blob_id string | repo_name string | path string | length_bytes int64 | score float64 | int_score int64 | text string | is_english bool |
|---|---|---|---|---|---|---|---|
95297b99949e123fbeaa89f84c7c4f14521b3187 | adhuliya/bin | /hidesplit | 2,288 | 4.125 | 4 | #!/usr/bin/env python3
"""
Splits a file into two chunks.
The first chunk is only few bytes.
"""
import sys
import os.path as osp
PREFIX = "hidesplit"
SIZE_OF_FIRST_CHUNK = 64 # bytes
BUFF_SIZE = 1 << 24
usageMsg = """
usage: hidesplit <filename>
note: file should be at least {} bytes.
It splits a file into two chunks.
The first chunk is only few bytes.
""".format(SIZE_OF_FIRST_CHUNK)
def genNextStrSeq(currStrSeq):
currIntSeq = int(currStrSeq)
nextIntSeq = currIntSeq+1
nextStrSeq = str(nextIntSeq)
if len(nextStrSeq) == 1:
nextStrSeq = "00" + nextStrSeq
elif len(nextStrSeq) == 2:
nextStrSeq = "0" + nextStrSeq
return nextStrSeq
def getExt(filename):
return filename.rsplit(".")[-1]
def main(filename):
if not osp.exists(filename):
print("hidesplit: error: file {} doesn't exists".format(filename),
file=sys.stderr)
exit(2)
currStrSeq = "001"
try:
file1 = open(filename, "rb")
except Exception as e:
print("hidesplit: error: cannot open file {}".format(filename),
file=sys.stderr)
exit(3)
try:
file2 = open(PREFIX+currStrSeq+"."+getExt(filename), "wb")
except Exception as e:
print("hidesplit: error: cannot create file {}".format(filename),
file=sys.stderr)
exit(4)
try:
bb = file1.read(SIZE_OF_FIRST_CHUNK)
# if len(bb) == SIZE_OF_FIRST_CHUNK:
file2.write(bb)
file2.close()
except:
print("hidesplit: error: cannot write to first chunk",
file=sys.stderr)
exit(5)
try:
file3 = open(PREFIX+genNextStrSeq(currStrSeq)+"."+getExt(filename), "wb")
while True:
bb = file1.read(BUFF_SIZE)
file3.write(bb)
if len(bb) != BUFF_SIZE:
file3.close()
file1.close()
break
except Exception as e:
print("hidesplit: error: cannot create/write-to second chunk",
file=sys.stderr)
if file1 is not None: file1.close()
if file3 is not None: file3.close()
exit(6)
if __name__ == "__main__":
if len(sys.argv) != 2:
print(usageMsg)
exit(1)
main(sys.argv[1])
| true |
4cff866c37cb8f29fad30abba17cfae985b4e72d | heecer/old | /day2/list.py | 336 | 4.125 | 4 | #!/usr/bin/env python
# -*- coding:utf-8 -*-
# Author:Haccer
list1=['a','b','c','d','e']
# print(list1[1])
# print(list1[1:3])
# print(list1[-1])
# print(list1[-3:-1])
# print(list1[-3:])
list1.append('f')#追加
list1.insert(-1,'f')#插入
list1.remove('f')
list1.pop(-1)
del list1[0]
print(list1)
Index= list1.index('d')
print(Index) | false |
3bd3eecf4fc50c4edca1f32bae401c39e6cd06a7 | deniscostadsc/becoming-a-better-programmer | /src/problems/cracking_the_code_interview/is_unique.py | 1,170 | 4.15625 | 4 | from typing import Dict
"""
Implement an algorithm to determine if a string has all unique characters. What
if you cannot use additional data structures?
"""
def is_unique(s: str) -> bool:
"""
Time: O(n)
Space: O(n)
"""
chars: Dict[str, int] = {}
for char in s:
if char in chars:
return False
else:
chars[char] = 1
return True
def is_unique_with_no_extra_space(s: str) -> bool:
"""
Time: O(n log n)
Space: O(1)
"""
if len(s) == 1:
return True
s = "".join(sorted(s))
for char in s:
begin = 0
end = len(s) - 1
while begin < end:
middle = begin + (end - begin) // 2
if s[middle] == char:
if middle >= 1 and s[middle - 1] == char:
return False
if middle <= len(s) - 2 and s[middle + 1] == char:
return False
break
elif s[middle] > char:
begin = middle + 1
else:
end = middle - 1
return True
__all__ = [
"is_unique",
"is_unique_with_no_extra_space",
]
| true |
07561b35e66d0ead5615059c3c56744da1e67280 | kamalkundal/PythonProjects | /prime.py | 214 | 4.125 | 4 | n = int(input("Enter any num"))
if n<2:
print("not a prime")
else:
for i in range (2,n):
if (n%i==0):
print("not a prime")
break
else:
print("num is prime")
| false |
f40b9ded5cd4c74fa5b67a77902b4b0778f0ff7e | anilasebastian-94/pythonprograms | /Functions/prime.py | 553 | 4.125 | 4 | num=int(input('enter number'))
if num>1:
for i in range(2,num):
if num%i==0 :
print('number is not prime')
break
else:
print('number is prime')
#..................print not used here because print continously executes in for loop............................#
# num=int(input('enter number'))
# flag=0
# if num>1:
# for i in range(2,num):
# if num%i==0 :
# print('number is not prime')
# break
# else:
# flag=1
# if flag==1:
# print('number is prime') | true |
3e49cf303d7e05de993dada0d63c803c04401b13 | anilasebastian-94/pythonprograms | /flowofcontrols/looping/factorial.py | 294 | 4.125 | 4 | num=int(input('enter number'))
fact=1
if num>0 :
for i in range(1,num+1) :
fact*=i
print(fact)
elif num == 0:
print('Factorial of zero is 1')
else :
print('factorial doesnt exist for negative number')
# i=1
# while i<=num :
# prdct*=i
# i+=1
# print(prdct)
| true |
12e0c0d33ee64f94b6624fcc6b67099cbe4a24a2 | jjgagne/learn-python | /ex20.py | 1,237 | 4.28125 | 4 | # Author: Justin Gagne
# Date: July 4, 2014
# File: ex20.py
# Usage: python ex20.py ex20_sample.txt
# Description: Use functions to print an entire file or a file line-by-line
# allow command line args
from sys import argv
# unpack args
script, input_file = argv
# print entire contents of file
def print_all(f):
print f.read()
# return to the beginning of the file
def rewind(f):
f.seek(0)
# print one line at a time; readline() reads the line from the current seek position (set to start of file using rewind function above)
# note: could add a comma after readline() to prevent \n\n from occurring
def print_a_line(line_count, f):
print line_count, f.readline()
# create file object
current_file = open(input_file)
print "First let's print the whole file:\n"
print_all(current_file)
print "Now let's rewind, kind of like a tape."
rewind(current_file)
print "Let's print three lines:"
# NOTE: current_line is just used for debugging purposes and is not actually passed to readline()
current_line = 1
print_a_line(current_line, current_file)
current_line += 1
print_a_line(current_line, current_file)
current_line += 1
print_a_line(current_line, current_file)
# don't forget to close the file afterwards
current_file.close() | true |
37279fce63c80eb1f39475815db70ef9a412f3f9 | jjgagne/learn-python | /ex15.py | 717 | 4.125 | 4 | # import argv, which allows user to pass in parameters from command line
from sys import argv
# unpack argv
script, filename = argv
# create file object from the filename file (ex15_sample.txt in this case)
txt = open(filename)
# print out the contents to the console by calling txt.read()
print "Here's your file %r:" % filename
print txt.read()
# since we are done with the file, we should close it
txt.close()
# grab filename from user input
print "Type the filename again:"
file_again = raw_input("> ")
# create new file object from user inputted filename
txt_again = open(file_again)
# then print its contents
print txt_again.read()
# since we are done with the file, we should close it
txt_again.close() | true |
cc29f8ab6be4e6148ddfcbd89bbe8e5e2c611dd5 | jshamsutdinova/TFP | /8_lab/task_4.py | 1,826 | 4.25 | 4 | #!/usr/bin/env python3
""" Laboratory work 8. Task 4 """
from abc import ABC, abstractmethod
class Edication():
"""
This class defines the interface of edication to client
"""
def __init__(self, edication_system):
self._edication_system = edication_system
@property
def edication_system(self):
return self._edication_system
@edication_system.setter
def set_edication_system(self, edication_system):
self._edication_system = edication_system
def get_edication_information(self):
""" Return information about edication system """
result = self._edication_system.system_interface()
return result
class EdicationSystem(ABC):
"""
This class declares an interface common to all suppoted systems
"""
@abstractmethod
def system_interface(self):
pass
class Schoolchild(EdicationSystem):
def system_interface(self):
print("Школьник учится 11 лет")
class StudentBachelor(EdicationSystem):
def system_interface(self):
print("Бакалавр длится 4 года")
class StudentMaster(EdicationSystem):
def system_interface(self):
print("Магистратура длится 5 лет")
if __name__ == "__main__":
edication = Edication(Schoolchild())
print("Сначала ребенок учится в школе:")
edication.get_edication_information()
print()
print("Далее поступает в университет:")
edication.set_edication_system = StudentBachelor()
edication.get_edication_information()
print()
print("Студент может получить магистерскую степень:")
edication.set_edication_system = StudentMaster()
edication.get_edication_information()
| true |
68f4607b1bc028ac64a29fc55b40f44bebef3b85 | creuter23/fs-tech-artist | /Staff/JConley/Scripts/Python Book Scripts/Chapter 2/trivia_script.py | 626 | 4.125 | 4 | #Trivia Script
#Minor exersice in using different types of data from user input
name = raw_input("Name? ")
age = int(raw_input("Age? "))
weight = int(raw_input("Weight?"))
print "\nHi, " + name
dog_years = age * 7
print "\nYou are ", dog_years , "in dog years."
seconds = age * 365 * 24 * 60 * 60
print "You are over ", seconds , "seconds old."
moon_weight = weight / 6.0
print "\nYour weight on the moon would be ", moon_weight , "pounds."
sun_weight = weight * 27.1
print "\nYour weight on the sun would be ", sun_weight , "pounds."
raw_input("\n\nPress the enter key to exit.")
| false |
59764f581af46fe1f43de830f3c8a09f641cd28b | edaniszewski/molecular_weight | /csv_reader.py | 802 | 4.1875 | 4 | import csv
class CSVReader():
"""
Implementation of a simple CSV reader. Contains a read method which operates on the filename which
the reader is instantiated with. This reader is tailored to read the resources/element_weights.csv
to create a dictionary, which is stored in the data member.
"""
def __init__(self, file_name):
self.file_name = file_name
self.data = None
def read(self):
"""
Read a .csv file, create a dictionary of data, and store the data within the CSVData.data variable
:return: None
"""
if not self.data:
with open(self.file_name, 'rb') as f:
reader = csv.reader(f, delimiter=',', quotechar='"')
self.data = {row[0]: float(row[1]) for row in reader} | true |
c233e6da46e7b2dc25d0e43c204745c15959b779 | hammam1311/age-calculator | /age_calculator.py | 971 | 4.21875 | 4 | from datetime import datetime
from datetime import date
def check_birthdate(year, month, day):
# write code here
today = date.today()
if int(year) > int(today.year):
return False
elif int(year) == int(today.year):
if int(month) > int(today.month):
return False
elif int(month) == int(today.month):
if int(day) > int(today.day):
return False
else:
return True
else:
return True
else:
return True
def calculate_age(year, month, day):
today = date.today()
ans = (int(today.year) - int(year)) + ((int(today.month) - int(month))/12) + ((int(today.day) - int(day))/365)
print ("you are "+str(int(ans//1))+"years old")
def main():
# write main code here
today = date.today()
y1 = input("Enter year of birth: ")
m1 = input("Enter month of birth: ")
d1 = input("Enter day of birth: ")
if check_birthdate(y1, m1, d1):
calculate_age(y1, m1, d1)
else :
print ("the birthdate is invalid.")
if __name__ == '__main__':
main()
| false |
efea2dfab3ed94a02d298ef65fbe913d8f669785 | charliedavidhoward/Learn-Python | /meanMedianMode.py | 1,244 | 4.40625 | 4 | # [Function] Mean, Median, and Mode
#
# Background
#
# In a set of numbers, the mean is the average, the mode is the number that occurs the most, and if you rearrange all the numbers numerically, the median is the number in the middle.
#
# Goal
#
# Create three functions that allow the user to find the mean, median, and mode of a list of numbers. If you have access or know of functions that already complete these tasks, do not use them.
#
# Subgoals
#
# In the mean function, give the user a way to select how many decimal places they want the answer to be rounded to.
#
# If there is an even number of numbers in the list, return both numbers that could be considered the median.
#
# If there are multiple modes, return all of them.
import math
input_numbers = input("your numbers separated by a space: ")
list = input_numbers.split(" ")
def mean_function(myList=[]):
sum = 0
for number in myList:
sum = sum + int(number)
mean = sum / len(myList)
return mean
def median_function(myList=[]):
print(myList)
myList.sort()
print(myList)
middle_number = myList[math.floor(len(myList) / 2)]
return middle_number
def mode_function(myList=[]):
return max(myList, key=myList.count)
| true |
8e1fc385e39c1e0c355a07238880bf087096ce5d | Gaurav-dawadi/Python-Assignment-III | /A/question1.py | 563 | 4.15625 | 4 | # Bubble Sort Algorithm
import timeit
start = timeit.default_timer()
def bubbleSort(array):
n = len(array)
for i in range(n):
already_sorted = True
for j in range(n - i - 1):
if array[j] > array[j + 1]:
array[j], array[j + 1] = array[j + 1], array[j]
already_sorted = False
if already_sorted:
break
return array
print("The sorted array is ", bubbleSort([8, 2, 6, 4, 5]))
end = timeit.default_timer()
print("The total time required: ", end-start, " seconds") | true |
4f57b9b429695ba7b4b84e52c822db4037d11288 | kanekko/python | /02-Introducción/Calculator.py | 1,253 | 4.15625 | 4 |
def operaciones(opcion, numero_a, numero_b):
if opcion == 1:
return numero_a + numero_b
elif opcion == 2:
return numero_a - numero_b
elif opcion == 3:
return numero_a * numero_b
elif opcion == 4:
return numero_a / numero_b
else:
print('Opción inválida')
print('Bienvenido a la calculadora')
while True:
print('Estas son las operaciones que puedes realizar:')
print('1 - Suma')
print('2 - Resta')
print('3 - Multiplicación')
print('3 - División')
try:
opcion = int(input('Introduce el número de operación que quieres realizar: '))
if opcion < 1 or opcion > 4:
print('Número de operación inválida\n')
continue
numero_a = int(input('Introduce el primer número: '))
numero_b = int(input('Introduce el segu1ndo número: '))
except ValueError:
print('Opción o número inválido')
else:
numero_c = operaciones(opcion, numero_a, numero_b)
print('El resultado es: ' + str(numero_c))
continuar = input("¿Deseas continuar? [si/no] ")
print('\n\n')
# if continuar == "no":
if continuar != "si":
break
print('¡Gracias!')
| false |
3e21d6d44a01c7d0129fb42cfcd76b93a9e5e171 | sojournexx/python | /Assignments/TanAndrew_assign4_problem1.py | 1,498 | 4.1875 | 4 | #Andrew Tan, 2/16, Section 010, Roll the Dice
from random import randint
result = False
while result == False:
s = int(input("How many sides on your dice? "))
#Check for valid data
if s < 3:
print("Sorry, that's not a valid size value. Please choose a positive number.")
