blob_id string | repo_name string | path string | length_bytes int64 | score float64 | int_score int64 | text string | is_english bool |
|---|---|---|---|---|---|---|---|
3c47d535501c877aae1e7c115534d309f9cfac32 | indexcardpills/python-labs | /15_generators/15_01_generators.py | 237 | 4.3125 | 4 | '''
Demonstrate how to create a generator object. Print the object to the console to see what you get.
Then iterate over the generator object and print out each item.
'''
daniel = (x+'f' for x in "hello")
for x in daniel:
print(x)
| true |
eb2128c3f33e18a27de998a58e00a0c1056d28d3 | indexcardpills/python-labs | /04_conditionals_loops/04_07_search.py | 575 | 4.15625 | 4 | '''
Receive a number between 0 and 1,000,000,000 from the user.
Use while loop to find the number - when the number is found exit the loop and print the number to the console.
'''
while True:
number = int(input("Enter a number between 0 and 1,000,000,000: "))
x = 7
if number < x:
print("no, higher")
if number > x:
print("no, lower")
if number == x:
print("correct, the number is 7")
break
#Is this what I was supposed to do? If not, I don't understand what is meant by "finding the number", if
#I'm inputting it myself
| true |
a4508650196850cea3aecd960dd0e9a06c98cbe3 | indexcardpills/python-labs | /10_testing/10_01_unittest.py | 690 | 4.34375 | 4 | '''
Demonstrate your knowledge of unittest by first creating a function with input parameters and a return value.
Once you have a function, write at least two tests for the function that use various assertions. The
test should pass.
Also include a test that does not pass.
'''
import unittest
def multiply(x, y):
return x*y-x
print(multiply(2, 4))
# import unittest
# x=2
# y=3
class Testtimes(unittest.TestCase):
def test_times(self):
self.assertEqual((multiply(2, 3)), (6))
self.assertEqual(multiply(3, 10), 30)
def test_times_two(self):
self.assertIs(multiply(5, 9), 45, "this failed")
#
if __name__ == '__main__':
unittest.main()
| true |
2ca18b8597a9d80d73fd7d1da52f7691664b5bf4 | indexcardpills/python-labs | /04_conditionals_loops/04_05_sum.py | 791 | 4.3125 | 4 | '''
Take two numbers from the user, one representing the start and one the end of a sequence.
Using a loop, sum all numbers from the first number through to the second number.
For example, if a user enters 1 and 100, the sequence would be all integer numbers from 1 to 100.
The output of your calculation should therefore look like this:
The sum is: 5050
'''
sum_number=0
first_number = 5
second_number = 9
range_numbers = range(first_number, second_number+1)
for x in range_numbers:
print(sum_number+first_number)
if sum_number == second_number+1:
break
# I think I'm getting closer? But now I don't know what else to do. I initialized a variable outside
# the for loop, but beyond that, I don't know what to do with it exactly. With this I just get 5 printed
# 5 times.
| true |
0ef559fbb64de7780dda09df151d2e4482bc1410 | tonynguyen99/python-stuff | /guess number game/main.py | 945 | 4.25 | 4 | import random
def guess(n):
random_number = random.randint(1, n)
guess = 0
while guess != random_number:
guess = int(input(f'Guess a number between 1 and {n}: '))
if guess > random_number:
print('Too high!')
elif guess < random_number:
print('Too low!')
print('You got it!')
def computer_guess(n):
lowestNumber = 1
highestNumber = n
feedback = ''
while feedback != 'c':
if lowestNumber != highestNumber:
guess = random.randint(lowestNumber, highestNumber)
else:
guess = lowestNumber
feedback = input(f'Is the computers guess {guess} correct (C), too high (H) or too low (L)? ')
if feedback.lower() == 'h':
highestNumber = guess - 1
elif feedback.lower() == 'l':
lowestNumber = guess + 1
print(f'The computer guessed your number {guess} correctly!')
computer_guess(10) | true |
b51af61c9960b18b8829708f6a0d3a8f18a5fd2f | sathvikg/if-else-statement-game | /GameWithBasics.py | 1,463 | 4.1875 | 4 | print("Welcome to your Game")
name = input("What is your name? ")
print("hi ",name)
age = int(input("What is your age? "))
#print(age)
#print(name,"you are good to go. As you are",age,"years old")
health = 10
print("you are starting with ",health,"health")
if age > 18:
print("you can continue the game.")
want_to_play = input("Dow you want to play?")
if want_to_play.lower() == 'yes':
print("lets play ")
choice = input("you wanna go left or right? ")
if choice.lower() == "left":
ans = input("you came the right way. Now you follow and reach the lake. Do you wanna swim or go around? ")
if ans.lower() == "around":
print("you went around and reached the other side of the lake.")
elif ans.lower() == "across":
print("you went around and lost 5 health.")
health-=5
ans2 = input("you notice a house and river. where do you go? ")
if ans2.lower() == 'house':
print("you went to the house and you are safe.")
else:
print('you couldnt swim now.')
health-=5
if health == 0:
print("you lose")
else:
print("you went in the wrong way")
else:
print("see you then")
else:
print("you cannot continue the game")
#print("The game begins") | true |
f642a041bcda31d628b16181016fae4cbb9c128d | nikonoff16/Simple_Number | /simple.py | 1,576 | 4.15625 | 4 | #Создаем переменную
quest_number = int(input("Введите число "))
# концепция проекта такая: если число простое, то при делению по модулю всегда будет остаток. Если посчитать эти остатки
# и сравнить их с самим числом за вычетом двух из него, то можно понять, простое оно или нет.
cycle_th = quest_number - 1
counter = 0
control_number = quest_number - 2
#Проводим базовый отвев числа
if quest_number == 2:
print('Простое число')
elif quest_number % 2 == 0:
print("Дурачок чтоли? посмотри на последнюю цифру в том, что ты нарисовал здесь!!!")
else:
#Проводим дальнейшую провеку на простоту
while cycle_th >= 2:
module_num = quest_number % cycle_th
cycle_th -= 1
if module_num > 0:
counter += 1
#
if control_number - counter == 0:
print("Это элементарно, Ватсон - число простое")
else:
print("Число не так просто, как кажется")
#Проверял таким образом то, какие результаты у меня в переменных хранились.
print("Проверка: Счетчик равен " + str(counter) + " итератор - " + str(cycle_th) + " контрольная сумма равна " + str(control_number)) | false |
d3246fddc314795c42ec10a19be60f2c0e026675 | GaborVarga/Exercises_in_Python | /46_exercises/8.py | 1,124 | 4.34375 | 4 | #!/usr/bin/env python
#######################
# Author: Gabor Varga #
#######################
# Exercise description :
# Define a function is_palindrome() that recognizes palindromes
# (i.e. words that look the same written backwards).
# For example, is_palindrome("radar") should return True.
#
# http://www.ling.gu.se/~lager/python_exercises.html
###########
# imports #
###########
from pip._vendor.distlib.compat import raw_input
from builtins import int
###############
# Function(s) #
###############
def ispalindrome(inputstring) :
if(len(inputstring)%2==0):
return inputstring[:int(len(inputstring)/2)] == \
reverse(inputstring[int(len(inputstring)/2):])
else :
return inputstring[:int((len(inputstring)-1)/2)]== \
reverse(inputstring[int(((int(len(inputstring))-1)/2)*-1):])
def reverse(inputstring):
result = ""
for karakter in inputstring:
result = karakter + result
return result
##########
# Script #
##########
inputstring = raw_input("Please enter a string: \n")
print("Is it a plaindrome? ",ispalindrome(inputstring)) | true |
e7e4117d2f583ceedce8c4cdba546523d9532ac8 | UCdrdlee/ScientificComputing_HW0 | /fibonacci.py | 2,824 | 4.5 | 4 | """
fibonacci
functions to compute fibonacci numbers
Complete problems 2 and 3 in this file.
"""
import time # to compute runtimes
from tqdm import tqdm # progress bar
# Question 2
def fibonacci_recursive(n):
if n == 0:
return 0
if n == 1:
return 1
else:
return fibonacci_recursive(n-1) + fibonacci_recursive(n-2)
pass
# Question 2
def fibonacci_iter(n):
if n == 0:
return 0
a=0
b=1
while n-1>0:
a,b = b, a+b
n-=1
return b
pass
print("The first 30 fibonacci numbers using fibonacci_recursive is: ")
for i in range(0,30):
print(fibonacci_recursive(i))
print("The first 30 fibonacci numbers using fibonacci_iter is: ")
for i in range(0,30):
print(fibonacci_iter(i))
import numpy as np
# Question 3
def fibonacci_power(n):
# The matrix A and vector x_1 are given and can be defined in NumPy arrays.
A = np.array([[1,1],[1,0]])
x_1 = np.array([1,0])
# We define what F_0 should be, which is 0.
if n == 0:
return 0
else:
def power(mat, i):
def isodd(i):
"""
returns True if n is odd
"""
return i & 0x1 == 1
if i == 0:
# when n=1, the output should be just x_1
return np.identity(2)
if isodd(i):
return power(mat@mat, i//2)@mat
else:
return power(mat@mat, i//2)
x_n = power(A,n-1)@x_1
# we need just the first element of x_n
return int(x_n[0])
for i in range(0,30):
print(fibonacci_power(i))
if __name__ == '__main__':
"""
this section of the code only executes when
this file is run as a script.
"""
def get_runtimes(ns, f):
"""
get runtimes for fibonacci(n)
e.g.
trecursive = get_runtimes(range(30), fibonacci_recusive)
will get the time to compute each fibonacci number up to 29
using fibonacci_recursive
"""
ts = []
for n in tqdm(ns):
t0 = time.time()
fn = f(n)
t1 = time.time()
ts.append(t1 - t0)
return ts
nrecursive = range(35)
trecursive = get_runtimes(nrecursive, fibonacci_recursive)
niter = range(10000)
titer = get_runtimes(niter, fibonacci_iter)
npower = range(10000)
tpower = get_runtimes(npower, fibonacci_power)
## write your code for problem 4 below...
import matplotlib.pyplot as plt
plt.loglog(nrecursive,trecursive, label=f"Recursive")
plt.loglog(niter,titer, label=f"Iteration")
plt.loglog(npower,tpower, label=f"Power")
plt.legend()
plt.xlabel("n")
plt.ylabel("run time")
plt.title("Fibonacci Algorithm Run Times (log-log)")
plt.show()
plt.savefig('Fibonacci_runtime.png')
| true |
837b4fe80020202ed9b37e0ac5dc0f868ec0aa8f | Yagomfh/holbertonschool-higher_level_programming | /0x07-python-test_driven_development/0-add_integer.py | 713 | 4.34375 | 4 | #!/usr/bin/python3
"""Module that that adds 2 integers.
Raises:
TypeError: if a or b are not int or floats
"""
def add_integer(a, b=98):
"""Function that adds 2 integers
Args:
a (int/float): first digit
b (int/float): second digit
Returns:
Int sum of both digits
"""
if type(a) != int and type(a) != float:
raise TypeError('a must be an integer')
if type(b) != int and type(b) != float:
raise TypeError('b must be an integer')
try:
a = int(a)
except Exception:
raise TypeError('a must be an integer')
try:
b = int(b)
except Exception:
raise TypeError('b must be an integer')
return a + b
| true |
c0a7efc9d8720a168c1677b413d2a86f8a494fe6 | kadahlin/GraphAlgorithms | /disjoint_set.py | 1,877 | 4.15625 | 4 | #Kyle Dahlin
#A disjoint set data structure. Support finding the set of a value, testing if
#values are of hte same set, merging sets, and how may sets are in the entire
#structure.
class Disjoint:
def __init__(self, value):
self.sets = []
self.create_set(value)
def create_set(self, value):
"""
Add a new grouping (set) that only contains this value
"""
self.sets.append(set(value))
def merge_sets(self, value1, value2):
"""
Merge the sets containing the two values if they are not already in the same set
"""
if not self.test_same_set(value1, value2):
set1 = self.find_set(value1)
set2 = self.find_set(value2)
new_set = set1.union(set2);
self.sets.remove(set2)
self.sets.remove(set1)
self.sets.append(new_set)
def find_set(self, value):
"""
Return the set that value belongs to
"""
for s in self.sets:
if value in s:
return s
def test_same_set(self, value1, value2):
"""
Return a boolean stating whether or not the two values are in the same set
"""
return self.find_set(value1) == self.find_set(value2)
def size(self):
"""
Return how many sets are in the structure
"""
return len(self.sets)
if __name__ == '__main__':
#Sanity test for myself when making this"""
d = Disjoint('A')
for char in ['B', 'C', 'D', 'E']:
d.create_set(char)
assert not d.test_same_set('A', 'B')
assert not d.test_same_set('C', 'D')
d.merge_sets('A', 'D')
d.merge_sets('B', 'C')
assert d.test_same_set('A', 'D')
assert d.test_same_set('C', 'B')
assert not d.test_same_set('A', 'C')
d.merge_sets('B', 'D')
assert d.test_same_set('A', 'C')
| true |
fb8318baba89169105e27fdc76b00e80888fb222 | Mone12/tictactoe-python | /.vscode/tictactoe.py | 1,176 | 4.1875 | 4 | import random
## Need to establish which player goes first through randomization
player = input("Please enter your name:")
cpu = "CPU"
p_order = [player,cpu]
if player:
random.shuffle(p_order)
if p_order[0] == player:
print(f"{player} you are Player 1. You go first!")
print(f"{cpu} is Player 2!")
elif p_order[0] == cpu:
print(f"{cpu} is Player 1. They go first!")
print(f"{player} your Player 2!")
