My list: ', my_list)
del my_list[-1]
print('My list after removing: ',my_list)
# step 6(Комплексные типы данных: словари)
my_dictionary = {
"city": "Москва",
"temperature": "20"
}
print('Name of city: ', my_dictionary.get('city'))
my_dictionary['temperature'] = int(my_dictionary['temperature']) - 5
print('Changed temperature: ', my_dictionary['temperature'])
print('Country key in My dictionary is: ', my_dictionary.get('country'))
print('Default value for country key: ', my_dictionary.get('country', 'Россия'))
my_dictionary['date'] = "27.05.2019"
print('Length of My dictionary: ', len(my_dictionary))
| false |
c22a3a2358c43fa2010303fa125199b77168d3b5 | colemai/coding-challenges | /leetcode_challenges/L267-PalindromeP2.py | 2,627 | 4.28125 | 4 | #!/usr/bin/env python3
"""
Author: Ian Coleman
Input:
Output:
Challenge:
Given a string s, return all the palindromic permutations (without duplicates) of it.
Return an empty list if no palindromic permutation could be form.
For example:
Given s = "aabb", return ["abba", "baab"].
Given s = "abc", return [].
"""
from sys import argv
import pdb
from itertools import permutations
from sympy.utilities.iterables import multiset_permutations
# 1. Test case x
# 2. Brute
# 3. Optimise
# 4. Run Edge cases
# 5. Write Tests
# TODO make sure even numbers of each letter x
# make sure palindromic perm possible (only one uneven count) x
# use is palandrome as test
# def make_permutations (s):
# """
# Input: STRING s
# Output: List of permutations of s
# """
# perms = permutations(s)
# perms = [''.join(x) for x in perms]
# return(list(perms))
def is_palindrome (s):
"""
Input: STRING s
Output: BOOLEAN, True if s is a palindrome
"""
if s[::-1] == s:
return True
else:
return False
def get_symmetrical_half (s):
"""
Input: STRING s, must have even numbers of each char (except optionally for one letter)
Output: STRING, Symmetrical half of s
"""
assert isinstance(s, str), 'Argument must be string'
even = len(s) % 2 == 0
letters = list(set(list(s))) # crudely get unique letters
counts = []
symm_half = ''
for letter in letters:
counts.append(s.count(letter))
# Actually let's get the odd letter here (in the case of uneven length)
odd_letter = ''
if not even:
# Check string valid (odd length):
if not len([x for x in counts if x % 2 != 0]) == 1: exit('[]')
odd_letter = [x for x in letters if s.count(x) % 2 != 0][0]
letters.remove(odd_letter)
counts = [x for x in counts if x % 2 == 0]
# Check valid string (even length):
if not len([x for x in counts if x % 2 != 0]) == 0: exit('[]')
for i in range(0, len(letters)):
symm_half += ((counts[i]//2) * letters[i])
return(symm_half, odd_letter)
def get_pal_perms (sym_half, odd_letter):
"""
Input: STRING sym_half - half the string as split by letter,
STRING odd_letter - the odd letter if there is one
Output: LIST of all palindromic permutations of s
"""
perms = list(multiset_permutations(list(sym_half)))
perms = [''.join(x) for x in perms]
pal_perms = [x + odd_letter + x[::-1] for x in perms]
return pal_perms
if __name__ == "__main__":
s = 'aabbc'
# Edge Cases
if len(s) == 1:
print(s)
exit()
elif len(s) == 0:
print('Error: must give non-empty string')
exit()
sym_half, odd_letter = get_symmetrical_half(s)
pal_perms = get_pal_perms(sym_half, odd_letter)
print(pal_perms)
| true |
afc794d1882bbfb28ea270f7b4a723762b3a246c | envisioncheng/python | /pyworkshop/1_intro_python/chapter5/exercise.py | 553 | 4.21875 | 4 | # Part 1
10 > 5
5 > 10
10 > 10
10 >= 10
5 < 10
5 < 5
5 <= 5
5 == 5
5 != 10
# Part 2
5 == True
# The number 5 does not equal True, but...
if 5:
print("The number 5 is truthy!")
# The number 5 is truthy for an if test!
# Part 3
1 == True
0 == False
# Part 4
True or False
[] or [1, 2, 3]
"Hello" or None
True and False
5 and 0
[1] and [1, 2, 3]
"Hello" and None
# Of course, you can use `and` and `or` aren't limited to two operands
a = False
b = False
c = False
a or b or c
b = True
a or b or c
a and b and c
a = True
c = True
a and b and c
| false |
0c0e23c15708325ec523a31bd9b7bea1044ad2e4 | prameya21/python_leetcode | /Valid_Palindrome.py | 800 | 4.25 | 4 | '''
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
Note: For the purpose of this problem, we define empty string as valid palindrome.