continue
#Roll the dice and display roll result
else:
print()
print("Thanks! Here we go ...")
print()
counter = 0
doubles = 0
t1 = 0
t2 = 0
while result == False:
counter += 1
d1 = randint(1, s)
d2 = randint(1, s)
t1 = t1 + d1
t2 = t2 + d2
print("%d. die number 1 is %d and die number 2 is %d." %(counter, d1, d2))
#Check for doubles
if d1 != d2:
continue
else:
doubles += 1
#Check for snake eyes
if d1 != 1 and d2 != 1:
continue
else:
print()
print("You got snake eyes! Finally! On try number %d!" %(counter))
print("Along the way you rolled doubles %d times" %(doubles))
print("The average roll for die #1 was %.2f" %(t1 / counter))
print("The average roll for die #2 was %.2f" %(t2 / counter))
result = True
| true |
da8c803202a2c3e2b428e4298dac3c19b80eecdd | sojournexx/python | /Assignments/TanAndrew_assign2_problem2.py | 1,188 | 4.28125 | 4 | #Andrew Tan, 2/2, Section 010, Grade Calculator
#Ask user for name and class
name = input("What is your name? ")
course = input("What class are you in? ")
print()
#Ask user for weightage and test scores
weight_test = float(input("How much are tests worth in this class (i.e. 0.40 for 40%): "))
test1 = float(input("Enter test score #1: "))
test2 = float(input("Enter test score #2: "))
test3 = float(input("Enter test score #3: "))
print()
#Calculate and display test avg
test_avg = (test1 + test2 + test3) / 3
print("Your test average is: %.2f" %(test_avg))
print()
#Ask user for weightage and hw scores
weight_hw = float(input("How much are homework assignments worth in this class (i.e. 0.60 for 60%): "))
hw1 = float(input("Enter homework score #1: "))
hw2 = float(input("Enter homework score #2: "))
hw3 = float(input("Enter homework score #3: "))
print()
#Calculate and display hw avg
hw_avg = (hw1 + hw2 + hw3) / 3
print("Your homework average is: %.1f" %(hw_avg))
print()
#Calculate and display final score
total = hw_avg * weight_hw + test_avg * weight_test
print("Thanks, %s. Your final score in %s is %.2f" %(name, course, total))
| true |
bffda2bd72fee372c85344b5274f7060e327aa47 | cecilmalone/PythonFundamentos | /Cap03/Lab02/calculadora_v1.py | 1,147 | 4.15625 | 4 | # Calculadora em Python
# Desenvolva uma calculadora em Python com tudo que você aprendeu nos capítulos 2 e 3.
# A solução será apresentada no próximo capítulo!
# Assista o vídeo com a execução do programa!
print("\n******************* Python Calculator *******************")
print()
print("""Selecione o número da operação desejada:
1 - Soma
2 - Subtração
3 - Multiplicação
4 - Divisão""")
print()
opcao = int(input("Digite sua opção (1/2/3/4): "))
print()
primeiro = int(input("Digite o primeiro número: "))
print()
segundo = int(input("Digite o segundo número: "))
print()
if opcao not in (1, 2, 3, 4):
print("Opção Inválida!")
elif opcao == 1:
resultado = primeiro + segundo
print("{} {} {} = {}".format(primeiro, '+', segundo, resultado))
elif opcao == 2:
resultado = primeiro - segundo
print("{} {} {} = {}".format(primeiro, '-', segundo, resultado))
elif opcao == 3:
resultado = primeiro * segundo
print("{} {} {} = {}".format(primeiro, '*', segundo, resultado))
elif opcao == 4:
resultado = primeiro / segundo
print("{} {} {} = {}".format(primeiro, '/', segundo, resultado)) | false |
760be593820d8aaf4ff443aa3892f1b96d577a9d | pulosez/Python-Crash-Course-2e-Basics | /#6: Dictionaries/favorite_numbers.py | 842 | 4.1875 | 4 | # 6.2.
favorite_numbers = {
'john': 1,
'anna': 7,
'edward': 22,
'jane': 5,
'kate': 3,
}
print(favorite_numbers)
num = favorite_numbers['john']
print(f"John's favorite number is {num}.")
num = favorite_numbers['anna']
print(f"Anna's favorite number is {num}.")
num = favorite_numbers['edward']
print(f"Edward's favorite number is {num}.")
num = favorite_numbers['jane']
print(f"Jane's favorite number is {num}.")
num = favorite_numbers['kate']
print(f"Kate's favorite number is {num}.")
print("\n")
# 6.10.
favorite_numbers = {
'john': [1, 11, 3],
'anna': [7, 5, 100],
'edward': [22, 42],
'jane': [5, 1, 10],
'kate': [3, 17],
}
for name, numbers in favorite_numbers.items():
print(f"\n{name.title()} likes the following numbers:")
for number in numbers:
print(f" {number}")
| false |
73682e4924c8d23279b3ef56c65ac5715bc2d7b2 | pulosez/Python-Crash-Course-2e-Basics | /#5: if Statements/favourite_fruits.py | 417 | 4.1875 | 4 | # 5.7.
favourite_fruits = ['orange', 'apple', 'tangerine']
if 'orange' in favourite_fruits:
print("You really like oranges!")
if 'apple' in favourite_fruits:
print("You really like apples!")
if 'tangerine' in favourite_fruits:
print("You really like tangerines!")
if 'banana' in favourite_fruits:
print("You really like bananas!")
if 'peach' in favourite_fruits:
print("You really like peaches!")
| true |
6c383b3dc9c39f8963ef2afac3eccfba06c378f4 | Jamesong7822/Python-Tutorials | /0) Introduction To Python/tutorials.py | 2,460 | 4.125 | 4 | # def myfunction(a,b,c):
# # Myfunction checks if any number from 0 - 99 is divisible by a, b and c
# ans_list = []
# for i in range(100):
# if i % a == 0 and i % b == 0 and i % c == 0:
# ans_list.append(i)
# return ans_list
# print(myfunction(1,2,3))
# D) Write a function that randomly chooses a number from 1 to 100. The player will have to guess \
# correctly within 5 tries for that number, with hints(too small or too big) given
# from random import randint
# def guess():
# # Let comp choose random number
# ans = randint(1, 101) # randomise from 1 to 100. Eg. 88
# max_num_of_tries = 5
# num_of_tries = 1
# while num_of_tries <= max_num_of_tries:
# print("Guess a Number from 1 to 100")
# user_guess = int(input("Try Num {}--> Tell me your guess>>".format(num_of_tries)))
# # Hints - conditionals
# if user_guess > ans:
# print("Guess is too HIGH")
# elif user_guess < ans:
# print("Guess is too LOW")
# else: # same as user_guess == ans
# print("Congrats You guess correctly in {} tries.".format(num_of_tries))
# return
# num_of_tries += 1
# print("HA U SUCK U DIDNT GUESS CORRECTLY")
# print("The correct ans is {} YOU FOOL!".format(ans))
# guess()
# Imports
from random import randint
def guess2(max_tries, range1, range2):
"""
This improved function allows the user to set parameters for his game
"""
# Randomise choice from range1 to range2
ans = randint(range1, range2+1)
# Set up variables to check/update during the game loop
try_count = 1
# Game loop
while try_count <= max_tries:
print("Guess a Number from {} to {}. You have {} tries left".format(range1, range2, max_tries - try_count + 1))
# Takes in input from user
user_guess = int(input("Try {} --> Guess: ".format(try_count)))
# Hints
if user_guess < ans:
print("Too Low")
elif user_guess > ans:
print("Too High")
else:
print("Congrats! That took {} tries!".format(try_count))
# Quit out of function
return
# Increment try_count by 1
try_count += 1
print("Boo, you SCRUB. My secret number is {}!!".format(ans))
# Call the function with user set params
max_tries = int(input("Max Tries: "))
range1 = int(input("Guess range from: "))
range2 = int(input("to: "))
guess2(max_tries, range1, range2)
| true |
90acca5899c8ecf3891bcc8339ad16a0fcb2fbca | strawsyz/straw | /ProgrammingQuestions/牛客/对称的二叉树.py | 928 | 4.15625 | 4 | # -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# 画个二叉树就能理解
# 一个节点左右节点的值相同
# 左节点的左节点要与右节点的右节点相同
# 左节点的右节点要与右节点的左节点相同
class Solution:
def isSymmetrical(self, pRoot):
# write code here
if not pRoot:
return True
return self.is_symmertrical(pRoot.left, pRoot.right)
def is_symmertrical(self, left, right):
if not left and not right:
# 如果两边都是空，说明两边相等
return True
if not left or not right:
return False
if left.val == right.val:
return True & self.is_symmertrical(left.left, right.right) & self.is_symmertrical(left.right, right.left)
else:
return False
| false |
4612293ef47d0ac146502d3ba6b5373b904fa951 | Metalscreame/Python-basics | /functions/zip.py | 368 | 4.25 | 4 | # В Pyhon функция zip позволяет пройтись одновременно по нескольким итерируемым объектам (спискам и др.):
a = [10, 20, 30, 40]
b = ['a', 'b', 'c', 'd', 'e']
for i, j in zip(a, b):
print(i, j) # (10, a)
zip_obj = zip(a, b)
list_zip = list(zip_obj) # list of tuples
print(list_zip)
| false |
8fd2ebf33486069e8e7939bd54465ceefe6370a3 | Metalscreame/Python-basics | /date.py | 2,670 | 4.125 | 4 | from datetime import datetime, timedelta
# parsing date
now = datetime.now()
some_date = '01/02/1903'
date_format = '%d/%m/%Y'
parsed_date = datetime.strptime(some_date, date_format)
print(parsed_date)
one_day = timedelta(
days=1,
minutes=2
)
new_date = parsed_date-one_day
print('Parsed minus one day: ', new_date)
print('Day: ' + str(now.day))
print('Month: ' + str(now.month))
print('Hour: ' + str(now.hour))
# added
some_date = '01/02/1903 12:30:12'
date_format = '%d/%m/%Y %H:%M:%S'
parsed_date = datetime.strptime(some_date, date_format)
print(parsed_date)
'''
Note: Examples are based on datetime.datetime(2013, 9, 30, 7, 6, 5)
Code Meaning Example
%a Weekday as locale’s abbreviated name. Mon
%A Weekday as locale’s full name. Monday
%w Weekday as a decimal number, where 0 is Sunday and 6 is Saturday. 1
%d Day of the month as a zero-padded decimal number. 30
%-d Day of the month as a decimal number. (Platform specific) 30
%b Month as locale’s abbreviated name. Sep
%B Month as locale’s full name. September
%m Month as a zero-padded decimal number. 09
%-m Month as a decimal number. (Platform specific) 9
%y Year without century as a zero-padded decimal number. 13
%Y Year with century as a decimal number. 2013
%H Hour (24-hour clock) as a zero-padded decimal number. 07
%-H Hour (24-hour clock) as a decimal number. (Platform specific) 7
%I Hour (12-hour clock) as a zero-padded decimal number. 07
%-I Hour (12-hour clock) as a decimal number. (Platform specific) 7
%p Locale’s equivalent of either AM or PM. AM
%M Minute as a zero-padded decimal number. 06
%-M Minute as a decimal number. (Platform specific) 6
%S Second as a zero-padded decimal number. 05
%-S Second as a decimal number. (Platform specific) 5
%f Microsecond as a decimal number, zero-padded on the left. 000000
%z UTC offset in the form +HHMM or -HHMM (empty string if the the object is naive).
%Z Time zone name (empty string if the object is naive).
%j Day of the year as a zero-padded decimal number. 273
%-j Day of the year as a decimal number. (Platform specific) 273
%U Week number of the year (Sunday as the first day of the week) as a zero padded decimal number. All days in a new year preceding the first Sunday are considered to be in week 0. 39
%W Week number of the year (Monday as the first day of the week) as a decimal number. All days in a new year preceding the first Monday are considered to be in week 0. 39
%c Locale’s appropriate date and time representation. Mon Sep 30 07:06:05 2013
%x Locale’s appropriate date representation. 09/30/13
%X Locale’s appropriate time representation. 07:06:05
%% A literal '%' character. %
''' | true |
2f8b4ee8c6145451ed2d9b588314bc1ee117fb9e | JomHuang/PythonStudy | /Day_2/Preview/OOP/person.py | 936 | 4.46875 | 4 | """
开始OOP学习
1.结构
2.封装
3.定制
定义一个类
"""
class Person:
# 相当与构造函数
def __init__(self, name, age, pay=0, job=None):
self.name = name;
self.age = age;
self.pay = pay;
self.job = job;
# 取lastname
def lastname(self):
return self.name.split()[-1];
# 加薪
def giveRaise(self, percent):
self.pay *= (1.0 + percent);
# 格式化输出 理解不了
def __str__(self):
return '<%s => %s : %s : %s>' % (self.__class__.__name__, self.name, self.job, self.pay);
if __name__ == "__main__":
bob = person('Bob Smith', 42, 40000, 'software');
sue = person('Sue Jones', 45, 50000, 'hardware');
print ('bob name:', bob.name, '\n sue pay:', sue
.pay);
print (bob.name.split()[-1]);
sue.pay += 200;
print (sue.pay);
print (bob.lastname());
sue.giveRaise(0.2);
print (sue.pay);
| false |
fb1158f7b2802185143717b1dfc4b1ebc28c7cbd | rahulvennapusa/PythonLearning | /SquareInt.py | 240 | 4.15625 | 4 | number = input("Enter the integer :")
if int(number) > 0:
iteration = int(number)
ans = 0
while iteration != 0:
ans = ans + int(number)
iteration = iteration - 1
print("Square of %s is %s :" % (number, ans))
| true |
33b4192303fd75f5cf8c60653820ec03589902cd | SuyeshBadge/Python-in-30-Days | /Day 1/DataType.py | 1,561 | 4.34375 | 4 | '''
Data type in python
int
float
string
list
tuple
dictionary
set
boolean (true or false)
'''
################################
'''
Iterable {list string tuples set dictionary}
non iterable {int float}
'''
#######################
# int float str
"""
int #integer for numbers 1,2,3,4
float #float for decimal numbers eg. 0.0 0.5
str #string for characters like "hello there"
"""
'''
integer stores all the numbers
non iterable
'''
a = 5
b = 10
print("a is "+str(a)+" and b is "+str(b)) # traditional method
print("a is {0} and b is {1}".format(a, b)) # string formatting
print(f'a is {a} and b is {b}') # f string method
# a is 5 and b is 10
# float
"""
Float stores all the decimal points
non iterable
"""
f = 37.5
print(f)
# string
c = 'himanshu'
c1 = 'h'
print(c)
# list
'''
mutable
iterable
'''
list1 = [1, 2, 3, 4, 5, 6, 7] # integer list
list2 = ['a', 'b', 'c'] # characters list
slist = [1, 1, 1, 1, 1.0, 1, 2, 1, 1]
# list1.pop()
print(list1[1])
print(slist)
# Tuples
'''
Tuples are immutable
iterable
'''
tup1 = (1, 2, 3, 4, 5)
# tup1[1]=5
# set
'''
address is not assinged to the block
only has distinct values
'''
s1 = {1, 2, 4, 3, 4, 100, 3, 4, 500, 4, 4}
s1list = set(slist)
# Dictionary
'''
dict_name={
key:value,
key:value,
key:value,
key:value
}
dict_name[key] => value
'''
himasnshu = [20, 'himanshu', 9384938493] # list
print(himasnshu[2])
himanshu = {
'Rollno': 20,
'name': 'himanshu',
'mobile': 9898938493
} # dictionary
print(himanshu['mobile'])
# boolean
ismale = True
isfemale = False
| false |
dc29468021664d1d55e42ac4d0121816655aeb85 | David-H-Afonso/python-basic-tests | /coinExchange.py | 818 | 4.21875 | 4 | # Function
def exchange(coin):
return coin / dollars
# Inputs
dollars = input("Write the amount of dollars you want to exchange: ")
dollars = float(dollars)
coin = input("""
Choose the coin that you want to exchange the value by typing the number. By default this value is "Euros"
1 - Euros
2 - Pesos argentinos
3 - Pesos colombianos
Type it here: """)
coin = int(coin)
# Coin values
euro_value = 1.19
peso_argentino = 95.89
peso_colombiano = 3834.74
# Coin comprobation
if coin == 1:
coin_value = euro_value
elif coin == 2:
coin_value = peso_argentino
elif coin == 3:
coin_value = peso_colombiano
else:
coin_value = euro_value
# Showing the data to the user
coin_value = exchange(coin_value)
coin_value = round(coin_value,2)
print(f"With {dollars}$ you have {coin_value} on the other coin")
| true |
e922fc7376a58caf2606224a864acd068c7e975a | SelimRejabd/Learn-python | /string method.py | 552 | 4.25 | 4 | # few useful method for string
name = "reja"
# lenth of string
print(len(name))
# finding a charecter
print(name.find("a"))
# Capitalize "reja" to "Reja"
print(name.capitalize())
# uppercase of a string
print(name.upper())
# lowecase of a string
print(name.lower())
# is string digit or not
print(name.isdigit())
# is string alphabatic or not
print(name.isalpha())
# count charecter in string
print(name.count("e"))
# replace charecter
print(name.replace("j", "z"))
# string print multiple time
print(name*2)
| true |
893ae9a7901c8ce7030ca0efdb4c81dba293c3ba | nicolealdurien/Assignments | /week-02/day-1/user_address.py | 1,673 | 4.4375 | 4 | # Week 2 Day 1 Activity - User and Address