## Need to draw board
a = [["-","-","-"],["-","-","-"],["-","-","-"]]
for i in range(len(a)):
for j in range(len(a[i])):
print(a[i][j], end=" ")
print()
## Need to assign X or O to each player
## Randomize who gets X or O
## Need to start game once Players and turns are established
## While loop in game
## Need to assign x and o to area on board player chooses
## CPU chooses randomly
## Game stops when a player gets 3 in a row
## Need to display winner when game is over
## Give score
## Be able to play again
## WHile Loop false
## Game ends when the first player reaches 3 points
## Increment points and display them
## Be able to start whole game over
## Second while loop false | true |
7182d6fa3958283a46dfdb5b9f4b6c5eea17055d | kontai/python | /面向對象/運算符重載/classＭethod.py | 970 | 4.25 | 4 | # Copyright (c) 2019.
# classＭethod.py
#
class FirstClass:
def setdata(self, value):
self.data = value
def display(self):
print(self.data)
class SecondClass(FirstClass):
def display(self):
print("Current data is %s" % self.data)
class ThirdClass(SecondClass):
def __init__(self, value):
self.data = value
# 重載'+'
def __add__(self, other):
return ThirdClass(self.data + other)
# 重載'str'
def __str__(self):
return '[ThirdClass: %s]' % self.data
# 重載'*'
def mul(self, other):
self.data *= other
if __name__ == '__main__':
x = SecondClass()
x.setdata("2nd class")
x.display() # 覆蓋FirstClass display方法
a = ThirdClass('abc')
a.display() # 繼承SecondClass display菲奧法
print(a) # __str__被調用
b = a + 'xyz' # __add__被調用
b.display()
print(b) # __str__被調用
a.mul(3)
print(a)
| false |
d867547f67b3c2697fffaabdccd39b4571ee473c | smukh93/Python_MITx | /iter_pow.py | 606 | 4.125 | 4 | def iterPower(base, exp):
'''
base: int or float.
exp: int >= 0
returns: int or float, base^exp
'''
# Your code here
prod = 1
if exp == 0:
return 1
else:
for x in range(exp):
prod *=base
return prod
def recurPower(base, exp):
'''
base: int or float.
exp: int >= 0
returns: int or float, base^exp
'''
# Your code here
if exp == 0:
return 1
else:
return base * recurPower(base,exp-1)
exp = int(input('Enter exp: '))
base = int(input('Enter base: '))
print(iterPower(base,exp)) | true |
a3a152ffc0a8fb2fe28def7b018ef54f4a65506d | gourav287/Codes | /Codechef-and-Hackerrank/NoOfStepsToReachAGivenNumber.py | 1,304 | 4.15625 | 4 | # -*- coding: utf-8 -*-
"""
Question Link:
https://practice.geeksforgeeks.org/problems/minimum-number-of-steps-to-reach-a-given-number5234/1
Given an infinite number line. You start at 0 and can go either to the left or to the right.
The condition is that in the ith move, youmust take i steps.
Given a destination D , find the minimum number of steps required to reach that destination.
"""
# Function to calculate minimum no of steps
def minSteps(self, D):
# Initializing source to track current place
source = 0
# i will track the length of next move
i = 1
# Keep going forward as it will minimize steps
while source < D:
source += i
i += 1
# Distance remaining to reach
diff = source - D
# If distance remaining is odd, we need to make it even
while diff % 2 != 0:
source += i
diff = source - D
i += 1
# If distance is even, it means altering a few steps taken will
# lead us to destination, hence this is the answer
return i - 1
# Driver Code
if __name__ == '__main__':
# No of test cases
for _ in range(int(input())):
# Destination
D = int(input())
#Calling the required function
print(minSteps(D)) | true |
089e2881ecc68c06e9fd9fe958d4ccd1414d55d0 | gourav287/Codes | /BinaryTree/TreeCreation/CreateTreeByInorderAndPreorder.py | 2,259 | 4.28125 | 4 | """
Implementation of a python program to create a binary tree
when the inOrder and PreOrder traversals of the tree are given.
"""
# Class to create a tree node
class Node:
def __init__(self, data):
# Contains data and link for both child nodes
self.data = data
self.left = None
self.right = None
# Function to print the inorder traversal of a tree, once created
def printInorder(start):
# if the tree exists
if start:
# Traverse left subtree
printInorder(start.left)
# print value
print(start.data, end = " ")
# Traverse right subtree
printInorder(start.right)
"""
Creation of the Tree from inOrder and preOrder traversals
"""
def createTree(inOrder, preOrder, left, right):
# A global variable to keep track of current preOrder element
global cur_ind
# If no element in the desired part of list
if left > right:
return None
# Create a tree node from element of preOrder traversal
# Increase the global variable's index value
tmp = Node(preOrder[cur_ind])
cur_ind += 1
# If left == right then the node is leaf
# Else work on to create child nodes as follows
if left != right:
# Getting the index in inOrder
# This tells us about left and right subtree
# of the current node
ind = inOrder.index(tmp.data)
# Recursively run the code for left subtree
tmp.left = createTree(inOrder, preOrder, left, ind - 1)
# Recursively run the code for right subtree
tmp.right = createTree(inOrder, preOrder, ind + 1, right)
# Returning the node to be added to the tree
return tmp
# The driver code
if __name__ == '__main__':
# inOrder traversal of the tree
inOrder = list(map(int, input().split())) #[1, 2, 3, 4, 5, 6, 7]
# preOrder traversal of the tree
preOrder = list(map(int, input().split())) #[4, 2, 1, 3, 6, 5, 7]
# Index to keep track of preOrder elements
cur_ind = 0
# Function to create the tree from inOrder and preOrder
# traversal given to us
start = createTree(inOrder, preOrder, 0, len(inOrder) - 1)
# printing the inOrder traversal of the tree created
printInorder(start)
| true |
263b8fdfdc3989979ec92e72110fd79be8ead6fd | gourav287/Codes | /Codechef-and-Hackerrank/Binary_to_decimal_recursive.py | 740 | 4.34375 | 4 | # -*- coding: utf-8 -*-
"""
Write a recursive code to convert binary string into decimal number
"""
# The working function
def bin2deci(binary, i = 0):
# Calculate length of the string
n = len(binary)
# If string has been traversed entirely, just return
if i == n - 1:
return int(binary[i])
# Left shifting i-th value to n-i-1 positions will actually multiply
# it with 2**(n-i-1) This will then be added to values at other i
# and final solution will be returned
return ((int(binary[i])<<(n-i-1)) + bin2deci(binary, i + 1))
# The driver code
if __name__ == "__main__":
# Input the string
binary = input()
# Print the output
print(bin2deci(binary)) | true |
ed43aa7eacb8f54d80102e3b05a2147451b636a8 | januarytw/Python_auto | /class_0605/0605作业.py | 1,702 | 4.15625 | 4 | # 1：创建一个名为 Restaurant 的类，其方法 init ()设置两个属性： restaurant_name 和 cooking_type。
# 创建一个名为 describe_restaurant()的方法和一个名为 open_restaurant()的方法，其中前者打印前述两项信息，
# 而后者打印一条消息， 指出餐馆正在营业。 根据这个类创建一个名为 restaurant 的实例，分别打印其两个属性，再调用前述两个方法。
class Restaurant():
def __init__(self,restaurant_name,cooking_type):
self.restaurant_name=restaurant_name
self.cooking_type=cooking_type
def describe_restaurant(self):
print("餐厅名字：%s 主营：%s"%(self.restaurant_name,self.cooking_type))
def open_restaurant(self):
print("餐厅正在营业！")
restaurant=Restaurant("蜀国演义","川菜")
restaurant.describe_restaurant()
restaurant.open_restaurant()
# 2:继承1 这个类，且添加函数：discount 打折扣用的 pay_money 支付餐费用 完成调用
class Restaurant_1(Restaurant):
def discount(self,pay_money):
print("您将享受8折优惠，优惠后价格：%s"%(int(pay_money)*0.8))
restaurant_1=Restaurant_1("海底捞","火锅")
restaurant_1.describe_restaurant()
restaurant_1.open_restaurant()
restaurant_1.discount(100)
# 3:超继承1这个类的open_restaurant方法，多加一个优惠信息宣传。
class Restaurant_2(Restaurant):
def open_restaurant(self):
super(Restaurant_2,self).open_restaurant()
print("餐厅正在营业，消费满100，可以享受8折优惠！")
restaurant_2=Restaurant_2("郭林家常菜","东北菜")
restaurant_2.describe_restaurant()
restaurant_2.open_restaurant() | false |
a056d2b012f5d2d3c8b0cb46cd26e5a960550970 | HANZ64/Codewars | /Python/8-kyu/07. Return Negative/index.py | 1,405 | 4.15625 | 4 | '''
Title:
Return Negative
Kata Link:
https://www.codewars.com/kata/return-negative
Instructions:
In this simple assignment you are given a number and have to make it negative. But maybe the number is already negative?
Example:
make_negative(1); # return -1
make_negative(-5); # return -5
make_negative(0); # return 0
Notes:
• The number can be negative already, in which case no change is required.
• Zero (0) is not checked for any specific sign. Negative zeros make no mathematical sense.
Problem:
def make_negative( number ):
# ...
'''
#=============================================#
# My solution:
def make_negative(number):
if (number < 0):
return number
return -number
#=============================================#
# 1. Alternative Solution:
def make_negative(number):
return -abs(number)
# 2. Alternative Solution:
def make_negative( number ):
return (-1) * abs(number)
# 3. Alternative Solution:
def make_negative( number ):
return -number if number>0 else number
# 4. Alternative Solution:
def make_negative(number):
if number >= 0:
return (0 - number)
else:
return number
# 5. Alternative Solution:
def make_negative( number ):
if number < 0:
return number
else:
return number * -1
#=============================================# | true |
a93e2fab8fe0972396945484bf75634184a2bf70 | albihasani94/PH526x | /week3/word_stats.py | 645 | 4.1875 | 4 | from week3.counting_words import count_words
from week3.read_book import read_book
def word_stats(word_counts):
"""Return number of unique words and their frequencies"""
num_unique = len(word_counts)
counts = word_counts.values()
return (num_unique, counts)
text = read_book("./resources/English/shakespeare/Romeo and Juliet.txt")
word_counts = count_words(text)
(num_unique, counts) = word_stats(word_counts)
print(num_unique, sum(counts))
text = read_book("./resources/German/shakespeare/Romeo und Julia.txt")
word_counts = count_words(text)
(num_unique, counts) = word_stats(word_counts)
# print(num_unique, sum(counts))
| true |
be8849979ce06c3cfa24cae9fb933e6673de5a3d | af94080/dailybyte | /next_greater.py | 1,066 | 4.21875 | 4 | """
This question is asked by Amazon. Given two arrays of numbers, where the first array is a subset of the second array, return an array containing all the next greater elements for each element in the first array, in the second array. If there is no greater element for any element, output -1 for that number.
Ex: Given the following arrays…
nums1 = [4,1,2], nums2 = [1,3,4,2], return [-1, 3, -1] because no element in nums2 is greater than 4, 3 is the first number in nums2 greater than 1, and no element in nums2 is greater than 2.
nums1 = [2,4], nums2 = [1,2,3,4], return [3, -1] because 3 is the first greater element that occurs in nums2 after 2 and no element is greater than 4.
"""
nums1 = [1,2,3,4]
nums2 = [2,4]
def next_greater_element(n1, n2):
if len(n1) < len(n2):
smaller, bigger = n1, n2
else:
smaller, bigger = n2, n1
out = []
for ele in smaller:
right_list = bigger[bigger.index(ele)+1:]
next_gt_ele = next((x for x in right_list if x > ele), -1)
out.append(next_gt_ele)
return out
print(next_greater_element(nums1, nums2))
| true |
3941ff68870c3b573885de1fb512c97953d29dcf | pjm8707/Python_Beginning | /py_basic_datatypes.py | 2,077 | 4.28125 | 4 | import sys
print("\npython data types -Numeric")
#create a variable with integer value
a=100
print("The type of variable having value", a, "is", type(a))
print("The maximum interger value", sys.maxsize)
print("The minimum interger value", -sys.maxsize-1)
#create a variable with float value
b=10.2345
print("The type of variable having value", b, "is", type(b))
#create a variable with complex value
c=100+3j
print("The type of variable having value", c, "is", type(c))
print("\npython data type - String")
d="string in a double quote"
e='string in a single quote'
print("The type of variable having value", d, "is", type(d))
print("The type of variable having value", e, "is", type(e))
#using ',' to concatenate the two or several strings
print(d, "concatenated with", e)
print(d+" concatenated with "+e)
print("\npython data typed - List")
#list of having only integer
f=[1,2,3,4,5]
print("list f is:",f)
#list of having only strings
g=["hey", "you", 1,2,3, "go"]
print("list g is:",g)
#index are 0 based. this will prnt a single character
print(g[1]) # this will print "you" in list g
print("\npython data type - Tuple")
#Tuple is another data type which is a sequence of data similar to list. But it is immutable.
