blob_id string | repo_name string | path string | length_bytes int64 | score float64 | int_score int64 | text string | is_english bool |
|---|---|---|---|---|---|---|---|
2fb8ce9f6ce719f4903127076c9582db1bab1034 | django-group/python-itvdn | /домашка/starter/lesson 3/Alex_Osmolowski/FuncValue.py | 228 | 4.125 | 4 | import math
x = input("Введите значение Х: ")
x = float(x)
if - math.pi <= x <= math.pi:
y = math.cos(3*x)
print("y = Cos(3x) = {}".format(y))
else:
y = x
print("y = x = {}".format(y)) | false |
21dda200a1ad54a3facb2892e4646967c87da327 | django-group/python-itvdn | /домашка/starter/lesson 6/Marukhniak Denys/Lesson_6_4.py | 435 | 4.25 | 4 | # Задание
# Напишите рекурсивную функцию, которая вычисляет сумму натуральных чисел, которые
# входят в заданный промежуток.
def foo(a, b, res):
res += a
a += 1
if a <= b:
return foo(a, b, res)
else:
return res
# a = int(input('From '))
# b = int(input('to '))
a = 3
b = 5
print(foo(a, b, res=0))
| false |
a35eb6b937034d3addc77726a50ac29e742c6820 | django-group/python-itvdn | /домашка/essential/lesson 3/Dmytro Marianchenko/t_2.py | 1,753 | 4.1875 | 4 | def operations(a, b):
"""
Operations with input numbers
"""
while True:
operation = input("What operation do you need:\n1. If add enter '+' \n2. If subtract enter '-' \n3. If "
"multiplication enter '*' \n4. If division enter '/'\n5. If exponentiation enter '^' \n>> ")
if operation == "+":
r = a + b
return r
elif operation == "-":
r = a - b
return r
elif operation == "*":
r = a * b
return r
elif operation == "/":
try:
r = a / b
return r
except ZeroDivisionError:
r = "no result"
return r
elif operation == "^":
try:
r = a ** b
return r
except ZeroDivisionError:
r = "no result"
return r
else:
print("Wrong operation choice... try again")
def inp_nums(x):
"""
Rule of initialisation inserted number
"""
while x is None:
try:
x = float(input("enter a number:\n>> "))
except ValueError:
print("It is not a number... try again")
return x
def main():
"""
Main program
"""
num1 = None
num2 = None
num1 = inp_nums(num1)
print("and second number")
num2 = inp_nums(num2)
result = operations(num1, num2)
result_end = result % 1
if result_end == 0:
print(round(result))
else:
print(result)
to_do = input("If you want to continue press 'Enter' or input 'exit':\n>> ")
if to_do == exit:
exit()
else:
main()
if __name__ == '__main__':
main()
| false |
2c9b58064ab9e14745f8be10ad96b7e98af65911 | django-group/python-itvdn | /домашка/starter/lesson 5/Alex_Osmolowski/hw_5_add.py | 1,234 | 4.375 | 4 | # Создайте программу, которая состоит из функции, которая принимает три числа и возвращает
# их среднее арифметическое, и главного цикла, спрашивающего у пользователя числа и
# вычисляющего их средние значения при помощи созданной функции.
def average3(x1, x2, x3):
return (x1 + x2 + x3)/3
def main():
while True:
op = input("Введите опреацию: ")
if op.lower() == "exit": # выход
return
elif op == "avg": # среднее арифметическое трёх чисел
x1 = input("Введите первое число x1 = ")
x1 = float(x1)
x2 = input("Введите второе число x2 = ")
x2 = float(x2)
x3 = input("Введите третье число x3 = ")
x3 = float(x3)
print("Среднее арифметическое трёх введённых чисел: {}".format(average3(x1, x2, x3)))
if __name__ == '__main__':
main() | false |
85cc9e7571335011818b5c12d4e06c0e74e0e2ff | django-group/python-itvdn | /домашка/starter/lesson 6/Dmytro Marianchenko/t_2.py | 680 | 4.25 | 4 | #это еще один способ кроме [::-1}, но правда сложнее, но еще один вариант)))
def palindrome(word="mom"):
if len(word) == 3:
if word[0] == word[2]:
print(f"Слово '{word}' полиндром")
else:
print(f"Слово '{word}' не полиндром")
elif word[0] == word[-1] and palindrome(word[1:-1]):
print(f"Слово '{word}' это полиндром =)")
else:
print(f"Слово '{word}' не полиндром")
word = input("Введи сло и я попробую угадать полиндром это или неи: ")
palindrome(word) | false |
a8c74c5a2cdfdbbf43de4fda2580bb87fc5694af | django-group/python-itvdn | /домашка/starter/lesson 5/Vitalii_K/foo_calc(task3).py | 1,746 | 4.1875 | 4 | def calc_add(a, b):
return a + b
def calc_sub(a, b):
return a - b
def calc_mult(a, b):
return a * b
def calc_div(a, b):
if b != 0:
return a / b
else:
return "Divisor must not be zero! OBVIOUSLY!"
while True:
print("Enter '1' if u want to calculate.",
"Enter '0' if u want to exit.", sep='\n')
j = input()
if j == "1":
op = input("Enter operation ('+', '-', '*', '/'): ")
if op != "+" and op != "-" and op != "*" and op != "/":
print("Read the parentheses part, plese...")
else:
x = input("Enter the number: ")
try:
x = float(x)
except ValueError:
print("You wanna try to calculate things other than numbers?"
"\nNot on my watch!")
continue
x = round(x)
y = input("Enter the number: ")
try:
y = float(y)
except ValueError:
print("You wanna try to calculate things other than numbers?"
"\nNot on my watch!")
continue
y = round(y)
if op == "+":
print(f"{x} + {y} = {calc_add(x, y)}")
elif op == "-":
print(f"{x} - {y} = {calc_sub(x, y)}")
elif op == "*":
print(f"{x} * {y} = {calc_mult(x, y)}")
elif op == "/":
if y != 0:
print(f"{x} / {y} = {calc_div(x, y)}")
else:
print(calc_div(x, y))
elif j == "0":
break
else:
print("It is a simple choise...\nDon’t overcomplicate things...")
input("Oh boy, that escalated quickly...")
| true |
ec6ab23e48d0e8c6d838e7a04dbb55e232f4a922 | django-group/python-itvdn | /домашка/essential/lesson 1/MaximKologrimov/Task Dop.py | 1,250 | 4.53125 | 5 | # Задание
# Создайте класс, описывающий автомобиль. Создайте класс автосалона, содержащий в себе список
# автомобилей, доступных для продажи, и функцию продажи заданного автомобиля.
class Cars:
def __init__(self, car_new, color_new, maxspeed_new):
self.car = car_new
self.color = color_new
self.maxspeed = maxspeed_new
def __str__(self):
return f'Марка Авто: {self.car}\n' \
f'Цвет: {self.color}\n' \
f'Максимальная скорость: {self.maxspeed}'
def __repr__(self):
return f'Марка Авто: {self.car}; ' \
f'Цвет: {self.color}; ' \
f'Максимальная скорость: {self.maxspeed}'
car1 = Cars('Ford', 'Black', 320)
car2 = Cars('Mazda', 'Red', 240)
print(car1)
print()
print(car2)
print()
class Market():
def price(self, price_new):
self.price = price_new
p1 = Market()
p1.price(30000)
p2 = Market()
p2.price(20000)
list = [car1.car, p1.price, car2.car, p2.price]
print(list) | false |
be5efa14fb63149c56ab803accce47330f5ae671 | django-group/python-itvdn | /домашка/starter/lesson 3/Dmytro Marianchenko/t_2.py | 293 | 4.25 | 4 | import math
print("Введите значение 'x':")
x = int(input(""))
if -math.pi <= x <=math.pi:
y = math.cos(3*x)
print(f"Ответ: {y}")
elif x < -math.pi or x > math.pi:
y = x
print(f"Ответ: {y}")
input("Для завершения нажмите 'Enter'...") | false |
a36704f61931057f842ba2cb0bc2a2d0b77dd7d8 | django-group/python-itvdn | /домашка/starter/lesson 3/Dmytro Marianchenko/t_4.py | 1,596 | 4.1875 | 4 | import math
print("Выберите пожалуйста НОМЕР МЕНЮ необходимой операции:")
print("1. '+'","2. '-'","3. '*'","4. '/'","5. '%'","6. Площадь круга", sep="\n")
oper = input("")
if oper == "1": #сложение
print("введите первое число:")
x = int(input(""))
y = int(input(""))
print(f"Ответ: {x + y}")
elif oper == "2": #разница
print("введите первое число:")
x = int(input(""))
y = int(input(""))
print(f"Ответ: {x - y}")
elif oper == "3": #умножение
print("введите первое число:")
x = int(input(""))
y = int(input(""))
print(f"Ответ: {x * y}")
elif oper == "4": #деление
print("введите первое число:")
x = int(input(""))
y = int(input(""))
print(f"Ответ: {x / y}")
elif oper == "5":
print("введите число от которого необходимо высчитать процент:")
x = int(input(""))
print("введите количество процентов в целым числом:")
y = int(input(""))
pers = x*(y/100)
print(f"Ответ: {round(pers,2)}")
elif oper == "6":
print("введите радиус:")
r = int(input(""))
s = math.pi*(r**2)
print(f"Ответ: {round(s,2)}")
else:
print(f"операция {oper}, при необходимости появится в следующей версии =)")
print("Для завершения нажмите кнопку 'Enter'...") | false |
21f9f280ef5d788c64b929a14ca1c1fc739016cf | django-group/python-itvdn | /домашка/starter/lesson 6/MaximKologrimov/Task 1.py | 1,685 | 4.1875 | 4 | # Задание 1
# Прочитайте в документации по языку Python информацию о перечисленных в резюме данного
# урока стандартных функциях. Проверьте их на практике.
def foo(a):
"""
Задание 1
Прочитайте в документации по языку Python информацию о перечисленных в резюме данного
урока стандартных функциях. Проверьте их на практике.
"""
a += 1
return a
x = 5
y = -5
real = -7
imag = 7
z = 7.1234567
s = 'Abraham'
print('abs(x) =', abs(x))
print('bin(x) =', bin(x))
print('bool(x) =', bool(x))
print('callable(f) =', callable(foo), callable(x))
print('chr(code) =', chr(1052))
print('complex(real, imag) =', complex(real, imag))
print('dir(obj) =', dir(print))
print('float(x) =', float(x))
print('format(x, fmt) =', '{}, {}'.format(y, x))
print('help(obj) =', help(print))
print('hex(x) =', hex(x))
print('id(obj) =', id(foo))
#input('who are you?')
print('int(x) =', int(x))
print('len(s)', len(s))
print('max(arg1, arg2, …) =', max(2, 77, 10))
print('min(arg1, arg2, …) =', min(2, 77, 10))
print('oct(x) =', oct(x))
print('ord(c) =', ord('m'))
print('pow(x, n) =', pow(x, imag))
print('print()')
print('range() =', min(range(77)),',', max(range(77)))
print('repr(obj) =', repr(help))
print('reversed(iterable):', list(reversed(s)))
for i in reversed(range(7)):
print(i)
print('round(number, ndigits) =', round(z,2))
print('sorted(iterable) =', sorted(s))
print('str(x) =', type(str(x)), type(x))
print(foo.__doc__)
| false |
1620a9cd97e0d9940eddad7852d682c54c4e8e51 | django-group/python-itvdn | /домашка/starter/lesson 7/Marukhniak Denys/Lesson_7_4.py | 482 | 4.34375 | 4 | # Создайте список, введите количество его элементов и сами значения, выведите эти значения на
# экран в обратном порядке.
cloud = []
element = 0
length = int(input('Enter length of list: '))
for i in range(length):
element = input(f'Enter {i+1} element of list: ')
cloud.append(element)
print('Your list with reversed elements:')
for i in cloud[::-1]:
print(i, end=' ')
| false |
eaa903dfea9fdb6184ce351ea87798c906504fec | django-group/python-itvdn | /домашка/starter/lesson 5/Marukhniak Denys/Lesson_5_2.py | 514 | 4.21875 | 4 | # Задание 2
# Создайте две функции, вычисляющие значения определённых алгебраических выражений.
# Выведите на экран таблицу значений этих функций от -5 до 5 с шагом 0.5.
def f1(x1=-5):
y = x1 ** 2
return y
def f2(x2=-5):
y = 1-x
return y
x = -5
print(f" x | y=x^2 | y=1-x")
while x < 5:
x += 0.5
print('{} | {:4.2f} | {:4.2f}'.format(x, f1(x), f2(x)))
| false |
8132d6fb46f2ee1b9edafe88addf3394289d88b5 | django-group/python-itvdn | /домашка/starter/lesson 4/KologrimovMaxim/Task dop.py | 423 | 4.3125 | 4 | #Задание
#Создайте программу, которая рисует на экране прямоугольник из звёздочек заданной
#пользователем ширины и высоты.
x = int(input('Введите высоту: '))
y = int(input('Введите ширину: '))
print()
for h in range(x):
for w in range(y):
print('*', end='')
print() | false |
b7984b9952e622516e17766668f8401c787fe308 | ericjiang107/Practice-Questions | /PermutationString.py | 1,256 | 4.1875 | 4 | '''
Problem Challenge:
Permutation in a String (hard)
Given a string and a pattern, find out if the string contains any permutation of the pattern.
Permutation is defined as the re-arranging of the characters of the string. For example, “abc” has the following six permutations:
abc
acb
bac
bca
cab
cba
If a string has ‘n’ distinct characters it will have n!n! permutations.
Example 1:
Input: String="oidbcaf", Pattern="abc"
Output: true
Explanation: The string contains "bca" which is a permutation of the given pattern.
Example 2:
Input: String="odicf", Pattern="dc"
Output: false
Explanation: No permutation of the pattern is present in the given string as a substring.
Example 3:
Input: String="bcdxabcdy", Pattern="bcdyabcdx"
Output: true
Explanation: Both the string and the pattern are a permutation of each other.
Example 4:
Input: String="aaacb", Pattern="abc"
Output: true
Explanation: The string contains "acb" which is a permutation of the given pattern.
'''
def permutation(String, Pattern):
empty_dic = {}
window_start = 0
match = 0
for letter in range(len(String)):
if String[letter] not in empty_dic:
empty_dic[String[letter]] = 1
else:
empty_dic[String[letter]] += 1
| true |
d98b6ee393e45ef9c6919442fe362f84e2e2024e | CoolM0nk3y/Functions | /pay.py | 1,022 | 4.15625 | 4 |
#calculate basic pay
def calculate_basic_pay(hours,rate):
total = hours * rate
return total
#calculate the over time pay
def calculate_overtime(hours,rate):
over_time = hours - 40
basic_pay = 40 * rate
overtime_pay = over_time * rate * 1.5
total = overtime_pay + basic_pay
return total
# the statment
def calculate_total_pay(hours,rate):
if hours <= 40:
total = calculate_basic_pay(hours,rate)
else:
total = calculate_overtime(hours,rate)
return total
# Get the data from the user
def get_data():
hours = int(input("Please enter the numbers of hours you do: "))
rate = int(input("Please enter the rate of pay you have: "))
return hours , rate
#calculate basic pay
def calculate_pay():
hours,rate = get_data ()
total_pay = calculate_total_pay (hours,rate)
dis_pay (total_pay)
#display the total
def dis_pay (total_pay):
print(total_pay)
#main program
calculate_pay()
| true |
e5466b52c74df1e3445d2753a97fc34d2c4bdf19 | Azarkasb/LearnPython | /UseLessProjects/Mosalas.py | 630 | 4.21875 | 4 | '''
Barname baraye tashkis noe mosalas bar asas azla
'''
# AZ
a = int(input('Pls enter the first side ... '))
b = int(input('Pls enter the second side ... '))
c = int(input('Pls enter the third side ... '))
if a + b <= c or a + c <= b or b + c <= a:
print('Impossible')
elif a == b and a == c:
print('Equilateral Triangle')
elif a == b or b == c:
if a**2 + b**2 == c or b**2 + c**2 == a**2 or a**2 + c**2 == b**2:
print('Isosceles Right Triangle')
else:
print('Isosceles Triangle')
elif a**2 + b**2 == c or b**2 + c**2 == a**2 or a**2 + c**2 == b**2:
print('Right Triangle')
else:
print('Normal Triangle')
| false |
bebc8bbdb3f3ea2e03da7f70663bc5d202b672b9 | jtcgen/PythonPuzzles | /balance.py | 2,566 | 4.125 | 4 | '''
Can we save them? Beta Rabbit is trying to break into a lab that contains the only known zombie cure - but there's an
obstacle. The door will only open if a challenge is solved correctly. The future of the zombified rabbit population
is at stake, so Beta reads the challenge: There is a scale with an object on the left-hand side, whose mass is given
in some number of units. Predictably, the task is to balance the two sides. But there is a catch: You only have this
peculiar weight set, having masses 1, 3, 9, 27, ... units. That is, one for each power of 3. Being a brilliant
mathematician, Beta Rabbit quickly discovers that any number of units of mass can be balanced exactly using this set.