Example 1:
Input: "A man, a plan, a canal: Panama"
Output: true
Example 2:
Input: "race a car"
Output: false
'''
class Solution:
def isPalindrome(self,s:str) -> bool:
s=s.lower()
l,r=0,len(s)-1
while l<=r:
if not s[l].isdigit() and not s[l].isalpha():
l+=1
elif not s[r].isdigit() and not s[r].isalpha():
r-=1
else:
if s[l]!=s[r]:
return False
l+=1;r-=1
return True
obj=Solution()
print(obj.isPalindrome("A man, a plan, a canal: Panama")) | true |
92cc8732044e9fc0a64a3d7de55ec4393b6b4beb | kangwonlee/16pfa_kangwonlee | /ex29_if/ex29.py | 714 | 4.1875 | 4 | # -*-coding:utf8
# http://learnpythonthehardway.org/book/
people = 10
cats = 30
dogs = 15
if people < cats:
print("Too many cats! The world is doomed!")
if people > cats:
print("No many cats! The world is saved!")
if people < dogs:
print("The world is drooled on!")
if people > dogs:
print("The world is dry!")
dogs += 5
if people >= dogs:
print("People are greater than or equal to dogs.")
if people <= dogs:
print("People are less than or equal to dogs.")
if people == dogs:
print("People are dogs.")
# 여기까지 입력 후 add, commit
# 각 행 주석 입력 후 commit
# 각자 Study drills 시도 후 필요시 commit
# 오류노트 에 각자 오류노트 작성
| true |
24dad66dd22395fc2a92f2a403de09dcb82fbfe4 | Maschenka77/100_Day_Python | /day_2_tip_project_calculator.py | 625 | 4.21875 | 4 | #If the bill was $150.00, split between 5 people, with 12% tip.
#Each person should pay (150.00 / 5) * 1.12 = 33.6
#Format the result to 2 decimal places = 33.60
#Tip: You might need to do some research in Google to figure out how to do this.
print("Welcome to the tip calculator!")
total_bill = float(input("What is the total bill? "))
percent = int(input("What percentage tip would you like to give? "))
percent_div = percent/100
split = int(input("How many people to split the bill? "))
bill_splitted = total_bill/split
with_tip = round(bill_splitted*(1+percent_div),2)
print(f"Each person should pay: ${with_tip}")
| true |
9d0dec604ab6ca5aefcec6aca0a3766082e1419d | Aravind-39679/tri.py | /triangle.py | 658 | 4.125 | 4 | t=True
while(t):
A=int(input('Enter first side of the triangle:'))
B=int(input('Enter second side of the triangle:'))
C=int(input('Enter third side of the triangle:'))
if(A+B>C and B+C>A and A+C>B):
print("It is a triangle")
if(A==B==C):
print('It is an equailateral triangle')
break
elif(A==B!=C or A==C!=B or B==C!=A):
print('It is isosceles triangle')
break
elif(A!=B!=C):
print('It is scalene triangle')
break
else:
print("It is not a triangle")
print('input correct inputs')
#t=True
| true |
cae0951dcb0aaed9aa287c1e924211365b6bcf80 | Lynch08/pands-problem-sheet | /squareroot.py | 1,061 | 4.3125 | 4 | #Programme that uses a function so when you input a number it will return the square root
#Author Enda Lynch
#REF https://medium.com/@sddkal/newton-square-root-method-in-python-270853e9185d
#REF https://www.youtube.com/watch?v=WsQQvHm4lSw - Understand Calculas
#REF https://www.homeschoolmath.net/teaching/square-root-algorithm.php
#REF https://www.geeksforgeeks.org/find-root-of-a-number-using-newtons-method/
#REF https://www.school-for-champions.com/algebra/square_root_approx.htm#.YD102-j7TDe
#REF Automate the Boring Stuff - Chapter 3 Functions
newt = input("Number to get square root of:" )
def sqrt(number, iterations = 100):
newt = float(number) # number to get square root of
for i in range(1, iterations): # iteration number
number = 0.5 * (newt / number + number) # √ number ≈ .5*(newt/number + number)
return round((number), 1) # return number in for loop and set to 1 decimal place
print ("Square number of", newt, "is approx:", (sqrt(float(newt))))
| true |
b96c473a1b366efebf30f888dcfe9aba5e53b2c6 | VladKrupin/DemoGit | /Calculator v1.0.py | 419 | 4.1875 | 4 | from math import pow
a = float(input("Enter the first number:"))
op = input("Input the operation:")
b = float(input("Enter the second number:"))
result = None
if op == "+":
result = a+b
elif op == "/":
result = a/b
elif op == "-":
result = a-b
elif op == "*":
result = a*b
elif op == "pow":
result=pow(a,b)
if not result:
print("Invalid input")
else:
print("Result: {:.2f}".format(result))
| false |
53b9d3683feff9adfa34c88be6c856db960072ad | sheelabhadra/LeetCode-Python | /1033_Moving_Stones_Until_Consecutive.py | 2,232 | 4.21875 | 4 | """PROBLEM:
Three stones are on a number line at positions a, b, and c.
Each turn, you pick up a stone at an endpoint (ie., either the lowest or highest position stone), and move it
to an unoccupied position between those endpoints. Formally, let's say the stones are currently at positions
x, y, z with x < y < z. You pick up the stone at either position x or position z, and move that stone to an
integer position k, with x < k < z and k != y.