#That means data in a tuple is write protected.
#Data in a tuple is wrriten using parenthesis and commas.
#tuple having only integer type of data
h=(1,2,3,4)
print("tuple h is:", h)
#tuple having multiple type of data
i=("hello", 1, 2, 3, "go")
print("tuple i is :", i)
#index of tuples are also 0 based
print("i[4] is ", i[4], "in tuple i")
print("\npython data type - Dictionary")
#Python Dictionary is an unordered sequence of data of key-value pair form.
#It is similar to the hash table type. Dictionaries are written within curly braces in the form key:vale.
#It is very useful to retrive data in an optimized way among large amount of data.
#a sample dictionary variable
j={1:"first name", 2:"second name", "age":3}
print(j[1])
print(j[2])
print(j["age"])
| true |
1f9bd3a166434e6f800e9f01938fd3bf615c5ec9 | inesjoly/toucan-data-sdk | /toucan_data_sdk/utils/postprocess/rename.py | 976 | 4.25 | 4 | def rename(df, values=None, columns=None, locale=None):
"""
Replaces data values and column names according to locale
Args:
df (pd.DataFrame): DataFrame to transform
values (dict):
- key (str): term to be replaced
- value (dict):
- key: locale
- value: term's translation
columns (dict):
- key (str): columns name to be replaced
- value (dict):
- key: locale
- value: column name's translation
locale (str): locale
"""
if values:
to_replace = list(values.keys())
value = [values[term][locale] for term in values]
df = df.replace(to_replace=to_replace, value=value)
if columns:
_keys = list(columns.keys())
_values = [column[locale] for column in columns.values()]
columns = dict(list(zip(_keys, _values)))
df = df.rename(columns=columns)
return df
| true |
6bee3da547307daf0e4245423a159b4196dcd025 | Vishal0442/Python-Excercises | /Guess_the_number.py | 480 | 4.21875 | 4 | #User is prompted to enter a guess. If the user guesses wrong then the prompt appears again until the guess is correct,
#on successful guess, user will get a "Well guessed!" message, and the program will exit
import random
while True:
a = random.randint(1,9)
b = int(input("Guess a number : "))
if a == b:
print ("Congrats!! Well Guessed...\n")
break
else:
print ("Oops...Wrong Guess!! Try again...\n")
| true |
3ecfd62848650e4bfe40cb5907d31eb9398e18c5 | nikhil-jayswal/6.001 | /psets/ps1b.py | 1,002 | 4.21875 | 4 | # Problem Set 1b
# Nikhil Jayswal
#
# Computing sum of logarithms of all primes from 2 to n
#
from math import * #import math to compute logarithms
n = int(raw_input('Enter a number (greater than 2): '))
start = 3 #the second prime; don't need this can do candidate = 3
log_sum = log(2) #sum of logarithms of primes, first prime is 2
candidate = start
while (candidate <= n): #upto n
divisor = 2
flag = 0
while (divisor < candidate):
if (candidate % divisor == 0): #if any number divides candidate, i.e.not prime
flag = 1
break
divisor = divisor + 1 #else, update divisor
if flag == 0: #if prime, update log_sum
log_sum += log(candidate)
candidate = candidate + 2 #update candidate
ratio = log_sum / n #log_sum should be float, hence division should be proper
print 'n = ', n
print 'Sum of logarithms of primes upto', n, 'is', log_sum
print 'Ratio of sum of logarithm of primes to n is', ratio
# Code works but results not verified
| true |
b6ccbbb9ab29a1cbe46ef64731dad597b94c337b | GaneshGoel/Basic-Python-programs | /Grades.py | 555 | 4.125 | 4 | #To print grades
x=int(input("Enter the marks of the student:"))
if(x<=100 or x>=0):
if(x>90 and x<=100):
print("O")
elif(x>80 and x<=90):
print("A+")
elif(x>70 and x<=80):
print("A")
elif(x>60 and x<=70):
print("B+")
elif(x>50 and x<=60):
print("B")
elif(x>40 and x<=50):
print("C")
elif(x>30 and x<=40):
print("D")
else:
print("Fail")
else:
print("Invalid input") | true |
ee5f0793f86f7396d38234c1ac621a387fa768f2 | kt00781/Grammarcite | /AddingWords.py | 826 | 4.1875 | 4 | from spellchecker import SpellChecker
spell = SpellChecker()
print("To exit, hit return without input!")
while True:
word = input("Input the word that you would like to add to the system: ")
if word == '':
break
word = word.lower()
if word in spell:
print ("Word ({}) already in Dictionary!".format(word))
else:
print ("Word ({}) is not currently in the Dictionary, would you like to add it?".format(word))
new_word = input("Input y if you would like to add this word other wise click return: ")
if word == '':
break
new_word = new_word.lower()
if new_word == 'y':
spell.word_frequency.add('{}'.format(word))
else:
print("Invalid input please click return to exit and rerun the program!")
break | true |
d03bc8035c16bdef2486f27e59d3e430b178790c | Leofariasrj25/simple-programming-problems | /elementary/python/sum_multiples(ex5).py | 258 | 4.34375 | 4 | print("We're going to print the sum of multiples of 3 and 5 for a provided n")
n = int(input("Inform n: "))
sum = 0
# range is 0 based so we add 1 to include n
for i in range(3, n + 1):
if i % 3 == 0 or i % 5 == 0:
sum += i
print(sum)
| true |
3042b228dba1572138e2216bcef1f3dac8f0368b | CTTruong/9-19Assignments | /Palindrome.py | 455 | 4.1875 | 4 | def process_text(text):
text = text.lower()
forbidden = ("!", "@", "#")
for i in forbidden:
text = text.replace(i, "")
return text
def reverse(text):
return text[::-1]
def is_palindrome(text):
new = process_text(text)
return process_text(text) == reverse(process_text(text))
something = input("Enter word: ")
if (is_palindrome(something)):
print("You got a palindrome")
else:
print("It is not a palindrome")
| true |
d0c094920da68f24db3d489cab6c1f2e37a17787 | yeshwanthmoota/Python_Basics | /lambda_and_map_filter/map_filter.py | 1,148 | 4.3125 | 4 | nums=[1,2,3,4,5,6,7,8,9]
def double(n):
return n*2
my_list=map(double,nums)
print(my_list)
def even1(n):
return n%2==0
def even2(n):
if(n%2==0):
return n
my_list=map(even2,nums) #This doesn't return a list it returns-
#-address of the genrators of the operation performed.
print(my_list)
my_list=list(map(even2,nums))
print(my_list)
# If we don't want to actually define a function just for
# single purpose we can use lamda functions.
my_list=list(map(lambda n: n*n,nums))
print(my_list)
# with map we cannot decide which to choose i mean see below
my_list=list(map(lambda n: n%2==0,nums))
print(my_list)
# The Output is "[False, True, False, True, False, True, False, True, False]"
# But that is not we wanted we wanted to print "[2, 4, 6, 8]"
# For this we have to select the elements for which n%2==0 is true for that purpose
# We have filter function it appends the element to the list if the
# condition in the function section comes out to be true.
my_list=list(filter(lambda n: n%2==0,nums))
print(my_list)
# Now the Output is "[2, 4, 6, 8]" as we expected.
my_list=list(map(lambda n: n*n,nums))
print(my_list) | true |
4abe4684d196821bdc39592e31e3b94d21f8e22b | yeshwanthmoota/Python_Basics | /comprehensions/zip_function.py | 433 | 4.125 | 4 | names=["Bruce","Clark","Peter","Logan","Wade"]
heroes=["Batman","Superman","Spiderman","Wolverine","Deadpool"]
# Now to use the zip function
Identity_list=list(zip(names,heroes))
print(Identity_list)
Identity_tuple=tuple(zip(names,heroes))
print(Identity_tuple)
Identity_dict=dict(zip(names,heroes)) # This is important names goes into key and
# Heroes goes into value of each key-value pair in the dictionary.
print(Identity_dict) | true |
866832f1f7e1429b4462491be12a4891f10934f7 | rivergillis/mit-6.00.1x | /midterm/flatten.py | 614 | 4.34375 | 4 | def flatten(aList):
'''
aList: a list
Returns a copy of aList, which is a flattened version of aList
[[1,'a',['cat'],2],[[[3]],'dog'],4,5] is flattened into [1,'a','cat',2,3,'dog',4,5]
'''
if aList == []:
return []
print(aList)
for index,elem in enumerate(aList):
print(elem)
if type(elem) == list:
print("Is a list")
return flatten(elem) + flatten(aList[index+1:])
else:
print("Is not a list")
return [elem] + flatten(aList[index+1:])
print(flatten([[1, 'a', ['cat'], 2], [[[3]], 'dog'], 4, 5]))
| true |
d5dae53d28dc9f6964c7adf0f83a5a5359309792 | Gaurav-dawadi/Python-Assignment | /Function/question19.py | 226 | 4.25 | 4 | """Write a Python program to create Fibonacci series upto n using Lambda."""
fib = lambda n: n if n<=1 else fib(n-1)+fib(n-2)
# print(fib(10))
print("The Fibonacci Series is :")
for i in range(10):
print(fib(i))
| false |
6406ba8d6fa1df233b6bd4ed010c5061bab39066 | Gaurav-dawadi/Python-Assignment | /Data Types/question21.py | 909 | 4.28125 | 4 | """Write a Python program to get a list, sorted in increasing order by the last
element in each tuple from a given list of non-empty tuples."""
def getListOfTuples():
takeList = []
num = int(input("Enter number of tuples you want in list: "))
for i in range(num):
takeTuples = ()
for j in range(2):
inputElement = int(input("Enter an element for a tuple: "))
takeTuples = takeTuples + (inputElement, )
takeList.append(takeTuples)
print("---------------------------------------------------")
print("The unsorted list of tuples is ",takeList)
print("---------------------------------------------------")
return takeList
def getSortedList():
unsortedList = getListOfTuples()
sortedList = sorted(unsortedList, key = lambda tup: tup[1])
return sortedList
print("The sorted List is ", getSortedList())
| true |
79a49b8caf4291c0c762e81a851212cc509e6c6a | Joyce-O/Register | /treehouse_python_basics.py/masterticket.py | 1,514 | 4.125 | 4 | TICKET_PRICE = 10
tickets_remaining = 100
#Run tickets untill its sold out
while tickets_remaining >= 1:
# Output how many tickets are remaining using the tickets remaining variable
print("There are {} tickets remaining.".format(tickets_remaining))
# Gather the user's name and assign it to a new variable
name = input("What is your name? ")
# Prompt the user by name and ask how many tickets they would loke
num_tickets = input("How many tickets would you like, {}? ".format(name))
# Catch error when a non number is inputed
try:
num_tickets = int(num_tickets) #Coarce it because input() always returns a string
print(num_tickets)
if num_tickets > tickets_remaining:
raise ValueError("There are only {} remaining".format(tickets_remaining))
except ValueError as err: # The as err keywords are used only when you have a raise keyword before the except
print("Sorry an error occured")
else:
# Calculate the price (number of tickets multiplied by the price) and assign it to a variable
amount_due = num_tickets * TICKET_PRICE
# Output the price in the screen
print("The total due is ${}".format(amount_due))
should_proceed = input("Do you want to proceed? Y/N ")
if should_proceed.lower == 'y':
print('SOLD')
tickets_remaining -= num_tickets
else:
print("Thank you {}".format(name))
print("Sorry tickets are sold out")
| true |
d5498151efd9677fe5247d12c6b4956822937f60 | erin-weeks/hello_world | /8_person.py | 777 | 4.46875 | 4 | #Building more complicated data structures
'''This exercise takes various parts of a name and then returns a dicionary.'''
def build_person(first_name, last_name, age = ''):
'''Return a dictionary of information about a person'''
person = {'first': first_name, 'last': last_name}
'''Because age is optional, we put it in an if statement just in case.'''
if age:
person['age'] = age
return person
musician = build_person('jimi','hendrix')
print(musician)
'''If we know the keys, can we format better?'''
print(musician['first'].title() + " " + musician['last'].title())
'''Testing with age'''
musician = build_person('jimi', 'hendrix', 37)
print(musician['first'].title() + " " + musician['last'].title() + ", " + str(musician['age']))
| true |
a2c94ca0ccf5569edfd3637bfeabc22c4ca84940 | Nelapati01/321810305034-list | /l1.py | 231 | 4.28125 | 4 | st=[]
num=int(input("how many numbers:"))
for n in range (num):
numbers=int(input("Enter the Number:"))
st.append(numbers)
print("Maximum elements in the list:",max(st),"\nMinimum element in the list is :",min(st))
| true |
adfaf200179e8e696cc6f1d74ae475a0beab43c5 | slovoshop/LightIT_test_task | /convert_roman_to_arabic.py | 1,303 | 4.34375 | 4 | # -*- coding: utf-8 -*-
""" Program in Python 3.6.4
Calculating the numeric value of a Roman numeral.