To help Beta get into the room, write a method called answer(x), which outputs a list of strings representing where
the weights should be placed, in order for the two sides to be balanced, assuming that weight on the left has mass x units.
The first element of the output list should correspond to the 1-unit weight, the second element to the 3-unit weight,
and so on. Each string is one of:
"L" : put weight on left-hand side
"R" : put weight on right-hand side
"-" : do not use weight
To ensure that the output is the smallest possible, the last element of the list must not be "-".
x will always be a positive integer, no larger than 1000000000.
Test cases
==========
Inputs:
(int) x = 2
Output:
(string list) ["L", "R"]
Inputs:
(int) x = 8
Output:
(string list) ["L", "-", "R"]
'''
from math import log, ceil, floor
import sys
def answer(x):
lgx = log(x,3)
# All powers of 3
if 3**floor(lgx) == x:
return ['-' if i != lgx else 'R' for i in range(int(lgx+1))]
# range of powers of 3 to look at
up_bnd = ceil(lgx)
low_bnd = floor(lgx)
branch = 3**up_bnd - x
# If diff between upper bnd and x is greater than x, answer will contain up to lower bnd
if branch > x:
ans = ['-'] * (low_bnd+1)
diff = answer(x-(3**low_bnd))
else:
# Answer will contain upper bnd
ans = ['-'] * (low_bnd+2)
diff = answer((3**up_bnd)-x)
# Populate result
for i in range(low_bnd+1):
if i < len(diff):
if branch > x:
ans[i] = '-' if diff[i] == '-' else 'R' if diff[i] == 'R' else 'L'
else:
ans[i] = '-' if diff[i] == '-' else 'L' if diff[i] == 'R' else 'R'
else:
ans[i] = '-'
ans[len(ans)-1] = 'R'
return ans
if __name__ == '__main__':
print(answer(sys.argv[1]))
| true |
6bfe2d8b1498b161c31f7d2ea809244a104e05e8 | Thea-J/python_practice | /tuple.py | 513 | 4.1875 | 4 | # Define a tuple & assign it to a variable
t = (1, 2, 3, 4)
# Use square-bracket notation to retrieve tuple element
# Note: tuple indices start at 0
x = t[0]
# Create a new list from a tuple
# via list reserved word
l = list(t)
# Define a range from 0 to inserted_number-1
# via range reserved word
r= range(7) #k = (0, 1, 2, 3, 4, 5, 6)
# Use square-bracket notation to retrieve range element
# Note: range indices start at 0
y= r[5]
# Create a new list from a range
# via list reserved word
z = list(r) | true |
fdab5914d1fb693b4ad7e7cdf549e76001b0e9db | portiaportia/csce204-assignment-answers | /12-file-writing/question-game-original.py | 1,769 | 4.25 | 4 | import random
FILE_NAME_TRIVIA = "assignments/12-file-writing/trivia.txt"
# Reach each line of the file into a dictionary. The dictionary will look like:
# Question -> Answer
# We use a convert to bool method to convert "true" to True
def getQuestions():
questions = {}
with open(FILE_NAME_TRIVIA) as file:
for line in file:
data = line.split(':')
question = data[0].strip()
answer = data[1].strip()
questions[question] = convertToBool(answer)
return questions
# Converts a string representation of a boolean to a Boolean
def convertToBool(answer):
if answer == "true":
return True
else:
return False
# Asks the user a random trivia question
# Gathers the users answer
# if their answer is the stored answer then return true (they won), otherwise return false
def playRound():
question = random.choice(list(questions.keys()))
userAns = input(f"True or False: {question} ").strip().lower()
compAns = questions.get(question)
if convertToBool(userAns) == compAns:
return True
else:
return False
# Let's play the game
print("Welcome to our Trivia Game")
# Load all the questions from the file
questions = getQuestions()
# Loop as long as the user wants to keep playing
while True:
command = input("(P)lay or (Q)uit: ").strip().lower()
# if they enter q, quit the game
if command == "q":
break
# if they enter an invalid command, indicate that
elif command != "p":
print("Invalid command")
continue
# otherwise play around, if playRound is true, they won
if playRound():
print("Yay, you got it!")
else:
print("Sorry, incorrect")
print("Goodbye")
| true |
c20ebef4edfce9a0dbb6eded6b23a2b9829b7f7c | sjriddle/CTCI | /2/a23.py | 446 | 4.125 | 4 | ## Implement an algorithm to delete a node in the middle (i.e any node but the
## first or last node, not necessarily the exact middle) of a singly linked list,
## given only access to that node
## EXAMPLE
## Input: The node c from linked list a->b->c->d->e->f
## Output: Nothing is returned, but the new linked list is a->b->d->e->f
# Memory: O(1)
# CPU: O(1)
def delete_node(node):
node.data = node.next.data
node.next = node.next.next
| true |
f45163ea2a7e32eb46f138321fe860c4ad356a12 | Yamikaze123/Nguyen-Tien-Hoang-c4t3-session-3-hw | /Fruit.py | 562 | 4.21875 | 4 | prices = {
"banana": 4,
"apple": 2,
"orange": 1.5,
"pear": 3
}
stock = {
"banana": 6,
"apple": 0,
"orange": 32,
"pear": 15
}
print()
print("Fruits' price and stock information:")
print()
for price in prices:
print(price)
print("price:", prices[price])
print("stock:", stock[price])
print()
print()
print("Money made:")
print()
total = 0
for price in prices:
print(price, end=": ")
multi = prices[price] * stock[price]
print(int(multi))
total = total + multi
print()
print("Total:", total)
| true |
c858740aca4412c22f714ae3e19d063c10383f54 | AsgardRodent/MyPy | /arguments.py | 1,143 | 4.1875 | 4 |
def print_name_age(name = "someone", age = "ancient"):
print("hi my name is ", name ,"i am ", age, " years old")
print_name_age("gaurav", 20)
print_name_age("regigas")
print_name_age("supahotfire", 28)
print_name_age(age=32)
# to solve the above error u can directly replace the line(4) with the following code which just [rints all the data as itself
# print("hi my name is ", name ,"i am ", age , " years old") ----> this is to be replaced
# as a back up the default line is provided below :D
# print("hi my name is " + name + " i am " + str(age) + " years old")--------> default line
# in line(9) only 32 is passed which will be replaced by the name since it as per the definition/the parameters provided to the function
# so the keyword "None" is used to skip a parameter hence the new line will be :D
# print_name_age(None,32) ------> new line
# print_name_age(32)---------> default line
# none --> bool
# this error can be sorted using key arguments like #print_name_age(age = 32)
# REMEMBER KEY ARGUMENTS CAN BE PASSED IN ANY MANNER AS YOU LIKE ,SINCE THEY ARE DEFINED TO A SPECIFIC VALUE
| true |
6d10d06f34d99747f20a97e50a7fd3ab6eba4c66 | moonlimb/scheme_to_js_translator | /refs/game/comp_guess.py | 1,468 | 4.28125 | 4 | import random
# Human will make random number
# Computer will make an initial constant guess: 50 constant. And print this number.
# The human will give feedback to computer if number guess is too hi or too low
# THe computer will then make a new guess, which is a middle number in the new range.
# the computer will repeat the process until it reaches the final number
# count the number of guesses
def human_feedback():
return raw_input("\nIs my guess 'too high', 'too low', or 'perfect'? ")
def smart_CS_guess():
CS_guess = 50
count = 1
# human_answer = a certain number
too_high = "too high"
too_low = "too low"
perfect = "perfect"
print "Hi. My name is Max the computer. I made my first guess: %d." %CS_guess
human_input = human_feedback()
#The computer does thinking below:
low_bound = 1
high_bound = 100
while (human_input != perfect):
if (human_input != too_high) and (human_input != too_low):
print "EPIC FAIL! Type in a sensible reply, you're not a clever human."
break
elif human_input == too_high:
high_bound = CS_guess - 1
print low_bound, high_bound
else: #human_input == too_low:
low_bound= CS_guess+1
print low_bound, high_bound
CS_guess = (low_bound + high_bound)/2
count+=1
print count
print "This is my new guess: %d." %CS_guess
human_input = human_feedback()
print "Congratulations Max! %d is the correct number. You got it in %d trials." %(CS_guess, count)
smart_CS_guess()
| true |
f7dae2376580f4e0e11a51fa9858e1bed7588e4a | bloobloons/codewars | /Thinkful - List Drills: Longest word.py | 250 | 4.1875 | 4 | # Write a function longest() that takes one argument, a list of words,
# and returns the length of the longest word in the list.
def longest(words):
lengths = []
for each in words:
lengths.append(len(each))
return max(lengths)
| true |
5a263cb2680ab8313b234d847dba5d09e23c2d47 | Nep-DC-Exercises/day-3 | /box.py | 481 | 4.1875 | 4 | # Given a height and width, input by the user, print a box consisting of * characters as its border.
width = int(input("Enter width of box > "))
height = int(input("Enter height of box > "))
star = "*"
i = 0
mid = width - 2
while i < height:
if i == 0: # top of box
print(width * star)
elif i == height - 1: # bottom of box
print(width * star)
elif i >= 1 or i < height - 1: # middle rows of box
print("*" + mid * " " + "*")
i += 1
| true |
08e7fa564f63818b5727ddc51905ec097e185d83 | Nep-DC-Exercises/day-3 | /blastoff4.py | 255 | 4.1875 | 4 | # Make sure user doesn't enter a number larger than 20
i = int(input("Enter the number to start counting from. >> "))
if i <= 20:
while i >= 0:
print(i)
i -= 1
else:
print("That number was too high. It can't be larger than 20.")
| true |
ed7e13bf6e91d2ab770f60b9c406d91663298947 | EricNg314/Code-Drills | /Python/day-01/04/challenge-prompt.py | 837 | 4.15625 | 4 | # Declare a variable of bobaShop with an input and a string of "Welcome to the Boba Shop!"
# Check if bobaShop is equal to true
# Write a print with a string of "Hello"
# Declare a variable of beverage with an input and a string of "What kind of boba drink would you like ?"
# Declare a variable of sweetness with an input and a string of "How sweet do you want your drink 0,50,100,200 ?"
# Now check
# if sweetness equals to 50 print "half sweetened"
# else if sweetness 100 print "normal sweet"
# else if sweetness 200 print "super sweet"
# else print with a string of "non-sweet"
# then print with a string of "your order of " variable beverage and a string of " boba with a sweet level of " and variable of sweetness
# and print string of "goodbye".
| true |
c180f9481c1e727b53060f5f6a018f829457f2d7 | YuJianMuNaiYi/Python | /day1/do_set.py | 441 | 4.15625 | 4 | s = set([1, 2, 3])
print(s)
# 重复元素在set中自动被过滤
t = set([1, 1, 2, 2, 3, 3, 4, 5, 6])
print(t)
# 添加元素
t.add(7)
t.add(3)
t.add(8)
t.add(7)
print(t)
# 移除元素
t.remove(3)
print(t)
# set可以看成数学意义上的无序和无重复元素的集合，因此，两个set可以做数学意义上的交集、并集等操作
print(s & t)
print(s |t)
#对于可变对象
a=['c', 'b', 'a']
a.sort()
print(a) | false |
e272790061c5be0d13216da2474c2dd89cc22e58 | EricaGuzman/PersonalProjects | /Python/payCheck/payCheck.py | 1,354 | 4.125 | 4 | #Erica Guzman
#This program calculates user's weekly paycheck
import math
#Define main
def main():
#Declare and initilaize float variables
weeklySalesAmount = commission = grossPay = netPay = socialSecurity = federalTax = 0.0
#constant for base pay
basePay = 400.00
#display Welcome message:
print("Welcome to your weekly paycheck generator!");
weeklySalesAmount = float(input("Enter the dollar amount of sales made for the week: "));
#calculations:
commission = weeklySalesAmount * .06
grossPay = weeklySalesAmount + basePay + commission
socialSecurity = grossPay * .06
federalTax = grossPay * .12
netPay = grossPay -(socialSecurity + federalTax)
#print paycheck
print ("*" * 40)
print("Base Pay:\t ${:,.2f}" .format (basePay));
print("Sales:\t\t ${:,.2f}" .format (weeklySalesAmount));
print("Commission:\t ${:,.2f}" .format (commission));
print("Gross Pay:\t ${:,.2f}" .format (grossPay));
print();
print("Deductions: ");
print ("*" * 40)
print("Social security: ${:,.2f}" .format (socialSecurity));
print("Federal Tax:\t ${:,.2f}" .format (federalTax));
print ("*" * 40)
print("Net Pay:\t ${:,.2f}" .format (netPay));
#display Goodbye
print();
print("Thank you for using the paycheck generator! \n Have a nice day!!");
main()
| true |
a5fe8b3a234987a1f13545bac3d096808d37108a | npraveen35/Python_Codes | /basic_codes/leap_year.py | 778 | 4.15625 | 4 | #!/usr/bin/python
'''
All the years that are perfectly divisible by 4 are called as Leap years except the century years.
Century year's means they end with 00 such as 1200, 1300, 2400, 2500 etc (Obviously they are divisible by 100).