# Create a User class and Address class, with a relationship between
# them such that a single user can have multiple addresses.
class User:
def __init__(self, first_name, last_name):
self.first_name = first_name
self.last_name = last_name
self.addresses = []
def get_first_name(self, first_name):
return self.first_name
def get_last_name(self, last_name):
return self.last_name
def add_address(self, address):
str_address = str(address)
self.addresses.append(str_address)
def display_addresses(self):
print(f"""Addresses listed for {self.first_name} {self.last_name}:\n {self.addresses}""")
def __str__(self):
return f"{str(self.first_name)} {str(self.last_name)}"
class Address:
def __init__(self, street, city, state, zip_code):
self.street = street
self.city = city
self. state = state
self.zip_code = zip_code
def get_street(self, street):
return self.street
def get_city(self, city):
return self.city
def get_state(self, state):
return self.state
def get_zip_code(self, zip_code):
return self.zip_code
def __str__(self):
return f"{str(self.street)}, {str(self.city)}, {str(self.state)} {str(self.zip_code)}"
mickey_mouse = User('Mickey', 'Mouse')
magic_kingdom = Address('1180 Seven Seas Dr', 'Lake Buena Vista', 'FL', '32836')
epcot = Address('200 Epcot Center Dr', 'Orlando', 'Florida', '32821')
mickey_mouse.add_address(magic_kingdom)
mickey_mouse.add_address(epcot)
mickey_mouse.display_addresses()
| true |
6011e9272ce81b903109d50f6c82fab57b46a60c | fpoppe/How-Many-Times-I-Have-to-Fold | /n_fold.py | 1,008 | 4.28125 | 4 | import math
def main ():
print("\n Welcome to 'HOW MANY TIMES DO I HAVE TO FOLD?' \n")
distance = float(input("\n Give me the distance (a number) \n"))
#usuário nao pode botar algo diferente de numero
unit = input("Give me its unit (km, m or mm) \n")
unit = unit.lower()
while unit != "km" and unit != "m" and unit != "mm":
print("\n Invalid Unit \n")
unit = input()
unit = unit.lower()
n_fold = calculate_dist_fold(distance,unit)
print("\n You have to fold your piece of paper " + str(n_fold) + " times to reach " + str(distance) + " " + unit + "\n\n")
def calculate_dist_fold(distance,unit):
if unit == "km":
thick = pow(10,-7)
n_fold = math.log((distance/thick),2)
elif unit == "m":
thick = pow(10,-4)
n_fold = math.log((distance/thick),2)
else:
thick = 0.1
n_fold = math.log((distance/thick),2)
n_fold = math.ceil(n_fold)
return n_fold
main()
| false |
980c11b94a9ee491ee2095727036a6ed2b966145 | sichkar-valentyn/Dictionaries_in_Python | /Dictionaries_in_Python.py | 1,597 | 4.75 | 5 | # File: Dictionaries_in_Python.py
# Description: How to create and use Dictionaries in Python
# Environment: PyCharm and Anaconda environment
#
# MIT License
# Copyright (c) 2018 Valentyn N Sichkar
# github.com/sichkar-valentyn
#
# Reference to:
# [1] Valentyn N Sichkar. Dictionaries in Python // GitHub platform [Electronic resource]. URL: https://github.com/sichkar-valentyn/Dictionaries_in_Python (date of access: XX.XX.XXXX)
# Creating an empty dictionary
a = dict()
d = {}
print(a)
print(d)
# Creating filled dictionary in form of Key --> Value
d = {'a': 101, 10: 202}
print(d)
print(d['a'])
print(d[10])
# Operations with dictionaries
print('a' in d)
print(11 not in d)
d['a'] = 102 # Reassigning value for the existing key
print(d)
# Getting the value of the element of the dictionary by the key
print(d[10]) # If there is no such key it will show mistake
print(d.get(11)) # If there is no such key it will show 'None'
# Deleting information from the dictionary
del d['a'] # It will delete the pair Key --> Value
print(d)
# Using loops for the dictionaries
a = {'A': 15, 'B': 23, 'C': 35, 'D': 43}
for x in a:
print(x, end=' ') # It will show keys
print()
for x in a.keys():
print(x, end=' ') # Again it will show keys
print()
for y in a.values():
print(y, end=' ') # It will show values
print()
for x, y in a.items():
print(x, y, end='; ') # It will show pairs Key --> Value
print()
# It is possible to store value as a list in the dictionary
print(a)
a['D'] = ['Diana', 'Rose']
print(a)
a['D'] += ['Sun']
print(a)
| true |
ca2e22bf764551e91fdaf13c469258ff3821462c | leinad520/Monty-Hall-Game | /mp_game_simulation.py | 1,867 | 4.125 | 4 | from random import randrange, choice
win_count = 0
def monty_python_game():
# print("You have three doors to choose from. Behind one door is the prize; two other doors have nothing. Which do you choose?")
choicesArr = ["1","2","3"]
prize = choice(choicesArr)
# your_choice = input("type 1 2 or 3\n> ")
your_choice = "1"
choicesArr.remove(prize)
if prize != your_choice:
choicesArr.remove(your_choice)
opendoor = choice(choicesArr)
choicesArr.remove(opendoor)
choicesArr2 = ["1","2","3"]
choicesArr2.remove(your_choice)
choicesArr2.remove(opendoor)
# print(f"You chose door {your_choice}. Now let me open one of the remaining two doors that does not have the prize, door {opendoor}.")
# print(f"*opens door {opendoor} to show there is nothing there.")
# print(f"You now have the option to stick with your original choice, door {your_choice}, or switch to the other remaining door, door {choicesArr2[0]}. Do you want to?")
# switch_decision = input("Type y or n \n> ")
switch_decision = "y"
if switch_decision == "y":
if your_choice == prize:
return
# print(f"You lose by switching to door {choicesArr[0]}! Your original door {your_choice} had the prize")
if your_choice != prize:
# print(f"You win by switching! You switched from door {your_choice} to door {prize} which had the prize")
global win_count
win_count += 1
if switch_decision == "n":
if your_choice == prize:
print(f"You win by not switching! Your original door {your_choice} had the prize")
win_count += 1
if your_choice != prize:
print(f"You lose by not switching! The other door, door {prize}, has the prize")
for i in range(10000):
monty_python_game()
print(win_count) | true |
1d0abd1db93be3e724570699313277d0c0970653 | rbdjur/Python | /Practice_Modules/Python_object_and_Data_Structure_basics/sets/set.py | 402 | 4.15625 | 4 | myset = set()
#add number 1 to set
myset.add(1)
print(myset)
#add number 2 to the set
myset.add(2)
#add number 2 again to the set
myset.add(2)
#see what the results of a set are
print(myset)
#Only {1,2} are in set despite adding two 2's. this is because a set only holds unique values. IF the value already exists, it wont take another number that is already identical to the number in the set. | true |
0a79efb7b2540019d6d572528ec99a71d41313a3 | rbdjur/Python | /Practice_Modules/Errors_handling_exceptions/try_except_else.py | 696 | 4.21875 | 4 | #The try block will execute and examine the code in this block
try:
result = 10 + 10
print(result)
except:
#if the code contains an error, the code in the except block will execute
print("Adding is not happening check type")
#The try block will execute and examine the code in this block
try:
# add = 5 + '5'
add = 5 + 5
print(add)
except:
#if the code contains an error, the code in the except block will execute
print("adding is not happening.")
else:
#If the code is clean and has no errors, then this code will run
print(add)
print("Went well.")
finally:
#This code runs regardless if there is an error or not
print("Thee End")
| true |
c427137c075c714607fc7b740437824b7fe4aae3 | rbdjur/Python | /Practice_Modules/Python_object_and_Data_Structure_basics/variable_assignments/variable_assignments.py | 235 | 4.125 | 4 | my_dogs = 2
my_dogs = ["Sammyy", "Frankie"]
#Above is an example of dynamic typing that allows user to assign different data type
a = 5
print(a)
a = 10
print(a)
a = a + a
print("should be 20", a)
print(type(a))
print(type(my_dogs)) | true |
98806536f6a7adea80fa82bdfaf9876ce8e57b00 | laszlokiraly/LearningAlgorithms | /ch04/linked.py | 1,740 | 4.40625 | 4 | """
Linked list implementation of priority queue structure.
Stores all values in descending.
"""
from ch04.linked_entry import LinkedEntry
class PQ:
"""Heap storage for a priority queue using linked lists."""
def __init__(self, size):
self.size = size
self.first = None
self.N = 0
def __len__(self):
"""Return number of values in priority queue."""
return self.N
def is_full(self):
"""If priority queue has run out of storage, return True."""
return self.size == self.N
def enqueue(self, v, p):
"""Enqueue (v, p) entry into priority queue."""
if self.N == self.size:
raise RuntimeError('Priority Queue is Full!')
self.N += 1
to_add = LinkedEntry(v, p)
if self.first:
# find first node SMALLER than key, and keep track of
# prev so we can insert properly
n = self.first
prev = None
while n:
if p > n.priority: # Stop once in the right place
if prev:
to_add.next = n
prev.next = to_add
else:
to_add.next = self.first
self.first = to_add
return
prev, n = n, n.next
prev.next = LinkedEntry(v, p)
else:
self.first = to_add
def dequeue(self):
"""Remove and return value with highest priority in priority queue."""
if self.first:
val = self.first.value
self.first = self.first.next
self.N -= 1
return val
raise RuntimeError('PriorityQueue is empty!')
| true |
f9e1c7f094f2104fa7975c4e843e8ca1cfc1e96d | arthurleemartinez/InterestTools | /high_interest_loan.py | 2,948 | 4.15625 | 4 | user_principle: float = float(input("How much is the principle amount of your loan?"))
def get_boolean_user_plans():
user_answer: str = input("Do you plan to pay off at least some of it soon? Answer 'yes' or 'no'.")
if user_answer != 'yes' or 'Yes' or 'YES' or 'y' or 'Y':
return False
else:
return True
user_plans:bool = get_boolean_user_plans()
def get_user_apr() -> float:
u_apr: float = float(input("How much is your APR? Use a percentage without a symbol."))
return u_apr
user_apr = get_user_apr()
print('Your interest rate is {}%.'.format(user_apr))
def get_user_daily_interest() -> float:
user_daily_interest: float = user_principle * (user_apr / 365 / 100)
return user_daily_interest
daily_interest: float = get_user_daily_interest()
print('You are spending ${} on interest every day.'.format(daily_interest))
user_pay_period: int = int(input("How many days long is your pay period?"))
def get_pay_check_cost() -> float:
weekly_interest_cost = daily_interest * 7
biweekly_interest_cost = daily_interest * 14
monthly_interest_cost = daily_interest * 30
if user_pay_period > 6:
pay_period_interest_cost = weekly_interest_cost
elif user_pay_period > 13:
pay_period_interest_cost = biweekly_interest_cost
elif monthly_interest_cost > 27:
pay_period_interest_cost = monthly_interest_cost
return round(pay_period_interest_cost, 2)
pay_check_cost: float = 2*round(get_pay_check_cost(),2)
user_pay_check_amount = float(input("How much do you get paid every pay period?"))
def get_paycheck_percentage():
pay_check_percentage1: float = round((pay_check_cost / user_pay_check_amount * 100), 2)
return pay_check_percentage1
pay_check_percentage = round(get_paycheck_percentage() , 2)
# determine how much money you will save on each minimum payment if you pay down a certain principle amount.
def pay_down_implications():
payment_more_than_minimum: float = float(str(input("How much do you want to pay toward the principle this pay-check?")))