Here is an example of calculating:
roman CXLIV
values [100, 10, 50, 1, 5]
values[:-1] [100, 10, 50, 1]
values[1:] [10, 50, 1, 5]
zip [(100,10), (10,50), (50,1), (1,5)]
sum 100 -10 50 -1 +5
result 144
"""
def convert_roman_to_arabic():
roman = input("Enter the roman number: ").upper()
"""Check valid input"""
trans = {'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000}
invalid = ['IIII', 'VV', 'XXXX', 'LL', 'CCCC', 'DD', 'MMMM']
if (any(sub in roman for sub in invalid) or
not any(sign in roman for sign in trans.keys())):
print('Invalid input format: {}. Reenter'.format(roman))
convert_roman_to_arabic()
return
"""Calculate the numeric value of a Roman numeral"""
values = [trans[sign] for sign in roman]
result = sum(
val if val >= next_val else -val
for val, next_val in zip(values[:-1], values[1:])
) + values[-1]
print('Numeric value of a Roman numeral: {}'.format(result))
convert_roman_to_arabic()
| false |
e049d49db2e69afaa2795331bdd9500cb0ebc2de | MilesBell/PythonI | /practicep1.py | 1,916 | 4.15625 | 4 | chapter="Introduction to PythonII"
print("original chapter name:")
print(chapter)
print("\nIn uppercase:")
print(chapter.upper())
print("\nIn 'swapcase':")
print(chapter.swapcase())
print("\nIn title format:")
print(chapter.title())
print("\nIn 'strip' format:")
print(chapter.strip())
print("\n\nPress the enter key to exit.")
print("My expenses are:")
car=input("insurance:")
gas=input("gas for my car:")
rent=input("my rent is:")
food=input("groceries and dining out:")
shopping=input("my personal shopping:")
total=car+gas+food+shopping
print("\nGrand total is:", total);
#this is good^, but adds the entered values together as a string instead of combining numeric values
#int(input( is better because it adds the integers
print("My expenses are:")
car=int(input("insurance:"))
gas=int(input("gas for my car:"))
rent=int(input("my rent is:"))
food=int(input("groceries and dining out:"))
shopping=int(input("my personal shopping:"))
total=car+gas+food+shopping
print("\nGrand total is:", total);
#below adds the entries together as a string, like 5+5+5+5 as 5555 instead of 5+5+5+5 as 20
print("My expenses are:")
car=str(input("insurance:"))
gas=str(input("gas for my car:"))
rent=str(input("my rent is:"))
food=str(input("groceries and dining out:"))
shopping=str(input("my personal shopping:"))
total=car+gas+food+shopping
print("\nGrand total is:", total);
#also adds together entries as integers, but always returns a decimal; i.e. 5+5+5+5=10.0
print("My expenses are:")
car=float(input("insurance:"))
gas=float(input("gas for my car:"))
rent=float(input("my rent is:"))
food=float(input("groceries and dining out:"))
shopping=float(input("my personal shopping:"))
total=car+gas+food+shopping
print("\nGrand total is:", total);
#practice01
#gets personal information from the user
print("\nUser\'s personal information:")
name=input("full name:")
age=input("age in years:")
weight=input("weight in lbs:")
| true |
2f32e330a2d5ce49f9687db2f826f1baeaef1089 | wkdghdwns199/Python_basic_and_study | /chap_4/22_dic_tion_ary.py | 352 | 4.15625 | 4 | dictionary={
"key1":"ant",
"key2":"bee",
"key3":"cake"
}
print("#딕셔너리의 items() 함수")
print("items():",dictionary.items())
print()
for key,element in dictionary.items():
print("dictionary[{}]= {}".format(key,element))
a_list=["1","2","3"]
b_list=["a","b","c"]
for i in range(len(a_list)):
print(a_list[i],b_list[i]) | false |
3410fdab20dd9e10e148c29a69b40ef8e10bc160 | HaythemBH2003/Object-Oriented-Programming | /OOP tut/Chapter4.py | 1,478 | 4.125 | 4 | ######## INTERACTION BETWEEN CLASSES ########
class Student:
def __init__(self, name, age, grade):
self.name = name
self.age = age
self.grade = grade
def get_grade(self):
return self.grade
def get_name(self):
return self.name
class Course:
def __init__(self, name, max_students):
self.name = name
self.max_students = max_students
self.students = []
# students -------> ATTRIBUTE
# NOT MENTIONED IN __init__()
def get_name(self):
return self.name
def add_student(self, student):
if len(self.students) < self.max_students:
self.students.append(student)
print(f"Student {student.get_name()} successfully added to the {self.get_name()} course.")
return True
else:
print(f"Maximum number of students reached. Unable to add {student.get_name()} to the {self.get_name()} course.")
return False
def get_average_grade(self):
sum = 0
for student in self.students:
sum += student.get_grade()
average = sum / len(self.students)
return average
s1 = Student("Tim", 16, 18)
s2 = Student("Bill", 18, 15)
s3 = Student("Joe", 19, 14)
course = Course("Computer Science", 2)
course.add_student(s1)
course.add_student(s2)
course.add_student(s3)
print(course.get_average_grade()) | true |
9d9e6c968ceabf5bceaeeeb848face6ddf0b24f1 | emberenot/pythoncourse | /lab1(py)/lab1_10.py | 526 | 4.28125 | 4 | #Напишите скрипт, позволяющий определить надежность вводимого пользователем пароля. Это задание является творческим: алгоритм
#определения надежности разработайте самостоятельно.
password = input("Введите пароль: ")
if len(password)>8 and not password.isdigit():
print("Пароль надёжный")
else:
print("Плохой пароль!")
| false |
59e52bd9931ca75082f8d2504722e8b2bc2c976a | alitsiya/InterviewPrepPreWork | /strings_arrays/maxSubArray.py | 670 | 4.21875 | 4 | # Find the contiguous subarray within an array (containing at least one number) which has the largest sum.
# For example:
# Given the array [-2,1,-3,4,-1,2,1,-5,4],
# the contiguous subarray [4,-1,2,1] has the largest sum = 6.
# For this problem, return the maximum sum.
def maxSubArray(A):
if len(A) == 0:
return []
if len(A) == 1:
return A
start = 0
end = 1
current = 1
max_sum = 0
while current < len(A):
cur_sum = sum(A[start:current])
if cur_sum > 0:
if cur_sum > max_sum:
max_sum = cur_sum
end = current
else:
start = current + 1
end = current + 1
current += 1
print max_sum
print maxSubArray([-2,1,-3,4,-1,2,1,-5,4]) | true |
dea9250a810a2f87be8ee0f6bbf720dfa963d144 | nmoore32/coursera-fundamentals-of-computing-work | /2 Principles of Computing/Week 4/dice_analysis.py | 1,893 | 4.21875 | 4 | """
Program to analyze expected value of simple dice game.
You pay $10 to play.
Roll three dice.
You win $200 if you roll triples, get your $10 back if doubles, and lose the $10 otherwise
"""
def gen_all_sequences(outcomes, length):
"""
Iterative function that enumerates the set of all sequences of
outcomes of given length
"""
ans = set([()])
for dummy_idx in range(length):
temp = set()
for seq in ans:
for item in outcomes:
new_seq = list(seq)
new_seq.append(item)
temp.add(tuple(new_seq))
ans = temp
return ans
def max_repeats(seq):
"""
Determines the maximum number of times any individual item occurs
in the sequence
"""
item_count = [seq.count(item) for item in seq]
return max(item_count)
def compute_expected_value():
"""
Compute the expected value (excluding the initial $10, so an answer of $10 corresponds to breaking even)
"""
all_rolls = gen_all_sequences([1, 2, 3, 4, 5, 6], 3)
results = [max_repeats(roll) for roll in all_rolls]
expected_value = 0
for result in results:
if result == 2:
# The probability for getting any given is 1/216 since there are 216 possible results
expected_value += 10 / 216
elif result == 3:
expected_value += 200 / 216
return expected_value
def run_test():
"""
Testing
"""
outcomes = set([1, 2, 3, 4, 5, 6])
print(
f"Number of possible outcomes for rolling three dice: {len(gen_all_sequences(outcomes, 3))}")
print(f"Max repeats for (3, 2, 1) is: {max_repeats([3, 2, 1])}")
print(f"Max repeats for (3, 3, 1) is: {max_repeats([3, 3, 1])}")
print(f"Max repeats for (3, 3, 3) is: {max_repeats([3, 3, 3])}")
print(f"Expected value is: {compute_expected_value()}")
run_test()
| true |
bff961baa068c0a52ce669edd31222aa2b45e928 | xxw1122/Leetcode-python-xxw | /283 Move Zeroes.py | 835 | 4.15625 | 4 | """
Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements.
For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].
Note:
You must do this in-place without making a copy of the array.
Minimize the total number of operations.
"""
class Solution(object):
def moveZeroes(self, nums):
"""
:type nums: List[int]
:rtype: void Do not return anything, modify nums in-place instead.
"""
num_zero = 0
while 1:
if 0 not in nums:
break
else:
index_0 = nums.index(0)
num_zero += 1
del nums[index_0]
for i in range(num_zero):
nums.append(0)
| true |
c0a4b16a1d9a2d072892909a4a50def04dc9cb68 | masterzht/note | /other/python/code/2_list.py | 709 | 4.3125 | 4 | # this is the code of list
word=['a','b','c','d','e','f','g']
a=word[2]
print " a is : " +a
b=word[1:3]
print b # index 1 and 2 elements of word.
c=word[:2]
print c # index 0 and 1 elements of word.
d=word[0:]
print "d is "
print d # All elements of word.
e=word[:2]+word[2:]
print "e is :"
print e # All elements of word.
f=word[-1]
print "f is :"
print f # the last elements of word.
g=word[-4:-2]
print " g is :"
print g # index 3 and 4 elements of word.
h=word[-2:]
print " h is :"
print h # the last two elements
i=word[:-2]
print "i is :"
print i # Everything except the last two characters
l=len(word)
print "Length of word is :" + str(l)
print " Adds new element ... "
word.append('h')
print word
| false |
72c69bbe24970624694106bf99d404fc77b21743 | SerioSticks/AprendiendoPython | /Aleatorio.py | 1,101 | 4.1875 | 4 | #Creador Jorge Alberto Flores Sánchez
#Matricula: 1622167 Grupo: 22
#Fecha de Creación : 18/09/2019
#En python existen muchas librerias para facilitar la programación,
#a estas se les denomina modulos (module).
#Para usar un modulo debe importarse y para esto se utiliza la,
#instrucción import
import random
#Definimos nuestra variables, que en esta ocasion sera de tipo,
#flotante (float) y le asignamos un valor.
numero1=float(10.5)
#En python las funciones son instrucciones que hacen una tarea,
#específica, como una unidad de ejecución.
#Se declaran con def.Y todo el codigo identado a la derecha de la ,
#definicion forma parte de la misma función.
def miFuncion():
#El numero aleatorio se convierte a float, este es generado por
#random.randrange() (este solo esta disponible si se importa,
# el modulo random)
numero2=float(random.randrange(1,10))
#Se utilizan sustituciones para mostrar resultados.
mensaje="La suma de {} y {} es {}"
print(mensaje.format(numero1,numero2,numero1+numero2))
#Se ejecuta la función definida en el código.
miFuncion() | false |
7fb3ca8717ca50c532ef518d2569aa39c43afef8 | gitjit/ehack | /py/emp.py | 993 | 4.25 | 4 | # This is a sample to demonstrate Python OOP features
class Employee(object):
raise_amount = 1.04 # class variable
num_emps = 0
def __init__(self, first, last, pay):
self._first = first
self._last = last
self._pay = pay
self._email = first + '.' + last + '@company.com'
Employee.num_emps += 1
def fullname(self):
return '{} {}'.format(self._first, self._last)
def pay(self):
return self.pay * self.raise_amount
def main():
emp1 = Employee('Jith', 'Menon', 300000)
emp2 = Employee('Ganga', 'Menon', 40000)
print(emp1.fullname(), Employee.fullname(emp2))
print(emp1.raise_amount)
print(Employee.raise_amount)
print (emp2.raise_amount)
emp1.raise_amount = 2.0
print(emp1.raise_amount)
print(Employee.raise_amount)
print (emp2.raise_amount)
print(emp1.__dict__)
print(Employee.__dict__)
print(emp2.__dict__)
print(Employee.num_emps)
main()
| false |
602a2952d43b2c43eab807be4a629af2a6fc4849 | feladie/info-206 | /hw3/BST.py | 2,481 | 4.46875 | 4 | #---------------------------------------------------------
# Anna Cho
# anna.cho@ischool.berkeley.edu
# Homework #3
# September 20, 2016
# BST.py
# BST
# ---------------------------------------------------------
class Node:
#Constructor Node() creates node
def __init__(self,word):
self.word = word
self.right = None
self.left = None
self.count = 1
class BSTree:
#Constructor BSTree() creates empty tree
def __init__(self, root=None):
self.root = root
#These are "external functions" that will be called by your main program - I have given these to you
#Find word in tree
def find(self, word):
return _find(self.root, word)
#Add node to tree with word
def add(self, word):
if not self.root:
self.root = Node(word)
return
_add(self.root, word)
#Print in order entire tree
def in_order_print(self):
_in_order_print(self.root)
def size(self):
return _size(self.root)
def height(self):
return _height(self.root)
#These are "internal functions" in the BSTree class - You need to create these!!!