For these century years we have to calculate further to check the Leap year:
If the century year is divisible by 400 then that year is a Leap year
If the century year is not divisible by 400 then that year is a Not Leap year
'''
#V1: This code reads input from user and test for leap year
num=int(input("Enter Year:"))
def leap_year(year):
return ( year % 4 == 0 and year % 100 != 0) or year % 400 == 0
print num,"is leap year" if leap_year(num) else "not a leap year"
#V2 use calender module
#import calendar
#print calendar.isleap(num)
| true |
02629ad2ac00dc8e0c8d274bd5c1c114ce1b62dc | npraveen35/Python_Codes | /basic_codes/words_in_line.py | 255 | 4.25 | 4 | #!/usr/bin/env python3
#v1 COUNTS NUMBER OF WORDS IN A SENTENCE:
def count_words(str):
return len(str.split(' ')) #splits the str with space & creates list
sentence = "An Apple a day keeps a Doctor away !!"
print (count_words(sentence)) #Output is 9
| true |
32de88abc8173d997907f1a87d23c8ba7bd0520b | LiamLead/to-do-list | /to-do-list.py | 1,755 | 4.25 | 4 | print("TO DO LIST\n")
class Queue(object):
def __init__(self):
self.storage = []
def enqueue(self, new_element):
self.storage.append([new_element])
def display_all(self):
return self.storage
def display_one(self):
return self.storage[-1]
def dequeue(self):
return self.storage.pop(0)
def delete_all(self):
return self.storage.clear()
def append(self):
add = str(input("Add: "))
self.storage.append([add])
q = Queue()
q.enqueue("Breakfast")
q.enqueue("Go to work")
q.enqueue("Lunch")
print(q.display_all())
drink_coffee = "Drink Coffee"
buy_chocolate = "Buy Chocolate"
drink_milk_tea = "Drink Milk Tea"
go_to_run = "Go To Run"
clear_list = "Empty the list"
show_list = "Check your list"
add_own_item = "Add your Own Item"
print("")
print("0: " + show_list)
print("1: " + drink_coffee)
print("2: " + buy_chocolate)
print("3: " + drink_milk_tea)
print("4: " + go_to_run)
print("5: " + clear_list)
print("6: " + add_own_item)
print("9: Exit the program")
while True:
choice = int(input("\nChoose one thing to add to the 'to do list': "))
if choice == 1:
q.enqueue(drink_coffee)
print(f"{q.display_one()}, Added to the list.")
elif choice == 2:
q.enqueue(buy_chocolate)
print(f"{q.display_one()}, Added to the list.")
elif choice == 3:
q.enqueue(drink_milk_tea)
print(f"{q.display_one()}, Added to the list.")
elif choice == 4:
q.enqueue(go_to_run)
print(f"{q.display_one()}, Added to the list.")
elif choice == 5:
q.delete_all()
elif choice == 0:
print(q.display_all())
elif choice == 6:
q.append()
elif choice == 9:
break | false |
bd840a9f621e4791a9440de95ecb1eb65a564d65 | amrutbadiger/python | /exercise1.py | 224 | 4.21875 | 4 | """
Write a Python program which accepts the user's first and last name and print them in reverse order with a space between them.
"""
a=str(input("Enter your First name:"))
b=str(input("Enter your Last name:"))
print (b,a)
| true |
c26ba526f16220c89265c688c06b0df838039302 | amrutbadiger/python | /exer4.py | 220 | 4.15625 | 4 | """
A program to calculate rate of interest.
"""
p=int(input("Enter the principal amount(P):"))
r=float(input("Enter the rate of interest(R):"))
t=int(input("Enter the period(T):"))
i=(p*r*t)/100
print("Interest is:",i)
| true |
5ba74f2ecef7a540f5f448eea2b8bcebbcb4b736 | amrutbadiger/python | /dict2.py | 903 | 4.5625 | 5 | '''
Assign a dictionary item's value to a variable using a
key that's a string. Then append that value to a list
'''
list_of_ages=[]
info={"first name": "Bob", "second name": "Mark", "age": 56}
age=info["age"]
list_of_ages.append(age)
print(list_of_ages)
'''
Assign a dictionary item's value to a variable
using a key that's a string. Then create a
dictionary that contains only that value.
'''
dict={"name": "Bob", "last name": "Mark", "address": "Parkinson Street"}
value=dict["name"]
new_dict={"my_name": value}
print(new_dict)
'''
Targeting the second item by its key, assign the
second item's value to a variable and display it.
'''
relatives ={"father": "Bob", "mother": "Lucy", "sister": "Rose",}
a_relative = relatives["mother"]
print(a_relative)
# In a single line of code,target the third item and display its value.
states = {1: "goa", 2: "assam", 3: "sikkim", 4: "kerala"}
print(states[3])
| true |
643a0c33d238121046ff0e2762afff7bb31579a8 | Elegant-Smile/PythonDailyQuestion | /Answer/Gorit的题解/2019-5-13/字符输出星期.py | 621 | 4.375 | 4 | #字符输入星期
str=input("请输入一个字符:\n")
if str is 'm' or str is 'M':
print("Monday")
elif str is 'T' or str is 't':
str1=input("请继续输入第二个字符:\n")
if str1 is 'u' or str1 is 'U':
print("Tuesday")
elif str1 is 'h' or str1 is 'H':
print("Thursday")
elif str is 'W' or str is 'w':
print("Wednesday")
elif str is 'f' or str is 'F':
print("Friday")
elif str is 's' or str is 'S':
str2=input("请继续输入第二个字符")
if str2 is 'A' or str2 is 'a':
print("Saturday")
elif str2 is 'u' or str2 is 'U':
print("Sunday")
| false |
7a1d91657e4d0c8be5286ad97f6d24e6aae53993 | Elegant-Smile/PythonDailyQuestion | /Answer/Yuzhou_1shu/20190524_2_hypotenuse.py | 363 | 4.21875 | 4 | import math
def hypotenuse(a, b):
'''
:param a: 直角边
:param b: 直角边
:return: 斜边长
'''
return math.sqrt(a ** 2 + b ** 2)
if __name__ == '__main__':
a = int(input("请输入一条直角边:"))
b = int(input("请输入另一条直角边:"))
print("The right triangle third side's length is", hypotenuse(a, b))
| false |
7bcce8de03e1d817be8ffcca8a42b1dd20bd595c | Elegant-Smile/PythonDailyQuestion | /Answer/weixiao/first_answer20190611.py | 1,175 | 4.125 | 4 | # 给定任意一个整数，打印出该整数的十进制、八进制、十六进制（大写）、二进制形式的字符串。
num = input('please input a number>>>')
dec_num = int(num)
print(dec_num)
print(oct(dec_num))
print(hex(dec_num).upper())
print(bin(dec_num))
# 给用户三次输入用户名和密码的机会，要求如下：
# 1）如输入第一行输入用户名为‘Kate’,第二行输入密码为‘666666’，输出‘登录成功！’，退出程序；
# 当一共有3次输入用户名或密码不正确输出“3次用户名或者密码均有误！退出程序。”
def check():
n = 0
while n < 3:
name = input('>>>')
password = input('>>>>')
if name == 'Kate' and password =='666666':
print('登录成功')
break
else:
n += 1
if __name__ == '__main__':
check()
| false |
c6e89c0feafa587a62fc6297c1f71d128bb1588c | NancyKou/python_lianxi | /python_oo/homework2/student.py | 1,248 | 4.125 | 4 | #定义私有属性，双下划线加变量名
#调用私有属性或私有方法，则双下划线加类型加 点，则可以找到私有方法或私有比属性
class Student: #定义一个Student类
def __init__(self,name,score): #初始化参数
self.name = name #定义实例变量name
self.score =score #定义实例变量score
def grade(self): #定义一个类方法grade
if self.score >= 90: #如果分数大于等于90分
print(self.name,',your grade is A！') #打印xx学生的等级为A
elif self.score >= 60: #如果分数大于等于60分
print(self.name,',your grade is B.') #打印xx学生的等级为B
else:
print(self.name,',your grade is C.') #否则打印xx学生的等级为C
def work(self): #定义一个类方法work
print(f'{self.name}学习很努力！') #打印XX学生学习很努力。
stu1 = Student('dannie',90) #实例化一个学生1，给init中定义的参数传入name和score参数
stu2 = Student('Alex',59) #实例化一个学生2，给init中定义的参数传入name和score参数
stu1.grade() #调用grade方法，打印学生1的等级
stu2.work() #调用work方法，打印学生2的努力程度
| false |
17c74c591e48ba4e8c575541b7c50677dba6dd84 | janekob/pp1 | /02-ControlStructures/zadania/18.py | 207 | 4.21875 | 4 | for x in range(31):
if(x%3==0 and not x%5==0):
print("three")
elif(x%5==0 and not x%3==0):
print("five")
elif(x%3==0 and x%5==0):
print('bingo')
else:
print(x) | false |
00f4cf0d311e3d87a4403b81a02818af6f9255b7 | celopesp/PROITEC | /criptotexto.py | 1,704 | 4.15625 | 4 | """César é um detetive que investiga uma série de roubos que acontecem em sua cidade.
Em todo lugar que um crime acontece, a pessoa que cometeu tal crime deixa uma mensagem escrita,
formada por letras maiúsculas e minúsculas.
César conseguiu achar um padrão nestas mensagens e agora extrai um texto oculto em cada mensagem e
pede a sua ajuda para tentar descobrir quem está cometendo tais crimes.
Entrada
A entrada é composta por vários casos de teste.
A primeira linha contém um número inteiro C (2 <= C <= 99) relativo ao número de casos de teste.
Nas C linhas seguintes, haverá mensagens codificadas, todas com um mesmo padrão em relação ao exemplo abaixo.
Saída
Para cada caso de teste de entrada do seu programa, você deve imprimir o texto extraído da mensagem original.
"""
valido = False
while(valido == False):
C = int(input("digite a quantidade de palavras que deseja descriptografar:"))
if(C>1 and C<100):
valido = True
lista = []
for contador in range (1, C+1):
palavra = input(f"digite a {contador}ª palavras que deseja descriptografar:")
tamanho_palavra = len(palavra)
palavra_ordenada = ""
reverso = -1
for contador_ordenar_palavra in range (tamanho_palavra-1, -1, -1):
palavra_ordenada += palavra[reverso]
reverso -= 1
lista.append(palavra_ordenada)
for contador in range (0, C):
tamanho_palavra = len(lista[contador])
palavra = lista[contador]
descriptografada = ""
for aux in range(0,tamanho_palavra):
if(palavra[aux].islower()):
descriptografada += palavra[aux]
print(f'descriptografada:{descriptografada}')
| false |
9799d78ef1457b5853596bb39994810970e77c47 | zedfoxus/stackoverflow-answers | /order-flights/flights-recursion.py | 845 | 4.15625 | 4 | def next_leg(flights, source):
"""Find and append next flight to trip"""
for leg in flights:
if leg[0] == source:
trip.append(leg)
source = leg[1]
flights.remove(leg)
next_leg(flights, source)
def previous_leg(flights, destination):
"""Find and prepend previous flight to trip"""
for leg in flights:
if leg[1] == destination:
trip.insert(0, leg)
destination = leg[0]
flights.remove(leg)
previous_leg(flights, destination)
flights = [
['Minneapolis', 'Las Vegas'],
['LA', 'Chicago'],
['Las Vegas', 'Seattle'],
['Chicago', 'Atlanta'],
['Atlanta', 'NY'],
['NY', 'Minneapolis'],
]
trip = [flights[0]]
flights.pop(0)
next_leg(flights, trip[0][1])
previous_leg(flights, trip[0][0])
print(trip)
| false |
64856f52540d3b445232b03820e0f02e7ac29eae | 00wendi00/Python-initiation | /algorithm/search_fibonacci.py | 1,704 | 4.21875 | 4 | #!/usr/bin/env python
# -*- coding: utf-8 -*-
# @File : search_fibonacci.py
# @Author: Wade Cheung
# @Date : 2018/7/3
# @Desc : 斐波拉契查找
from algorithm.Fibonacci import fibonacci_create
def fibonacci_search(lis, key):
"""斐波拉契查找, 平均性能，要优于二分查找。但是在最坏情况下，比如这里如果key为1，则始终处于左侧半区查找，此时其效率要低于二分查找
:param lis:
:param key:
:return:
"""
fib = fibonacci_create(100)
low = 0
high = len(lis) - 1
# 为了使得查找表满足斐波那契特性，在表的最后添加几个同样的值
# 这个值是原查找表的最后那个元素的值
# 添加的个数由fib[k]-i-1决定
k = 0
while high > fib[k] - 1:
k += 1
print(k)
i = high
while fib[k] - 1 > i:
lis.append(lis[high])
i += 1
print(lis)
times = 0
while low <= high:
times += 1
# 为了防止fib列表下标溢出
if k < 2:
mid = low
else:
mid = low + fib[k - 1] - 1
print('low=%s, mid=%s, high=%s' % (low, mid, high))
if key < lis[mid]:
high = mid - 1
k -= 1
elif key > lis[mid]:
low = mid + 1
k -= 2
else:
if mid <= high:
print('times: %s' % times)
return mid
else:
print('times: %s' % times)
return high
print('times: %s' % times)
return False
if __name__ == '__main__':
LIST = [1, 5, 7, 8, 22, 54, 99, 123, 200, 222, 444]
result = fibonacci_search(LIST, 222)
print(result)
| false |
0a49e0d5e5eeda698c11c100a4a3fb8f8cc6a9d6 | dhruvsharma95/intro-to-python | /day3/recap.py | 2,891 | 4.28125 | 4 | """
program to check if the number (say 'n') given as input is a prime
a prime no. is the one which has only 1 and the numbers itself as its divisors
1 is not a prime no.
eg. n = 2, 3, 5, 7, ...
idea: start a loop from 2 to n/2, if n is divisble by the divisor => not prime
because a no. cannot be divided by a no. > n/2 except n
"""
n = int(input())
checked = 0
for divisor in range(2, n//2 + 1):
if (n % divisor == 0):
print("No is not prime!")
checked = 1
# no need checking further, can break the loop here itself
break
# the above code doesn't say that 1 is not a prime because loop won't run even once
# to handle that case, we can add the extra condition:
if (n == 1):
print("No is not prime!")
checked = 1
# in cases where checked is false or equal to 0, can say it is prime!
if (not checked):
print("No is prime!")
"""
program to print the following pattern:
*
* *
* * *
* * * *
...
idea: As you can observer, we are looping in two directions, row wise and column wise
1st row has 1 star
2nd row has 2 stars
3rd row has 3 stars and so on...
the outer loop would represent looping over the rows
and in the inner loop, we will loop over the elements in the rows, i.e. columns
Always better to use loop variable 'r' or 'row' for looping on rows
and the loop variable 'c' or 'col' to loop on columns
for easy understanding and debugging of your code !
"""
num_rows = n
for row in range(1, num_rows + 1):
# r_th row will have r stars
for col in range(1, row + 1):
print('*', end=' ')
# endline
print('')
# is it possible to do the same thing using one loop, think?
# idea is the same: print 1 char in 1st row, 2 chars in 2nd row, ...
line_num = 1
printed_on_cur_row = 0
while (line_num <= num_rows):
if (printed_on_cur_row < line_num):
print('*', end=' ')
printed_on_cur_row += 1
else:
print('')
# reset count of stars on cur_row to be zero
printed_on_cur_row = 0
# increment the line_num, i.e. move to next row
line_num += 1
"""
function to return maximum of three numbers
idea: let numbers be a, b, c
when can a be the max? when a >= b and a >= c
similarly for b and c
so, write three conditions, and return the answer if that condition is true
also, you can use if elif else as well!
"""
def maximum_of_three(num_one, num_two, num_three):
if (num_one >= num_two and num_one >= num_three):
return num_one
if (num_two >= num_one and num_two >= num_three):
return num_two
if (num_three >= num_one and num_three >= num_two):
return num_three
return -1
n1 = 3
n2 = 7
n3 = 1
res = maximum_of_three(n1, n2, n3)
print(res)
# idea: assign max = a, now check if b >= max, max = b, again if c >= max, max = c
def maximum_of_three_simple(n1, n2, n3):
maxi = n1
if (n2 >= maxi):
maxi = n2
if (n3 >= maxi):
maxi = n3
return maxi
res = maximum_of_three(n1, n2, n3)
print(res) | true |
e8566d6e9626c1f5d04d524071ca158cb535677e | Vahi-vai/python-bootcamp | /task13 - all exercises.py | 890 | 4.40625 | 4 | # Exercise 1
import re
def is_allowed_specific_char(string):
charRe = re.compile(r'[^a-zA-Z0-9.]')
string = charRe.search(string)
return not (string)
c = input("Enter anything")
print(is_allowed_specific_char(c))
#Exercise 2
import re
a = input("Enter any word")
b = re.search("ab",a)
print(b)
# Exercse 3
j = input("Enter a sentence")
res= re.finditer(r"([0-9]{1,3})", j)
print(" words end with")
for i in res:
print(i.group(0))
# Exercise 4
import re
results = re.finditer(r"([0-9]{1,3})", "i got 93 in maths, 96 in chemistry and 100 in physics")
print("Number of length 1 to 3")
for n in results:
print(n.group(0))
# Exercise 6
import re
def text_match(text):
if re.search("^[A-Z]*$", text):
return ("all the letters in the entered word is uppercase")
o = input("Enter a word")
print( text_match(o)) | true |
562fc5da1d34291a12e1e32ca7b179166b10c2c4 | javierunix/ca_computerscience | /CS101/functions/challenges.py | 2,576 | 4.21875 | 4 | # Create a function named divisible_by_ten() that takes a list of numbers named nums as a parameter.