The game ends when you cannot make any more moves, ie. the stones are in consecutive positions.
When the game ends, what is the minimum and maximum number of moves that you could have made?
Return the answer as an length 2 array: answer = [minimum_moves, maximum_moves]
"""
"""SOLUTION:
The maximum number of moves would be the total number of empty spots between the stones.
The minimum number of moves can be obtained by analyzing the following 4 cases:
1. If the number of empty spots between any one of the pairs of successive stones is 1,
e.g. [1, 3, 5], and [3, 5, 10], then only 1 move is required which fills the empty spot between the pair of stones.
2. If there are no empty spots between both the successive pair of stones, e.g. [1, 2, 3], then no move is required.
3. If there is no empty spot between any one of the pairs of successive stones,
e.g. [2, 10, 11] or [2, 3, 10], then only 1 move is required.
4. For all the other possible configurations, at least 2 moves are required.
"""
class Solution:
def numMovesStones(self, a: int, b: int, c: int) -> List[int]:
stones = sorted([a, b, c])
max_moves = (stones[1] - stones[0] - 1) + (stones[2] - stones[1] - 1) # points between the stones
# add cases for min_moves
if stones[1] - stones[0] == 2 or stones[2] - stones[1] == 2:
# [1, 3, 5]
min_moves = 1
elif stones[1] - stones[0] == 1 and stones[2] - stones[1] == 1:
# [1, 2, 3]
min_moves = 0
elif stones[1] - stones[0] == 1 or stones[2] - stones[1] == 1:
# [2, 10, 11] or [2, 3, 10]
min_moves = 1
else:
# all other cases
min_moves = 2
return [min_moves, max_moves]
| true |
22e6457a4791af6292a17bbb294566aefdb9834e | sheelabhadra/LeetCode-Python | /728_Self_Dividing_Numbers.py | 1,425 | 4.125 | 4 | #A self-dividing number is a number that is divisible by every digit it contains.
# For example, 128 is a self-dividing number because 128 % 1 == 0, 128 % 2 == 0, and 128 % 8 == 0.
# Also, a self-dividing number is not allowed to contain the digit zero.
# Given a lower and upper number bound, output a list of every possible self dividing number, including the bounds if possible.
# Example 1:
# Input:
# left = 1, right = 22
# Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22]
class Solution(object):
def selfDividingNumbers(self, left, right):
"""
:type left: int
:type right: int
:rtype: List[int]
"""
res = []
for num in range(left,right+1):
digits = []
temp = num
print(num)
while temp != 0:
lsd = temp%10
flag1 = 0
if lsd == 0:
print("Contains 0")
flag1 = 1
break
digits.append(lsd)
temp = temp//10
if flag1:
continue
for d in digits:
flag2 = 0
if num%d != 0:
flag2 = 1
break
if flag2:
continue
else:
res.append(num)
print(res)
return res
| true |
8f977472fd05fb250e180589c6545a545805a2ac | zahraaassaad/holbertonschool-machine_learning | /math/0x06-multivariate_prob/0-mean_cov.py | 854 | 4.28125 | 4 | #!/usr/bin/env python3
""" calculates the mean and covariance of a data set """
import numpy as np
def mean_cov(X):
"""
X is a numpy.ndarray of shape (n, d) containing the data set:
n is the number of data points
d is the number of dimensions in each data point
Returns: mean, cov:
mean numpy.ndarray of shape (1, d) containing the mean of the data set
cov numpy.ndarray of shape (d, d) containing the covariance matrix
of the data set
"""
if not isinstance(X, np.ndarray) or len(X.shape) != 2:
raise TypeError("X must be a 2D numpy.ndarray")
if X.shape[0] < 2:
raise ValueError("X must contain multiple data points")
d = X.shape[1]
mean = np.mean(X, axis=0)
a = X - mean
c = np.matmul(a.T, a)
cov = c/(X.shape[0] - 1)
return mean.reshape((1, d)), cov
| true |
e4a322de533519abc12a1819ef9bcea84113c488 | kwierman/Cards | /Source/Shuffle.py | 1,032 | 4.125 | 4 | """ @package Shuffle
Performs shuffling operations on decks
"""
import Deck
## splits the deck into two decks
# @param deck The deck to be split
# @param split_point how many cards from the top of the deck to split it
def split_deck(deck, split_point=26):
deck_1 = Deck.Deck()
deck_2 = Deck.Deck()
for x, card in enumerate(deck.private.cards):
if(x<split_point):
deck_1.add_card(card)
else:
deck_2.add_card(card)
return (deck_1, deck_2)
## performs a tent shuffle
def tent_shuffle(deck, split_point=26, groups=[], shuffle_left=True):
#first split it into two decks
deck1, deck2 = split_deck( deck, split_point )
if(shuffle_left):
deck3=deck1
deck1=deck2
deck2=deck3
#for each of the groups of cards
result_deck=Deck.Deck()
for x,i in enumerate(groups):
tent_group=Deck.Deck()
if(x%2==0 ):
tent_group,deck1 = split_deck(deck1, i)
else:
tent_group,deck2 = split_deck(deck2, i)
result_deck.add_deck(tent_group)
result_deck.add_deck(deck1)
result_deck.add_deck(deck2)
return result_deck
| true |
5fd87282bd2429a43a83feb00c542b07da1a4259 | simiyub/python | /interview/practice/code/arrays/SortedSquareFromSortedArray.py | 642 | 4.28125 | 4 | """
Given an array of sorted integers, this function returns a sorted array of the square of the numbers
"""
def sorted_square_array(array):
result = [0 for _ in array]
smaller_value_index = 0
larger_value_index = len(array) - 1
for index in reversed(range(len(array))):
smaller_value = array[smaller_value_index]
larger_value = array[larger_value_index]
if (abs(smaller_value) > abs(larger_value)):
result[index] = smaller_value ** 2
smaller_value_index += 1
else:
result[index] = larger_value ** 2
larger_value_index += 1
return result
| true |
6e713b61bedbe73121c1326b8d17ae7cd5d1ef06 | simiyub/python | /interview/practice/code/recursion/Permutations.py | 1,724 | 4.15625 | 4 | """
Requirement:This function will receive an array of integers and
create an array of unique permutations or ordering of integers
that can be created from the integers in the array.