#Function to add node with word as word attribute
def _add(root, word):
if root.word == word:
root.count +=1
return
if root.word > word:
if root.left == None:
root.left = Node(word)
else:
_add(root.left, word)
else:
if root.right == None:
root.right = Node(word)
else:
_add(root.right, word)
#Function to find word in tree
def _find(root, word):
if root is None: # word is not in the tree
return 0
if root.word == word: # word has been found
return root.count
if root.word > word:
return _find(root.left, word)
if root.word < word:
return _find(root.right, word)
#Get number of nodes in tree
def _size(root):
if root is None: # Leaf nodes
return 0
else: # Root node and its children
return 1 + _size(root.left) + _size(root.right)
#Get height of tree
def _height(root):
if root is None:
return 0 # Leaf node
else:
return max(_height(root.left) + 1, _height(root.right) + 1) # Returns the longest path
#Function to print tree in order
def _in_order_print(root):
if not root:
return
_in_order_print(root.left)
print(root.word)
print(root.count)
_in_order_print(root.right)
| true |
9ce73959107cd05593c971af07ac82530e9b6a35 | chilango-o/python_learning | /seconds_calculator.py | 739 | 4.34375 | 4 | def calcSec():
""" A function that calculates a given number of seconds (user_seconds)
and outputs that value in Hour-minute-second format
"""
user_seconds = int(input('cuantos segundos? '))
hours = user_seconds // 3600 #integer division between the given seconds and the number of seconds in 1 hour (calculation of the whole hrs in the given seconds)
reminder_hours = (user_seconds % 3600) # The remainder of seconds of the hour division (modulus operation)
minutes = reminder_hours // 60 # the whole minutes in the seconds given
reminder_seconds = reminder_hours % 60 # the number of whole seconds of the total given
print('hours:', hours, 'minutes:', minutes, 'seconds:', reminder_seconds)
calcSec() | true |
21026c68e2fb2a211d72eaf48ab67edc023ffe32 | MulengaKangwa/PythonCoreAdvanced_CS-ControlStatements | /53GradingApplication.py | 790 | 4.15625 | 4 |
m=int(input("Enter your maths score (as an integer):"))
if m >=59:
p = int(input("Enter your physics score(as an integer):"))
if p >= 59:
c = int(input("Enter your chemistry score(as an integer):"))
if c >= 59:
if (p + m + c) / 3 <= 59:
print("You secured a C grade")
elif (p + m + c) / 3 <= 69:
print("You secured a B grade")
elif (p + m + c) / 3 > 69:
print("You secured an A grade")
else:
print("Unfortunately, you have not passed the examination.")
else: print("You have not passed the Chemistry examination")
else: print("You have not passed the Physics examination")
else: print("You have not passed the the Mathematics examination")
| true |
3b0e84aa74c12e1a1771755e73d37993d27e1ba6 | push-95/Project-Euler | /p001.py | 279 | 4.375 | 4 | # Problem 1 - Multiples of 3 and 5
# Pushyami Shandilya
# http://github.com/push-95
def calc_sum(N):
'''
Function to find the sum of all the multiples of 3 or 5 below a number N.
'''
return sum(i for i in range(N) if (i%3==0 or i%5==0))
if __name__ == '__main__':
print calc_sum(1000) | false |
8d8e09d5241dcf6a3055e72355029a84967aa0af | ellisgeek/linux1 | /chapter7/scripts/month.py | 332 | 4.34375 | 4 | #!/usr/bin/env python
#define the months of the year
months = ["January", "Febuary", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"]
print "Traditional order:"
for month in months:
print month
print "\n\nAlphabetical order:"
for month in sorted(months):
print month
| true |
eec1de729ffc2a62859746605198505a3f8a994a | tamjidimtiaz/CodeWars | /<8 Kyu> Logical calculator.py | 1,216 | 4.34375 | 4 | '''
Your task is to calculate logical value of boolean array. Test arrays are one-dimensional and their size is in the range 1-50.
Links referring to logical operations: AND, OR and XOR.
You should begin at the first value, and repeatedly apply the logical operation across the remaining elements
in the array sequentially.
First Example:
Input: true, true, false, operator: AND
Steps: true AND true -> true, true AND false -> false
Output: false
Second Example:
Input: true, true, false, operator: OR
Steps: true OR true -> true, true OR false -> true
Output: true
Third Example:
Input: true, true, false, operator: XOR
Steps: true XOR true -> false, false XOR false -> false
Output: false
Input:
boolean array, string with operator' s name: 'AND', 'OR', 'XOR'.
Output:
calculated boolean
'''
def logical_calc(array, op):
if op=='AND':
if False in array:
result = False
else:
result = True
elif op=="OR":
if True in array:
result = True
else:
result = False
else:
result = array[0]
for i in range(1,len(array)):
result = result ^ array[i]
#boolean
return result
| true |
80ce0b6a98ffad061eeb45a97c81851afe42a216 | Venkatesh123-dev/100-Days-of-code | /Day_06_2.py | 1,860 | 4.1875 | 4 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Thu Aug 13 19:28:24 2020
@author: venky
There are two kangaroos on a number line ready to jump in the positive direction (i.e, toward positive infinity). The first kangaroo starts at location and moves at a rate of meters per jump. The second kangaroo starts at location and moves at a rate of
meters per jump. Given the starting locations and movement rates for each kangaroo, can you determine if they’ll ever land at the same location at the same time?
Input Format
A single line of four space-separated integers denoting the respective values of , , , and .
Constraints
Output Format
Print YES if they can land on the same location at the same time; otherwise, print NO.
Note: The two kangaroos must land at the same location after making the same number of jumps.
Sample Input 0
0 3 4 2
Sample Output 0
YES
Explanation 0
The two kangaroos jump through the following sequence of locations:
Thus, the kangaroos meet after jumps and we print YES.
Sample Input 1
0 2 5 3
Sample Output 1
NO
Explanation 1
The second kangaroo has a starting location that is ahead (further to the right) of the first kangaroo’s starting location (i.e., ). Because the second kangaroo moves at a faster rate (meaning ) and is already ahead of the first kangaroo, the first kangaroo will never be able to catch up. Thus, we print NO.
"""
#!/bin/python3
import math
import os
import random
import re
import sys
# Complete the kangaroo function below.
def kangaroo(x1, v1, x2, v2):
for n in range(10000):
if((x1+v1)==(x2+v2)):
return "YES"
x1+=v1
x2+=v2
return "NO"
if __name__ == '__main__':
x1, v1, x2, v2 = raw_input().strip().split(' ')
x1, v1, x2, v2 = [int(x1), int(v1), int(x2), int(v2)]
result = kangaroo(x1, v1, x2, v2)
print(result)
| true |
2fc9577d7073827d48449eb4c3988998e55da166 | DizzyMelo/python-tutorial | /numbers.py | 288 | 4.25 | 4 | # There are three numeric types in python
# int, float and complex
import random
x = 1 # int
y = 2.8 # float
z = 1j # complex
print(complex(x))
print(int(y))
print(type(z))
# the last number is not included, then, numbers from 1 to 9 will show up
print(random.randrange(1, 10))
| true |
115e1cc9db1fc292b214b1f31ff5b23b62dbd684 | DizzyMelo/python-tutorial | /ifelse.py | 1,263 | 4.375 | 4 | # Python Conditions and If statements
# Python supports the usual logical conditions from mathematics:
# Equals: a == b
# Not Equals: a != b
# Less than: a < b
# Less than or equal to: a <= b
# Greater than: a > b
# Greater than or equal to: a >= b
# These conditions can be used in several ways, most commonly in "if statements" and loops.
a = 44
b = 100
if a > b:
# Remember that python relies on identation
print('a is greater than b')
elif a == b:
print('a is equals to b')
else:
print('b is greater than a')
if b > a:
print('b is greater than a')
a = 2
b = 330
print('1') if a > b else print('2')
a = 3
b = 4
print('A') if a > b else print('==') if a == b else print('B')
# The and logical operator
a = 220
b = 33
c = 500
if a > b and c > a:
print('both conditions are true')
if a > b or a > c:
print('at least one of the conditions are true')
# Nested If
a = 41
if a > 10:
print('Above 10')
if a > 20:
print('Above 20')
if a > 40:
print('Above 40')
else:
print('But not above 40')
else:
print('But not above 20')
# Pass statement can be used if for some reason you need an empty if statement, which by default will raise an error
if 10 > 5:
pass
| true |
6869f7ccbbe5124779270653201b7bfeb5b12ba5 | DizzyMelo/python-tutorial | /trycatch.py | 1,830 | 4.375 | 4 | # The try block lets you test a block of code for errors.
# The except block lets you handle the error.
# The finally block lets you execute code, regardless of the result of the try- and except blocks.
# Exception Handling
# When an error occurs, or exception as we call it, Python will normally stop and generate an error message.
# These exceptions can be handled using the try statement:
# Example
# The try block will generate an exception, because x is not defined:
try:
print('ola')
except NameError:
print('An exception occured')
except:
print('Another error')
# Else
# You can use the else keyword to define a block of code to be executed if no errors were raised:
try:
print('Hello World')
except NameError:
print('An exception occured')
except:
print('Another error')
else:
print('Everything went OK')
# Finally
# The finally block, if specified, will be executed regardless if the try block raises an error or not.
try:
print('asdf')
except NameError:
print('Name error occured')
finally:
print('End of try')
# This can be useful to close objects and clean up resources:
# Example
# Try to open and write to a file that is not writable:
try:
f = open("demofile.txt")
f.write("Lorum Ipsum")
except:
print("Something went wrong when writing to the file")
finally:
# f.close()
print('close')
# Raise an exception
# As a Python developer you can choose to throw an exception if a condition occurs.
# To throw (or raise) an exception, use the raise keyword.
x = -1
# if x < 0:
# raise Exception('No numbers below 0 allowed')
# The raise keyword is used to raise an exception.
# You can define what kind of error to raise, and the text to print to the user.
x = 'Hello'
if not type(x) is int:
raise TypeError('It has to be a number')
| true |
abbeac5ff24ee0715d8f0e788244239d4740de5a | danieldiniz1/blue | /aula 15 21.05/exercicio 1.py | 323 | 4.15625 | 4 | numeros = []
for nm in range(5):
numero = int(input(f"Digite o {nm+1}º número: "))
for chave, valor in enumerate(numeros):
if numero < valor:
numeros.insert(chave, numero)
break
else:
numeros.append(numero)
print("Lista: ", numeros)
for nmrs in numeros:
print(nmrs) | false |
79dd6ddfe911e50f068d60014f5713afd3572345 | danieldiniz1/blue | /aula 14.05/aula 14.05.py | 515 | 4.125 | 4 | # exercício de while
opc = 1
while opc == 1:
numero = float(input("digite um numero: "))
print()
if (numero == 0):
print(f'o numero digitado é: {numero}')
print()
elif (numero > 0):
print(f'o numero {numero:.2f} é positivo')
print()
else:
print(f'o numero {numero:.2f} é negativo')
print()
resposta = input("você deseja continuar? (sim/não) ")
print()
if resposta =="não":
opc = 0
print("programa finalizado") | false |
0798b0374d82ddc94292b856402b782474d8c891 | chaithanyasubramanyam/pythonfiles | /linkedlist.py | 992 | 4.15625 | 4 | class Node:
def __init__(self,data):
self.data = data
self.next = None
class Linkedlist:
def __init__(self):
self.head = None
def push(self,new_data):
new_node = Node(new_data)
new_node.next = self.head
self.head = new_node
def insert(self,previous_node,new_data):
if previous_node == None:
print('no')
new_node = Node(new_data)
new_node.next = previous_node.next
previous_node.next = new_node
def append(self,new_data):
new_node = Node(new_data)
if self.head == None:
self.head = new_node
else:
n = self.head
while n.next != None:
n = n.next
n.next = new_node
def printlist(self):
n = self.head
while n != None:
print(n.data,end = ' ')
n = n.next
l = [1,2,3,4]
a = Linkedlist()
for i in l:
a.append(i)
a.printlist()
| false |
16ad2f811290b82e99ad541713ca0b48adc99f87 | gauravlahoti80/Project-97-Number-Guessing-Game | /guessing Game/main.py | 1,899 | 4.21875 | 4 | #importing modules
import random
import pyttsx3
#welcome screen
input_speak = pyttsx3.init()
input_speak.say("Enter your name: ")
input_speak.runAndWait()
#taking name input from the user
user_name = input("Enter your name: ")
#showing hello to the user
input_speak.say(f"Hello,{user_name}")
input_speak.runAndWait()
print(f'Hello , {user_name}.')
#chances
chances = 5
print(f"You have {chances} chances to win the game.")