# Return the count of how many numbers in the list are divisible by 10.
def divisible_by_ten(nums):
counter = 0 # define and initialize counter
for num in nums: # iterate over list
if num % 10 == 0: # if number is divisible by ten
counter += 1 # increment counter
return counter
# print(divisible_by_ten([20, 25, 30, 35, 40]))
# Create a function named add_greetings() which takes a list of strings named names as a parameter.
# In the function, create an empty list that will contain each greeting. Add the string 'Hello, ' in front of each name in names and append the greeting to the list.
# Return the new list containing the greetings.
def add_greetings(names):
greetings = []
for name in names:
greetings.append("Hello, %s" %name)
return greetings
# print(add_greetings(["Owen", "Max", "Sophie"]))
#Write a function called delete_starting_evens() that has a parameter named lst.
# The function should remove elements from the front of lst until the front of the list is not even. The function should then return lst.
# For example if lst started as [4, 8, 10, 11, 12, 15], then delete_starting_evens(lst) should return [11, 12, 15].
def delete_starting_evens(lst):
while (len(lst) > 0 and lst[0] % 2 == 0):
lst = lst[1:]
return(lst)
# print(delete_starting_evens([4, 8, 10, 11, 12, 15]))
# print(delete_starting_evens([4, 8, 10, 14, 18, 20]))
# Create a function named odd_indices() that has one parameter named lst.
# The function should create a new empty list and add every element from lst that has an odd index.
# The function should then return this new list.
def odd_indices(lst):
new_list = []
for i in range(1, len(lst), 2): # iterate from index 1, in 2 position steps
new_list.append(lst[i])
return new_list
# print(odd_indices([4, 3, 7, 10, 11, -2]))
# Create a function named exponents() that takes two lists as parameters named bases and powers.
# Return a new list containing every number in bases raised to every number in powers.
def exponents(bases, powers): # the function accepts as arguments two lists: bases and powers
new_list = [] # new list to store the exponents
for base in bases: # iterate through bases
for power in powers: # iterate through powers
new_list.append(base ** power) # append to new list the exponential number
return new_list
print(exponents([2, 3, 4], [1, 2, 3])
)
| true |
e28a3cdba85b68d2a09da8486c745c22ce4ea9e8 | javierunix/ca_computerscience | /CS102/BinarySearch/binary_iterative.py | 821 | 4.1875 | 4 | def binary_search(sorted_list, target):
left_pointer = 0
right_pointer = len(sorted_list)
# fill in the condition for the while loop
while left_pointer < right_pointer:
# calculate the middle index using the two pointers
mid_idx = (left_pointer + right_pointer) // 2
mid_val = sorted_list[mid_idx]
if mid_val == target:
return mid_idx
if target < mid_val:
# set the right_pointer to the appropriate value
right_pointer = mid_idx
if target > mid_val:
# set the left_pointer to the appropriate value
left_pointer = mid_idx + 1
return "Value not in list"
# test cases
print(binary_search([5,6,7,8,9], 9))
print(binary_search([5,6,7,8,9], 10))
print(binary_search([5,6,7,8,9], 8))
print(binary_search([5,6,7,8,9], 4))
print(binary_search([5,6,7,8,9], 6)) | true |
4aecc602de2fc81ee18ef45eaf70639c99c8f268 | javierunix/ca_computerscience | /CS102/DepthFirstTrees/TreeNode.py | 1,918 | 4.15625 | 4 | from collections import deque
class TreeNode:
def __init__(self, value):
self.value = value # data
self.children = [] # references to other nodes
def __repr__(self):
return self.value
def add_child(self, child_node):
# creates parent-child relationship
print("Adding " + child_node.value)
self.children.append(child_node)
def remove_child(self, child_node):
# removes parent-child relationship
print("Removing " + child_node.value + " from " + self.value)
self.children = [child for child in self.children
if child is not child_node]
def traverse(self):
# moves through each node referenced from self downwards
nodes_to_visit = [self]
while len(nodes_to_visit) > 0:
current_node = nodes_to_visit.pop()
print(current_node.value)
nodes_to_visit += current_node.children
def print_tree(root):
stack = deque()
stack.append([root, 0])
level_str = "\n"
prev_level = 0
level = 0
while len(stack) > 0:
prev_level = level
node, level = stack.pop()
if level > 0 and len(stack) > 0 and level <= stack[-1][1]:
level_str += " "*(level-1)+ "├─"
elif level > 0:
level_str += " "*(level-1)+ "└─"
level_str += str(node.value)
level_str += "\n"
level+=1
for child in node.children:
stack.append([child, level])
print(level_str)
def print_path(path):
# If path is None, no path was found
if path == None:
print("No paths found!")
# If a path was found, print path
else:
print("Path found:")
for node in path:
print(node.value)
sample_root_node = TreeNode("A")
two = TreeNode("B")
three = TreeNode("C")
sample_root_node.children = [three, two]
four = TreeNode("D")
five = TreeNode("E")
six = TreeNode("F")
seven = TreeNode("G")
two.children = [five, four]
three.children = [seven, six] | true |
832d4063787cbe9ed48ac8a72aa27be8b89aea83 | swaroop9ai9/problemsolving | /generator.py | 2,301 | 4.46875 | 4 | # Python generators are the functions that return the traversal object and used to create iterators.
# It traverses the entire items at once
# The generato can also be an expression in which syntac is similar to list comprehension
# There is a lot of complexity in creating iteration in python, we need to implement __iter__() and __next__() method to keep track of internal states
# Generator plays an essential role in simplifying this process.
# If no value it raises "StopIteration" exception
# Syntax to create a generator function
# Similar to normal function, but uses 'yeild' keyword instead of return
import time
def simple_even():
for i in range(10):
print("gen_func")
time.sleep(2)
if(i%2==0):
yield i
# Successive function call using for loop, this returns the even number list
for i in simple_even():
print("loop")
time.sleep(2)
print(i)
# The yeild keyword, is responsible for controlling the flow of the generator function.
# It pauses the function execution by saving all states and yielded to the caller,
# Later resumes execution, when a successive function is called.
# But, a return statement returns a value and terminates the whole function, (only one return can be used).
time.sleep(3)
print("Multiple Yield's")
def multiple_yield():
str1="First String"
yield str1
str2 = "second string"
yield str2
str3="Third String"
yield str3
obj = multiple_yield()
print(next(obj))
time.sleep(2)
print(next(obj))
time.sleep(2)
print(next(obj))
# In Generator functions are called, the normal function is paused immediately and control transferred to the caller
# Local variable and their states are remembered between succesive calls
# StopIteration exception is raised automatically when the function terminates
# Generator Expression: Could be created easily without useer-defined function.
# Similar to Lambda function, which created an anonymous generator function
# Very similar to list comprehension, but square braces are replaced with ()
print("Generator Expression")
lis = [1,2,3,4,5,78]
a = (x**3 for x in lis)
print(a)
print(next(a))
print(next(a))
# Advantages of Generators
# 1) Easy to implement
# 2) Memory Efficient
# 3) Pipelining with Generators
# 4) Generate Infinite Sequence
| true |
2d143b66b04ee61acdcc848f7b6e51c60aacc993 | vinkrish/ml-jupyter-notebook | /DS & Algorithm/Stack/StackArraySize.py | 1,632 | 4.15625 | 4 | class EmptyStackError(Exception):
pass
class StackFullError(Exception):
pass
class Stack:
def __init__(self, max_size = 10):
self.items = [None] * max_size
self.count = 0
def is_empty(self):
return self.count == 0
def is_full(self):
return self.count == len(self.items)
def size(self):
return self.count
def push(self, item):
if self.is_full():
raise StackFullError('Stack is full, can\'t push')
self.items[self.count] = item
self.count += 1
def pop(self):
if self.is_empty():
raise EmptyStackError('Stack is empty, can\'t pop')
x = self.items[self.count-1]
self.items[self.count-1] = None
self.count -+ 1
return x
def peek(self):
if self.is_empty():
raise EmptyStackError('Stack is empty, can\'t peek')
return self.items[-1]
def display(self):
print(self.items)
stack = Stack()
print('Check if stack is empty :')
check_stack = stack.is_empty()
print(check_stack)
print('Pushing element to stack :')
stack.push(1)
stack.push(2)
stack.push(3)
stack.push(4)
stack.push(5)
stack.push(6)
stack.push(7)
stack.push(8)
stack.push(9)
stack.push(10)
stack.display()
print('Is stack full :')
full = stack.is_full()
print(full)
print('Pop element from stack :')
pop = stack.pop()
print(pop)
stack.display()
print('Peek at stack :')
peek = stack.peek()
print(peek)
stack.display()
print('Size of stack :')
size = stack.size()
print(size)
print('Check if stack is empty :')
check_stack = stack.is_empty()
print(check_stack) | false |
5573c2210cc86fc41e421581e462961320cc7439 | HDLahlou/pythonTetris | /playground/initWindow.py | 1,532 | 4.15625 | 4 | #!/usr/bin/python3
# -*- coding: utf-8 -*-
#imports TK=k and Frame class, and the constant BOTH
#Tk is used to create a root window
#Frame is container for other widgets
from tkinter import Tk, BOTH
from tkinter.ttk import Frame
#Example class inherits Frame container widget
#calls constructor of our inherited class in the _init_()
class Example(Frame):
def __init__(self):
super().__init__()
#delegates creation of the user interface to the initUI()
self.initUI()
def initUI(self):
#Set title of window, master attribute gives access to the root window Tk
self.master.title("Simple")
#pack() method one of 3 geometry managers in Tkinter
#organizes widgets into horizontal and vertical boxes, here goes the frame widget
#Frame accessed through self attribute in Tk root window
#expands in both directions, taking whole client space of the root window
self.pack(fill=BOTH, expand=1)
def main():
#root window created, main app window, must be created before any widgets
root = Tk()
#sets size and positions it on the screen 1) width 2) height 3) X 4) Y coord
root.geometry("250x150+300+300")
#Creates instance of application class
app = Example()
#main loop, event handling starts from this point, receives events from the window system
#dispatches them to application widgets
root.mainloop()
if __name__ == '__main__':
main() | true |
635490925bbf77cc19972bfff875466095b590c0 | xuedagong/hello | /leetcode/python/BinaryTreeLevelOrderTraversal_II.py | 1,252 | 4.15625 | 4 | '''
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
'''
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def levelOrderBottom(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
my_lst=self.make_q([root], [])
my_lst.reverse()
return my_lst
def make_q(self,root_lst,global_lst):
one_list=[]
new_root_lst=[]
for item in root_lst:
if item is not None:
one_list.append(item.val)
new_root_lst.append(item.left)
new_root_lst.append(item.right)
if len(one_list)>0:
global_lst.append(one_list)
if len(new_root_lst)>0:
return self.make_q(new_root_lst,global_lst)
else:
return global_lst
if __name__ == '__main__':
print 1 | true |
d122d474b7de57df01d4e8fe789d5f3e7cab137a | carvalhopedro22/Programas-em-python-cursos-e-geral- | /Programas do Curso/Desafio 9.py | 228 | 4.125 | 4 | n = int(input('Digite um numero: '))
print('Tabuada do numero digitado a seguir\n:')
print(n * 0)
print(n * 1)
print(n * 2)
print(n * 3)
print(n * 4)
print(n * 5)
print(n * 6)
print(n * 7)
print(n * 8)
print(n * 9)
print(n * 10) | false |
d9a7a6d5c685b4de80f54c86d563408bb7cbedc3 | fwang1395/LeetCode | /python/LinkList/DeleteNode.py | 1,656 | 4.28125 | 4 | #!/usr/bin/python
# _*_ coding=UTF_* _*_
'''
Delete Node in a Linked List
Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.
Given linked list -- head = [4,5,1,9], which looks like following:
4 -> 5 -> 1 -> 9
Example 1:
Input: head = [4,5,1,9], node = 5
Output: [4,1,9]
Explanation: You are given the second node with value 5, the linked list
should become 4 -> 1 -> 9 after calling your function.
Example 2:
Input: head = [4,5,1,9], node = 1
Output: [4,5,9]
Explanation: You are given the third node with value 1, the linked list
should become 4 -> 5 -> 9 after calling your function.
Note:
The linked list will have at least two elements.
All of the nodes' values will be unique.
The given node will not be the tail and it will always be a valid node of the linked list.
Do not return anything from your function.
'''
class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None
# def setNext(self,y):
# self.next = y
# def getNext(self):
# return self.next
# def getValue(self):
# return self.val
class Solution(object):
def deleteNode(self, node):
"""
:type node: ListNode
:rtype: void Do not return anything, modify node in-place instead.
"""
if node.next is not None:
nextNode = node.next
node.val = nextNode.val
node.next = nextNode.next
if __name__ == '__main__':
node3,node2,node1 = ListNode(3),ListNode(2),ListNode(1)
node1.setNext(node2)
node2.setNext(node3)
| true |
8813688d22baf6497f9a255a679dd82d4a5adeed | 405620294/python_classwork | /basic grammar/計算bmi.py | 525 | 4.125 | 4 | # 計算BMI BMI：weight/height(m)^2
#型態：數字(整數、小數)、字串
#要注意回傳值得型態，可以回傳數字英文等等，或是有意義的0
height = float(input("請輸入你的身高"))
weight = float(input("請輸入你的體重"))
print("身高是：", height)
print("體重是：", weight)
print("bmi：", weight / (height / 100) ** 2)
BMI = weight / (height / 100) ** 2
if BMI > 25:
print("過重")
print("少吃多動")
elif 25 > BMI >18:
print("正常")
else:
print("過輕") | false |
8482f36a3647203c03c4ccb954ce726f79bf236b | mehedi2258/Pythonist-DP01 | /2ndClass.py | 2,090 | 4.59375 | 5 | #Python Class2 (28-02-2017)
'''a = 12
b = 5.5
sum = a+b
print (sum)
print(type(a))
print(type(b))
print(type(sum))'''
'''a = 12
b = 5.5
print(a)
print(float(b))
print(b)
print(int(b))'''
'''c = 6.5
print(round(c))
d = 6.4
print(round(d))
e = -5.5
print(round(e))
f= 5.5
print(round(f))
#How to Ceil
print(math.ceil(c)) #ceil
print(math.ceil(d)) #ceil
print(math.ceil(e)) #ceil
print(math.ceil(f)) #ceil
#How to Floor
print(math.floor(c)) #floor
print(math.floor(d)) #floor
print(math.floor(e)) #floor
print(math.floor(f)) #floor'''
#print double or more lines
'''a= """Mehedi
Hasan
Milon
"""
print(a)'''
#Add int & String
'''a= 'Python programming class. '
b = 'We use pycharm 3.5'
print(a+b)
a= 'Python programming class.'
b = ' We use pycharm 3.5'
print(a+b)
a= 'Python programming class.'