Implementation: We build all permutations for the array within the array in place using pointers.
We select the first element in the array and use it as a starting element of all the combinations
of permutations that can be formed from the rest of the elements in the array.
We swap the first element in the remaining array of elements with the current anchor index.
Then we iteratively call the helper method a second time with the next element in the array of
remaining elements, the array and the array of permutations.
When we are done with this call, we swap back the elements to how they were before
and move on to the next index after the anchor index. When the index is the last one in the array,
we copy the array into the array of permutations.
Complexity: O(n*n!) T and O(n.n!) S For each element in the array,
we make calls to the helper equivalent to the position of the element in the array
which means the last one will have n.n! calls.
"""
def __swap(array, i, j):
array[i], array[j] = array[j], array[i]
def __permutations_helper(anchor_index, array, all_permutations):
if anchor_index == len(array) - 1:
all_permutations.append(array[:])
else:
for j in range(anchor_index, len(array)):
__swap(array, anchor_index, j)
__permutations_helper(anchor_index + 1, array, all_permutations)
__swap(array, anchor_index, j)
def permutations(array):
permutations_found = []
__permutations_helper(0, array, permutations_found)
return permutations_found
| true |
61066f828ffa00dd2652b07ea8e546e255682cf8 | simiyub/python | /interview/practice/code/linked_lists/ReverseLinkedList.py | 1,248 | 4.15625 | 4 | """
Requirement: Given a linked list where each node points to the next node,
this function will return the linked list in reverse order
Implementation: Conceptually, a node is reversed when the next pointer points to the previous node.
However, as this is a singly linked list, we do not have a pointer to the previous node.
So we need to create one. We do this using a temporary node that we initialise to None
as we start traversing the linked list from head.
We also need to keep a reference to the next node after the current node because
if we move the next pointer of the current node to the previous node,
we would lose connection to the rest of the linked list.
Once we have these two temporary references in place,
we start moving our pointers along recursively until the current node is null
and at that point the linked list will be reversed.
Complexity: O(n) T O(1) S as we have to recursively traverse the whole linked list.
However, we do not use any additional space than that of the linked list.
"""
def reverse(head):
previous = head
current = head
while current is not None:
next_node = current.next
current.next = previous
previous = current
current = next_node
return previous
| true |
fe1c5e5cea9e34e329b658982acd5bd7fd8f23af | shorya97/AddSkill | /DS & Algo/Week 2/Stack/Implement Stack using Queues.py | 1,420 | 4.15625 | 4 |
from queue import Queue
class MyStack:
def __init__(self):
"""
Initialize your data structure here.
"""
self.q1 = Queue() #inbuilt queues
self.q2 = Queue()
self.cur_size = 0 #contains number of elements
def push(self, x: int) -> None:
"""
Push element x onto stack.
"""
self.cur_size+=1
self.q2.put(x)
while (not self.q1.empty()):
self.q2.put(self.q1.queue[0])
self.q1.get()
self.q = self.q1
self.q1 = self.q2
self.q2 = self.q
def pop(self) -> int:
"""
Removes the element on top of the stack and returns that element.
"""
if (self.q1.empty()):
return
return self.q1.get()
self.cur_size -= 1
def top(self) -> int:
"""
Get the top element.
"""
if self.q1.empty():
return -1
return self.q1.queue[0]
def empty(self) -> bool:
"""
Returns whether the stack is empty.
"""
if self.q1.empty():
return True
else:
return False
# Your MyStack object will be instantiated and called as such:
# obj = MyStack()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.top()
# param_4 = obj.empty()
| true |
be1b56dc341fc45d7c0d017bc979c4d1426f7afa | prabhu30/Python | /Strings/Set - 1/maketrans_translate.py | 743 | 4.5625 | 5 | """
maketrans() :-
It is used to map the contents of string 1 with string 2 with respective indices to be translated later using translate().
translate() :-
This is used to swap the string elements mapped with the help of maketrans().