#random number
max_number = 10
min_number = 1
random_number = random.randint(min_number , max_number)
#taking input for number
input_speak.say("Enter any number from 1 to 10: ")
input_speak.runAndWait()
guessed_number = int(input("Enter any number from 1 to 10: "))
while True:
#checking the numbers
if guessed_number == random_number:
input_speak.say(f"Congrats, {user_name} you have guessed the number in {chances} chances.")
input_speak.runAndWait()
print(f"Congrats, {user_name} you have guessed the number in {chances} chances")
break
#if the user looses
elif chances == 1:
input_speak.say(f"sorry you were not able to guess the number. The number is {random_number}")
input_speak.runAndWait()
print(f"sorry you were not able to guess the number. The number is {random_number}")
break
#checking the numbers for high anf low
else:
if random_number > guessed_number:
input_speak.say("that was too low.")
input_speak.runAndWait()
print('That was too low.')
else:
input_speak.say("that was too high.")
input_speak.runAndWait()
print('That was too high.')
#taking the input again on wrong guessed
chances -= 1
print(f'You have {chances} chances left.')
guessed_number = int(input('Guess again: ')) | true |
bf5e398738221308a2c5fa8462c32ff1d8de1cbb | emmanuelaboah/Data-Structures-and-Algorithms | /Solved Problems/Data structures/Recursion/String-Permutations.py | 1,790 | 4.125 | 4 | # Problem Statement
# Given an input string, return all permutations of the string in an array.
# Example 1:
# string = 'ab'
# output = ['ab', 'ba']
# Example 2:
# string = 'abc'
# output = ['abc', 'bac', 'bca', 'acb', 'cab', 'cba']
# Note - Strings are Immutable
# Recursive Solution
"""
Param - input string
Return - compound object: list of all permutations of the input string
"""
def permutations(string):
return return_permutations(string, 0)
def return_permutations(string, index):
# output to be returned
output = list()
# Terminaiton / Base condition
if index >= len(string):
return [""]
# Recursive function call
small_output = return_permutations(string, index + 1)
# Pick a character
current_char = string[index]
# Iterate over each sub-string available in the list returned from previous call
for subString in small_output:
# place the current character at different indices of the sub-string
for index in range(len(small_output[0]) + 1):
# Make use of the sub-string of previous output, to create a new sub-string.
new_subString = subString[0: index] + current_char + subString[index:]
output.append(new_subString)
return output
# Test - function
def test_function(test_case):
string = test_case[0]
solution = test_case[1]
output = permutations(string)
output.sort()
solution.sort()
if output == solution:
print("Pass")
else:
print("Fail")
string = 'ab'
solution = ['ab', 'ba']
test_case = [string, solution]
test_function(test_case)
string = 'abc'
output = ['abc', 'bac', 'bca', 'acb', 'cab', 'cba']
test_case = [string, output]
test_function(test_case)
| true |
2eff58b50c69128a2999948651acc3cb11701e98 | emmanuelaboah/Data-Structures-and-Algorithms | /Solved Problems/Data structures/Recursion/Reverse _string_input.py | 1,389 | 4.53125 | 5 |
def reverse_string(input):
"""
Return reversed input string
Examples:
reverse_string("abc") returns "cba"
Args:
input(str): string to be reversed
Returns:
a string that is the reverse of input
"""
# (Recursion) Termination condition / Base condition
if len(input) == 0:
return ""
else:
first_char = input[0]
the_rest = slice(1, None) # `the_rest` is an object of type 'slice' class
sub_string = input[the_rest] # convert the `slice` object into a list
# Recursive call
reversed_substring = reverse_string(sub_string)
return reversed_substring + first_char
# Test Cases
print ("Pass" if ("" == reverse_string("")) else "Fail")
print ("Pass" if ("cba" == reverse_string("abc")) else "Fail")
#-------------------------------------------------#
'''
**Time and Space Complexity Analysis**
Each recursive call to the `reverse_string()` function will create
a new set of local variables - first_char, the_rest, sub_string, and reversed_substring.
Therefore, the space complexity of a recursive function would always be proportional to the
maximum depth of recursion stack.
The time complexity for this function will be O(k*n), where k is a constant and n is the
number of characters in the string (depth of recursion stack).
'''
| true |
f10c03cbd6d9a175258290a51135d1ab0609281b | ParthRoot/Basic-Python-Programms | /list Python program to print even numbers in a list.py | 444 | 4.15625 | 4 | # Programme by parth.py
# Python program to print even numbers in a list
def even(lst):
for i in lst:
if i % 2 == 0:
print(i, end=" ")
def odd(lst):
for i in lst:
if i % 2 != 0:
print(i, end=" ")
print("Even Number is-:")
even([1,2,3,4,5,6,7,8,9])
print("\nodd Number is:-")
odd([1,2,3,4,5,6,7,8,9])
"""
output:
Even Number is-:
2 4 6 8
odd Number is:-
1 3 5 7 9
"""
| false |
0424290f9aa1009976c74c19bfc41f65da59146e | ParthRoot/Basic-Python-Programms | /Fectorial.py | 285 | 4.1875 | 4 | num=int(input("enter the num"))
factorial=1
if num < 0:
print("sorry factorial doesnot exit negative num")
elif num == 0:
print("factorial always start 1")
else:
for i in range(1,num+1):
factorial = factorial * i
print("the factorial of",num,factorial) | true |
bb328aa070b6889b12d9dac1b3508801da268fd6 | daniel-laurival-comp19/slider-game-1o-bimestre | /getStartingBoard.py | 574 | 4.25 | 4 | def getStartingBoard():
''' Return a board data structure with tiles in the solved state. For example, if BOARDWIDTH and BOARDHEIGHT are both 3, this function returns [[1, 4, 7], [2, 5, 8], [3, 6, BLANK]]''' counter = 1 board = []
for x in range(BOARDWIDTH):
column = []
for y in range(BOARDHEIGHT):
column.append(counter)
counter += BOARDWIDTH
board.append(column)
counter -= BOARDWIDTH * (BOARDHEIGHT - 1) + BOARDWIDTH - 1
board[BOARDWIDTH-1][BOARDHEIGHT-1] = BLANK
return board | true |
d3911c904e2e9cae6da70563f707270a07aae8f6 | jnegro/knockout | /knockout_steps/knockout_step3.py | 2,339 | 4.46875 | 4 | #!/usr/bin/python
# STEP 3 - Code the 'main' function to play the game
# In this step we replace the 'main' function with code that runs the game
from __future__ import print_function
# We need to import the 'random' Python library in order to generate random numbers
import random
class Boxer(object):
"""
An Object to describe a Boxer and the actions he/she takes
"""
def __init__(self, name):
"""Create a new boxer with a name and set starting health"""
self.name = name
self.health = 100
@property
def standing(self):
"""A property we can check at any time to see if the boxer is still standing"""
if self.health > 0:
return True
else:
return False
def punch(self):
"""Boxer throws a punch. We randomly choose damage between 1 and 10"""
damage = random.randint(1, 10)
return damage
def punched(self, damage):
"""Boxer gets punched and receives damage"""
self.health = self.health - damage
def main():
"""This function starts the program"""
# Ask the player for their name
name = raw_input("Please enter your name: ")
# Create the player as a new Boxer
player = Boxer(name)
# Create the Computer boxer
computer = Boxer("Computer")
# Print out the initial health stats
print("{0}: {1}".format(player.name, player.health))
print("{0}: {1}".format(computer.name, computer.health))
print("")
# Start our game loop and keep repeating until someone falls down
while player.standing and computer.standing:
# Pause and wait for the player to hit enter to punch
raw_input("Press enter to punch")
# Computer punches the player
player.punched(computer.punch())
# Player punches the computer
computer.punched(player.punch())
# Print out the stats after they punch each other
print("{0}: {1}".format(player.name, player.health))
print("{0}: {1}".format(computer.name, computer.health))
print("")
# Once someone falls down the loop stops
if player.standing:
print("KNOCKOUT! YOU WON!")
print("")
else:
print("You lost. Too bad")
print("")
if __name__ == "__main__":
# run the main function
main()
| true |
0ae1e723b9a16cd4c9a5bad9d756158888ccfe27 | SabiqulHassan13/python3-programiz-try | /NumberIsOddOrEven.py | 290 | 4.46875 | 4 | # python program to check a number is even or odd
#take input a number to check from the user
num = float(input("Enter a number: "))
# checking a number is odd or even
if num % 2 == 0:
print("{} is even number".format(num))
elif num % 2 == 1:
print("{} is odd number".format(num)) | true |
886979bbd25e5a15e02eef4cad6519d56f710c6c | SabiqulHassan13/python3-programiz-try | /FindFactorsOfANumber.py | 251 | 4.4375 | 4 | # python program to find the factors of a number
# take input a number
num = int(input("Enter a positive integer: "))
# print the factors
print("The factors of {} are: ".format(num))
for i in range(1, num + 1):
if num % i == 0:
print(i)
| true |
56353728f6e3f62a021e580e8e73cb4231c5e41a | Sohan11Sarkar/Basic-Calculator.github.io- | /main.py | 1,011 | 4.1875 | 4 | print('*********** WELCOME TO MY CALCULATOR ************')
while True:
print()
print("Please select the operation you want to perform : \n\n1.Addition\n2.Subtraction\n3.Multiplication\n4.Division\n5.Exit\n")
option = int(input("Enter pption here -> "))
if option==1:
no1=int(input("Enter First Number -> "))
no2=int(input("Enter Second Number -> "))
print(no1,' + ',no2,' = ',no1+no2)
elif option==2:
no1=int(input("Enter First Number -> "))
no2=int(input("Enter Second Number -> "))
print(no1,' - ',no2,' = ',no1-no2)
elif option==3:
no1=int(input("Enter First Number -> "))
no2=int(input("Enter Second Number -> "))
print(no1,' * ',no2,' = ',no1*no2)
elif option==4:
no1=int(input("Enter First Number -> "))
no2=int(input("Enter Second Number -> "))
print(no1,' / ',no2,' = ',no1/no2)
elif option==5:
print("Thanks for using my calculator!")
break
else:
print("""Operation is not available....Try something else!""") | false |
ed5cd23e31fcabc16b3ec9a5822e666bfa66b667 | utolee90/Python3_Jupyter | /workspace2/10-collection/exam1_.py | 730 | 4.1875 | 4 | # 문자열 여러개 한꺼번에 저장
str_list = ['국어', '영어', '수학', '사회', '한국사']
print(str_list)
# 인덱싱 : 데이터 1개 처리
print(str_list[0])
print(str_list[3])
# 슬라이싱 : 데이터 여러개 처리
print(str_list[1:4])
print(str_list[:4])
print(str_list[1:])
print(str_list[:])
print(str_list[::2])
print(str_list[::-1])
print('-' * 30)
# 정수 저장
num_list = [1, 2, 3, 4, 5]
print(num_list)
print('-' * 30)
# 실수 저장
float_list = [5.5, 3.14, 7.7]
print(float_list)
print('-' * 30)
# boolean값 저장
bool_list = [True, False, False]
print(bool_list)
print('-' * 30)
# 여러 데이터 저장
list1 = [5, 7.7, True, "Hello"]
print(list1)
print('-' * 30)
| false |
a59bc8107619ffaa5468ee7d5df1e9cf6cc3ebee | utolee90/Python3_Jupyter | /workspace2/06-operator/exam6.py | 646 | 4.25 | 4 | '''
논리 연산자 : 수학의 집합 기호를 명령어로 만들어 놓은 것
=> boolean 연산
<진리표>
x y x and y x or y not x
true true true true false
true false false true false
false true false true true
false false false false true
'''
a = 100
b = 200
x = 5
y = 3
r1 = a >= b
r2 = x >= y
print(r1, r2)
print('=' * 30)
result = r1 and r2
print(result)
print('=' * 30)
result = r1 or r2
print(result)
print('=' * 30)
result = not r1
print(result)
print('=' * 30) | false |
55d460794565caf77b2f5c357a550f86049a7c9e | gawalivaibhav/Vg-python | /08-usefull datastructures/07-sets.py | 370 | 4.15625 | 4 | #Sets :- collection of uniqu elements
#Union "|" #intersection "^" #diffrence "-"
my_set = set(['one','two','three','one'])
#print(my_set)
my_set1 = set(['two','three','four'])
a = my_set - my_set1
print(a <= my_set) #subset
my_set1.add('five')
print(my_set1)
#print(my_set|my_set1) #Union
#print(my_set ^ my_set1) #intersection
#print(my_set - my_set1) #Diffrence
| false |
2a14971a28aee88ed93ee3aefb06dc1ec2fc0151 | rickyqiao/data-structure-sample | /binary_tree.py | 1,401 | 4.125 | 4 | class Node:
def __init__(self, data):
self.left = None
self.right = None
self.parent = None
self.data = data
self.height = 0
class BinaryTree:
def __init__(self):
self.root = None
def __str__(self, node = 0, depth = 0, direction_label = ""):
"The tree structure in string form, to be used in str(my_node) or print(my_node)."
if node == 0:
node = self.root
if node:
height_info = "(H"+str(node.height)+")" if node.height > 0 else ""
return depth * "\t" + direction_label + height_info + str(node.data) + "\n" + \
self.__str__(node.left, depth+1, "L:") + self.__str__(node.right, depth+1, "R:")
else:
return ""
def inorder(self, node = 0, result = None):
if result is None:
result = []
if node == 0:
node = self.root
if node:
self.inorder(node.left, result)
result.append(node.data)
self.inorder(node.right, result)
return result
if __name__ == "__main__":
tree = BinaryTree()
tree.root = Node("0")
tree.root.left = Node("1")
tree.root.right = Node("2")
tree.root.left.right = Node("3")
print(tree)
print(tree.inorder())
tree = BinaryTree()
tree.root = Node("0")
print(tree)
print(tree.inorder())
| true |
9670dc0c87a0205c248e98154aa92186b48addba | arinmsn/My-Lab | /Books/PythonCrashCourse/Ch8/8-6_CityNames.py | 720 | 4.53125 | 5 | # Write a function called city_country() that takes in the name
# of a city and its country. The function should return a string formatted like this:
# "Santiago, Chile"
# Call your function with at least three city-country pairs, and print the value
# that’s returned.
def city_country(city, country):
message = city + ', ' + country
return message
while True:
print("Type 'q' at any time to quit ")
city_name = input("Enter a city name: ")
if city_name == 'q':
break
country_name = input("Enter name of that city's country: ")
if country_name == 'q':
break
formatted_message = city_country(city_name, country_name)
print(f'{formatted_message}')
city_country() | true |
de0f6049fdf79eddf97e30129d1c449999dc4cab | slayer96/codewars_tasks | /pattern_craft_decorator.py | 1,691 | 4.5 | 4 | """
The Decorator Design Pattern can be used, for example, in the StarCraft game to manage upgrades.