b = 'We use pycharm 3.5'
print(a+' '+b)
'''
#Add String & Int
'''
x = 'Python Programming Class'
y = 2
print(x+str(y))
x = 1.0
y = ' Python Programming Class'
print(str(x)+(y))
a = '10'
b = 10
print(int(a)+b)
print(float(a)+b)
'''
#Print Output
'''
a = 7
b = 3
sum = a+b
print(sum)
print(str(a) + '+' + str( b) +' = ' +str(sum))
print(a,'+',b,'=',sum)
print('{}+{}={}'.format(a,b,sum))
'''
'''a = input('Enter the 1st number: ')
b = input('Enter the second number: ')
sum = int(a) + int(b)
print(sum)'''
'''a = int(input('Enter the 1st number: '))
b = int(input('Enter the second number: '))
sum = (a) + (b)
print('sum = ',sum,end='')'''
'''a = int(input('Enter 1st number: '))
b = int(input('Enter 2nd number: '))
print(a+b)'''
'''a = input('Enter 1st number: ')
b = input('Enter 2nd number: ')
print(a+b)'''
'''a = int(input('Enter 1st number: '))
b = int(input('Enter 2nd number: '))
print(a*b)'''
'''a = int(input('Enter 1st number: '))
b = int(input('Enter 2nd number: '))
print(a/b)'''
'''a = int(input('Enter 1st number: '))
b = int(input('Enter 2nd number: '))
print(a%b)'''
| false |
54138dd34573193910a644a42cd518d8ccb5941b | clevercarl/udemy.ds | /preprocessing_data.py | 2,003 | 4.34375 | 4 | # Data Preprocessing
# Importing the Libraries
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
# Importing the Dataset
dataset = pd.read_csv('Data.csv')
x = dataset.iloc[:,:-1].values
y = dataset.iloc[:,3].values
# x are independent variables
# y is the dependent variable
# Missing Data: Taking care of it
from sklearn.preprocessing import Imputer
# sklearn is a giant library with helpful ML tools
# preprocessing is package that helps with preprocessing datasets
# Imputer is a class that helps to take care of missing data
imputer = Imputer(missing_values = 'NaN', strategy = 'mean', axis = 0)
imputer = imputer.fit(x[:,1:3]) #using imputer on the second & third column, upperbound is excluded that is why its 3 instead of 2 (ending at two and not going to 3)
x[:,1:3] = imputer.transform(x[:,1:3])
# Encoding Categorical data
from sklearn.preprocessing import LabelEncoder, OneHotEncoder
# LabelEncoder is a class that changes the categorical variables into a hierarchy of numeric values
# OneHotEncoder is a class that creates dummy variables
# Encode country column
labelencoder_x = LabelEncoder()
x[:,0] = labelencoder_x.fit_transform(x[:,0])
onehotencoder = OneHotEncoder(categorical_features = [0]) # 0 represents the index for the country column
x = onehotencoder.fit_transform(x).toarray()
# Encode purchase column
labelencoder_y = LabelEncoder()
y = labelencoder_y.fit_transform(y)
# Splitting dataset into test and train
from sklearn.model_selection import train_test_split
x_train, x_test, y_train, y_test = train_test_split(x, y, test_size = 0.2, random_state = 0)
# random_state is like setting the seed
# test_size 0.2 means that the test set contains 20% of the dataset
# Feature Scaling
from sklearn.preprocessing import StandardScaler
sc_x = StandardScaler()
x_train = sc_x.fit_transform(x_train)
x_test = sc_x.transform(x_test)
| true |
b3f6be1cba245f468e37e4e28d2e55076e6d9111 | konjoinfinity/python-checkpoint | /oop.py | 2,747 | 4.1875 | 4 | # 1: Define a Vehicle class with the following properties and methods:
# - `vehicle_type`
# - `wheel_count`
# - `name`
# - `mpg`, a 'dict', with the following properties:
# - `city`
# - `highway`
# - `combined`
# - `get_vehicle_type` should return the `vehicle_type`
# - `get_vehicle_drive` if the `wheel_count` for that class is "no wheels!" then
# it should return "no wheels send it back to the shop"
# otherwise it should return "I have {self.wheel_count} wheel drive" as a formatted string
#
# Your Vehicle class should take one extra argument in the __init__ method (a `dict`) with the above
# attributes. Define the properties on the class from the dict that is passed in.
#
# Here's an example of the dict that will be passed in to your class:
#
# vehicle_dict_vehicle = {
# "vehicle_type": "Vehicle",
# "wheel_count": 'no wheels!',
# "mpg": {
# "city": 19,
# "highway": 30,
# "combined": 27
# },
# "name": "Unidentified Flying Object",
# }
class Vehicle:
def __init__(self, vehicle_type, wheel_count, mpg, name):
self.vehicle_type = vehicle_type
self.wheel_count = wheel_count
self.name = name
self.mpg = {city, highway, combined}
def get_vehicle_type(self):
print(self.vehicle_type)
def get_vehicle_drive(self):
if self.wheel_count == "no wheels!":
print("no wheels send it back to the shop")
else:
print(f"I have {self.wheel_count} wheel drive")
# #2: Create a Motorcycle class that inherits from the Vehicle class and has the
# following properties and methods:
# - all the properties inherited from the Vehicle class
# - method: `pop_wheelie` if `wheel_count` is not equal to 2 then it should return False
# otherwise return "popped a wheelie!"
class Motorcycle(Vehicle):
def __init__(self, wheel_count):
self.vehicle_type = vehicle_type
self.wheel_count = wheel_count
self.name = name
self.mpg = {city, highway, combined}
def pop_wheelie(self):
if self.wheel_count != 2:
print(False)
else:
print("popped a wheelie!")
# #3: Define a Car class that inherits from the Vehicle class with the following properties and methods:
# - all the properties inherited from the Vehicle class
# - property: `wheel_count` defaults to 4
# - method: `can_drive` that should return 'Vrrooooom Vroooom'
# #4: Define a Truck class that inherits from the Vehicle class with the following properties and methods:
# - all the properties inherited from the Vehicle class
# - method: `rev_engine` that should return a string 'rreevv!'
# Commit when you finish working on these questions!
| true |
551010ad5f5ecc55e90a9d7dea65343d62f7ab15 | marvage/Practice-Python | /problem10.py | 1,498 | 4.3125 | 4 | #List Overlap Comprehensions
#Exercise 10 (and my Solution) from www.practicepython.com
#This week’s exercise is going to be revisiting an old exercise (see Exercise 5), except require the solution in a different way.
#Take two lists, say for example these two:
a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
#and write a program that returns a list that contains only the elements that are common between the lists (without duplicates). Make sure your program works on two lists of different sizes. Write this in one line of Python using at least one list comprehension. (Hint: Remember list comprehensions from Exercise 7).
#The original formulation of this exercise said to write the solution using one line of Python, but a few readers pointed out that this was impossible to do without using sets that I had not yet discussed on the blog, so you can either choose to use the original directive and read about the set command in Python 3.3, or try to implement this on your own and use at least one list comprehension in the solution.
#Extra:
#Randomly generate two lists to test this
import random
#generate random lists
randomlist1 = random.sample(range(10),5)
randomlist2 = random.sample(range(10), 5)
#list comparisons/comprehensions
results = list(set(i for i in randomlist1 if i in randomlist2))
print "These are the random lists"
print randomlist1, randomlist2
print "This is the overlap with duplicates removed:"
print results
| true |
3581ebe6bca3cb2713e4d6601af4cce3f7a53bf9 | fishofweb/python-programs | /thesaurus.py | 993 | 4.375 | 4 | print("Welcome to the Thesaurus App!\n")
print("Choose a word from the thesaurus and I will give you the synonym.\n")
my_dict = {"hot": ["warm", "fire"],
"cold": ["chilly","cool"],
"happy": ["content","glad"],
"sad": ["gloomy","bad mood"],
}
print("Here are the words in the thesaurus:")
for key in my_dict:
print("\t\t- "+key)
word = input("What word would you like a synonym for: ").lower().strip()
if word not in my_dict.keys():
print(word,"is not in thesaurus")
quit()
for key,value in my_dict.items():
if word == key:
print("Synonym of", key, "is",value[0])
choice = input("Would you like to see the whole thesaurus (yes/no): ").lower().strip()
if choice == "yes":
for k , v in my_dict.items():
print("Synonyms of",k,"are:")
for val in v:
print("\t\t-"+val)
else:
print("Thank you for using this app , Good Bye")
quit()
print("Thank you for using this app , Good Bye")
| false |
149d21cefd2181335ed048e0b881498668cc5421 | fishofweb/python-programs | /database-admin.py | 1,685 | 4.1875 | 4 | print("Welcome to the Database Admin Program.\n")
db = {
"admin":"12345678",
"shanza": "shanza1000",
"sadaf": "sadaf1000",
"ammara": "ammara1000",
"mahnoor": "mano1000",
"saniya": "saniya1000",
"maryam": "maryam1000",
"sonia": "sonia1000",
}
username = input("Enter your username: ").lower().strip()
password = input("Enter your password: ").lower().strip()
if username not in db.keys():
print("incorrect username")
quit()
if password not in db.values():
print("incorrect password")
quit()
if username == 'admin' and password == '12345678':
print("welcome", username, "here is you database\n\n")
for k , v in db.items():
print(k)
print("user:",k,"\tpassword:",v)
for user,passwd in db.items():
if user == username and password == passwd and username != 'admin' and password != '12345678':
print("Hello",username,"You are logged in!")
print("\n")
change_pwd = input("Would you like to change your password(y/n): ").lower().strip()
if change_pwd == 'y':
new = input("Type your new password: ")
if len(new)<8:
print("Password is too short , type atleast 8 characters.")
break
new_confirm = input("Type your new password again: ")
if new == new_confirm:
db[username] = new_confirm
print("your password is changed to", new_confirm)
else:
print("password mismatched")
break
else:
print("ok your password is still", passwd)
break
print("\tUsername:",username,"\t\tpassword:",password) | false |
94adddb0057e94366e19429f008a064a82bb88bf | krishols/pedal | /pedal/utilities/text.py | 1,844 | 4.5 | 4 | """
Utilities for handling English Language rules.
"""
def add_indefinite_article(phrase):
"""
Given a word, choose the appropriate indefinite article (e.g., "a" or "an")
and add it the front of the word. Matches the original word's
capitalization.
Args:
phrase (str): The phrase to get the appropriate definite article
for.
Returns:
str: Either "an" or "a".
"""
if phrase[0] in "aeiou":
return "an "+phrase
elif phrase[0] in "AEIOU":
return "An "+phrase
elif phrase[0].lower() == phrase[0]:
return "a "+phrase
else:
return "A "+phrase
def escape_curly_braces(result):
""" Replaces all occurrences of curly braces with double curly braces. """
return result.replace("{", "{{").replace("}", "}}")
def safe_repr(obj, short=False, max_length=80) -> str:
"""
Create a string representation of this object using :py:func:`repr`. If the
object doesn't have a repr defined, then falls back onto the default
:py:func:`object.__repr__`. Finally, if ``short`` is true, then the string
will be truncated to the max_length value.
Args:
obj (Any): Any object to repr.
short (bool): Whether or not to truncate to the max_length.
max_length (int): How many maximum characters to use, if short is True.
"""
try:
result = repr(obj)
except Exception:
result = object.__repr__(obj)
if short and len(result) >= max_length:
result = result[:max_length] + ' [truncated]...'
result = result
return result
def chomp(text):
""" Removes the trailing newline character, if there is one. """
if not text:
return text
if text[-2:] == '\n\r':
return text[:-2]
if text[-1] == '\n':
return text[:-1]
return text
| true |
e802ccdf928c5da395ff846aba1f29338cc3321c | puneeth-techie/python-example | /cel_fan_To_fan_cel.py | 484 | 4.21875 | 4 | #Converting Celsius to Fahrenheit
#Formula Fahrenheit = (Celsius * 1.8) + 32
Celsius = float(input('Enter the Celsius: '))
Fahrenheit = (Celsius * 1.8) + 32
print('%0.2f Celsius is equal to %0.2f Fahrenheit\n'%(Celsius, Fahrenheit))
#Converting Fahrenheit to Celsius
#Formula Celsius = (Fahrenheit - 32) / 1.8
Fahrenheit = float(input('Enter the Fahrenheit value: '))
Celsius = (Fahrenheit - 32) / 1.8
print('%0.2f Fahrenheit is equal to %0.2f Farhenheit'%(Fahrenheit, Celsius))
| false |
ed2dee91f663f2dcd34b7b4c1f428bd7d1e918cb | LightbulbProductions/thinkpython | /function_demo.py | 785 | 4.15625 | 4 | def subtractor(a, b):
"""I subtract b from a and return the result"""
print("I'm a function. My name is {}".format(subtractor.__name__))
print("I'm about to subtract {} and {}\n\n".format(a,b))
return a - b # i output a value by using the return statement
if __name__ == '__main__':
subtractor(3, 2)
def function3(a=1, b=1):
"""I'm a function that calls other functions """
print("I'm {} and I'm about to call subtractor function".format(function3.__name__))
total = subtractor(a,b)
print("I'm {} and I'm about to call print_function".format(function3.__name__))
print_function()
print("I'm {} and I'm about return total".format(function3.__name__))
return total
if __name__ == '__main__':
total = function3()
print("total is {}".format(total)) | true |
d0a9f3d4b33d1685a62bc506c2dc071f824d06d5 | PrzemyslawMichalski/bootcamp_python | /różne/arytmetyka.py | 647 | 4.15625 | 4 | # 1. przypisz do zmiennych x i y
# jakieś wartości liczbowe
# 2. korzystajac z funkcji print
# wypisz na ekranie podstawowe dzialania
# arytmetyczne (dodawanie, odejmowanie, dzielenie
# mnożenie, dzielenie całkowitem, reszta z dzielenia
# potęgowanie )
# 3. uruchom program kilka razy dla różnych wartośći x i y
"""
dsd
"""
x = 10
y = 15
print("x=", x, "y=", y) # wywolanie print
print("dodawanie:", x + y)
print("odejmowanie:", x - y)
print("mnożenie:", x * y)
print("dzielenie:", x / y)
print("dzielenie całkowite:", x // y)
print("dzielenie modulo:", x % y)
print("potęgowanie:", x ** y)
# ctrl + alt + l
| false |
3bde2884de5102feae48b349ee66bf47213e368a | S1nEat3r/PythonJourney | /script.py | 2,874 | 4.21875 | 4 | #!/bin/python3
#Variables And Methods
quote = "All is fair in love and war."
print(quote)
print(quote.upper()) #uppercase
print(quote.lower()) #lowercase
print(quote.title()) #title case
print(len(quote))
name = "Simon" #string
age = 37 #int(30)
height = 5.10 #float float(5.9)
print (int(age))
print (int(30.9))
print ("My name is " + name + "I am " + str(age) + " years old")
age += 1
print (age)
birthday = 1
age += birthday
print(age)
print('\n')
#Functions
print ("Here is an example Function:")
def who_am_i(): #this is a function
name = "Simon"
age = 37
print ("My Name is " + name + " I am " + str(age) + " years old.")
who_am_i()
#adding parameters
def add_one_hundred(num):
print(num + 100)
add_one_hundred(100)
#multiple parameters
def add(x,y):
print (x + y)
add(7,7)
def square_root(x):
print(x ** .5)
square_root(64)
print (square_root)
def nl():
print('\n')
nl()
#Boolean expressions
print ("Boolean expressions:")
bool1 = True
bool2 = 3*3 == 9
bool3 = False
bool4 = 3*3 != 9
print (bool1,bool2,bool3,bool4)
print(type(bool1))
nl()
#Relational And Boolean Operators
greater_than = 7 > 5
less_than = 5 < 7
greater_than_equal_to = 7 >= 7
less_than_equal_to = 7 <= 7
test_and = (7 >5) and (5 > 7) #True
test_and2 = (7 > 5) and (5 > 7) #False
test_or = (7 > 5) or (5 < 7) #True
test_or2 = (7 > 5) or (5 > 7) #True
test_not = not True #False
test_not2 = not False #True
nl()
#Conditional Statements
def drink(money):
if money >= 2:
return "You've got yourself a drink"
else:
return "NO drink for you!"
print(drink(3))
print(drink(1))
def alcohol(age,money):
if (age >= 21) and (money >= 5):
return "We're getting a drink!"
elif (age >= 21) and (money < 5):
return "Come back with more money"
elif (age < 21) and (money >= 5):
return "Nice try kid, come back when your 21"
else:
return "Your too poor and too young kid"
print(alcohol(21,10))
print(alcohol(21,4))
print(alcohol(20,4))
nl()
#Lists - Have brackets []
movies = ["The Hangover", "Halloween", "The Rock", "The Exorcist", "Top Gun"]
print(movies[1]) #returns the second item
print(movies[0]) #returns the first item in the list
print(movies[1:4]) #returns all items from 2 to 4
print(movies[1:]) # returns all items in the list from second item
print(movies[:2]) # returns all item in the list up to the 3rd item
print(movies[-1]) # returns last item in the list
print(len(movies))
movies.append("Jaws")
print(movies)
movies.pop(0)
print(movies)
nl()
#Tuples - Do not change, ()
grades = ("a", "b", "c", "d", "f")
print(grades[1])
nl()
#Looping
#For Loops - start to finish of an iterate
vegetables = ["cucumber", "spinach", "cabbage"]
for x in vegetables:
print(x)
#While loops - Execute as long as true
i = 1
while i < 10:
print(i)
i += 1
| false |
95dbb6c0cd20090caacfda04e7ba966059bf228c | dpk3d/HackerRank | /ParenthesisChecker.py | 1,958 | 4.375 | 4 | """
https://practice.geeksforgeeks.org/problems/parenthesis-checker2744/1
Given an expression string x. Examine whether the pairs and the orders of {,},(,),[,] are correct in exp.