"""
# Python code to demonstrate working of
# maketrans() and translate()
from string import maketrans # for maketrans()
str = "geeksforgeeks"
str1 = "gfo"
str2 = "abc"
# using maktrans() to map elements of str2 with str1
mapped = maketrans( str1, str2 )
# using translate() to translate using the mapping
print "The string after translation using mapped elements is : "
print str.translate(mapped)
"""
Output :
The string after translation using mapped elements is :
aeeksbcraeeks
"""
| true |
bdbfdd41268b98866a0eb26d1d335267b1a4f0b9 | prabhu30/Python | /Lists/Set - 1/indexing_negative.py | 666 | 4.375 | 4 | """
Negative indexing :-
In Python, negative sequence indexes represent positions from the end of the array.
Instead of having to compute the offset as in List[len(List)-3], it is enough to just write List[-3].
Negative indexing means beginning from the end, -1 refers to the last item, -2 refers to the second-last item, etc.
"""
List = [1, 2, 'Geeks', 4, 'For', 6, 'Geeks']
# accessing a element using
# negative indexing
print("Accessing element using negative indexing")
# print the last element of list
print(List[-1])
# print the third last element of list
print(List[-3])
"""
Output:
Accessing element using negative indexing
Geeks
For
"""
| true |
e8bc8d4d3b21a4e6525a0e23a3e462f4feabb9a0 | Manish674/Cs50-problem-sets | /pset6/mario/mario.py | 201 | 4.15625 | 4 | height = int(input("Height: "))
while height < 0:
height = int(input("Height: "))
col = 0
row = 0
while row <= height:
while col <= row:
print("#"*col)
col += 1
row += 1 | false |
df60640085e6f84ebd72aee33d3dd29dfc4d950e | TrevorSpitzley/PersonalRepo | /PythonProjects/listCheck.py | 920 | 4.25 | 4 | #!/usr/bin/env python
# This file is used for comparing two lists.
# If there is an element in the input_list,
# and it is not in the black_list, then it
# will be added to the out_list and printed
def list_check(input_list, black_list):
out_list = []
for element in input_list:
if element not in black_list:
out_list.append(element)
return out_list
chars = ['a', 'b', 'c', 'd']
chars2 = ['e', 'f', 'g', 'h']
chars3 = ['a', 'c']
chars4 = ['e', 'h']
ints = [1, 2, 3, 4]
ints2 = [5, 6, 7, 8]
ints3 = [1, 2, 3]
ints4 = [5, 6, 7, 8]
# Should return ['b', 'd']
print(list_check(chars, chars3))
# Should return ['f', 'g']
print(list_check(chars2, chars4))
# Should return [4]
print(list_check(ints, ints3))
# Should return []
print(list_check(ints2, ints4))
# Should return ['e', 'f', 'g', 'h']
print(list_check(chars2, ints))
# Should return [1, 2, 3]
print(list_check(ints3, chars3))
| true |
fba0793e6a6508be9717c0db93287d455ed13d5d | nathan-create/assignment-problems | /linked_list.py | 1,601 | 4.15625 | 4 | class Node:
def __init__(self, data):
self.data = data
self.next = None
self.index = 0
class LinkedList:
def __init__(self, data):
self.head = Node(data)
def append(self, data):
current_node = self.head
node_index = 0
while current_node.next != None:
current_node = current_node.next
node_index += 1
current_node.next = Node(data)
current_node.next.index = current_node.index + 1
def print(self):
current_node = self.head
while current_node != None:
print(current_node.data)
current_node = current_node.next
def length(self):
current_node = self.head
nodes_visited = 1
while current_node.next != None:
current_node = current_node.next
nodes_visited += 1
return nodes_visited
def push(self, data):
former_head = self.head
self.head = Node(data)
self.head.next = former_head
current_node = former_head
while current_node != None:
current_node.index += 1
linked_list = LinkedList('b')
linked_list.append('e')
linked_list.append('f')
linked_list.push('a')
assert linked_list.length() == 4, linked_list.length()
assert linked_list.head.index == 0, linked_list.head.index
assert linked_list.head.next.index == 1
assert linked_list.head.next.next.index == 2
assert linked_list.head.next.next.next.index == 3, linked_list.head.next.next.next.index
assert linked_list.get_node(0) == 'a', linked_list.get_node(0) | false |
62615472dad8b7cbd47655acab1ee1dcafcfc4f3 | cbolson13/Pyautogui_CO | /Personality_Survey_C.py | 1,876 | 4.21875 | 4 | print("What's your name?")
name = input().title()
if name == "Connor":
print("Same here!")
else:
print(name + " is a pretty cool name.")
print("What's your favorite sport")
sport = input().title()
if sport == "Soccer":
print("Cool, what's your favorite team!")
soccerteam = input().title()
if soccerteam == "Liverpool":
print("That is my favorite team also!")
elif soccerteam == "Manchester United" or soccerteam == "Everton":
print("Oh no, I like Liverpool and not " + soccerteam)
else:
print(soccerteam + " is not my favorite team, but I don't mind them")
elif sport == "Tenis":
print("Cool, sometimes I like to play tenis too!")