new_daily_interest = (user_principle - payment_more_than_minimum)/365/100*user_apr
new_pay_check_percentage: float = round((100*(new_daily_interest * user_pay_period) / user_pay_check_amount), 2)
new_pay_check_cost:float = round((new_pay_check_percentage/100*user_pay_check_amount) , 2)
message = 'If you make a payment of ${} before the next due date, this loan will change to taking up %{} of your pay-checks. This means that your next payment would be ${}'.format(payment_more_than_minimum, new_pay_check_percentage, new_pay_check_cost)
return message
print('Your pay period is {} days long'.format(user_pay_period))
print('You are spending approximately ${} per pay-check on this loan'.format(pay_check_cost))
print('Your high-interest loan will eat up %{} of your paycheck this pay period.'.format(pay_check_percentage))
print(pay_down_implications())
| true |
6e4fd23cb4a7129f0a9b5ca3ffb6fb1647f0e321 | nawaraj-b5/random_python_problems | /Python Basics/datatype_conversion.py | 391 | 4.5 | 4 | #Find the length of the text python and convert the value to float and convert it to string
length_of_python = ( len('python') )
length_of_python_in_float = float(length_of_python)
length_of_python_in_string = str( length_of_python )
print ('Length of the python in integer is {}, float is {}, and string is {} '.format(length_of_python,length_of_python_in_float,length_of_python_in_string)) | true |
67d38a1b0a4d052650402639c71eb1b6807f8d00 | heba-ali2030/examples | /largest_number.py | 423 | 4.40625 | 4 | #Python Program to Find the Largest Among Three Numbers
num1= int(input('choose first number: '))
num2= int(input('choose second number: '))
num3= int(input('choose third number: '))
if num1 > num2 and num1 > num3 :
print('num 1 is the largest')
elif num2 > num1 and num2 > num3 :
print('num 2 is the largest')
else:
print('num 3 is the largest')
#how to do this for more than 3 samples
| false |
a78fae1eb24fbb75cf328d20e83cec97d051356d | oluwafenyi/code-challenges | /CodeWars/6Kyu/Valid Braces/valid_braces.py | 737 | 4.3125 | 4 |
# https://www.codewars.com/kata/valid-braces
# Write a function that takes a string of braces, and determines if the order
# of the braces is valid. It should return true if the string is valid,
# and false if it's invalid.
def validBraces(string):
while '[]' in string or '{}' in string or '()' in string:
string_list = list(string)
if '[]' in string:
string_list.remove('[')
string_list.remove(']')
elif '{}' in string:
string_list.remove('{')
string_list.remove('}')
elif '()' in string:
string_list.remove('(')
string_list.remove(')')
string = ''.join(string_list)
return not bool(string)
| true |
1db206efd070dfa46493a2cdabf2f268e8d7cad0 | oluwafenyi/code-challenges | /CodeWars/6Kyu/Find the Missing Letter/find_missing_letter.py | 294 | 4.25 | 4 |
#https://www.codewars.com/kata/find-the-missing-letter
from string import ascii_letters
def find_missing_letter(chars):
ind = ascii_letters.index(chars[0])
corr_seq = list(ascii_letters[ind:ind+len(chars)+1])
return [char for char in corr_seq if char not in chars][0]
| true |
345c2f179becc434547007db6863d0cd1b1226cf | l0neaadil/Python_for_beginner | /10_if_elif_else.py | 860 | 4.21875 | 4 | # if...elif ...else
x = float(input("enter a no.: "))
if x == 0:
print("no. is neither positive nor negative")
elif x < 0:
print("no. is negative")
else:
print("no. is positive")
print("Done")
# Nested if_else
x = int(input("enter any integer: "))
if x == 0:
print("no. is neither positive nor negative")
elif (x > 0) and x <= 100:
print("no. is positive but very small")
if (x % 2) == 0:
print("even")
else:
print("odd")
elif (x < 0) and x >= -100:
print("no. is negative but very small")
if (x % 2) == 0:
print("even")
else:
print("odd")
elif 100 < x < 10000 or -10000 < x < -100:
print("no. contains three or more digits")
if (x % 2) == 0:
print("even")
else:
print("odd")
else:
print("only smaller integral values are to be entered")
print("Done")
| false |
3a38adf993db23d3b0dca7820c04e4e57ec4db9d | l0neaadil/Python_for_beginner | /14_for_loop.py | 552 | 4.15625 | 4 | # For Loop
string = "3456789"
list = [3, 4, 5, 6, 7, 8, 9]
tuple = (3, 4, 5, 6, 7, 8, 9)
set = {3, 4, 5, 6, 7, 8, 9}
dictionary = {1: 'a', 2: 'b', 3: 'c'}
print(string, list, tuple, set, dictionary)
for element in string:
print(element)
for element in list:
print(element)
for element in tuple:
print(element)
for element in set:
print(element)
for element in dictionary.keys():
print(element)
for element in dictionary.values():
print(element)
for element in range(0, 31, 2):
print(element % 2)
print(element / 2)
| false |
f78d289529201648db63dbc4981b5ab121b3c61f | pathakamaresh86/python_class_prgms | /day5_assign.py | 1,119 | 4.28125 | 4 | #!/usr/bin/python
num1=input("Enter number")
print "Entered number ", num1
# if LSB(last bit) is zero then even if 1 then odd
if num1&1 == 0:
print str(num1) + " is even number"
else:
print str(num1) + " is odd number"
num1=input("Enter number")
print "Entered number ", num1
if num1&15 == 0:
print str(num1) + " is divisible by 16"
else:
print str(num1) + " is not divisible by 16"
num1,num2,num3=input("Enter lenght of three sides of triangle")
print "Entered side1", num1
print "Entered side2", num2
print "Entered side3", num3
if num1>num2 and num1>num3 :
side3=num1
temp=num2 + num3
elif num2>num3 :
side3=num2
temp=num1 + num3
else :
side3=num3
temp=num1 + num2
if temp > side3:
print "It is triangle"
else:
print "not triangle"
'''
D:\F DATA\python_class>python day5_assign.py
Enter number12
Entered number 12
12 is even number
Enter number144
Entered number 144
144 is divisible by 16
Enter lenght of three sides of triangle5,6,7
Entered side1 5
Entered side2 6
Entered side3 7
It is triangle
''' | true |
e48a1567b7c529cb7a6ff50ab7cf22844dc2b19f | pathakamaresh86/python_class_prgms | /day4_assign.py | 1,011 | 4.21875 | 4 | #!/usr/bin/python
str=input("Enter string")
print "Entered string ", str
print "first tow and last two char string ", str[1:3:1]+str[-1:-3:-1]
str1=input("Enter string")
print "Entered string ", str1
print "occurance replaced string", str1[:1]+str1[1:].replace("b", "*")
str1,str2=input("Enter two strings")
print "str1=", str1
print "str2=", str2
print "swapped first two character strings", str2[:2] + str1[2:], str1[:2] + str2[2:]
str1,str2=input("Enter two strings")
print "str1=", str1
print "str2=", str2
size1=len(str1)
size2=len(str2)
if size1 != size2:
print "2nd string is not right rotation of 1st"
temp=str1+str1
#if (temp.count(str2)> 0):
if (str2 in temp):
print "2nd string is right rotation of 1st"
else:
print "2nd string is not right rotation of 1st"
str1=input("Enter string")
print "Entered string ", str1
str2="not bad"
if (str2 in str1):
print str1.replace("not bad", "good")
else:
print "string not contain not bad"
| true |
4e012bf0e6c051ee2585c2c27d33cd92c9babbe0 | krishnagoli/python | /Harikah/test5.py | 836 | 4.1875 | 4 | my_string="Hello world"
print(my_string.isalpha())
Output : False
str1="HelloWorld"
print(str1.isalpha())
Output : True
str="hfdgkjfdhg"
print(str.isdigit())
Output : False
str="hfdgkjfdhg123"
print(str.isdigit())
Output : False
str="45435435435h"
print(str.isdigit())
Output : False
str="45435435435"
print(str.isdigit())
Output : True
a="Hello world"
print(a.isalnum())
Output : False
str2="356546"
print(str2.isalnum())
Output : True
str1="HELLO"
print(str1.isupper())
Output : True
str5="H78"
print(str5.isupper())
Output : True
str="Hello"
print(str.isupper())
Output : False
a='hello'
print(a.islower())
Output : True
a='Hello'
print(a.islower())
Output : False
a="hkh6767"
print(a.islower())
Output : True
b='Hello World'
print(b.endswith('d'))
Output : True
b='Hello World'
print(b.startswith('H'))
Output : True
| false |
08ab6206785466e01758fc3df211b289d73e4fb6 | MunsuDC/visbrain | /examples/signal/02_3d_signals.py | 1,665 | 4.15625 | 4 | """
Plot a 3D array of data
=======================
Plot and inspect a 3D array of data.
This example is an extension to the previous one (01_2d_signals.py). This time,
instead of automatically re-organizing the 2D grid, the program use the number
of channels for the number of rows in the grid and the number of trials for the
number of columns.
To illustrate this point, we generate a random EEG dataset composed with 20
channels, 10 trials of 4000 points each. The 2D grid will have a shape of
(20 rows, 10 columns).
.. image:: ../../picture/picsignal/ex_3d_signal.png
"""
from visbrain import Signal
from visbrain.utils import generate_eeg
sf = 512. # sampling frequency
n_pts = 4000 # number of time points
n_channels = 20 # number of EEG channels
n_trials = 10 # number of trials in the dataset
"""Generate a random EEG dataset of shape (n_channels, n_trials, n_pts).
Also get the associated time vector with the same length as the data.
"""
data, time = generate_eeg(sf=sf, n_pts=n_pts, n_trials=n_trials,
n_channels=n_channels, smooth=200, noise=1000)
time += 8. # force the time vector to start at 8 seconds
time *= 1000. # millisecond conversion
"""The data have a shape of (20, 10, 4000). Hence, the time dimension is
defined as the last axis i.e `axis=2`
"""
axis = 2 # localization of the time axis
"""Add a label to the x-axis (xlabel), y-axis (ylabel) and a title
"""
xlabel = 'Time (ms)'
ylabel = 'Amplitude (uV)'
title = 'Plot of a 1-d signal'
Signal(data, sf=sf, axis=axis, time=time, xlabel=xlabel, ylabel=ylabel,
title=title, bgcolor=(.1, .1, .1), axis_color='wlite',
color='white').show()
| true |
82d1b9ee2f5a78cd095960a36a58e19b6239409e | zoeyouyu/GetAhead2020 | /Q3 - solution.py | 1,240 | 4.25 | 4 | # Longest path in the Tree
class Tree:
def __init__(self, value, *children):
self.value = value
self.children = children
# We walk the tree by iterating throught it,
# yielding the length of the path ending at each node we encounter,
# then take the max of that.
def longest_path(tree):
def rec(current, parent_value = 0, parent_path_length = 0):
# Length of the longest chain this node is part of.
current_path_length = (parent_path_length + 1
if current.value == parent_value + 1 else 1)
# Emit the length of this node.
yield current_path_length
# Recurse into the children
for child in current.children:
# For each of the descendant nodes, emit their lengths as well
for value in rec(child, current.value, current_path_length):
yield value
# Take the overall maximum length.
return max(rec(tree))
# Tests
if __name__ == "__main__":
assert longest_path(Tree(1)) == 1
assert longest_path(
Tree(1,
Tree(2,
Tree(4)),
Tree(3))
) == 2
assert longest_path(
Tree(5,
Tree(6),
Tree(7,
Tree(8,
Tree(9,
Tree(15),
Tree(10))),
Tree(12)))
) == 4
| true |
d54ce4cf2c186d149f51a6a37de807fbabdca25f | ritomar/ifpi-404-2017 | /Simulado01-Q02.py | 297 | 4.15625 | 4 | quantidade = 0
soma = 0
n = int(input("Digite um valor qualquer, zero para terminar: "))
while n != 0:
quantidade += 1
soma += n
n = int(input("Digite um valor qualquer, zero para terminar: "))
print("Quantidade:", quantidade)
print("Soma:", soma)
print("Média:", soma/quantidade)
| false |
d386a27c06a6252f2331beff919a290887cd4825 | saikumarsandra/My-python-practice | /Day-5/variableLengthArg.py | 418 | 4.25 | 4 | # when * is added before any argument then t act as the variable length argument
#it act as a parameter that accepts N number of values to one argument act as a tuple
def myFun(a,*b):
print (a,b)
myFun("sai",2.0)
#type 2
myFun("this is 'a'",2.0,1,2,3,4,5,6,[1,2,3],(1,2,3),{9,8,10})
def avg(*vals):
Average= sum(vals)/len(vals)
print("avarage of the values",Average)
avg(1,2,3,4,5,6,9,8,7,) | true |
684e219a422c35b538a6acfe6a780d3d04c06b89 | xpony/LearnPython | /def_function.py | 1,777 | 4.28125 | 4 | #自定义一个求绝对值的my_abs函数为例：\
# 方式 依次写 def 函数名 括号(及参数) ：
def my_abs(x):
if x >= 0:
return x #return 返回函数值
else:
return -x
print(my_abs(-3))
#注意，函数体内部的语句在执行时，一旦执行到return时，函数就执行完毕，并将结果返回。
#如果没有return语句，函数执行完毕后也会返回结果，只是结果为None。return None可以简写为return。
#如果你已经把my_abs()的函数定义保存为abstest.py文件了，那么，
#可以在该文件的当前目录下启动Python解释器，用from abstest import my_abs来导入my_abs()函数，
#注意abstest是文件名
#空函数 如果想定义一个什么事也不做的空函数，可以用pass语句：
def nop():
pass
#pass可以用来作为占位符，比如现在还没想好怎么写函数的代码，
#就可以先放一个pass，让代码能运行起来。
# 检查参数 用内置函数 isinstace()
def my_Newabs(x):
if not isinstance(int, float):
raise TypeError('bad operand type')
if x >= 0:
return x
else:
return -x
#print(my_Newabs('12'))
#返回多个值
import math #导入math包，后续代码就可以使用math包里的sin、cos等函数了
def move(x, y, step, angle = 0):
nx = x + step * math.cos(angle)
ny = y + step * math.sin(angle)
return nx, ny
print(move(100, 100, 60, math.pi/6))
#但其实这只是一种假象，Python函数返回的仍然是单一值
r = move(100, 100, 60, math.pi/6)
print(r)
# 返回值其实是一个tuple (151.96152422706632, 130.0)
#在语法上，返回一个tuple可以省略括号，而多个变量可以同时接收一个tuple，按位置赋给对应的值，
#list，set，dict 都类似可以这样赋值给变量
| false |
9a756fc8eac6f99a6e95a7cbe9e99c04ce777eb0 | xpony/LearnPython | /filter.py | 2,078 | 4.25 | 4 | #filter( )函数 用于过滤序列。同样接收一个函数和一个序列，把传入的函数依次作用于每个元素，
#然后根据返回值是True还是False决定保留还是丢弃该元素。和map()一样返回的是Iterator
#例如，在一个list中，删掉偶数，只保留奇数，可以这么写：
def is_odd(n):
return n % 2 == 1 # 判断是否为奇数
num = filter(is_odd, [1, 2, 3, 4, 5, 6, 10])
print(list(num))
#把一个序列中的空字符串删掉，可以这么写：
def not_empty(s):
return s and s.strip() #strip()去除字符串首尾空格，对单个空格作用后，bool值为False
s = filter(not_empty, ['A', 'B', '', None, ' ']) # None的bool值为False
print(list(s))
#用filter求素数
#构造一个从3开始的奇数序列
def _odd_iter():
n = 1
while True:
n = n + 2
yield n #这是个一个生成器函数
#生成器函数可以用for循环来不断输入结果，也可以生成个对象，然后用next()函数返回下个值
#g = _odd_iter()
# print(next(g))
#定义一个帅选函数
def _not_divisible(n):
return lambda x: x % n > 0
# 再定义一个生成器，不断返回下一个素数
def primes():
yield 2
it = _odd_iter() #生成一个对象，其包括大于3的奇数
while True:
n = next(it) #返回序列的第一个数
yield n #返回素数
it = filter(_not_divisible, it) # 构造新序列
for n in primes(): # 打印100以内的素数:
if n < 100:
print(n)
else:
break
# 练习 数是指从左向右读和从右向左读都是一样的数，例如12321，909。请利用filter()筛选出回数：
#教训！： 命名变量时千万不要用一些内置函数名，否则找找bug找到你哭
# 自然正整数生成函数
def _add_nums():
n = 10
while True:
yield n
n = n + 1
# 判断是否为回数
def number_h(n):
return str(n) == str(n)[::-1]
numbers = _add_nums() #生成自然是想序列的对象
nums = filter(number_h, numbers) # 生成回数序列
for n in nums: #打印出1000以内的回数
if n < 1000:
print(n)
else:
break | false |
790c398b7699707dd0881780f58dd63961956cf5 | xpony/LearnPython | /slice.py | 1,795 | 4.125 | 4 | #切片（Slice）：
#取一个list或tuple的部分元素是非常常见的操作。比如，取前n个元素，我们可以用循环，但是太麻烦了。如果用切片就会非常简单：
L = [1, 2, 3, 4, 5]
print(L[0:2]) #取前两个元素
print(L[:2]) # 从索引零开始，可以省略
print(L[1:]) # 前
print(L[2:5]) #可以从索引2开始取，取到第五个
print(L[2:]) #默认取完
#既然Python支持L[-1]取倒数第一个元素，那么它同样支持倒数切片
print(L[-2:]) #从倒数第二个开始，取完
print(L[-2:-1])
print(L[1:3])
print(L[1:2])
#切片操作十分有用。我们先创建一个0-99的数列：
L = list(range(100)) #range(100)表示从0开始生成含100个整数的序列
print(L[:10]) #取出前十个元素
print(L[-10:]) #取出后十个元素
print(L[10:20]) #前十一个到第二十个元素
print(L[:10:2]) #前十个每两个取一个
print(L[::5]) #所有数每五个取一个
print(L[:]) #原模原样复制出一个
print('---------------------------------------------------------------')
#tuple也是一种list，唯一区别是tuple不可变。因此，tuple也可以用切片操作，只是操作的结果仍是tuple：
t = (1,2,3,4,5)
print(t[:3]) #取出前三个
#字符串'xxx'也可以看成是一种list，每个元素就是一个字符。因此，字符串也可以用切片操作，只是操作结果仍是字符串：
s = 'mahongwei'
print(s[:2]) #取出前两个
### 练习 切除字符串首尾的空格
def trim(str):
if len(str) == 0:
print('请输入有效的字符串')
return str
elif str[0] == ' ':
str = str[1:]
return trim(str)
elif str[-1] == ' ':
str = str[0:-1]
return trim(str)
return str, len(str)
print(trim(' ma '))
## if、 elif 后边都要跟条件语句 | false |
0eb90832223c089e86b7ca5c0eda3f576f7b414f | mondon11/leetcode-top100 | /0021mergeTwoLists.py | 1,585 | 4.125 | 4 | #!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time : 2021/6/14 下午9:06