For example, the function should return 'true' for exp = [()]{}{[()()]()} and 'false' for exp = [(]).
Note: The drive code prints "balanced" if function return true, otherwise it prints "not balanced".
Example 1:
Input:
{([])}
Output:
true
Explanation:
{ ( [ ] ) }. Same colored brackets can form
balanced pairs, with 0 number of
unbalanced bracket.
Example 2:
Input:
()
Output:
true
Explanation:
(). Same bracket can form balanced pairs,
and here only 1 type of bracket is
present and in balanced way.
Example 3:
Input:
([]
Output:
false
Explanation:
([]. Here square bracket is balanced but
the small bracket is not balanced and
Hence , the output will be unbalanced.
"""
# Create an empty stack to keep track of opening parentheses.
# Loop through each character in the input string.
# If the character is an opening parentheses, push it onto the stack.
# If the character is a closing parentheses, pop the top element from the stack and
# check whether it matches the corresponding opening parentheses for the current closing parentheses.
# If the stack is empty and all parentheses have been matched, return True. Otherwise, return False.
def isValid(self, s: str) -> bool:
stack = []
for char in s:
if char == '(' or char == '{' or char == '[':
stack.append(char)
else:
if not stack:
return False
if char == ')' and stack[-1] == '(':
stack.pop()
elif char == '}' and stack[-1] == '{':
stack.pop()
elif char == ']' and stack[-1] == '[':
stack.pop()
else:
return False
return not stack
| true |
5815114592824cfae3a2ae06d555085925fef282 | dpk3d/HackerRank | /AmazonBinaryDigit.py | 969 | 4.28125 | 4 | """
Given an array of binary digits, 0 and 1, sort the array so that all 0 are at one end and
all the 1 are at other end, which end doesn't matter. To sort the array swap any two adjacent
elements. Determine the minimum Number of swap to sort the array.
Example:
arr=[0,1,0,1] -> with one move switching element 1 and 2 yields [0,0,1,1]
Complete the function minMoves below.
minMoves has following parameters :
int arr[n] : An array of binary digits
Return : Minimum number of moves required.
"""
def minMoves(arr):
count = 0
numOfUnplacedZero = 0
for moves in range(len(arr)):
if arr[moves] == 0:
numOfUnplacedZero += 1
else:
count += numOfUnplacedZero
return count
digits = [1, 1, 1, 1, 0, 1, 0, 1]
print("Minimum Number of moves required to sort array is ", minMoves(digits))
# Minimum Number of moves required to sort array is 3
# This is not optimal solution but pases 12 test case out of 13.
| true |
5c21807e96ea9f9afde868c7c999df24772a71b4 | dpk3d/HackerRank | /NumberOfPairs.py | 2,251 | 4.1875 | 4 | """
Given two arrays X and Y of positive integers, find the number of pairs such that
x^y > y^x (raised to power of) where x is an element from X and y is an element from Y.
"""
# Time Complexity O( n * m )
def simpleApproach(arr1, len1, arr2, len2):
pairs = 0
for a in range(len1):
for b in range(len2):
if pow(arr1[a], arr2[b]) > pow(arr2[b], arr1[a]):
print("Pairs are ===>", arr1[a], arr2[b])
pairs += 1
return pairs
arr1 = [2, 1, 6]
arr2 = [1, 5]
print("Number of Pairs count ==>", simpleApproach(arr1, 3, arr2, 2))
## TODO : Need to work on this to handle exception and corner cases
def binarySearchToGetIndex(arr, n, element):
low = 0
high = n - 1
pairs = -1
while low <= high:
mid = low + high // 2
if arr[mid] > element:
pairs = mid
high = mid - 1
else:
low = mid + 1
print("Pairs === >>", pairs)
return pairs
# Time Complexity O ( n + m)
# To Handle Exception and corner cases , need to call binarySearchToGetIndex method in below method.
def numberOfPairs(deepArray, moniArray, deepLength, moniLength):
zeros = 0
one = 0
two = 0
three = 0
four = 0
deepArray.sort()
moniArray.sort()
for x in range(moniLength):
if moniArray[x] == 0:
zeros += 1
if moniArray[x] == 1:
one += 1
if moniArray[x] == 2:
two += 1
if moniArray[x] == 3:
three += 1
if moniArray[x] == 0:
four += 1
# Traversing in First Array
pairs = 0
for x in range(deepLength):
if deepArray[x] == 0:
continue
elif deepArray[x] == 1:
pairs += 1
elif deepArray[x] == 2:
pairs += one + zeros
pairs -= three
pairs -= four
else:
pairs += one + zeros
return pairs
deepArray = [2, 1, 6, 8, 9, 0, 10, 11]
moniArray = [1, 5, 0, 13, 2, 7, 9, 98]
print("Number of Pairs count 2nd Approach ==>", numberOfPairs(deepArray, moniArray, 8, 8))
"""
Output:
Pairs are ===> 2 1
Pairs are ===> 2 5
Pairs are ===> 6 1
Number of Pairs count ==> 3
Number of Pairs count 2nd Approach ==> 12
"""
| true |
7d51cee9a200c7c2f40fc3657844ca8314e7eba1 | dpk3d/HackerRank | /MinimumPlatforms.py | 1,353 | 4.46875 | 4 | """
Given arrival and departure times of all trains that reach a railway station.
Find the minimum number of platforms required for the railway station so that no train is kept waiting.
Consider that all the trains arrive on the same day and leave on the same day.
Arrival and departure time can never be the same for a train but we can have arrival time of one train equal to departure time of the other.
At any given instance of time, same platform can not be used for both departure of a train and arrival of another train.
In such cases, we need different platforms,
"""
# Time Complexity O (n + nlogn)
def minimumPlatform(arrival, departure, n):
platform = 1
required_platform = 1
arrival.sort
departure.sort
x = 1
y = 0
# Iterating in both array
while (x < n and y < n):
if arrival[x] <= departure[y]:
platform += 1
x += 1
elif arrival[x] > departure[y]:
platform -= 1
y += 1
if platform > required_platform:
required_platform = platform
return required_platform
n = 6
arrival = [900, 940, 950, 1100, 1500, 1800]
departure = [910, 1200, 1120, 1130, 1900, 2000]
print("Platform Required is ===> ", minimumPlatform(arrival, departure, 6))
"""
Output:
Platform Required is ===> 3
"""
| true |
a2b77186cc914d93ddb699895ea9678cf1655cb5 | dpk3d/HackerRank | /MajorityElement.py | 1,758 | 4.15625 | 4 | """
Given an array of size n, find all elements in array that appear more than n/k times.
For example, if the input arrays is {3, 1, 2, 2, 1, 2, 3, 3} and k is 4, then the output should be [2, 3].
Note that size of array is 8 (or n = 8), so we need to find all elements that appear more than 2 (or 8/4) times.
There are two elements that appear more than two times, 2 and 3.
https://www.geeksforgeeks.org/given-an-array-of-of-size-n-finds-all-the-elements-that-appear-more-than-nk-times/
"""
def moreThanNbyK(arr, k):
number = len(arr) // k
unorderedMap = {} # unordered_map initialization
for x in range(len(arr)):
if arr[x] in unorderedMap:
unorderedMap[arr[x]] += 1
else:
unorderedMap[arr[x]] = 1
# Traversing the unorderedMap
for y in unorderedMap:
# Checking if value of a key-value pair is greater than x (where x = len(array) // k)
if unorderedMap[y] > number:
print(y , end= " ")
arr = [1, 1, 2, 2, 3, 5, 4, 2, 2, 3, 1, 1, 1]
k = 4
moreThanNbyK(arr, k)
# Using Boyer-Moore Majority Vote Algorithm - Time complexity - O(N)
def majorityElement(arr, k):
if not arr:
return []
count1, count2, candidate1, candidate2 = 0, 0, 0, 0
for x in arr:
if x == candidate1:
count1 += 1
elif x == candidate2:
count2 += 1
elif count1 == 0:
candidate1, count1 = x, 1
elif count2 == 0:
candidate2, count2 = x, 1
else:
count1, count2 = count1 - 1, count2 - 2
return [x for x in (candidate1, candidate2) if arr.count(x) > len(arr) // k]
arr = [3, 1, 2, 2, 1, 2, 3, 3]
k = 4
print("Elements appearing more than n/k is : ", majorityElement(arr, k))
| true |
04c18bf08827463218ebe45699de708766c6f435 | dpk3d/HackerRank | /InversionCount.py | 2,889 | 4.15625 | 4 | """
Given an array of integers. Find the Inversion Count in the array.
Inversion Count: For an array, inversion count indicates how far (or close) the array is from being sorted.
If array is already sorted then the inversion count is 0. If an array is sorted in the reverse order then
the inversion count is the maximum.
Formally, two elements a[i] and a[j] form an inversion if a[i] > a[j] and i < j.
"""
############################################################################
def SimpleInversionCount(arr):
count = 0
for x in range(len(arr)):
for y in range(x + 1, len(arr)):
if arr[x] > arr[y]:
count += 1
return count
arr = [2, 0, 4, 3, 9, 8, 9, 7, 10]
print("Inversion count is ==>", SimpleInversionCount(arr))
############################################################################
def InversionCountMergeSort(input_arr, n):
# A resulting_arr is temporary array to store sorted array of merge func.
resulting_arr = [0] * n
return mergeSort(input_arr, resulting_arr, 0, n - 1)
# This Function will use MergeSort to count inversions
def mergeSort(input_arr, resulting_arr, left, right):
inversion_count = 0
# Making recursive call only if more than one elements are there
if left < right:
# mid is dividing array into 2 sub arrays
mid = (left + right) // 2
# Calculating inversion counts in the left sub array
inversion_count += mergeSort(input_arr, resulting_arr, left, mid)
# Calculating inversion counts in right sub array
inversion_count += mergeSort(input_arr, resulting_arr, mid + 1, right)
# Merge 2 sub arrays into a sorted one
inversion_count += final_merge(input_arr, resulting_arr, left, mid, right)
return inversion_count
# Merge two sub arrays into a single sorted sub array
def final_merge(input_arr, resulting_arr, left, mid, right):
a = left
b = mid + 1
c = left
inv_count = 0
# No Inversion here
while a <= mid and b <= right:
if input_arr[a] <= input_arr[b]:
resulting_arr[c] = input_arr[a]
c += 1
a += 1
else:
# Inversion
resulting_arr[c] = input_arr[b]
inv_count += (mid - a + 1)
c += 1
b += 1
# Copying remaining array to left sub array
while a <= mid:
resulting_arr[c] = input_arr[a]
c += 1
a += 1
# Copying remaining array to right sub array
while b <= right:
resulting_arr[c] = input_arr[b]
c += 1
b += 1
# Copying the sorted array into final array
for x in range(left, right + 1):
input_arr[x] = resulting_arr[x]
return inv_count
arr = [1, 3, 4, 8, 20, 0, 11, 20, 6, 4, 5]
n = len(arr)
result = InversionCountMergeSort(arr, n)
print("Number of inversions are", result)
| true |
40380ebad4b1725941f360569bda5e2d4b67cab0 | dpk3d/HackerRank | /WordBreak.py | 2,943 | 4.21875 | 4 | """
https://practice.geeksforgeeks.org/problems/word-break1352/1
Given a string A and a dictionary of n words B, find out if A can be segmented into a space-separated sequence of dictionary words.
Note: From the dictionary B each word can be taken any number of times and in any order.
Example 1:
Input:
n = 12
B = { "i", "like", "sam",
"sung", "samsung", "mobile",
"ice","cream", "icecream",
"man", "go", "mango" }
A = "ilike"
Output:
1
Explanation:
The string can be segmented as "i like".
Example 2:
Input:
n = 12
B = { "i", "like", "sam",
"sung", "samsung", "mobile",
"ice","cream", "icecream",
"man", "go", "mango" }
A = "ilikesamsung"
Output:
1
Explanation:
The string can be segmented as
"i like samsung" or "i like sam sung".
Expected time complexity: O(s2)
Expected auxiliary space: O(s) , where s = length of string A
"""
def wordBreakDP(given_string, given_dictionary):
# dp_arr initialized to all FALSE
dp_arr = [False] * (len(given_string) + 1)
# Setting to True as a Base case
dp_arr[0] = True
# This loop will be ahead of inner loop
for x in range(1, len(given_string) + 1):
# this loop will start at input_string[0] and increment up to x
# Checking each time if input_string[y:x] matches in input_dictionary
for y in range(x):
# dp_arr[y] tells us if we have successfully created a word up to that index.
# Checking dp_arr[y] helps us ensure that we're only marking tracker True when words are adjacent.
if dp_arr[y] and given_string[y:x] in given_dictionary:
dp_arr[x] = True
break # control goes to outer loop to increment x
return dp_arr[-1]
input_dictionary = {"i", "like", "sam",
"sung", "samsung", "mobile",
"ice", "cream", "icecream",
"man", "go", "mango"}
input_string = "ilikesamsung"
print("We can able to break the word : ", wordBreakDP(input_string, input_dictionary))
# We can break the word : True
def wordBreakDPAnother(given_string, given_dictionary):
# dp_array initialized to all FALSE
dp_array = [False] * (len(given_string) + 1)
# Setting to True as a Base case
dp_array[0] = True
for x in range(1, len(given_string) + 1):
# check each word in wordDict rather than iterating from 0 to i
for word in given_dictionary:
if len(word) > x:
continue
if given_string[x - len(word):x] == word and dp_array[x - len(word)]:
dp_array[x] = dp_array[x - len(word)]
break
return dp_array[len(given_string)]
input_dictionary = {"i", "like", "sam",
"sung", "samsung", "mobile",
"ice", "cream", "icecream",
"man", "go", "mango"}
input_string = "ilikesamsung"
print("We can able to break the word : ", wordBreakDPAnother(input_string, input_dictionary))
| true |
d68bf8c7f1d677d1e5794b4be0da1b2cc37ae015 | dpk3d/HackerRank | /sortWithoutUsingFunction.py | 911 | 4.4375 | 4 | """
Sort A Given List/Array.
data_list = [-5, -23, 5, 0, 23, -6, 23, 67]
"""
data_list = [-5, -23, 5, 0, 23, -6, 23, 2, 67]
# Using 2 for loops to sort list not recommended solution
# Complexity of this approach is n * n
for x in range(0, len(data_list)):
for y in range(x + 1, len(data_list)):
if data_list[x] > data_list[y]:
data_list[x], data_list[y] = data_list[y], data_list[x]
print("For Loop : Sorted List is ==>", data_list)
"""
For Loop : Sorted List is ==> [-23, -6, -5, 0, 2, 5, 23, 23, 67]
"""
# While Loop Another Approach better than using two for loops.
new_list = []
while data_list:
lowest = data_list[0]
for x in data_list:
if x < lowest:
lowest = x
new_list.append(lowest)
data_list.remove(lowest)
print(" While Loop : Sorted List is ==> ", new_list)
"""
For Loop : Sorted List is ==> [-23, -6, -5, 0, 2, 5, 23, 23, 67]
"""
| true |
ffae53eaa3cc09c504781c26f80e4adc3a55e40c | TobyChen320/CSPT15_HashTables_II | /src/demos/demo2.py | 1,694 | 4.5 | 4 | """
Given a string, sort it in decreasing order based on the frequency of characters.