else:
print("Wow, " + sport + " sound fun!")
print("What's your favorite TV Show?")
tvshow = input().title()
if tvshow == "Flash" or tvshow == "Rick And Morty" or tvshow == "The Flash":
print("I love " + tvshow + " too! Who is your favorite character?")
character = input().title()
if character == "Cisco" or character == "Barry" or character == "The Flash" or character == "Joe" or character == "Wally" or character == "Kid Flash" or character == "HR" or character == "Rick" or character == "Morty" or character == "Beth" or character == "Jerry" or character == "Summer" or character == "Mr.Meeseeks":
print("Awesome, " + character + " is one of my favorite character too!")
elif character == "Iris":
print("I don't really like " + character + " that much.")
else:
print(character + " isn't my favorite, but I don't mind them.")
elif tvshow == "Leverage" or tvshow == "Psych" or tvshow == "Wizards Of Waverly Place":
print("I used to watch " + tvshow + " too!")
else:
print("I don't know about " + tvshow + " but it sounds good!")
| true |
e120d6a0b5708af15a2a50b654ef68837a12c03e | Abhiaish/GUVI | /ZEN class/day 1/fare.py | 236 | 4.15625 | 4 | distance=int(input("enter the distance"))
peaktime=int(input("enter the peaktime"))
if(distance>5):
distance=distance-5
else:
print("fare is 100")
fare=int(distance*8+100)
if(peaktime==1):
fare=fare+0.25*fare
print(fare)
| true |
bbdb68232f6a2d04d342d4cc274faa2d3d083aeb | renzhiliang/python_work | /chapter4/4-10.py | 293 | 4.34375 | 4 | #numbers=range(3,31,3)
numbers=[number**3 for number in range(1,11)]
print(numbers)
print("The first three items in the list are:")
print(numbers[0:3])
print("Three items from the middle of the list are:")
print(numbers[5:8])
print("The last three items in the list are:")
print(numbers[-3:])
| true |
cabc815a05896d9d72e0f904eac1aede308438e7 | thu4nvd/project_euler | /prob014.py | 1,184 | 4.15625 | 4 | #!/usr/bin/env python3
'''
Longest Collatz sequence
Problem 14
The following iterative sequence is defined for the set of positive integers:
n → n/2 (n is even)
n → 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
Which starting number, under one million, produces the longest chain?
NOTE: Once the chain starts the terms are allowed to go above one million.
23.73s user
0.00s system
99% cpu
23.734 total
https://projecteuler.net/problem=14
'''
def collatz_seq(n):
length = 1
while n != 1:
if n % 2 == 0 : n = n // 2
else: n = 3 * n + 1
length += 1
return length
def solve(N):
max_length = 1
for i in range(2, N):
if max_length < collatz_seq(i):
max_length = collatz_seq(i)
max_i = i
return max_i
def main():
print(solve(1000000))
if __name__ == "__main__":
main()
| true |
a3b53793279b63e1facca3a5b4bef20146aff93d | leemanni/receiving-GV_Python | /6.30/stackModule.py | 1,583 | 4.1875 | 4 | #!/usr/bin/env python
# coding: utf-8
# In[27]:
class Stack :
# constructor
def __init__(self, size = 5) :
self.size = size
self.stack = list()
self.top = 0
# push
def push(self, data):
if data not in self.stack :
if self.size > self.top:
self.stack.append(data)
print('데이터 저장 완료')
self.top += 1
else:
print('overflow')
else :
print('중복!...{} 데이터는 받을 수 없습니다'.format(data))
self.view()
#pop
def pop(self):
if self.top == 0 :
print('제거할 데이터 없음')
else :
data = self.stack.pop()
self.top -= 1
# pop() 은 제거할 데이터를 가지고 있다가 없애는 것을 이용
print('{} 가 제거되었습니다'.format(data))
self.view()
#view
def view(self) :
if self.top != 0 :
print('stack==> ',end ='')
for i in range(self.top):
print(self.stack[i],end=' ')
else : # 데이터가 없을 때
print('underflow!! 데이터가 없어용')
print()
# In[34]:
if __name__ == '__main__' :
st = Stack()
st.push(25)
st.push(34)
st.push(33)
st.push(34)
st.push(1)
st.push(2)
st.push(33)
print('='*35)
st.pop()
st.pop()
st.pop()
st.pop()
st.pop()
st.pop()
| false |
12abd510bd551ccafd2101bb6392c35357bdbe47 | mtmccarthy/langlearn | /bottomupcs/Types.py | 1,530 | 4.71875 | 5 | from typing import NewType, Dict
"""
In Chapter 1, we learned that 'Everything is a File!'. Here we've created a new type, 'FileType' which models a file
as its path in the filesystem.
"""
FileType = NewType('FileType', str)
"""
In Chapter 1, we learned about FileDescriptors, which are essentially integer indexes into a table
stored by the kernel called the file descriptor table (shocking!)
Since a file descriptor is essentially just an integer, we can model it as such.