# @Author : jt_hou
# @Email : 949241101@qq.com
# @File : 0021mergeTwoLists.py
# Definition for singly-linked list.
class ListNode(object):
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution(object):
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
res = ListNode()
cur = res
while(l1 is not None and l2 is not None):
if l1.val < l2.val:
cur.next = ListNode(l1.val)
cur = cur.next
l1 = l1.next
else:
cur.next = ListNode(l2.val)
cur = cur.next
l2 = l2.next
'''
while(l1 is not None):
cur.next = ListNode(l1.val)
cur = cur.next
l1 = l1.next
while(l2 is not None):
cur.next = ListNode(l2.val)
cur = cur.next
l2 = l2.next
'''
if l1 is not None:
cur.next = l1
if l2 is not None:
cur.next = l2
return res.next
def genListNode(l):
start = ListNode()
cur = start
for i in l:
cur.next = ListNode(i)
cur = cur.next
return start.next
if __name__ == '__main__':
sol = Solution()
l1 = [1, 2, 4]
l2 = [1, 3, 4]
l1_node = genListNode(l1)
l2_node = genListNode(l2)
res = sol.mergeTwoLists(l1_node,l2_node)
print(res)
| false |
03edbcdf7a43db403c91d9499f8c7c90a1afdcbe | tsoporan/exercises_and_algos | /python/observer.py | 1,096 | 4.28125 | 4 | """
An example of the Observer pattern in Python.
A Subject keeps track of its observers and has a method of notifying them.
"""
class Subject(object):
def __init__(self):
self.observers = []
def register_observer(self, observer):
self.observers.append(observer)
def unregister_observer(self, observer):
self.observers.remove(observer)
def notify_observers(self, *args, **kwargs):
for obs in self.observers:
obs.notify(*args, **kwargs)
class Observer(object):
def __init__(self, name):
self.name = name
def notify(self, *args, **kwargs):
print("Called with args: %s %s" % (args, kwargs))
def __repr__(self):
return 'Observer: %s' % self.name
if __name__ == '__main__':
sub = Subject()
obs1 = Observer('1');
obs2 = Observer('2');
obs3 = Observer('3');
sub.register_observer(obs1)
sub.register_observer(obs2)
print(sub.observers, len(sub.observers) == 2)
sub.unregister_observer(obs2)
print(sub.observers, len(sub.observers) == 1)
sub.register_observer(obs3)
sub.notify_observers('notifying!')
| true |
f0407174a9908c1a73ee250296d334ae73bb178a | tsoporan/exercises_and_algos | /python/utopian_tree.py | 762 | 4.34375 | 4 | """
Utopian tree goes through 2 cycles of growth a YEAR.
First cycle in spring: doubles in height
Second cycle in summer: height increases by 1
Tree is planted onset of spring with height 1.
Find the height of tree after N growth cycles.
"""
def growthAfterCycles(lst):
out = []
for cycle in lst:
height = 1
if cycle == 0:
out.append(height)
else:
for run in range(cycle):
if run % 2 == 0:
height += height
else:
height += 1
out.append(height)
return out
if __name__ == '__main__':
r1, r2, r3, r4 = growthAfterCycles([3,4,0,1])
print(r1 == 6)
print(r2 == 7)
print(r3 == 1)
print(r4 == 2)
| true |
bb8bb6ca07ad5d3793b8e24fdee189d83657e294 | wprudencio97/csc121 | /wprudencio_lab6-8.py | 1,759 | 4.40625 | 4 | #William Prudencio, Chapter 6 - Lab 8, 8/18/19
''' This program reads the random numbers that were generated into
randomNumbers.txt and will display the following: 1)Count of the
numbers in the file, 2)Total of the numbers in the file, 3)Average of
the numbers in the file, 4)Largest and smallest number in the file. '''
#-----------------Define the main function---------------------
def main():
randomNumbers = open("randomNumbers.txt", "r") #open randomNumbers.txt
readFile(randomNumbers) #send the file to the file reader function
randomNumbers.close() #close randomNumbers.txt
#-----------------Define the file reader function--------------
def readFile(randomNumbers):
#start the counter and sum of the numbers
counter, total = 0, 0
#declare the variables to store the biggests and smallest numbers
smallestNumber, biggestNumber = 500, 0
print("\nNumbers read from file: \n") #print title
for line in randomNumbers: #read lines on file
line = int(line.rstrip("\n"))#remove new line character
print(line) #print the lines
#increase the counter and add to the total
counter += 1
total += line
#compare numbers to find biggest/smallest value
if smallestNumber > line:
smallestNumber = line
if biggestNumber < line:
biggestNumber = line
#Print the Results
print("\nNumbers read from the file: ", counter)
print("Sum of the numbers read from the file: ", total)
print("Average of the numbers read from the file: ", total/counter)
print("Smallest number on the file: ", smallestNumber)
print("Biggest number on the file: ", biggestNumber)
#-----------call the main function to run the program-----------
main() | true |
315af1bee922074ca17a8b8002f87057224921e6 | deanjingshui/Algorithm-Python | /17_位运算/231. 2的幂.py | 1,735 | 4.3125 | 4 | """
给定一个整数，编写一个函数来判断它是否是 2 的幂次方。
示例 1:
输入: 1
输出: true
解释: 2^0 = 1
示例 2:
输入: 16
输出: true
解释: 2^4 = 16
示例 3:
输入: 218
输出: false
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/power-of-two
"""
class Solution_iterate:
"""
date:2020.9.18
author:fenghao
思路：
从0开始尝试，计算2的m次幂是否等于n，当计算结果大于n说明无法满足
时间复杂度：O(log n)
"""
def isPowerOfTwo(self, n: int) -> bool:
m = 0
while 2**m < n:
m += 1
if 2**m == n:
return True
else:
return False
class Solution_bit_manipulation:
"""
date:2020.9.18
author:labuladong
思路：
位运算：去除二进制中最右边的 1
n&(n-1) 这个操作是算法中常见的，作用是消除数字 n 的二进制表示中的最后一个 1。
一个整数如果是2的幂，则其二进制表达只有一个1
反证法：num = 2^n + 2^m = 2^n(1 + 2^(m-n)) 括号中的树为基数，则num不可能为2的幂
时间复杂度：O(1)
"""
def isPowerOfTwo(self, n: int) -> bool:
if n <= 0: # 2的幂不可能为负数、零
return True
result = True if n & (n-1) == 0 else False
return result
class Solution_bit_manipulation_simplify_code:
"""
date:2020.9.18
author:labuladong
思路：
精简代码
时间复杂度：O(1)
"""
def isPowerOfTwo(self, n: int) -> bool:
return n > 0 and n & (n-1) == 0
n = 16
my_sol =Solution_bit_manipulation()
print(my_sol.isPowerOfTwo(n))
| false |
a2989a38d363fa9517600d06844a71e590ae38c4 | deanjingshui/Algorithm-Python | /5_二叉树/1. 二叉树的前序遍历.py | 2,253 | 4.375 | 4 | """
给定一个二叉树，返回它的 前序 遍历。
示例:
输入: [1,null,2,3]
1
\
2
/
3
输出: [1,2,3]
进阶: 递归算法很简单，你可以通过迭代算法完成吗？
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/binary-tree-preorder-traversal
"""
from typing import List
# Definition for a binary tree node.
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution_recursive:
"""
date:2020.9.23
author:fenghao
思路：递归
时间复杂度：O(2^n) n为二叉树的高度
空间复杂度：O()
"""
def preorderTraversal(self, root: TreeNode) -> List[int]:
if not root:
return []
left = right = []
if root.left:
left = self.preorderTraversal(root.left)
if root.right:
right = self.preorderTraversal(root.right)
return [root.val] + left + right
class Solution_iterate_stack:
"""
date:2020.9.27
author:fenghao
思路：迭代
维护一个数据结构（栈），每次弹出栈顶的节点(暂时留着右节点)，并将该节点的值存进结果，然后先后压入这个节点的右节点、左节点（注意入栈的顺序）
不断重复，直到这个数据结构(栈)为空
[root]
[root.left, root.right]
[root.left.left, root.left.right, root.right]
这是深度优先遍历DFS
时间复杂度：O(n) n为二叉树的节点个数
空间复杂度：O(h) h为二叉树的高度
"""
def preorderTraversal(self, root: TreeNode) -> List[int]:
if not root:
return []
result = []
nodes_stack = [root]
while nodes_stack:
node = nodes_stack.pop(0)
result.append(node.val)
if node.right:
nodes_stack.insert(0, node.right)
if node.left:
nodes_stack.insert(0, node.left)
return result
node_1 = TreeNode(1)
node_2 = TreeNode(2)
node_3 = TreeNode(3)
node_1.right = node_2
node_2.left = node_3
my_sol = Solution_iterate_stack()
print(my_sol.preorderTraversal(node_1))
| false |
c841dfbd88298c7f0a27e3a18d0e1533aad0b6e2 | deanjingshui/Algorithm-Python | /1_双指针/11. 盛最多水的容器.py | 1,411 | 4.125 | 4 | """
给你 n 个非负整数 a1，a2，...，an，每个数代表坐标中的一个点 (i, ai) 。
在坐标内画 n 条垂直线，垂直线 i 的两个端点分别为 (i, ai) 和 (i, 0) 。
找出其中的两条线，使得它们与 x 轴共同构成的容器可以容纳最多的水。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/container-with-most-water
"""
from typing import List
class Solution_force:
"""
author:fenghao
date:2021.4.5
思路：暴力法
两层循环穷举所有可能
时间：O(n^2)
空间：O(1)
"""
def maxArea(self, height: List[int]) -> int:
pass
class Solution_two_pointer:
"""
author:力扣+fenghao
date:2021.4.5
思路：双指针
难点：左右指针移动的条件，应该移动左or右指针？
时间：O(n)
空间：O(1)
"""
def maxArea(self, height: List[int]) -> int:
left = 0
right = len(height) - 1
maxArea = 0
while left < right:
tmp = min(height[left], height[right])*(right-left)
maxArea = max(maxArea, tmp)
if height[left] <= height[right]:
left += 1
else:
right -= 1
return maxArea
height = [1,8,6,2,5,4,8,3,7]
my_sol = Solution()
print(my_sol.maxArea(height)) | false |
ff9d8d7b4f3b4cba24822850d4fb1d78f52b9439 | Gyol/Multicampus_Cloud | /common_ground/python_programming_stu/mycode/11_Lambda/lambda01.py | 1,407 | 4.125 | 4 | # 보통 함수
def add(x, y):
return x + y
print(add(10, 20))
# lambda
add2 = lambda x, y : x + y
print(add2(100, 200))
# 제곱승, 곱하기, 나누기를 람다 함수로 정의해서 호출
print()
multi = lambda x, y : x ** y
print(multi(2, 4))
mul = lambda x, y : x * y
print(mul(18, 2))
dev = lambda x, y : x / y
print(dev(20, 4))
mulmul = lambda x: x * 2
print(mulmul(3))\
print()
llist = [1, 2, 3, 4, 5]
result = map(lambda x: x * 2, llist)
print(list(result))
result = list(map(lambda x: x * 2, llist))
print(result)
# [1, 2, 3, 4, 5] + [1, 2, 3, 4, 5]
# add(1, 1) add(2, 2) add(3, 3)
print()
ddist = [13, 14, 15, 16, 17]
f_add = lambda x, y: x + y
print(f_add(1, 88))
result = list(map(f_add, llist, ddist))
print(result)
# llist 값을 제곱승해서 다른 리스트에 저장
# lambda 함수랑 map()함수 사용
other = lambda x: x ** 2
# 얘네는 한번에
# result = list(map(other, llist))
# print(result)
# 이러면 하나씩
result = map(other, llist)
print(next(result))
print(next(result))
print(next(result))
print(next(result))
print(next(result))
# filter 함수
print()
result = list(filter(lambda x: x > 2, llist))
print(result)
for a in filter(lambda x: x > 2, llist):
print(a)
# reduce
# functools.py 라는 모듈 안에 있는 reduce함수를 불러오기
print()
from functools import reduce
result = reduce(lambda x, y: x + y, llist)
print(result) | false |
126540b591dd05de840b78447860cc2130c03326 | svetlana-strokova/GB_BigData_1055_Python | /3/3 les 2page.py | 817 | 4.21875 | 4 | # 2 Задание Реализовать функцию, принимающую несколько параметров
# переменные
name = input('Введите имя - ')
surname = input('Введите фамилию - ')
year = int(input('Введите год рождения - '))
city = input('Введите город проживания - ')
email = input('Введите email - ')
telephone = input('Введите телефон - ')
#функция с возвращением кортежа
def my_func (name, surname, year, city, email, telephone):
return ' '.join([name, surname, year, city, email, telephone])
print(my_func(name = 'Иван', surname = 'Иванов', year = '1988', city = 'Санкт-Петербург', email = 'email@mail.ru', telephone = '8-800-100-98-88'))
| false |
0a889f350a8c1b1efd46316f5786a8191250d934 | svetlana-strokova/GB_BigData_1055_Python | /1lesson/1les 2page.py | 1,289 | 4.125 | 4 | # 2 Задание. Переведение секунд в минуты и часы с форматированием строк
time = int(input('Введите время в секундах - '))
hours = time // 3600
#целочисленное деление
minutes = (time // 60) - (hours * 60)
# Из целого остатка часов остаток - минуты
seconds = time % 60
# Остаток от деления часов и минут - секунды. 60 - потому что в мин - 60 сек
print(f'{hours:02}:{minutes:02}:{seconds:02}')
# Вариант, где в формат "часы минуты и секунды" введенное число переводится отдельно, минуты в часы не переводятся и тд
#sek = int(input('Введите число секунд -'))
#min = sek / 60
#hour = min / 60 or sek / 3600
#print(hour,'часов', ':', min, 'минут', ':', sek, 'секунд')
# второй вариант, уже через форматирование строк
#sek1 = int(input('Введите число секунд -'))
#min1 = sek1 / 60
#hour1 = min1 / 60 or sek1 / 3600
#text: str = '''{0} часов : {1} минут : {2} секунд'''.format(hour1, min1, sek1)
#print(text)
| false |
a9caaccc5a5ade933817c50c8a8daf912b3c0255 | Tripl3Six/Rock-paper-scissors-lizard-spock | /rpsls.py | 2,206 | 4.1875 | 4 | # Rock-paper-scissors-lizard-Spock
# The key idea of this program is to equate the strings
# "rock", "paper", "scissors", "lizard", "Spock" to numbers
# as follows:
#
# 0 - rock
# 1 - Spock
# 2 - paper
# 3 - lizard
# 4 - scissors
import random as rand
# helper functions
def name_to_number(name):
if name == 'rock':
number = 0
elif name == 'Spock':
number = 1
elif name == 'paper':
number = 2
elif name == 'lizard':
number = 3
elif name == 'scissors':
number = 4
else:
print "Error - enter a correct name inside quotation marks."