Example 1:
```plaintext
Input:
"free"
Output:
"eefr"
Explanation:
'e' appears twice while 'f' and 'r' appear once.
So 'e' must appear before 'f' and 'r'. Therefore, "eerf" is also a valid answer.
```
Example 2:
```plaintext
Input:
"dddbbb"
Output:
"dddbbb"
Explanation:
Both 'd' and 'b' appear three times, so "bbbddd" is also a valid answer.
Note that "dbdbdb" is incorrect, as the same characters must be together.
```
Example 3:
```plaintext
Input:
"Bbcc"
Output:
"ccBb"
Explanation:
"ccbB" is also a valid answer, but "Bbcc" is incorrect.
Note that 'B' and 'b' are treated as two different characters.
```
"""
# use the counter from collections to save us some time
import collections
def frequency_sort(s: str) -> str:
"""
Inputs:
s -> str
Output:
str
"""
# Your code here
# count up all of the occurrences of each letter
# ref: https://docs.python.org/3/library/collections.html#collections.Counter
letters_count = collections.Counter(s)
# create a list to build the string
string_build = list() # []
# iterate over the sorted counts (using the ".most_common()" method) extracting the letter and frequency key value pair
for letter, frequency in letters_count.most_common():
# letter * frequency
# "V" * 5 -> "VVVVV"
# append the letter * frequency to the string_build list
string_build.append(letter * frequency)
# turn the list back in to a string and return it (use a join?)
return "".join(string_build)
print(frequency_sort("free")) # => "eefr"
print(frequency_sort("dddbbb")) # => "dddbbb" | true |
52bdaae5ae41eb2045cd11d4df386534a09c5165 | mauabe/python-thw | /ex3.py | 506 | 4.125 | 4 | print ("I will now count my chickens")
print ("Hens", 25 + 30.0 / 6)
print ("Roosters", 100 - 25 * 3 % 4)
print ("Now I will count the eggs:")
print (3.0 + 2 + 1 - 5.0 + 4 % 2 - 1 / 4.0 + 6.0)
print ("Is it true that 3 + 2 < 5 - 7?")
print (3 + 2 < 5 - 7)
print ("What is 3 + 2?", 3 + 2)
print ("What is 5 - 7?", 5 - 7)
print ("Oh, that is wha it's False.")
print ("How about some more.")
print ("Is it greater?", 5 > -2)
print ("Is it greater or equal?", 5 >= -2)
print ("Is it less or equal?", 5 <= -2)
| true |
c048d4b596c44f713a76df3686b5ed44cc9275cb | UnSi/2_GeekBrains_courses_algorithms | /Lesson 3/hw/task5.py | 822 | 4.40625 | 4 | # 5. В массиве найти максимальный отрицательный элемент. Вывести на экран его значение и позицию в массиве.
# Примечание к задаче: пожалуйста не путайте «минимальный» и «максимальный отрицательный».
# Это два абсолютно разных значения.
from random import randint
MIN_VALUE = -100
rand_array = [randint(MIN_VALUE, 100) for _ in range(50)]
print(rand_array)
max_number = MIN_VALUE
for i, number in enumerate(rand_array):
if 0 > number > max_number:
max_number = number
pos = i
print(f"Максимальное отрицательное число: {max_number}, индекс {pos}, позиция: {pos+1}")
| false |
26896d0a35149e576c7e163e23c331af5794880b | The-Wayvy/Intro_to_CS_MIT_OCW | /pset_02/hangman.py | 2,395 | 4.25 | 4 | # 6.00 Problem Set 3
#
# Hangman
#
# -----------------------------------
import random
import string
WORDLIST_FILENAME = "words.txt"
def load_words():
"""
Returns a list of valid words. Words are strings of lowercase letters.
Depending on the size of the word list, this function may
take a while to finish.
"""
print "Loading word list from file..."
# inFile: file
inFile = open(WORDLIST_FILENAME, 'r', 0)
# line: string
line = inFile.readline()
# wordlist: list of strings
wordlist = string.split(line)
print " ", len(wordlist), "words loaded."
return wordlist
def choose_word(wordlist):
"""
wordlist (list): list of words (strings)
Returns a word from wordlist at random
"""
return random.choice(wordlist)
def print_word(word,guessed_letters):
to_print = ""
for letter in word:
if letter in guessed_letters:
to_print += letter
else:
to_print += "?"
return to_print
def finished(word,guessed_letters):
for letter in word:
if letter not in guessed_letters:
return False
else:
return True
# actually load the dictionary of words and point to it with
# the wordlist variable so that it can be accessed from anywhere
# in the program
wordlist = load_words()
secret_word = choose_word(wordlist)
guesses_remaining = 7
guessed_letters = []
print "Welcome to hangman!"
while guesses_remaining > 0:
print "=========================================="
print "you have",guesses_remaining,"guesses left"
print "you've guessed the following letters",guessed_letters
print "the word so far:",print_word(secret_word,guessed_letters)
guess = raw_input("guess a letter: ")
while guess in guessed_letters:
print "you already guessed that!"
guess = raw_input("guess a letter: ")
if guess in secret_word:
print "good guess"
guessed_letters.append(guess)
if finished(secret_word,guessed_letters):
print "=================================="
print "You win!"
print "the word was",secret_word,"!!"
break
else:
print "sorry"
guesses_remaining -= 1
guessed_letters.append(guess)
else:
print "==============================="
print "sorry, you lost"
print "the word was",secret_word | true |
0508f833481927c25028946b78825f3f8736986d | philipteu/Fibonacci-Sequence | /Fibonacci Sequence.py | 353 | 4.125 | 4 |
#Fibonacci Sequence
def Fibonacci(n):
if n < 0:
print('Invalid Input')
elif n <= 1:
return n
else:
return (Fibonacci(n-1)+ Fibonacci(n - 2 )) #F= Fn-1 + Fn-2
n= int(input('This is Fibonacci Sequence. Enter the iteration: '))
for i in range(n):
print(Fibonacci(i))
#0,1,1,2,3,5,8,13,21,34,55 | false |
46cdb00040fece861462422e46590373a8983cc4 | Psandoval93/HigherLowerGame | /HigherLowerGame.py | 1,380 | 4.21875 | 4 | # Import random class
import random
# Receiving/assigning variables & seed value
seed_value = int(input("What seed should be used? "))
random.seed(seed_value)
lower = int(input('What is the lower bound?'))
upper = int(input('What is the upper bound?'))
# Creating loop statement & condition
while lower > upper:
print('Lower bound must be less than upper bound.')
# Receiving user input and assigning variable
lower = int(input('What is the lower bound?'))
upper = int(input('What is the upper bound?'))
# Assigning variable for number to win game
rand_num = random.randint(lower, upper)
# Assigning variable in order to quit game
user_quits = 00
# Receive user input and assigning to variable for guess at winning number
user_guess = int(input('What is your guess?'))
# Creating loop statement & conditions
while user_guess != user_quits:
# Creating if, if else, and if else statements with conditions for output
if user_guess < rand_num:
print('Nope, too low.')
user_guess = int(input('What is your guess? '))
elif user_guess > rand_num:
print('Nope, too high.')
user_guess = int(input('What is your guess? '))
elif user_guess == rand_num:
print('You got it!')
else:
# user_guess == rand_num
print('You have quit the game. Thank you for playing')
| true |
2649ef9301c51fda72a8138bc5be11dd52692f1b | Maryanushka/MITx-6.00.1-2017- | /week2/Problem 2 other solution.py | 1,798 | 4.375 | 4 | # ----------------------------another solution------------------------
# PROBLEM 2: PAYING DEBT OFF IN A YEAR (15 points possible)
# Now write a program that calculates the minimum fixed monthly payment needed in order pay off a credit card balance within
# 12 months. By a fixed monthly payment, we mean a single number which does not change each month, but instead is a constant
# amount that will be paid each month.
# In this problem, we will not be dealing with a minimum monthly payment rate.
# The following variables contain values as described below:
# balance = 3329
#
# annualInterestRate = 0.2
#
# # The program should print out one line: the lowest monthly payment that will pay off all debt in under 1 year, for example:
#
# lowestPayment = 0
# finding = True
#
#
# def payment(balance, annualInterestRate, lowestPayment):
# month = 1
# while month <= 12:
#
# # Monthly interest rate= (Annual interest rate) / 12.0
# monthlyIntRate = annualInterestRate / 12.0
#
# # Monthly unpaid balance = (Previous balance) - (Minimum monthly payment)
# balance -= lowestPayment
#
# # Updated balance each month = (Monthly unpaid balance) + (Monthly interest rate x Monthly unpaid balance)
# balance += (monthlyIntRate * balance)
#
# month += 1
#
# if balance <= 0:
# return lowestPayment
# else:
# return False
#
#
# while finding == True:
# if payment(balance, annualInterestRate, lowestPayment):
# finding = False
# print
# 'Lowest Payment:', payment(balance, annualInterestRate, lowestPayment)
# else:
# lowestPayment += 10
# payment(balance, annualInterestRate, lowestPayment)
# ------------------another solution-------------------------------- | true |
fa8b56d81d03077d615920df4df46669cfca8b0d | blane612/-variables | /example_code.py | 2,062 | 4.75 | 5 | # author: elia deppe
# date: 6/6/21
#
# description: example code for variable exercises
# Saving a String to a variable.
name = 'elia'
# ----- Printing the contents of the variable.
print('my name is', name) # using regular strings
print(f'my name is {name}') # using f-strings
# When inserting a variable into an f-string, simply surround the name of the variable with {}
# Notice that the color of the curly braces and the variable aren't the same color as the string. This helps us
# identify what is and what isn't a string. Notice also that the curly braces aren't printed either, only the contents
# of the variable.
# ----- Saving a number to a variable.
num1 = 7
num2 = 11.5
num3 = -21
print('here are a few numbers:', num1, num2, num3) # regular string
print(f'here are a few numbers: {num1}, {num2}, {num3}') # f-string
# ----- Overwriting Variables (Re-Use)
num1 = 121
num2 = 4.422
num3 = -31.23 # Variables can switch types at any point, so feel free to save whatever you want to any variable!
print('here are the same variables, but now with different values:', num1, num2, num3)
print(f'here are the same variables, but now with different values: {num1}, {num2}, {num3}')
# ----- Saving the Result of an Operation to a Variable
num1 = 15 - 2
num2 = 12 / 3
num3 = -2 * 4.5
print('nums:', num1, num2, num3)
print(f'nums: {num1}, {num2}, {num3}')
# ----- Using Variables in Operations
num1 = num2 + num3
num2 = num1 - 15
num3 = num3 * 2 # You can use a variable in an operation that is assigning to itself!
print(f'nums: {num1}, {num2}, {num3}')
# ----- String Duplication
# String duplication is where you duplicate strings using multiplication.
# For example, if I wanted to duplicate 'cat' 5 times then:
print('cat' * 5)
# Yes, you're allowed to do operations in functions, remember that the result is what matters!
print('dog' * 3, 'gone' * 3)
# If you want to duplicate a string inside a variable, simply multiply the variable by a number
cash = '$'
hashtag = '#'
print(cash * 5, hashtag * 3, cash * 5) | true |
6ebc6ad98fa1996c1c20ce7bbdb792c6d1d61d48 | RayOct18/effective_python | /46_property_decorator.py | 2,058 | 4.125 | 4 |
class Grade2:
def __init__(self):
self._value = 0
def __get__(self):
return self._value
def __set__(self, value):
if not (0 <= value <= 100):
raise ValueError('Grade must be between 0 and 100.')
self._value = value
class Exam2:
def __init__(self):
# create different object when Exam2 is created
self.math_grade = Grade2()
self.writing_grade = Grade2()
self.science_grade = Grade2()
class Grade:
# use dictionary to save different instance's value
def __init__(self):
self._value = {}
def __get__(self, instance, instance_type):
if instance is None:
return self
return self._value.get(instance, 0)
def __set__(self, instance, value):
if not (0 <= value <= 100):
raise ValueError('Grade must be between 0 and 100.')
self._value[instance] = value
class Exam:
# use same object
math_grade = Grade()
writing_grade = Grade()
science_grade = Grade()
if __name__ == '__main__':
first_exam = Exam()
first_exam.writing_grade = 82
first_exam.science_grade = 99
print('Writing', first_exam.writing_grade)
print('Science', first_exam.science_grade)
second_exam = Exam()
second_exam.writing_grade = 88
second_exam.science_grade = 95
print('Writing', second_exam.writing_grade)
print('Science', second_exam.science_grade)
print('Writing', first_exam.writing_grade)
print('Science', first_exam.science_grade)
print('=' * 20)
first_exam = Exam2()
first_exam.writing_grade = 82
first_exam.science_grade = 99
print('Writing', first_exam.writing_grade)
print('Science', first_exam.science_grade)
second_exam = Exam2()
second_exam.writing_grade = 88
second_exam.science_grade = 95
print('Writing', second_exam.writing_grade)
print('Science', second_exam.science_grade)
print('Writing', first_exam.writing_grade)
print('Science', first_exam.science_grade) | true |
628820783ca7a797023d5d7d8bfee7fb5b8cd9d5 | joaomendonca-py/tip-calculator-start | /main.py | 1,525 | 4.125 | 4 | #If the bill was $150.00, split between 5 people, with 12% tip.
#Each person should pay (150.00 / 5) * 1.12 = 33.6
#Format the result to 2 decimal places = 33.60
#Tip: There are 2 ways to round a number. You might have to do some Googling to solve this.💪
#HINT 1: https://www.google.com/search?q=how+to+round+number+to+2+decimal+places+python&oq=how+to+round+number+to+2+decimal
#HINT 2: https://www.kite.com/python/answers/how-to-limit-a-float-to-two-decimal-places-in-python
print("Welcome to the tip calculator.")
bill = input("What was the total bill? ")
percentage = input("What percentage tip would you like to give? 10, 12 or 15? ")
manyPeople = input("How many people to split the bill? ")
bill_int = float(bill)
percentage_int = float(percentage)
manyPeople_int = int(manyPeople)
result = round((bill_int * (1 + (percentage_int / 100))) / manyPeople_int, 2)
pay = (f"Each person should pay: {result}")
print(pay)
"""
print("Welcome to the tip calculator!")
bill = float(input("What was the total bill? $"))
tip = int(input("How much tip would you like to give? 10, 12, or 15? "))
people = int(input("How many people to split the bill?"))
tip_as_percent = tip / 100
total_tip_amount = bill * tip_as_percent
total_bill = bill + total_tip_amount
bill_per_person = total_bill / people
final_amount = round(bill_per_person, 2)
#FAQ: How to round to 2 decimal places?