"""
FileDescriptorType = NewType('FileDescriptorType', int)
"""
We can model the table as a dictionary with key file descriptor and value file.
The fd table is generated for every process and holds a pointer to the system fd table, but the model still holds up.
"""
FileDescriptorTableType = Dict[FileDescriptorType, FileType]
"""
In Chapter 1, we learned about Device Drivers are software which provide an abstraction between the kernel and
external devices such as a keyboard, mouse, monitor etc.
"""
DeviceDriverType = NewType('DeviceDriverType', None)
"""
Programs are files that can also be executed. We can model that like this:
"""
ProgramType = NewType('ProgramType', FileType)
"""
In Chapter 1, the kernel was mentioned. The kernel is essentially a program executed when the system boots.
The kernel is responsible for processing low level system calls, examples include input/output and memory management.
Since the kernel is just a special type of program, we can model it like this:
"""
KernelType = NewType('KernelType', ProgramType)
| true |
6bb5fbe258c60e039c2380dcbbf50dc6b8e5d803 | mtmccarthy/langlearn | /dailyredditor/easy/imgurlinks_352.py | 1,574 | 4.25 | 4 | from common import *
from typing import Union
"""
Finds the greatest exponent of n that is less than or equal to another number m
Example: if n = 2 and m = 17 return 4
"""
def get_base62_character(index: int) -> Union[str, None]:
capital_letters = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
lower_case = 'abcdefghijklmnopqrstuvwxyz'
if index < 10:
return str(index)
elif 10 <= index < 36:
return lower_case[index - 10]
elif 36 <= index < 62:
return capital_letters[index - 36]
else:
return None
def find_greatest_exponent_of_n_loe_than_m(n: int, m: int) -> int:
current_exponent = 0
while n ** current_exponent <= m:
current_exponent += 1
return current_exponent - 1
def convert_from_base10ton(baseN: int, num: int) -> str:
conversion = []
power = find_greatest_exponent_of_n_loe_than_m(baseN, num)
while power > 0:
place_value = baseN ** power
digit = 0
while num - place_value > 0:
digit += 1
num -= place_value
conversion.append(digit)
power -= 1
conversion.append(num) # One's place
return ''.join(list(map(lambda digit: get_base62_character(digit), conversion)))
def generate_output(path: str) -> str:
lines = get_file_lines(path)
output = ""
for line in lines:
output += convert_from_base10ton(62, int(line.rstrip())) + '\n'
return output
def main():
path = 'easytests/imgurlinks_352_challenge_input.txt'
print(generate_output(path))
if __name__ == '__main__':
main()
| true |
17ce6e5c354abe914abb538356201ed5cfc55a25 | lamyai-norn/algorithsm-week11 | /week11/thursday1.py | 279 | 4.15625 | 4 | def sumFromTo(start,end):
result=0
if start>end:
return 0
else:
for index in range(start,end+1):
result=result+index
return result
startValue=int(input("Start value : "))
endValue=int(input("End value : "))
print(sumFromTo(startValue,endValue)) | true |
3ce463b0b5860ae48129db242597bb45d7620ecc | NikolasSchildberg/guesser | /guesser_simple.py | 1,218 | 4.21875 | 4 | """
Welcome to the number guesser in python!
The logic is to use the bissection method to find a number chosen by the user, only
providing guesses and asking to try a bigger or smaller one next time.
"""
# Initial message
print("Welcome to the number guesser!\nThink of a number between 1 and 1000 (including them).I'll take only 10 guesses (or less) to find out which number you picked.\nYou'll just have to tell me if your number is smaler or bigger than my guess. Wanna try it?")
print("Please enter '0' if I'm already right, '1' if your number is smaller, and '2' if your number is bigger.\n")
right = False
smaller = 0
bigger = 1000
guess = int((smaller + bigger)/2)
count_guesses = 0
while(right != True):
whattodo = input("Is it "+str(guess)+"?\n>>> " )
if whattodo == '0':
right = True
elif whattodo == '1':
bigger = guess
guess = int((smaller + bigger)/2)
elif whattodo == '2':
smaller = guess
guess = int((smaller + bigger)/2)
else: print("Try again...")
count_guesses += 1
else:
if(count_guesses == 1):
print("Gotcha! Needed",count_guesses,"guesse!\nBye!")
else:
print("Gotcha! Needed",count_guesses,"guesses!\nBye!") | true |
6150528fcd6ecb0f9a68655746318a2f680ee546 | Antonio24ch/BEDU-281120 | /01_tabla_multiplicar.py | 485 | 4.15625 | 4 | #Preguntar el numero que el usuario quiere multiplicar
numero = int(input('Que número quieres multiplicar?\t'))
print(f'A continuacion se muestra la tabla del {numero}')
print('------------------------------------------------')
#Esta función hace esto:
# n * 1 = n
# n * 2 = n
# n * 3 = n
# ...
for n in range(10):
indice = n + 1 #Sirve para que si queremos multiplicar cada 2, 3, etc. solo se cambie el numero(1)
r = numero * indice
print(f'{numero} * {indice} = {r}') | false |
826f03f9f1dc41fe4aca2c70b505d7381ba091bf | EagleRock1313/Python | /circle.py | 523 | 4.40625 | 4 | # Load the Math Library (giving us access to math.pi and other values)
import math
# Ask for the radius of a circle that will be entered as a decimal number
radius = eval(input(("Enter the radius of the circle: ")))
# Computer circumference and area of this circle
circumference = 2 * math.pi * radius
area = math.pi * (radius ** 2)
# Results
print()
print("Circumference is", circumference)
print("Area is", area)
print("\nRounded: ")
print("Circumference is", round(circumference,2))
print("Area is", round(area,2)) | true |
ebcf7ae783f11ff6e8908df29b9c0c46cd0b54af | petemulholland/learning | /Python/ThinkPython/Chapter18/MyCard.py | 1,364 | 4.15625 | 4 | class Card(object):
""" Represents a playing card"""
#suit_names = ['Clubs', 'Diamonds', 'Hearts', 'Spades']
#rank_names = [None, 'Ace', '2', '3', '4', '5', '6', '7',
# '8', '9', '10', 'Jack', 'Queen', 'King']
suit_names = ('C', 'D', 'H', 'S')
rank_names = (None, 'A', '2', '3', '4', '5', '6', '7',
'8', '9', '10', 'J', 'Q', 'K')
def __init__(self, suit=0, rank=2):
self.suit = suit
self.rank = rank
# class variables
def __str__(self):
#return '%s of %s' % (Card.rank_names[self.rank],
return '%s%s' % (Card.rank_names[self.rank],
Card.suit_names[self.suit])
# for this implementation suit is more important than rank,
# so all Spaces outrank all diamonds etc.
def __cmp__(self, other):
c1 = self.suit, self.rank
c2 = other.suit, other.rank
return cmp(c1, c2)
def cmp_to_string(res):
if res > 0:
return ">"
if res < 0:
return "<"
return "="
if __name__ == '__main__':
card1 = Card(1, 11)
card2 = Card(1, 11)
card3 = Card(2, 8)
print card1, cmp_to_string(cmp(card1, card2)), card2
print card1, cmp_to_string(cmp(card1, card3)), card3
print card3, cmp_to_string(cmp(card3, card2)), card2
print card1 | false |
fdeecfeec4bbea1b1c5bd1738bbadf905dcb3772 | Kapilmundra/Complete-Python-for-Lerner | /4 set.py | 691 | 4.21875 | 4 | #Create set
print("Create set")
s={"teena","meena","reena"}
print(s) # it can print in unorder way
#pop the element from the set
print("\npop the element from the set")
s.pop() # it can pop the element in unorder way
print(s)
#Add element in the set
print("\nAdd element in the set")
s.add("priya") # it can add the new value in unorder way
print(s)
s.add("priya") # distinct(not return same value in set)
print(s)
#Operation
print("\nOperation")
p={"rajesh","manish","reena"}
print(p)
print("Difference")
x=s.difference(p)
print(x)
print("Intersection")
z=s.intersection(p)
print(z)
print("Difference_update")
y=s.difference_update(p)
print(y)
| true |
d86d7fdc1e4fb8c3e6c328b2d4141f5b2e078f66 | das-jishu/data-structures-basics-leetcode | /Leetcode/hard/regular-expression-matching.py | 2,506 | 4.28125 | 4 | """
# REGULAR EXPRESSION MATCHING
Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*' where:
'.' Matches any single character.
'*' Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).
Example 1:
Input: s = "aa", p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".
Example 2:
Input: s = "aa", p = "a*"
Output: true
Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".
Example 3:
Input: s = "ab", p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".
Example 4:
Input: s = "aab", p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore, it matches "aab".
Example 5:
Input: s = "mississippi", p = "mis*is*p*."
Output: false
Constraints:
0 <= s.length <= 20
0 <= p.length <= 30
s contains only lowercase English letters.
p contains only lowercase English letters, '.', and '*'.
It is guaranteed for each appearance of the character '*', there will be a previous valid character to match.
"""
class Solution:
def isMatch(self, s: str, p: str) -> bool:
table = [[None for _ in range(len(p) + 1)] for _ in range(len(s) + 1)]
table[0][0] = True
for row in range(1, len(s) + 1):
table[row][0] = False
for col in range(1, len(p) + 1):
if p[col - 1] == "*":
table[0][col] = table[0][col - 2]
else:
table[0][col] = False
for row in range(1, len(s) + 1):
for col in range(1, len(p) + 1):
if p[col - 1] == ".":
table[row][col] = table[row - 1][col - 1]
elif p[col - 1] != "*":
table[row][col] = table[row - 1][col - 1] and s[row - 1] == p[col - 1]
else:
if p[col - 2] == ".":
table[row][col] = table[row][col - 1] or table[row][col - 2] or table[row - 1][col - 1] or table[row - 1][col]
else:
table[row][col] = table[row][col - 1] or table[row][col - 2] or (table[row - 1][col - 1] and p[col - 2] == s[row - 1])
for row in range(len(s) + 1):
print(table[row])
return table[-1][-1] | true |
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