return number
# convert name to number using if/elif/else
def number_to_name(number):
if number == 0:
name = 'rock'
elif number == 1:
name = 'Spock'
elif number == 2:
name = 'paper'
elif number == 3:
name = 'lizard'
elif number == 4:
name = 'scissors'
else:
print "Error number not in the range 0 to 4 inclusive"
return name
# convert number to a name using if/elif/else
def rpsls(player_choice):
# print a blank line to separate consecutive games
print
# print out the message for the player's choice
print 'Player choice: ' + str(player_choice)
# convert the player's choice to player_number using the function name_to_number()
player_number = name_to_number(player_choice)
# compute random guess for comp_number using random.randrange()
comp_number = rand.randrange(0, 5)
# convert comp_number to comp_choice using the function number_to_name()
comp_choice = number_to_name(comp_number)
# print out the message for computer's choice
print 'Computer choice: ' + str(comp_choice)
# compute difference of comp_number and player_number modulo five
diff = (player_number - comp_number) % 5
# use if/elif/else to determine winner, print winner message
if diff == 1 or diff == 2:
print "Player wins!!"
elif diff == 3 or diff == 4:
print "Computer wins"
else:
print "Tie"
player = input("Enter 'rock', 'paper',
'scissors', 'lizard', or 'spock': ")
rpsls(player)
| true |
af76dc1189d8b1985cd069e250ca17ee1bde6dff | MahjubeJami/Python-Programming-01 | /python essentials/mod03-exe03.py | 994 | 4.5625 | 5 | """
3. You are trying to build a program that will ask the user the following:
First Name
Temperature
Based on the user's entries, the program will recommend the user to wear
a T-shirt if the temperature is over or equal to 70º or bring a sweater if it is less than 70º.
Console Output
What shall I wear today?
Please Enter Your First Name: John
What is Today's Temperature: 33
Hi John , You should probably bring a sweater
____________________________________________
Console Output
What shall I wear today?
Please Enter Your First Name: Mike
What is Today's Temperature: 75
Hi Mike , It will be a warm day , T-shirt time!
"""
print("What shall I wear today? \n")
name = input("Please Enter Your First Name: ")
temp = int(input("What is Today's Temperature: "))
print("\n")
if temp >= 70:
print("Hi" , name, ", It will be a warm day , T-shirt time!")
elif temp < 70:
print("Hi" , name , ", You should probably bring a sweater")
| true |
1201f750d7ccc0c120fe4773ec50337bce5dc8c2 | MahjubeJami/Python-Programming-01 | /python strings/mod06-exe_6.7.py | 1,890 | 4.21875 | 4 | # Using the variable famous_list, write a program to check
# if a famous individual is in the list above, if they are
# then print: Sorry, the individual did not make the top 20 cut!
# Otherwise print: Yup, the individual did make the top 20 cut.
#
# Console:
#
# Please Enter the name of the famous individual? Albert Einstein
# Yup, Albert Einstein did make the Top 20 cut!
#
# Please Enter the name of the famous individual? leonardo Da vinci
# Sorry, Leonardo Da Vinci did not make the Top 20 cut!
famous_list = ''' \
Marilyn Monroe (1926 – 1962) American actress, singer, model
Abraham Lincoln (1809 – 1865) US President during American civil war
Nelson Mandela (1918 – 2013) South African President anti-apartheid campaigner
John F. Kennedy (1917 – 1963) US President 1961 – 1963
Martin Luther King (1929 – 1968) American civil rights campaigner
Queen Elizabeth II (1926 – ) British monarch since 1954
Winston Churchill (1874 – 1965) British Prime Minister during WWII
Donald Trump (1946 – ) Businessman, US President.
Bill Gates (1955 – ) American businessman, founder of Microsoft
Muhammad Ali (1942 – 2016) American Boxer and civil rights campaigner
Mahatma Gandhi (1869 – 1948) Leader of Indian independence movement
Margaret Thatcher (1925 – 2013) British Prime Minister 1979 – 1990
Mother Teresa (1910 – 1997) Macedonian Catholic missionary nun
Christopher Columbus (1451 – 1506) Italian explorer
Charles Darwin (1809 – 1882) British scientist, theory of evolution
Elvis Presley (1935 – 1977) American musician
Albert Einstein (1879 – 1955) German scientist, theory of relativity
Paul McCartney (1942 – ) British musician, member of Beatles
Queen Victoria ( 1819 – 1901) British monarch 1837 – 1901
Pope Francis (1936 – ) First pope from the Americas
'''
fullname = input("Please Enter the name of the famous individual? ")
if fullname in famous_list:
print("Yup,", fullname, "did make the Top 20 cut!")
else:
print("Sorry,", fullname, "did not make the Top 20 cut!")
| true |
3ab92c4162bb89747c786a4a6975a083f06dd368 | MahjubeJami/Python-Programming-01 | /python strings/mod06-exe_6.4.py | 834 | 4.4375 | 4 | """
4. Write a Python function to create the HTML
string with tags around the word(s).
Sample function and result are shown below:
add_html_tags('h1', 'My First Page')
<h1>My First Page</h1>
add_html_tags('p', 'This is my first page.')
<p>This is my first page.</p>
add_html_tags('h2', 'A secondary header.')
<h2>A secondary header.</h2>
add_html_tags('p', 'Some more text.')
<p>Some more text.</p>
"""
# input from Console to have didffernt entrance for tags and contents
input1 = input("Enter The Tag please: ")
input2 = input("Enter the contents please: ").title()
def add_html_tags(tag, content):
print('<'+ tag + '>'+ content +'</' +tag +'>')
# insted of repeated method call, I used input and call the function once
add_html_tags(input1, input2)
| true |
be23e3bb52a6a191041264d2430fde3a4a1ebdd6 | richardOlson/Intro-Python-I | /src/13_file_io.py | 1,172 | 4.375 | 4 | """
Python makes performing file I/O simple. Take a look
at how to read and write to files here:
https://docs.python.org/3/tutorial/inputoutput.html#reading-and-writing-files
"""
# importing the os to get the path to find the file
import os
fooPath = os.path.join(os.path.dirname(__file__), "foo.txt")
# Open up the "foo.txt" file (which already exists) for reading
# Print all the contents of the file, then close the file
# Note: pay close attention to your current directory when trying to open "foo.txt"
# opening the file
f = open(fooPath)
print(f.read())
f.close()
# Open up a file called "bar.txt" (which doesn't exist yet) for
# writing. Write three lines of arbitrary content to that file,
# then close the file. Open up "bar.txt" and inspect it to make
# sure that it contains what you expect it to contain
# making the path for the bar file
barPath = os.path.join(os.path.dirname(__file__), "bar.txt")
# opening the path
theFile = open(barPath, "w")
theFile.write("Hey this is the first line in the bar file.\n")
theFile.write("This is the second line of the file, yeah!\n")
theFile.write("This is now the third line of the file!\n")
theFile.close()
| true |
399553cdf2290f687ec9466a0ccf1353c2a1e2b9 | pratikshyad32/pratikshya | /rock paper scissor game.py | 1,833 | 4.1875 | 4 | name=input("Enter your name")
while name.isalpha()==False or len(name)<6:
name=input("your name is wrong please enter again")
else:
print("Hello",name)
age=(input("Enter your age"))
while age.isdigit()==False :
age=input("your age is wrong please enter again")
else:
print("your age is accepted",age)
password=int(input("Enter your password"))
password=int(1234567)
while True:
if password ==1234567:
print("your password is accepted")
break
else:
password("your password is wrong please enter again")
count = 0
choice=input(" \n 1.Play Game \n 2.Exit\n")
while choice != "2":
if choice =="1":
print("Game begins...")
import random
game_word =["rock","paper","scissor"]
pc_result = random.choice(game_word)
user_word = str(input("Enter in small letter type :'rock','paper','scissor':"))
game_word = ['rock','paper','scissor']
pc_result = random.choice(game_word)
if pc_result =='rock' and user_word =='paper' :
print('you win')
count= count + 1
elif pc_result == 'paper' and user_word == 'scissor':
print("you win")
count=count + 1
elif pc_result == 'scissor' and user_word == 'rock':
print("you win")
count=count + 1
elif pc_result == 'rock' and user_word == 'scissor':
print("you lose")
elif pc_result =='paper' and user_word == 'rock':
print("you lose")
elif pc_result == 'scissor' and user_word == 'paper':
print("you lose")
else:
print("MATCH IS DRAW...")
print("you have entered :",user_word)
print("computer entered :",pc_result)
print("your score is ",count)
print("\n you wanna play again")
if choice == "2":
print("exit")
choice = input("\n 1. play game \n 2.exit\n")
| true |
a2caca192e370b9e3565139dddb0515cff9ea421 | SaiSudhaV/TrainingPractice | /ArraysI/array_square.py | 312 | 4.1875 | 4 | # 3 Given an integer array nums sorted in non-decreasing order, return an array of the squares of each number sorted in non-decreasing order.
def squares(ar):
return sorted([i ** 2 for i in ar])
if __name__ == "__main__":
n = int(input())
ar = list(map(int, input().split()))
print(squares(ar)) | true |
7e60df36ee9effdd2a5b732fb34ad6b33b33b89b | Ap6pack/PythonProjects | /UserMenu | 2,436 | 4.34375 | 4 | #!/usr/bin/env python3
### What and where the information is being stored
userTuples = ["Name", "Age", "Sex"]
userList = [ ]
age2 = []
sex2 = []
### I have used def to define my menu so I can use this format over and over.
def userDirectory():
print "1. Add your information to the list"
print "2. Find out the number of males and the number of females and the average age of each in the list"
print "3. Print the list"
print "4. Quit"
print
### calling the Directory and setting the number of options
userDirectory()
userDirectory = 0
while userDirectory != 4:
### Printing user prompt and requesting users information
userDirectory = input("Pick a number from the menu: ")
print
if userDirectory == 1:
name = raw_input("What is your name? ")
userList.append(name)
userTuples = tuple(userList)
age2 = raw_input("How old are you " + name + "? ")
userList.append(age)
age.append(age)
userTuples = tuple(userList)
sex2 = raw_input("What sex do you identify with Male or Female " + name + "? ")
userList.append(sex)
sex.append(sex)
userTuples = tuple(userList)
print
print "Your name is " + name + ", you are " + age + " years old" + "and you identfy as a " + sex
print
### Seconed user option asking if they would like to find out the number of males and the number of females and the average age of each in the list.
elif userDirectory == 2:
print "Number if Males in the list ", sex.count("male") + sex.count("Male") + sex.count("m") + sex.count("M")
print "Number of Females in the list ", sex.count("female") + sex.count("Female") + sex.count("f") + sex.count("F")
print "The average age of the Males ",'\n' ### still trying to solve this question
print "The average age of the Females ", ### still trying to solve this question
print
print
### Third user option where they can print the information that has been inputed.
elif userDirectory == 3:
print userTuples
print
### Fourth options is where the user can end the loop.
elif userDirectory == 4:
print "Thank you, Have a nice day"
break
| true |
c045170367b61dbc0554b69f6c63ccd8caf176fc | chloebeth/codex-celerate | /benchmarks/code_output_4.py | 542 | 4.1875 | 4 | # - Time complexity: O(n)
# - Space complexity: O(n)
def myPow(x, n):
if n == 0:
return 1
if n < 0:
return 1 / myPow(x, -n)
if n % 2:
return x * myPow(x, n - 1)
return myPow(x * x, n / 2)
# Optimize the space complexity of the above code
#
# - Time complexity: O(n)
# - Space complexity: O(1)
def myPow(x, n):
if n == 0:
return 1
if n < 0:
return 1 / myPow(x, -n)
res = 1
while n > 0:
if n % 2:
res *= x
x *= x
n //= 2
return res | false |
d311436ad2902a7ccc5ea809669a280ab0aea17a | oekeur/MinProg_DataToolkit | /Homework/Week 1/exercise.py | 2,028 | 4.25 | 4 | # Name : Oscar Keur
# Student number : 11122102
'''
This module contains an implementation of split_string.
'''
# You are not allowed to use the standard string.split() function, use of the
# regular expression module, however, is allowed.
# To test your implementation use the test-exercise.py script.
# A note about the proper programming style in Python:
#
# Python uses indentation to define blocks and thus is sensitive to the
# whitespace you use. It is convention to use 4 spaces to indent your
# code. Never, ever mix tabs and spaces - that is a source of bugs and
# failures in Python programs.
def split_string(source, separators):
'''
Split a string <source> on any of the characters in <separators>.
The ouput of this function should be a list of strings split at the
positions of each of the separator characters.
'''
output = []
j = 0
k = 0
# loop over the source
while j in range(len(source)):
k = 0
if source[j] not in separators:
k = j + 1
# as long as k is not the list of separators, check for the next letter
while k in range(len(source)) and source[k] not in separators:
k += 1
# when a separator is found, add the characters between the separators to the output
output.append(source[j:k])
# if a word was found, skip over the length of that word
if k > 1:
j += len(output[-1])
# else go to the next letter
else:
j += 1
# return the outputlist
return output
if __name__ == '__main__':
# You can try to run your implementation here, that will not affect the
# automated tests.
# should print: ['c', 'd', 'r']
print split_string('abacadabra', 'ab')
# should print: ['abc']
print split_string('abc', '')
# should print: ['aba', 'adaba', 'a']
print split_string('abacadabra', 'cr')
# should print: ['a', 'acada']
print split_string('abacadabra', 'b')
| true |
9f8efa161d8ef1b8bf35075c4723e4453cc265fc | licup/interview-problem-solving | /Module 7/kBackspace.py | 838 | 4.125 | 4 | '''
K Backspaces
The backspace key is broken. Every time the backspace key is pressed, instead of deleting the last
(non-backspace) character, a '<' is entered.
Given a string typed with the broken backspace key, write a program that outputs the
intended string i.e what the keyboard output should be when the backspace key works properly.
Input
One line containing the string that was written in the text editor.
Only contains lowercase letters from the English alphabet as well as the character '<'.
'<' will not be the first character.
'''
def k_backspace(inputString):
stack = []
for character in inputString:
if character == "<":
stack.pop()
else:
stack.append(character)
return "".join(stack)
# Test Case
testInput = 'a<bc<'
actualOutput = k_backspace(testInput)
print(actualOutput) #returns 'b'
| true |
968f323dc9381b444f1341813449e458dcc78e62 | licup/interview-problem-solving | /Module 7/sya.py | 1,342 | 4.15625 | 4 | '''
Reverse polish notation is a postfix notation for mathematical expressions.
For example, the infix expression (1 + 2) / 3 would become 1 2 + 3 /.
More detailed explanation here: https://en.wikipedia.org/wiki/Reverse_Polish_notation
Task:
Given a mathematical expression in reverse polish notation, represented by an array of strings,
find the answer to this expression. Operators consist only of +, -, *, /,
and all numbers are integer values. When performing a division on two numbers,
use python's integer division operator (//). Your output should be a single integer,
which is the value of the expression when evaluated. Each expression is guaranteed to be valid.