The pattern consists in "incrementing" your base class with extra functionality.
A decorator will receive an instance of the base class and use it to create a new instance with the new things you want "added on it".
Your Task
Complete the code so that when a Marine gets a WeaponUpgrade it increases the damage by 1, and if it is a ArmorUpgrade then increase the armor by 1.
You have 3 classes:
Marine: has a damage and an armor properties
MarineWeaponUpgrade and MarineArmorUpgrade: upgrades to apply on marine. Accepts a Marine in the constructor and has the same properties as the Marine
"""
import test_framework as Test
class Marine:
def __init__(self, damage, armor):
self.damage = damage
self.armor = armor
class Marine_weapon_upgrade:
def __init__(self, marine):
self.damage = marine.damage + 1
self.armor = marine.armor
class Marine_armor_upgrade:
def __init__(self, marine):
self.damage = marine.damage
self.armor = marine.armor + 1
Test.describe('Basic tests')
Test.it('Single upgrade')
marine = Marine(10, 1)
Test.assert_equals(Marine_weapon_upgrade(marine).damage, 11)
Test.assert_equals(Marine_weapon_upgrade(marine).damage, 11)
Test.it('Double upgrade')
marine = Marine(15, 1)
marine = Marine_weapon_upgrade(marine)
marine = Marine_weapon_upgrade(marine)
Test.assert_equals(marine.damage, 17)
Test.it('Triple upgrade')
marine = Marine(20, 1)
marine = Marine_weapon_upgrade(marine)
marine = Marine_weapon_upgrade(marine)
marine = Marine_weapon_upgrade(marine)
Test.assert_equals(marine.damage, 23)
| true |
5a31b5a75acc5bba977ad3117973e674063ffcc8 | joseph-guidry/dot | /mywork/ch5_ex/ex7.py | 636 | 4.125 | 4 | #! /usr/bin/env python3
def update_value(dictionary, num):
"""Take a dictionary and number to add values. Returns an updated
dictionary"""
for key in dictionary:
dictionary[key] += num #with no return
dict_value = {"cats":0, "dogs":0}
#get input and convert to int
number = int(input("What value do you want to increase by?\n>"))
update_value(dict_value, number) #no need to use a return of function()
#dict_value is updated in place.
animal = list(dict_value.keys())
for thing in animal:
print("there are " + str(dict_value[thing]) + " " + thing +" in the zoo")
| true |
5380d9d84232b3c71cac9aa3e622a86d34dd1939 | dayna-j/Python | /codingbat/not_string.py | 238 | 4.375 | 4 | # Given a string, return a new string where "not " has been added
# to the front. However, if the string already begins with "not", return the string unchanged.
def not_string(str):
if str[0:3]=='not':
return str
return 'not '+str
| true |
aa5611dd17a02c0642afabcb5d37e5816079631d | deepakgd/python-exercises | /class7.py | 719 | 4.1875 | 4 | # multiple inheritance init call analysis
class A:
def __init__(self):
print("init of A")
def feature1(self):
print("Feature 1-A")
def featurea(self):
print("Feature A")
class B:
def __init__(self):
print("init of B")
def feature1(self):
print("Feature 1-B")
def featureb(self):
print("Feature B")
class C(A, B):
# what happen if there is no init in sub class. it will call super call init but here we have
# two super class. it will call A because it is in first. Left to right order
# check class8.py
def feature1(self):
print("Feature 1-C")
def featureb(self):
print("Feature C")
c = C() | true |
f0c63a412162c445546ebdb2b5f12239f85e2212 | pxblx/programacionPython | /practica03/E01Cuadrado.py | 925 | 4.15625 | 4 | """
Ejercicio 1 de clases
Implementa en Python las clases GatoSimple, Cubo y Cuadrado vistas en el libro "Aprende Java con Ejercicios" y sus
respectivos programas de prueba.
"""
class Cuadrado:
def __init__(self, lado):
self.__lado = lado
def __str__(self):
resultado = ""
c = 0
while c < self.__lado:
resultado += "##"
c += 1
resultado += "\n"
c = 1
while c < self.__lado - 1:
resultado += "##"
espacios = 1
while espacios < self.__lado - 1:
resultado += " "
espacios += 1
resultado += "##\n"
c += 1
c = 0
while c < self.__lado:
resultado += "##"
c += 1
resultado += "\n"
return resultado
# Principal
if __name__ == "__main__":
miCuadradito = Cuadrado(5)
print(miCuadradito)
| false |
db71200cb44b9a230d57338187a30ad8045b7912 | pxblx/programacionPython | /practica05/E10HashMap.py | 1,025 | 4.1875 | 4 | """
Ejercicio 10 de POO4
Crea un mini-diccionario español-inglés que contenga, al menos, 20 palabras
(con su correspondiente traducción). Utiliza un objeto de la clase HashMap para
almacenar las parejas de palabras. El programa pedirá una palabra en español
y dará la correspondiente traducción en inglés.
"""
diccionario = {
"hola": "hello",
"adiós": "goodbye",
"por favor": "please",
"nombre": "name",
"hoy": "today",
"ayer": "yesterday",
"semana": "week",
"día": "day",
"mes": "month",
"lunes": "monday",
"martes": "tuesday",
"miércoles": "wednesday",
"jueves": "thursday",
"viernes": "friday",
"sábado": "satuday",
"domingo": "sunday",
"enero": "january",
"febrero": "february",
"marzo": "march",
"abril": "april"
}
palabra = input("Introduce una palabra en español: ")
if palabra in diccionario:
print("La traducción en inglés es: " + diccionario.get(palabra))
else:
print("La palabra no está en el diccionario.")
| false |
6d51a33cce5ed2c713c298e7a0ae483a3c8ebedf | pxblx/programacionPython | /practica01/repetitivas/E05Repetitivas.py | 1,166 | 4.25 | 4 | """
Ejercicio 5 de repetitivas
Escribe un programa que pida el limite inferior y superior de un intervalo. Si el limite inferior es mayor que el
superior lo tiene que volver a pedir. A continuacion se van introduciendo numeros hasta que introduzcamos el 0. Cuando
termine el programa dara las siguientes informaciones:
- La suma de los numeros que estan dentro del intervalo (intervalo abierto).
- Cuantos numeros estan fuera del intervalo.
- Informa si hemos introducido algun numero igual a los limites del intervalo.
"""
# Pedir valores
while True:
desde = int(input("Desde: "))
hasta = int(input("Hasta: "))
if desde < hasta:
break
numero = int(input("Introduce un numero: "))
# Resultado
suma = 0
fuera = 0
igual = 0
while True:
suma = suma + numero
if numero < desde or numero > hasta:
fuera += 1
if numero == desde or numero == hasta:
igual += 1
numero = int(input("Introduce un numero: "))
if numero == 0:
break
print("Has introducido " + str(fuera) + " numeros fuera del rango, " + str(igual) + " numeros iguales a los limites y la suma de todos es igual a " + str(suma) + ".")
| false |
3cf82f53cf591934fb01755beabf9904c3c1ffdf | Aobie/Project-Euler | /Euler9.py | 1,170 | 4.1875 | 4 | #Euler 9
#Pythagorean triplet is a set of 3 natural numbers which can define a right triangle
#a ** 2 + b ** 2 = c ** 2, where a < b < c
import math
import time
def find_product():
# since a, b, and c are natural numbers and a < b < c
# the smallest set possible is a = 1, b = 2, c = 3
# this also means that the largest a is 1000 / 3 - 1 = 332
# since this would lead to a = 332, b = 333, c = 334
for a in range(1, 332):
# since b > a, start the inner loop at a + 1
# since b < c, and a + b + c = 1000, set a = 1 to find largest b
# leads to b + c = 999, so largest b = floor (999 / 2)
for b in range (a + 1, 499):
if (math.sqrt (a ** 2 + b ** 2)).is_integer():
c = int(math.sqrt(a ** 2 + b ** 2))
#print ("A: ", a, "B: ", b, "C: ", c)
if a + b + c == 1000:
return a * b * c
if __name__ == "__main__":
print("Problem 9")
print("Find the product of the one Pythagorean triplet for which a + b + c = 1000.")
t1 = time.time()
answer = find_product()
t2 = time.time()
print((t2-t1)*1000.0, " ms")
print(answer)
| false |
f87730c01046eddf792665e051463ff59fde7aa0 | LilMelt/Python-Codes | /Rock_Paper_Scissors.py | 1,880 | 4.1875 | 4 | import random
def opponent():
choice = random.choice(["rock", "paper", "scissors"])
print("Your opponent chose " + choice)
return choice
def main():
game = True
score = 0
opponent_score = 0
while game:
# input user move
user_move = input("Type \"rock\", \"paper\" or \"scissors\" then press enter: ")
# check attack is valid
while user_move != 'rock' and user_move != 'paper' and user_move != 'scissors':
print("Please enter a valid move")
user_move = input("Type \"rock\", \"paper\" or \"scissors\" then press enter: ")
opponent_move = opponent()
if user_move == 'rock' and opponent_move == 'scissors':
score += 1
print("You won! Nice job!")
elif user_move == 'paper' and opponent_move == 'rock':
score += 1
print("You won! Nice job!")
elif user_move == 'scissors' and opponent_move == 'paper':
score += 1
print("You won! Nice job!")
elif user_move == opponent_move:
print("You tied!")
else:
opponent_score += 1
print("You lose!")
# ask user if they want to play again
play = input("The score is [" + str(score) + "," + str(opponent_score) + "]. Do you want to play another "
"game? Type \"yes\" or \"no\": ")
# ensure this is a valid input
while play != 'yes' and play != 'no':
print("Please enter a valid answer")
play = input("Do you want to play another game? Type \"yes\" or \"no\": ")
if play == 'no':
print("Ok, lets play again soon!")
game = False
else:
print("Ok, lets play another game!")
if __name__ == "__main__":
main()
| true |
0a749ccfccd2d7b18168455fca9c7701f26177fd | Pankhuriumich/EECS-182-System-folder | /Homeworks/HW5/printmovie.py | 994 | 4.25 | 4 | '''TASK: Fill in the code to generate the HTML-formatted string
from the fields of a movie. See the README.txt for the
format
'''
def print_movie_in_html(movieid, movie_title, moviedate, movieurl):
'''STUB code. Needs to change so that the return value
contains HTML tags, as explained in README.txt.'''
resulthtml = movieid + " " + movie_title + " " + moviedate + " " + movieurl + "\n";
# Output resulthtml to a file
print_to_file(movieid + ".html", resulthtml);
return resulthtml;
'''
You do not have to change this function. It writes the string s
to the specified filename. In this program, s will be a raw HTML
string. By writing the HTML to a file, you can view
the HTML file in a browser. This is called from main.cpp.
'''
def print_to_file(filename, s):
f = open(filename, "w");
f.write(s);
f.close();
def main():
result = print_movie_in_html("1", "Sample movie", "1-1-2005", "http://imdb.com");
print result;
main();
| true |
30e6b14d953e87fb0f37066e53f5222755743ddb | Pankhuriumich/EECS-182-System-folder | /Lecture_Code/Recursion/stone_to_dust_simulation/stonetodust.py | 797 | 4.21875 | 4 | def turn_stone_into_dust(stone):
if (isdust(stone)):
print "We got a dust piece! Nothing more to do on the piece!";
return;
else:
(piece1, piece2) = strike_hammer(stone);
turn_stone_into_dust(piece1);
turn_stone_into_dust(piece2);
def isdust(stone):
# Stones of size 1 or smaller are considered to be dust
if (stone <= 1):
return True;
else:
return False;
def strike_hammer(stone):
print "Striking hammer on a stone of size:", stone;
# Divide stone into two smaller pieces.
piece1 = stone / 2;
piece2 = stone - piece1;
print "Created stones of size:", piece1, "and", piece2;
return (piece1, piece2);
def main():
# Turn a stone of volume 7 units into dust.
turn_stone_into_dust(7);
main();
| true |
92d1c7f8015d4d552bb11fa8c818f81840d49646 | lavenderLatte/89926_py | /homework/helperfunctions.py | 536 | 4.34375 | 4 | """
Create a python module helperfunctions.py with the following functions.
add - returns the sum of two numbers
diff - returns the difference between two numbers
product - returns the product of two numbers
greatest - returns the greatest of two numbers.