'''
def evaluate_expression(expression):
stack = []
for element in expression:
if element.isnumeric():
stack.append(int(element))
else:
if element == '+':
n = stack.pop()
n1 = stack.pop()
stack.append(n + n1)
elif element == '-':
n = stack.pop()
n1 = stack.pop()
stack.append(n1 - n)
elif element == '/':
n = stack.pop()
n1 = stack.pop()
stack.append(n1 // n)
elif element == '*':
n = stack.pop()
n1 = stack.pop()
stack.append(n1 * n)
return stack.pop()
print(evaluate_expression(["3","4","+","5","-"])) #( 3 + 4 ) - 5 = 2
| true |
de6836359ebbac88fe38613153af669a9af33902 | thisislola/Tutorials | /temp_calculator.py | 1,825 | 4.28125 | 4 | # Temperature Calculator by L. Carthy
import time
def intro_options():
""" Takes the option and returns the fuction
that correlates """
option = int(input("1 for Fahrenheit to Celsius \n"
"2 for Celcius to Fahrenheit \n"
"3 for Fahrenheit to Kelvin: "))
if option == 1:
ftoc_calc()
elif option == 2:
ctof_calc()
elif option == 3:
ftk_calc()
else:
print("That is not an option. Try again.")
time.sleep(2)
intro_options()
def ftoc_calc():
"""Calculates from Fahrenheit to Celsius.
Returns the value of the calculation """
try:
ftc_input = int(input("Enter the Fahrenheit value: "))
ftc_conversion = (ftc_input - 32) * 5/9
print("Your answer in Celsius is: ", ftc_conversion)
except ValueError:
print("Error: Your input is not a number. Try again.")
time.sleep(2)
ftoc_calc()
def ctof_calc():
"""Calculates from Celsius to Fahrenheit.
Returns the value of the calculation """
try:
ctf_input = int(input("Enter the Celsius value: "))
ctf_conversion = ctf_input * 9/5 + 32
print("Your answer in Fahrenheit is: ", ctf_conversion)
except ValueError:
print("Error: Your input is not a number. Try again.")
time.sleep(2)
ftoc_calc()
def ftk_calc():
"""Calculates from Fahrenheit to Kelvin.
Returns the value of the calculation """
try:
ftc_input = int(input("Enter the Fahrenheit value: "))
ftc_conversion = (ftc_input + 459.67) * 5/9
print("Your answer in Kelvin is: ", ftc_conversion)
except ValueError:
print("Error: Your input is not a number. Try again.")
time.sleep(2)
ftoc_calc()
intro_options()
| true |
db54e4a06079cbba0c755e0df03afbe2892eb739 | satishkr39/MyFlask | /Python_Demo/Class_Demo.py | 1,415 | 4.46875 | 4 | class Sample:
pass
x = Sample() # creating x of type Sample
print(type(x))
class Dog:
# def __init__(self, breed):
# self.breed = breed
# CLASS OBJECT ATTRIBUTE IT will always be same for all object of this class
species = 'Mammal'
# INIT METHOD CALLED EVERY TIME AN OBJECT IS CREATED
def __init__(self, breed, name):
self.breed = breed
self.name = name
# repr returns the string representation of class object. when we print an object this repr is called.
def __repr__(self):
return f" Breed: {self.breed}, name: {self.name}"
# it used to return the len() method
def __len__(self):
return self.name
# myDog = Dog('Lab')
# print("Breed of myDog is " + myDog.breed)
myDog = Dog('Lab', 'MyDogName')
print(myDog.breed, myDog.name)
print("Printing objects calls repr method: ", myDog)
print(myDog.species)
print(len(myDog))
# CIRCLE CLASS DEMO
class Circle:
pi = 3.14
def __init__(self, radius=1):
self.radius = radius
def area(self, radius):
return self.pi * self.radius * self.radius
def circumference(self, radius):
return 2 * self.pi * self.radius
myCircle = Circle()
print("Circle with default value : "+ str(myCircle.area(1)))
myCircle2 = Circle()
print("Circle with radius 10 : "+ str(myCircle2.area(10)))
print("Circumference with radius 10: "+ str(myCircle2.circumference(10)))
| true |
c9685fe1d887a5f52952043f559ce6e71108ef27 | anand-ryuzagi/Data-Structue-and-Algorithm | /Sorting/merge-sort.py | 1,123 | 4.25 | 4 | # mergesort = it is divide and conquer method in which a single problem is divided in to small problem of same type and after solving each small problem combine the solution.
# Time complexity : O(nlogn)
# Space complexity : O(n)
# algorithms :
# 1. divide the array into two equal halves
# 2. recursively divide each sub arrays into two halves until length of last subarray is equal to 1.
# 3. after that apply merge function to join the solution of all the problems.
def merge(a,b,arr):
len_a= len(a)
len_b = len(b)
i = 0
j = 0
k = 0
while i<len_a and j<len_b:
if a[i] <= b[j]:
arr[k] = a[i]
i +=1
else:
arr[k] = b[j]
j+=1
k += 1
while i<len_a:
arr[k] = a[i]
i += 1
k += 1
while j<len_b:
arr[k] = b[j]
j += 1
k += 1
def mergeSort(arr):
if len(arr) <= 1:
return
mid = len(arr)//2
left = arr[:mid]
right = arr[mid:]
mergeSort(left)
mergeSort(right)
merge(left, right, arr)
arr =[1,5,2,3,9,7,0]
mergeSort(arr)
print(arr)
| true |
4691857dd272f5ed6cd0bb040dfe7157ad429407 | micaris/Data-Structures-and-Algorithms | /Sorting.py | 1,715 | 4.125 | 4 | #Bubble sort
def bubble_sort(a_list):
for pass_num in range(len(a_list) - 1, 0, -1):
for i in range(pass_num):
if a_list[i] > a_list[i + 1]:
temp = a_list[i]
a_list[i] = a_list[i + 1]
a_list[i + 1] = temp
def selection_sort(a_list):
for fill_slot in range(len(a_list) -1 , 0, 1):
pos_of_max = 0
for location in range(1, fill_slot + 1):
if a_list[location] > a_list[pos_of_max]:
pos_of_max = location
temp = a_list[fill_slot]
a_list[fill_slot] = a_list[pos_of_max]
a_list[pos_of_max] = temp
def insertion_sort(a_list):
for index in range(1, len(a_list)):
current_value = a_list[index]
position = index
while position > 0 and a_list[position - 1] > current_value:
a_list[position] = a_list[position - 1]
position = position - 1
a_list[position] = current_value
def shell_sort(a_list):
sublist_count = len(a_list) // 2
while sublist_count > 0:
for start_position in range(sublist_count):
gap_insertion_sort(a_list, start_position, sublist_count)
print("After increments of size", sublist_count, "The list is", a_list)
sublist_count = sublist_count // 2
def gap_insertion_sort(a_list, start, gap) :
for i in range(start + gap, len(a_list), gap):
current_value = a_list[i]
position = i
while position >= gap and a_list[position - gap] > current_value:
a_list[position] = a_list[position - gap]
position = position - gap
a_list[position ] = current_value
def merge_sort(a_list):
pass | false |
7b498522f914cf3f88d686578da8201adaf2bb10 | devathul/prepCode | /DP/boxStacking.py | 2,661 | 4.21875 | 4 | """
boxStacking
Assume that we are given a set of N types 3D boxes; the dimensions are defined for i'th box as:
h[i]= height of the i'th box
w[i]= width of the i'th box
d[i]= depth of the i'th box
We need to stack them one above the other and print the tallest height that can be achieved. One type of box can be used multiple times.
Condition of the stacking is:
1. The box can be placed on top of another box, iff the lower box's top surface area is strictly larger than the upper box's base area
2. Strictly larger area is where the both length and width are higher
n = 4
height[] = {4,1,4,10}
width[] = {6,2,5,12}
length[] = {7,3,6,32}
Output: 60
Explanation: One way of placing the boxes is
as follows in the bottom to top manner:
(Denoting the boxes in (l, w, h) manner)
{ 2 , 1 , 3 }
{ 3 , 2 , 1 }
{ 5 , 4 , 6 }
{ 6 , 5 , 4 }
{ 7 , 6 , 4 }
{ 12 , 10 , 32 }
{ 32 , 12 , 10 }
Hence, the total height of this stack is
10 + 32 + 4 + 4 + 6 + 1 + 3 = 60.
No other combination of boxes produces a height
greater than this.
"""
class Box:
def __init__(self,length,width,height):
self.length=length
self.width=width
self.height=height
self.area=length*width
def printbox(self):
print("{",self.length,",",self.width,",",self.height,"}")
def allBoxesWithDimensions(box):
if box is None:
return None
return [Box(max(box.length,box.width),min(box.length,box.width),box.height),Box(max(box.length,box.height),min(box.length,box.height),box.width),Box(max(box.width,box.height),min(box.width,box.height),box.length)]
def stackBoxes(length,width,height):
if length is None or width is None or height is None or len(length)!=len(width) or len(length)!=len(height):
return None
n=len(height)
boxes=[]
for i in range(n):
boxes.append(Box(length[i],width[i],height[i]))
boxAllType=[]
for i in range(n):
boxAllType.extend(allBoxesWithDimensions(boxes[i]))
boxAllType=sorted(boxAllType,key=lambda x:x.area)[::-1]
#for i in boxAllType:
# i.printbox()
m=len(boxAllType)
maxTillNow=[i.height for i in boxAllType]
result=[i for i in range(m)]
for i in range(1,m):
for j in range(i):
if boxAllType[i].length<boxAllType[j].length and boxAllType[i].width<boxAllType[j].width:
maxTillNow[i]=maxTillNow[j]+boxAllType[i].height
result[i]=j
maxVal=max(maxTillNow)
print("Max Value:",maxVal)
maxIndex=maxTillNow.index(maxVal)
i=maxIndex
su=0
while i!=result[i]:
boxAllType[i].printbox()
su+=boxAllType[i].height
i=result[i]
boxAllType[i].printbox()
height = [4,1,4,10]
width = [6,2,5,12]
length = [7,3,6,32]
stackBoxes(length,width,height)
height=[4,5]
width=[2,2]
length=[1,3]
stackBoxes(length,width,height) | true |
92bcb1209ffeea2320590bd05a442c73bf492dfe | connorholm/Cyber | /Lab2-master/Magic8Ball.py | 960 | 4.125 | 4 | #Magic8Ball.py
#Name:
#Date:
#Assignment:
#We will need random for this program, import to use this package.
import random
def main():
#Create a list of your responses.
options = [
"As I see it, yes.",
"Ask again later.",
"Better not tell you now.",
"Cannot predict now.",
"Concentrate and ask again.",
"Don’t count on it.",
"It is certain.",
"It is decidedly so.",
"Most likely.",
"My reply is no.",
"My sources say no.",
"Outlook not so good.",
"Outlook good.",
"Reply hazy, try again.",
"Signs point to yes.",
"Very doubtful.",
"Without a doubt.",
"Yes.",
"Yes – definitely.",
"You may rely on it."
]
#Prompt the user for their question.
response = random.choice(options)
# Answer question randomly with one of the options from your earlier list.
input("Ask the Magic 8 Ball something: ")
print(response)
main()
| true |
115595d836d0f9202a444db48c5abb71020eeb00 | xMijumaru/GaddisPythonPractice | /Chapter 2/Gaddis_Chap2_ProgEX10_Ingredient/main.py | 486 | 4.125 | 4 | #this program will run the ingredients needed to produce cookies
amount=int(input('How many cookie(s) do you wish to make: '))
#this is the amount that makes 48 cookies
sugar=1.5/48.0
butter=1/48.0
flour=2.75/48.0
#reminder that // is int division and / is float division
print("Amount to make ",amount, " cookie(s)")
print("Sugar: ", format(sugar*amount, ',.02f'), "cups")
print("Butter: ", format(butter*amount,',.02f'), "cups")
print("Flour: ", format(flour*amount,',.02f'), "cups") | true |
fe5d02ecda6c3b78d3e3b418211ad508a60c000f | yms000/python_primer | /rm_whitespace_for_file.py | 820 | 4.15625 | 4 | #! /usr/bin/python
# -*- coding: utf-8 -*-
# rm_whitespace_for_file.py
# author: robot527
# created at 2016-11-6
'''
Remove trailing whitespaces in each line for specified file.
'''
def rm_whitespace(file_name):
'''Remove trailing whitespaces for a text file.'''
try:
with open(file_name, 'r+') as code:
line_list = [item.rstrip() + '\n' for item in code]
code.seek(0)
code.truncate(0)
code.writelines(line_list)
print 'Removed trailing whitespaces for file: ' + file_name
except IOError as err:
print 'File error: ' + str(err)
if __name__ == '__main__':
import sys
argc = len(sys.argv)
if argc < 2:
print 'Usage: python rm_whitespace_for_file.py /path/to/file'
else:
rm_whitespace(sys.argv[1])
| true |
02c841ac52ca32b61b87f71502f072c047ac38ea | BrothSrinivasan/leet_code | /cycle graph/solution.py | 1,534 | 4.1875 | 4 | # Author: Barath Srinivasan
# Given an unweighted undirected graph, return true if it is a Cycle graph;
# else return false. A Cycle Graph is one that contains a single cycle and every
# node belongs to that cycle (it looks like a circle).
# Notes:
# a cycle is only defined on 3 or more nodes.
# adj_matrix is an n-by-n list of lists representing the adjacency matrix of a simple graph. adj_matrix[i][i] = 0 for all i from 0 to n-1 (inclusive).
# Explanation of Code
"""
My reasoning was that for the graph to be a single cycle is if each node has exactly two paths. For example consider this valid example: 4<-1<->2<->3<->4->1. Every node is surrounded by a path foward and a path backward. Using this reasoning, I just run through the matrix to see if row has two paths.
"""
# Analysis of Code
"""
The time complexity of this code is O(E^2). This is because a single row has enough elements to correspond to every node in the graph, thus V equals E. Since we we do a count function for each row, the complexity is O(E^2). Space complexity is O(1) since I don't use any additional space for my computation.