Import this module in your python program and use the functions your created on any two numbers and print the result.
"""
def add(a, b):
return a + b
def diff(a, b):
return a - b
def product(a, b):
return a * b
def greatest(a, b):
if a >= b:
return a
else:
return b
| true |
54bc71f87b9de5e9956881744c63d1501c24ddd4 | saragregory/hear-me-code | /pbj_while.py | 1,801 | 4.28125 | 4 | # Difficulty level: Beginner
# Goal #1: Write a new version of the PB&J program that uses a while loop. Print "Making sandwich #" and the number of the sandwich until you are out of bread, peanut butter, or jelly.
# Example:
# bread = 4
# peanut_butter = 3
# jelly = 10
# Output:
# Making sandwich #1
# Making sandwich #2
# All done; only had enough bread for 2 sandwiches.
bread = 6
pb = 4
jelly = 10
sandwiches = 0
while bread >= 2 and pb > 0 and jelly > 0:
bread = bread - 2
pb = pb - 1
jelly = jelly - 1
sandwiches = sandwiches + 1
print "I just made sandwich #{0}.".format(sandwiches)
# print "I have enough bread for {0} more sandwiches, enough peanut butter for {1} more and enough jelly for {2} more".format(bread/2,pb,jelly)
# Goal #2: Modify that program to say how many sandwiches-worth of each ingredient remains.
# Example 2:
# bread = 10
# peanut_butter = 10
# jelly = 4
# Output:
# Making sandwich #1
# I have enough bread for 4 more sandwiches, enough peanut butter for 9 more, and enough jelly for 3 more.
# Making sandwich #2
# I have enough bread for 3 more sandwiches, enough peanut butter for 8 more, and enough jelly for 2 more.
# Making sandwich #3
# I have enough bread for 2 more sandwiches, enough peanut butter for 7 more, and enough jelly for 1 more.
# Making sandwich #4
# All done; I ran out of jelly.
if bread == 0 and pb == 0 and jelly == 0:
print "All done, I ran out of ingredients."
elif bread <= 1 and pb >= 1 and jelly >= 1:
print "All done, I ran out of bread."
elif pb <= 1 and bread >= 2 and jelly >= 1:
print "All done, I ran out of peanut butter."
elif jelly <= 1 and bread >= 2 and pb >= 1:
print "I ran out of jelly but I can make a peanut butter sandwich."
print "I had enough ingredients to make {0} sandwiches.".format(sandwiches)
| true |
a9d2ac3204cf92eb968c39fcc51ba887805b9645 | coolguy-kr/questions | /py-multiple_inheritance.py | 826 | 4.4375 | 4 | # The following code is an example. The tree structure of class inheritance relationships is displayed as a list on the console.
class X:
pass
class Y:
pass
class Z:
pass
class A(X, Y):
pass
class B(Y, Z):
pass
class M(B, A, Z):
pass
print(M.mro())
# So, It displays a result as a list.
""" Console Output
[<class '__main__.M'>, <class '__main__.B'>, <class '__main__.A'>, <class '__main__.X'>, <class '__main__.Y'>, <class '__main__.Z'>, <class 'object'>]
"""
# However, I expected a result of below to be printed.
""" The result that I expected
[<class '__main__.M'>, <class '__main__.B'>, <class '__main__.Y'>, <class '__main__.Z'>, <class '__main__.A'>, <class '__main__.X'>, <class 'object'>]
"""
# Q. If we look at them in order, the expected result should be output. Shouldn't it?
| true |
87bf49b4a94a28bd4ae46e00b09ef44e3ddfa977 | the-brainiac/twoc-problems | /day6/program_14.py | 699 | 4.15625 | 4 | #to solve this question i used geeksforgeeks{in anticlockwise} as a reference
# https://www.geeksforgeeks.org/inplace-rotate-square-matrix-by-90-degrees/
#
N = 4
def rotateMatrix(mat):
for x in range(0, int(N / 2)):
for y in range(x, N-x-1):
temp = mat[N-1-y][x]
mat[N-1-y][x]=mat[N-1-x][N-1-y]
mat[N-1-x][N-1-y]=mat[y][N-1-x]
mat[y][N-1-x]=mat[x][y]
mat[x][y]=temp
# Function to print the matrix
def displayMatrix( mat ):
for i in range(0, N):
for j in range(0, N):
print (mat[i][j], end = ' ')
print ("")
mat = [ [1, 2, 3, 4 ],
[5, 6, 7, 8 ],
[9, 10, 11, 12 ],
[13, 14, 15, 16 ] ]
rotateMatrix(mat)
displayMatrix(mat) | false |
8d37ab6a0ee58ee4f45551e06ac99603c239f64b | suboqi/-python | /ex3.py | 794 | 4.53125 | 5 | #现在数一数我的鸡
print("I will now count my chickens:")
#hens有30只鸡，打印出来
print("Hens",25+30/6)
#roosters有97只鸡，打印出来
print("Roosters",100-25*3%4)
#现在数一数我的蛋
print("Now i will count the eggs:")
#一共有以下几只蛋，并打印出来
num1=print(3+2+1-5+4%2-1/4+6)
#这是一个判断
print("Is it true that 3 + 2 <5 - 7?")
#判断真假
print(3+2<5-7)
#这是一个判断
print("What is 3 + 2?",3+2)
#这是一个判断
print("What is 5 - 7?",5-7)
#哦！为什么它是假的
print("oh!,thst's why it's False")
#要不，再点一点例子
print("How about some more")
#判断是否为真
print("Is it greater?",5 > 2)
#判断是否为真
print("Is it greater or equal?",5 >= -2)
#判断是否为假
print("Is it less or equal?",5 <= -2) | false |
511f378d45e5474c7a7eb679a78a14af4bac5cbf | asikurr/Python-learning-And-Problem-Solving | /3.Chapter three/elseif_stste.py | 350 | 4.125 | 4 | age = input("Input Your Age : ")
age = int(age)
if 0<age<=3:
print("Ticket is free for you. ")
elif 3<age<=10:
print("Ticket price is 150 tk.")
elif 10<age<=20:
print("Ticket price is 250 tk.")
elif 20<age<=150:
print("Ticket Price is 350 tk.")
elif age == 0 or 0>age or age>150:
print("Sorry... You are not exist in the World") | true |
1ba6cdf92840fef7f2deef408520de98826d61e9 | asikurr/Python-learning-And-Problem-Solving | /8.Chapter eight SET/set.py | 499 | 4.28125 | 4 | # Set is unordered collection of unique data item
# why we user set method
# Because it contain unique data
#but we cannot user 'list' and 'dictionary' in set
# what is we do by set functon
# list data unique and tuple data unique
s = {1,2,3}
# print(s)
list1 = [6,3,4,5,4,3,5,4,4,3,3,4,6,8]
l = list(set(list1)) # Removed Duplicate data from list using set method
# print(l)
s.add(4)# add data in set
#s.remove(2)# Remove dat from set
#s.discard(4)
# s.clear()
s1 = s.copy()#copy set
print(s) | true |
34585338db8d706740fe31e9c42ba2dd934580e6 | asikurr/Python-learning-And-Problem-Solving | /2.Chapter two/exercise3.py | 517 | 4.21875 | 4 | #String case insensetive
name , char = input("Enter a Name and a character separate by coma : ").split(",")
print(f"the length of your name : {len(name)}")
print(f"Number of Character in your name case sensetive: {name.count(char)}")#case sensetive
print(f"Number of Character in your name case insensetive: {name.lower().count(char)}")
#remove spacese
# name.strip().lower().count(char.strip().lower())
print(f"Number of Character in your name case insensetive: {name.strip().lower().count(char.strip().lower())}") | false |
832e49a939e333b9d95193acd6aae656167855df | asikurr/Python-learning-And-Problem-Solving | /6.Chapter six/about_tuple.py | 667 | 4.34375 | 4 | #looping in tuple
#tuple in one element
# tuploe without paranthesis
# tuple unpacking
# list inside tuple
# some function that you can use tuple
mix = (2,1,4,1,4,5,6,32)
# i = 0
# while i<len(mix):
# print(mix[i])
# i+=1
# for i in mix:
# print(i)
#Single element with tuple
n = (1,)
s = ('asikur',)
# print(type(n))
# print(type(s))
#Tuple without paranthesis
name = 'asikur', 'rahaman asikur', 'rahaman'
# print(type(name))
#Tuple upaking
n1,n2,n3 = (name)
# print(n1)
#List inside tuple
nam = ('asikur','rahaman asikur',['rahaman','fayez'])
# nam[2].pop()
nam[2].append("Abdur Rahaman")
# print(nam)
# some function using tuple max-min-sum-len-count
print(sum(mix)) | false |
6e632c820c3eda237542b2035be429ead1069019 | asikurr/Python-learning-And-Problem-Solving | /8.Chapter eight SET/more_set.py | 343 | 4.125 | 4 | # more about set method
# Union and Intersection methode
s = {'a','b','v', 'd'}
# if 'as' in s:
# print('Present')
# else:
# print('Not present')
# for i in s:
# print(i)
# Union operation
s1 = {1,2,3,3,4}
s2 = s | s1
# print(s2)
#Intersection
s3 = {1,2,3,4,5,6,7,8,9}
s4 = {1,2,3,45,6,7,8,9,10,11,12}
s5 = s3 & s4
print(s5)
| false |
674fb6ebc81e0fca4a75301afcd0ec17922d5051 | chamzheng/python-cookbook | /c01_data_structures_algorithms/p09_find_commonalities_in_dicts.py | 1,943 | 4.1875 | 4 | # -*- coding:utf-8 -*-
# 问题
# 怎样在两个字典中寻寻找相同点(比如相同的键、相同的值等等)？
# 解决方案
# 考虑下面两个字典：
a = {
'x': 1,
'y': 2,
'z': 3
}
b = {
'w': 10,
'x': 11,
'y': 2
}
# 为了寻找两个字典的相同点，可以简单的在两字典的 keys() 或者 items() 方法返回结果上执行集合操作。比如：
# Find keys in common
print(a.keys() & b.keys()) # {'y', 'x'}
# Find keys in a that are not in b
print(a.keys() - b.keys()) # {'z'}
# Find (key, value) pairs in common
print(a.items() & b.items()) # {('y', 2)}
# 这些操作也可以用于修改或者过滤字典元素。 比如，假如你想以现有字典构造一个排除几个指定键的新字典。
# 下面利用字典推导来实现这样的需求：
# Make a new dictionary with certain keys removed
c = {key: a[key] for key in a.keys() - {'z', 'w'}}
print(c) # {'x': 1, 'y': 2}
# 讨论
# 一个字典就是一个键集合与值集合的映射关系。
# 字典的 keys() 方法返回一个展现键集合的键视图对象。
# 键视图的一个很少被了解的特性就是它们也支持集合操作，比如集合并、交、差运算。
# 所以，如果你想对集合的键执行一些普通的集合操作，可以直接使用键视图对象而不用先将它们转换成一个set。
# 字典的 items() 方法返回一个包含(键，值)对的元素视图对象。
# 这个对象同样也支持集合操作，并且可以被用来查找两个字典有哪些相同的键值对。
# 尽管字典的 values() 方法也是类似，但是它并不支持这里介绍的集合操作。
# 某种程度上是因为值视图不能保证所有的值互不相同，这样会导致某些集合操作会出现问题。
# 不过，如果你硬要在值上面执行这些集合操作的话，你可以先将值集合转换成set，然后再执行集合运算就行了。
| false |
edb125836089b25155cd4c46222393d502e7881e | wangmengyu/pythontutorial27 | /ch04/unchangeable.py | 625 | 4.125 | 4 | # -*- coding: utf-8 -*
# 参数默认值
i = 5
def my_fun(num=i):
"""默认值只被赋值一次，传递了不可变的对象"""
print num
i = 6
my_fun()
def append_list(a, l=[]):
"""默认值只被赋值一次,传递了可变的对象"""
l.append(a)
return l
print append_list(1)
print append_list(2)
print append_list(3)
def append_list_2(a, l=None):
"""将传递的可变的对象换成NOne,函数内部进行初始化，可避免每次调用会影响该对象"""
if l is None:
l = []
l.append(a)
return l
print append_list_2(1)
print append_list_2(2)
print append_list_2(3)
| false |
e433f895bb2fc8d0cf6607b9e5801bd8f61f57c4 | erikaosgue/python_challenges | /easy_challenges/7-write_a_function.py | 349 | 4.1875 | 4 | #!/usr/bin/python3
def is_leap(year):
leap = False
# leap year have to be divided by 4 and not by 100, or
# leap year have to be divided by 400
if (year % 4 == 0 and year % 100 != 0) or year % 400 == 0:
leap = True
return leap
if __name__ == "__main__":
year = int(input())
print(is_leap(year)) | false |
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