"""
def is_cycle(adj_matrix):
if adj_matrix is not None: # checks if we can param is null
for c in range(0, len(adj_matrix)): # goes through all rows of the matrix
if adj_matrix[c].count(1) is not 2: # checks the number of connections in the node
return False; # is there is not exactly 2 connections, then the cycle doesn't exist
return True; # exactly 2 connections for all Nodes
return False; # param was null
| true |
fa5ee033646904e358b3f504d3597831a5701336 | Tajveez/intermediate-python | /lists.py | 629 | 4.125 | 4 | myList = ["banana", "apple", "cherry"]
print(myList)
myList2 = list()
print(myList2)
myList3 = [5, True, "apple", "apple"]
print(myList3)
print(myList[2])
print(myList[-1])
for x in myList:
print(x)
if "banana" in myList:
print('Yes')
else:
print('No')
print(len(myList3))
myList.append("lemon")
myList.insert(1, "blueberry")
print(myList)
# removing all element
myList.pop()
myList.remove("cherry")
# Delete all item
myList.clear()
myList3.reverse()
myList.sort()
new_list = sorted(myList) #sorted method
myListNum = [0] * 5
print(myListNum)
myListNum2 = [1, 2, 3]
new_num_list = myListNum + myListNum2 | false |
a85e19425d0c1ca0e7d9907fb2d7421acdbfa71e | avneetkaur1103/workbook | /DataStructure/BinaryTree/tree_diameter.py | 1,342 | 4.1875 | 4 | """ Print the longest leaf to leaf path in a Binary tree. """
class Node:
def __init__(self, value):
self.key = value
self.left = self.right = None
def diameter_util(root):
if not root:
return 0, []
left_height, left_subtree_path = diameter_util(root.left)
right_height, right_subtree_path = diameter_util(root.right)
if left_height + right_height + 1 > diameter_util.max_width:
diameter_util.max_width = left_height + right_height + 1
diameter_util.final_path.clear()
diameter_util.final_path += left_subtree_path + [root.key] + list(reversed(right_subtree_path))
if left_height > right_height:
return 1+left_height, left_subtree_path + [root.key]
return 1+right_height, right_subtree_path + [root.key]
diameter_util.max_width = float('-inf')
diameter_util.final_path = list()
def diameter(root):
diameter_util(root)
print('Ans = ', diameter_util.final_path)
if __name__ == '__main__':
# Enter the binary tree ...
# 1
# / \
# 2 3
# / \
# 4 5
# \ / \
# 8 6 7
# /
# 9
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.left.right.left = Node(6)
root.left.right.right = Node(7)
root.left.left.right = Node(8)
root.left.left.right.left = Node(9)
diameter(root) | true |
3fd71c526ccbfcf4ab615d8615084924088c0c3e | natanisaitejasswini/Python-Programs | /OOP/underscoreduplicte.py | 997 | 4.5625 | 5 | """
map => Take a list and a function, and return the list you get by applying that function to every item in the list
filter => Take a list and return only the values when a given function is true
my_filter([1,2,3,4,5], lambda x: x%2==0) => [2,4]
reject => The exact opposite of filter
my_reject([1,2,3,4,5], lambda x: x%2==0) => [1,3,5]
find => Return the first value in the list where the function is true
my_find([1,2,3,4,5], lambda x: x%2==0) => 2
all => Return True if the function is true for every item in the list, false otherwise
my_all([1,2,3,4,5], lambda x: x%2==0) => False
any => Return True if the function is true for any item in the list, false otherwise
my_any([1,2,3,4,5], lambda x: x%2==0) => True
"""
def my_map(lst, func):
output = []
for item in lst:
# print("for loop")
output.append(func(item))
# print(item)
# print(func(item))
return output
def square(num):
return num**2
a = [1,2,3,4,5]
print(my_map(a, square))
print(a)
Add Comment | true |
4107b89ed8a9320e97038724aff412aad3bc84e4 | LVargasE/CS21 | /Extracurricular/OOP-Door-Example.py | 1,044 | 4.15625 | 4 | """ OOP example using doors
"""
class Door:
color = 'brown'
def __init__(self, number, status):
self.number = number
self.status = status
@classmethod
def knock(cls):
print("Knock!")
@classmethod
def paint(cls, color):
cls.color = color
def open(self):
self.status = 'open'
def close(self):
self.status = 'closed'
class SecurityDoor(Door):
locked = True
def __init__(self, number, status):
self.door = Door(number, status)
def open(self):
if self.locked:
return
self.door.open()
def __getattr__(self, attr):
return getattr(self.door, attr)
class ComposedDoor:
def __init__(self, number, status):
self.door = Door(number, status)
def __getattr__(self, attr):
return getattr(self.door, attr)
class Room:
def __init__(self, door):
self.door = door
def open(self):
self.door.open()
| false |
75b5e1c59aa553349af121218c3281442fa60a78 | shivakarthikd/practise-python | /single_linked_list.py | 1,615 | 4.28125 | 4 | class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class LinkedList:
# Function to initialize head
def __init__(self):
self.head = None
# This function prints contents of linked list
# starting from head
def printList(self):
temp = self.head
while (temp):
print(temp.val)
temp = temp.next
class Solution:
def mergeKLists(self, lists:ListNode) :
l=list()
for i in lists:
temp=i.head
while (temp):
l.append(ListNode(temp.val))
temp=temp.next
for i in l.sort():
print(i.val)
# Code execution starts here
if __name__ == '__main__':
# Start with the empty list
ls=list()
llist = LinkedList()
llist.head = ListNode(1)
second = ListNode(2)
third = ListNode(3)
llist.head.next = second # Link first node with second
second.next = third; # Link second node with the third node
ls.append(llist)
llist1 = LinkedList()
llist1.head = ListNode(1)
second1 = ListNode(2)
third1 = ListNode(3)
llist1.head.next = second1 ; # Link first node with second
second1.next = third1
ls.append(llist1)
llist2 = LinkedList()
llist2.head = ListNode(1)
second2 = ListNode(2)
third2 = ListNode(3)
llist2.head.next = second2 # Link first node with second
second2.next = third2
ls.append(llist2)
#llist.printList()
#llist1.printList()
#llist2.printList()
sol=Solution()
sol.mergeKLists(ls) | true |
bddacb0f55e6d1b2332f6a48471ad8b0b31c51b6 | StRobertCHSCS/ics2o1-201819-igothackked | /life_hacks/life_hack_practice_1.3.py | 527 | 4.34375 | 4 |
'''
-------------------------------------------------------------------------------
Name: minutes_to_days.py
Purpose: converts minutes into days, hours, minutes
Author: Cho.J
Created: date in 22/02/2019
------------------------------------------------------------------------------
'''
#PRINT
minute= int(input("How many minutes: "))
#COMPUTE
days= minute//1440
hours_left= minute%1440
hours= hours_left//60
days_left= minute%60
#PRINT
print(minute, "minutes=", days, "days, ", hours, "hours, ", days_left, "minutes")
| true |
f680f5deb0653aa00c080a84d992f8bacb58d6d0 | StRobertCHSCS/ics2o1-201819-igothackked | /unit_2/2_6_4_while_loops.py | 276 | 4.34375 | 4 |
numerator = int(input("Enter a numerator: "))
denominator = int(input("Enter denominator: "))
if numerator // denominator :
print("Divides evenly!")
else:
print("Doesn't divide evenly.")
while denominator == 0:
denominator = int(input("Enter denominator: "))
| true |
7d05afce4800e027970c6523cb3f83abf677a020 | mihaivalentistoica/Python-Fundamentals | /python-fundamentals-master/09-flow-control/while-loop-exercise.py | 1,012 | 4.3125 | 4 | user_input = ""
# b. If the input is equal to “exit”, program terminates printing out provided input and “Done.”.
while user_input != "exit":
# a. Asks user for an input in a loop and prints it out.
user_input = input("Provide input: ")
# c. If the input is equal to “exit-no-print”, program terminates without printing out anything.
if user_input == "exit-no-print":
break
# d. If the input is equal to “no-print”, program moves to next loop iteration without printing anything.
if user_input == "no-print":
continue
# a. Asks user for an input in a loop and prints it out.
# b. If the input is equal to “exit”, program terminates printing out provided input and “Done.”.
# e. If the input is different than “exit”, “exit-no-print” and “no-print”, program repeats.
print(user_input)
else:
# b. If the input is equal to “exit”, program terminates printing out provided input and “Done.”.
print("Done.")
| true |
5f37a37f5b8d2ac55c0bdc1b6be249058c8b411e | ShivangiNigam123/Python-Programs | /regexsum.py | 381 | 4.1875 | 4 | #read through and parse a file with text and numbers. You will extract all the numbers in the file and compute the sum of the numbers.
import re
name = input ("enter file :")
sum = 0
fhandle = open(name)
numlist = list()
for line in fhandle:
line = line.rstrip()
numbers = re.findall('[0-9]+',line)
for number in numbers :
sum = sum + int(number)
print(sum)
| true |
5396067aa7a5580b864123469f232d1c7fce7369 | danbeyer1337/Python-Drill--item-36 | /item36Drill.py | 1,963 | 4.125 | 4 | #Assign an integer to a variable
number = 4
#Assign a string to a variable
string = 'This is a string'
#Assign a float to a variable
x = float (25.0/6.0 )
#Use the print function and .format() notation to print out the variable you assigned
print 'The variables are {0}, {1}, {2}'.format(number, string, x)
#Use each of these operators +, - , * , / , +=, = , %
print 4+2, 3-2, 5*5, 36/6, 9%2
y = 3
y += 2
print y
#Use of logical operators: and, or, not
if 3 < 4 and 5 > 4:
print('condition met')
if 10 < 11 or 15 < 223:
print('condition met')
#Use of conditional statements: if, elif, else
x = 10
if x == 10:
print'x = 10'
elif x == 9:
print'x = 9'
else:
print 'x does not equal 9 or 10'
#Use of a while loop
counter = 0
while counter < 5:
print counter
counter += 1
#Use of a for loop
for counter in range(0,5):
print counter
#Create a list and iterate through that list using a for loop to print each item out on a new line
favorite_games_list = ['Overwatch',
'PUBG',
'Hearthstone',
'Destiny']
for game in favorite_games_list:
print game
#11. Create a tuple and iterate through it using a for loop to print each item out on a new line
candy = ('apple', 'orange', 'banana', 'chocolate')
print ('\n Flavors of candy that we have', candy)
print ('Your items: ')
for item in candy:
print (item)
#12. Define a function that returns a string variable
#Function that determines if a girl is too young for you or not
def allowed_dating_age(my_age):
girls_age = my_age/2 +7
return girls_age
dans_limit = allowed_dating_age(24)
carls_limit = allowed_dating_age(27)
glens_limit = allowed_dating_age(40)
print('Dan can date girls', dans_limit, 'or older')
print('Carl can date girls', carls_limit, 'or older')
print('Glen can date girls', glens_limit, 'or older')
| true |
3506f85dcd5b448b4f0c009e1fa9a239dacc45ba | IamBikramPurkait/100DaysOfAlgo | /Day 3/Merge_Meetings.py | 1,451 | 4.25 | 4 | '''Write a function merge_ranges() that takes a list of multiple meeting time ranges and returns a list of condensed
ranges.Meeting is represented as a list having tuples in form of (start time , end time)'''
# Time complexity is O(nlogn)
def merge_meetings_time(meetinglist):
# Sort the meetings by start time
meetings_sorted = sorted(meetinglist)
# Initialize with the first element of meetings_sorted
merged_meetings = [meetings_sorted[0]]
for current_meeting_start , current_meeting_end in meetings_sorted[1:]:
last_merged_meeting_start , last_merged_meeting_end = merged_meetings[-1] #Last merged will always be the last item in list
# First case: Where the two meetings overlap
if current_meeting_start <= last_merged_meeting_end:
print('Merger done between', (current_meeting_start , current_meeting_end) ,'and' ,(last_merged_meeting_start , last_merged_meeting_end))
merged_meetings[-1] = (last_merged_meeting_start , max(current_meeting_end ,last_merged_meeting_end))
else: # If they don't overlap
merged_meetings.append((current_meeting_start, current_meeting_end))
return merged_meetings
sample_input = [(0, 1), (3, 5), (4, 8), (10, 12), (9, 10)]
print('Given schedule is :', sample_input)
print('After compaction the schedule is : ', end = '')
print(merge_meetings_time(sample_input)) | true |
41e0d6d733fba92c65018cb3df37ed00baf1eff3 | IamBikramPurkait/100DaysOfAlgo | /Day 4/FirstComeFirstServe.py | 1,258 | 4.21875 | 4 | # Recursive Approach
# Problem statement : Write a function to check if a restaurant serves first come , first serve basis
# Assumptions: 1)There are three lists dine_in_orders , take_out_orders , served_orders
# 2)The orders number will not be ordered and are randomly assigned
# 3)They are given as lists
''' Input format : Three lists
Output Format: True/False
Edge cases : 1) When one or both the lists are empty
2) If few orders were take away and not marked in the being served list'''
def is_first_come_first_serve(take_out_orders , dine_in_orders , served_orders):
if len(served_orders) == 0:
return True
if len(take_out_orders) and take_out_orders[0] == served_orders[0]:
return is_first_come_first_serve(take_out_orders[1:],dine_in_orders,served_orders[1:])
elif len(dine_in_orders) and dine_in_orders[0] == served_orders[0]:
return is_first_come_first_serve(take_out_orders ,dine_in_orders[1:],served_orders[1:])
else:
return False
take_out_sample = [17 ,8 ,24]
dine_in_sample = [12 ,19 ,2]
served_sample = [17 ,8 ,12 ,19 ,24 ,2]
print(is_first_come_first_serve(take_out_sample , dine_in_sample ,served_sample))
| true |
1e4d7b0f98b1d63b4bf82cb8f1b353176cc71c42 | IamBikramPurkait/100DaysOfAlgo | /Day1/Tower of Hanoi.py | 1,367 | 4.25 | 4 | ''' Problem Tower of Hanoi:
About: Tower of Hanoi is a mathematical puzzle where we have three rods and n disks. The objective of the puzzle is to move the entire stack to another rod, obeying the following simple rules:
1) Only one disk can be moved at a time.
2) Each move consists of taking the upper disk from one of the stacks and placing it on top of another stack i.e. a disk can only be moved if it is the uppermost disk on a stack.
3) No disk may be placed on top of a smaller disk
4) That is we cant have a bigger disk over smaller disk'''
'''Sources: https://www.geeksforgeeks.org/c-program-for-tower-of-hanoi/
Sources: https://www.khanacademy.org/computing/computer-science/algorithms/towers-of-hanoi/e/move-three-disks-in-towers-of-hanoi'''
# Khan Academy link provides excellent visualization of problem
# Used a recursive approach
def tower_of_hanoi(from_disk , to_disk , aux_disk , n):
if(n==1) : #Base case
print('Move disk 1 from',from_disk,'To',to_disk)
return
tower_of_hanoi(from_disk , aux_disk ,to_disk , n-1) #Recursive calling with n-1 disks
print('Move disk',n,'from',from_disk ,'To',to_disk)
tower_of_hanoi(aux_disk ,to_disk , from_disk, n-1)
# Sample inputs
print('Enter number of disks :',end='')
disks = input() # Sample input
tower_of_hanoi('A','B','C',int(disks)) #Passing value to the function
| true |
c62c719c3585ce32b829726429725e03a650c4a5 | G1998P/SEII-GuilhermePeresSilva | /Semana02/prog018.py | 617 | 4.15625 | 4 | '''
ordenacao de lista
'''
# metodos sort e sorted
#sorted cria uma nova lista
#sort organiza a propria lista
l = [1,4,3,2,10,65,3]
l2 = sorted(l)
print(l2)
print(l)#l nao for ordenada
# l.sort()
# print(l)# l foi ordenada
t = tuple(l)
# tuiples nao possui o metodo sort
print(sorted(t))
# pode ser passado uma funcao para atribuir um valor ordenavel a um objeto, util para organizar dicionarios
d = []
for i in range(len(l)):
d.append({'name':f'pessoa{i}','age':l[i]})
# dd = sorted(d)
dd2 = sorted(d,key=lambda a:a['age'])
for pessoa in dd2:
nome , idade = pessoa.values()
print(nome,idade)
| false |
f93e10068776b3ab4e298bf55eae39598a38ec5d | indexcardpills/python-labs | /02_basic_datatypes/2_strings/02_09_vowel.py | 444 | 4.25 | 4 | '''
Write a script that prints the total number of vowels that are used in a user-inputted string.
CHALLENGE: Can you change the script so that it counts the occurrence of each individual vowel
in the string and print a count for each of them?
'''
#string=input("write a sentence: ")
string="the dog ate the cat"
a=string.count("a")
e=string.count("e")
i=string.count("i")
o=string.count("o")
u=string.count("u")
print(a+e+i+o+u)
| true |
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