#https://www.udemy.com/course/100-days-of-code/learn/lecture/17965132#questions/13315048
print(f"Each person should pay: ${final_amount}")""" | true |
58c62ed2048d5255e99700d2f0e43ae701849bc6 | kuan1/test-python | /30days/day04_string/02_method.py | 847 | 4.4375 | 4 | # Accessing Characters in Strings by Index
print('Python'[2])
# Slicing Python Strings
print('Python'[0:2])
print('Python'[-3:])
print('Python'[1:])
# Reversing a String
print('Python'[::-1])
# Skipping Characters While Slicing
print('Python'[0::2])
# capitalize
print('python'.capitalize())
# count
print('python python'.count('p'))
# endswith
print('Python'.endswith('on'))
# startswith
print('Python'.startswith('P'))
# find
print('thirty days of python'.find('y')) # 5
print('thirty days of python'.find('th')) # 0
print('a'.find('b'))
# rfind
print('thirty days of python'.rfind('y')) # 16
print('thirty days of python'.rfind('th')) # 17
# index
print('abc'.index('a'))
print('abc'.index('c'))
# print('abc'.index('d'))
# join
print('|'.join(["a", "b", "c"]))
# split
print('1,2,3,4'.split(','))
print(type('1,2,3,4'.split(','))) | false |
a945f0f5dfd7632cd84b894bde2cc233cd1a71c1 | milinilla/Practice | /are_anagrams.py | 1,122 | 4.375 | 4 | def are_anagrams(s1, s2, s3):
"""
we convert function arguments into the list, as it would be easier
to deal with it later on, as every argument needs to be the subject
of the same actions
"""
arr = [s1, s2, s3]
"""
we check if the length of all arguments doesn't exceed the provided
limit of 5 characters
"""
if not all(len(item)<= 5 for item in arr):
return("Sorry, all \"are_anagrams\" function arguments shall be "\
"at most 5 characters long.")
"""
first, we unify the formating of input data by lowering the case,
next we convert every string into the list of characters, which is
then sorted in alphabetical order to allow the verification if every
string consists of the same characters
"""
arr = [sorted(list(x.lower())) for x in arr]
"""
first idea was to just compare every element of the list
(arr[0]==arr[1]==arr[2]), but to make the code scalable we compare
if the first element of the list appears the same number of times as
the number of all list elements
"""
return(arr.count(arr[0])==len(arr)) | true |
b7ca6eb34a68da479904f6b93a9dc692f336850e | YvesIraguha/pythonlearning | /simple_game/calculator.py | 2,534 | 4.40625 | 4 |
def calculator():
print "Welcome to caclcualator \n you can calculate any number you want"
print "Here there are many operations."
print "write add to add two numbers"
print "Write multiply to multiply two numbers"
print "Write divide to make division of two numbers"
print "Write substract to substract one number from another"
print "Write remainder to know the \n remainder of division of one number with another"
print "Write square to square a number. "
def addition():
first_number=float(raw_input("Enter the first number: "))
second_number = float(raw_input("Enter the second number: "))
result = first_number + second_number
print "From the calculation the result is %s "%result
def substraction():
first_number=float(raw_input("Enter the first number: "))
second_number = float(raw_input("Enter the second number: "))
result = first_number - second_number
print "From the calculation the result is %s "%result
def multiplication ():
first_number=float(raw_input("Enter the first number: "))
second_number = float(raw_input("Enter the second number: "))
result = first_number * second_number
print "From the calculation the result is %s "%result
def division ():
first_number=float(raw_input("Enter the first number: "))
second_number = float(raw_input("Enter the second number: "))
result = first_number / second_number
print "From the calculation the result is %s "%result
def squaring():
first_number=float(raw_input("Enter the first number: "))
second_number = float(raw_input("Enter the second number: "))
result = first_number ** second_number
print "From the calculation the result is %s "%result
def remainder():
first_number=float(raw_input("Enter the first number: "))
second_number = float(raw_input("Enter the second number: "))
result = first_number % second_number
print "From the calculation the result is %s "%result
def calculation():
user_command = str(raw_input("Can you write your operation: "))
if user_command== "add":
addition()
elif user_command=="multiply":
multiplication()
elif user_command== "divide":
division()
elif user_command=="substract":
substraction()
elif user_command=="remainder":
remainder()
elif user_command=="square":
squaring()
else:
print "Incorrect operation!"
move_out = str(raw_input("Do you want to make more calculation: "))
if move_out == "yes":
calculation()
else:
print "Thank you for using this awesome calculator"
calculation()
calculator()
| true |
4c508bf06a0d87676b0168988eba6ed00ed47bfa | Andrei-Kulik/python_for_dqe | /Module_1.py | 1,869 | 4.21875 | 4 | # importing 'randint' method for creating new list with random numbers
from random import randint
# declaring size of list 'n'
n = 100
# declaring new empty list 'a'
a = []
# loop for creating our new list
for i in range(n):
# 'append' method for adding new elements to list
# 'randint' method for creating a random number from 0 to 1000
a.append(randint(1, 1000))
# sorting algorithm (I used bubble sorting)
# index 'i' we will use for further reducing our scope of unsorted numbers
for i in range(len(a) - 1):
# index 'j' we will use for list elements iterations
# we need to find maximum form the list and put it at the end
# after we decrease our range() on 'i', because we no longer need to compare sorted items at the end
for j in range(len(a) - 1 - i):
# comparing element with next element
if a[j] > a[j + 1]:
# replacing elements with each other
a[j], a[j + 1] = a[j + 1], a[j]
# declaration variables for counting average values (sum / count = average)
even_sum, even_count = 0, 0
odd_sum, odd_count = 0, 0
# loop for counting sum and count for even and odd elements
for elem in a:
# sum and count for even elements (even % 2 == 0)
if elem % 2 == 0:
# increasing the sum on element value
even_sum += elem
# increasing the count on one element
even_count += 1
# sum and count for odd elements
else:
# increasing the sum on element value
odd_sum += elem
# increasing the count on one element
odd_count += 1
# printing final results
# added converting to the 'int' type to round float values (* optional)
# also added converting average values to the 'str' type for concatenating with the string
print('Even average: ', str(int(even_sum / even_count)))
print(' Odd average: ', str(int(odd_sum / odd_count)))
| true |
77be81fd9138f3ef5eed7432a1733f16553f1124 | jharp9077/CTI110 | /P2HW1_PoundsKilograms_JordanHarper.py | 521 | 4.1875 | 4 | # Calculates the conversion of pounds to kilograms
# 4 Febuary 2019
# CTI-110 P2HW1 - Pounds to Kilograms Converter
# Jordan Harper
#
# Get pounds
# convert total kilograms
# display kilograms
# Get the total kilograms
pounds = float(input('Enter the number of pounds: '))
# calculate pounds to kilograms by doing the equation pounds times 1kg devided by 2.21 equals total KG
kilograms = pounds * 1/2.2046
# display the total amount of kilograms
print ('Pounds equals number of kilograms ', kilograms, 'kg')
| true |
9175767b47d02926234c300f97bc12e256351aad | jharp9077/CTI110 | /P4HW2_ PoundsKilos _JordanHarper.py | 636 | 4.1875 | 4 | # Calculates the conversion of pounds to kilograms
# 5 March 2019
# CTI-110 P4HW2 - Pounds to Kilograms table
# Jordan Harper
#
# Get pounds
# convert total kilograms
# display kilograms
# Get the total kilograms
def kilo_table():
# Defined pounds in the range using start stop step format
for pounds in range(100, 301, 10):
# Enter the formula held in a variable format
kilograms = pounds / 2.2046
# Displays result specifically calling for 2 decimal places.
print("{:.2f}".format(kilograms), 'kg')
kilo_table()
| true |
a98dc9f4ad246805c593b5ed0582f7fc8ef6b482 | jcattanach/python-practice | /practice6.py | 526 | 4.46875 | 4 | # Ask the user for a string and print out whether
# this string is a palindrome or not.
user_string = input("Enter a word: ").lower()
def reverse(string):
new_string = ""
for index in range(len(string) -1, -1, -1):
new_string = new_string + string[index]
return new_string
def is_palindrome(reversed):
if(reversed == user_string):
print("{0} is a palindrome.".format(user_string))
else:
print("{0} is NOT a palindrome.".format(user_string))
is_palindrome(reverse(user_string))
| true |
c3c8e7f98315dd8c316f084f1517d3f583b8d5b8 | truptikolte/Learn-Python | /datatype.py | 226 | 4.125 | 4 | str = 'Good Morning' # string data type
num = 123 #integer data type
fnum = 3.14 # float data type
print(str,num,fnum)
print(type(str))
print(type(num))
print(type(fnum)) # type function return the data type of any variable
| true |
50bf2e4bc5e0e0f47378e12a34b4ef23b35c7ab5 | Saeed-Jalal/Basics-of-python | /Basics of python.py | 1,206 | 4.34375 | 4 | #Arithmetic Operators
num1=60
num2=45
print(num1+num2)
print(num1-num2)
print(num1*num2)
print(num1//num2)
print(num1%num2)
print(7**3)
#Assignemt Operator
num3= num1+num2
print(num3)
num3+=num2
print(num3)
#Comparsion Operator
print(num3>num2)
print(num2==num3)
print(num1!=num2)
#Logical Operators
A=True
B=False
print(A and B)
print(A or B)
print(not A)
#Identity Operators
X = 5
X is 5
X = 5
X is not 5
#Membership Operators
ABC=[1,2,3,4,5,6,7,8,9]
4 in ABC
6 not in ABC
#Python has two types of Data 1. Immutable (Numbers, Strings, Tuples) 2. Mutable (Lists, Dictionaries, Sets)
#Numbers(Integer,Float and Complex)
#Strings
name1="My name is Saeed."
name2="I am genius."
#Concatenation
print(name1+name2)
#Repetition
print(name1*2)
#Slicing
print(name2[2:7])
#find()
print(name1.find('S'))
# replace()
print(name2.replace("am","can be"))
# split()
job="Don"
print(job.split(","))
# count()
print(name1.count("a"))
#upper
print(name1.upper())
#max
print(max(name2))
#min
str="HOME"
print(min(str))
#isalpha
print(name1.isalpha())
#Tuples
myTup = ("Home","Office","Car","Computer")
#Concatenation
print(myTup+("Copter","Ship"))
#Reprtition
print(myTup*2)
#Slicing
print(myTup[1:3])
#
| true |
bccd0f8e1a1ede2fc7ba425da057723a1d4219fd | StarbzYT/Sorting-Algorithms | /Merge_Sort/merge.py | 773 | 4.21875 | 4 | import merging_arrays # merge func to merge two sorted arrays
# Pseudocode
# Break up the array into halves until you have arrays that are empty or have one element
# Once you have smaller sorted arrays, merge those arrays with other
# sorted arrays until you are back at the full length of the array
# once the array has been merged back together, return the merged (and sorted!) array
def merge(arr):
if len(arr) <= 1: # base case
return arr
mid = len(arr) // 2 # floored
left = merge(arr[0:mid]) # 0 to mid (exclusive)
right = merge(arr[mid::]) # mid to end
return merging_arrays.merge(left, right) # keep merging left and right sorted arrays
print(merge([1, 18, 6, 41, 3, 100])) # [1, 3, 6, 18, 41, 100]
| true |
610519db921e71a98636862398b2351bd2041b61 | dilayercelik/CSE160-Data-Programming-UW | /Lists Check-in/make_out_word.py | 748 | 4.15625 | 4 | # -*- coding: utf-8 -*-
"""
Created on Mon May 18 19:45:20 2020
@author: dilayerc
"""
# Practice
# Given an "out" string length 4, such as "<<>>", and a word,
# return a new string where the word is in the middle of the out string,
# e.g. "<<word>>".
# Examples:
## make_out_word('<<>>', 'Yay') → '<<Yay>>'
## make_out_word('<<>>', 'WooHoo') → '<<WooHoo>>'
## make_out_word('[[]]', 'word') → '[[word]]'
# Answer
def make_out_word(out, word):
new_string = out[:2] + word + out[2:]
return new_string
# Tests
print(make_out_word('<<>>', 'Yay')) # correct output
print(make_out_word('<<>>', 'WooHoo')) # correct output
print(make_out_word('[[]]', 'word')) # correct output
| true |
3c456176c6968092afb32d106ba629d9243d8487 | dilayercelik/CSE160-Data-Programming-UW | /Loops, If, & Functions Check-in/sum_odds.py | 740 | 4.25 | 4 | # -*- coding: utf-8 -*-
"""
Created on Thu May 14 23:11:57 2020
@author: Dilay Ercelik
"""
# Return the sum of the odd ints in the given list.
# Note: the % "mod" operator computes the remainder, e.g. 5 % 2 is 1.
# Examples:
## sum_odds([5, 2, 6, 3, 4]) → 8
## sum_odds([3, 6, 11, 2, 5]) → 19
## sum_odds([]) → 0
# Answer:
def sum_odds(nums):
sum_odd = 0
for num in nums:
if num % 2 == 1:
sum_odd += num
return sum_odd
# Tests:
print(sum_odds([5, 2, 6, 3, 4])) # correct output
print(sum_odds([3, 6, 11, 2, 5])) # correct output
print(sum_odds([])) # correct ouput
| true |
9b1a98d704ca48daabc4380d415f88e60371ecb6 | dilayercelik/CSE160-Data-Programming-UW | /Lists Check-in/front_back.py | 532 | 4.25 | 4 | # Practice
# Given a string, return a new string where the first and last chars have been exchanged.
# Examples:
## front_back('code') → 'eodc'
## front_back('a') → 'a'
## front_back('ab') → 'ba'
# Answer
def front_back(str):
if len(str) > 1:
new_string = str[-1] + str[1:-1] + str[0]
else:
new_string = str
return new_string
# Tests
print(front_back('code')) # correct output
print(front_back('a')) # correct output
print(front_back('ab')) # correct output
| true |
eef4aa43fb7507de91bd76870d2bb0bdfeb77a18 | Kaminagiyou123/python_practice | /Games/randomnumber.py | 802 | 4.28125 | 4 | import random
def guess(x):
random_number=random.randint(1,x);
guess=0;
while guess!=random_number:
guess=int(input(f"guess the number between 1 and {x}: "));
if guess<random_number:
print("sorry, the guess is too low")
elif guess>random_number:
print("sorry the guess is too high")
print(f"yay congrats, you got the number {random_number}")
def computer_guess(x):
low=1;
high=x;
feedback=""
while feedback!="c":
if low!=high:
guess=random.randint(low,high)
else:
guess=low;
feedback =input(f"Is {guess} to high (H),too low (L), or correct(C)").lower();
if feedback=="h":
high=guess-1;
elif feedback=="l":
low=guess+1;
print(f"yay, computer got the number {guess}, correctly")
computer_guess(10)
| true |
ccbcc6245e68930e7d81f6b4f326347f9ed622e8 | Kaminagiyou123/python_practice | /Second/Database/app copy.py | 1,074 | 4.1875 | 4 | from utils import database
USER_CHOICE="""
Enter:
-'a' to add a book
-'l' list all books
-'r' to mark a book as complete
-'d' to delete a book
-'q' to quit
Your choice:"""
def add_book():
name_input=input("Enter the name of the book: ")
author_input=input("Enter the author of the book: ")
database.add_book(name_input,author_input)
def list_books():
books=database.get_all_books()
for book in books:
print(book)
def mark_read():
name_input=input("Enter the name of the book you read: ")
books=database.get_all_books()
for book in books:
if book['name']==name_input:
database.markread(name_input)
def delete_book():
name_input=input("Enter the name of the book you want to delete: ")
database.delete_book(name_input)
def menu():
user_input=input(USER_CHOICE)
database.create_book_table()
while user_input!='q':
if user_input=='a':
add_book()
elif user_input=='l':
list_books()
elif user_input=='r':
mark_read()
elif user_input=='d':
delete_book()
user_input=input(USER_CHOICE)
menu() | true |
ad7fa474866597dba60464b706c262d4f69af017 | ananthramas/pythonBasics | /Day12/fillColor.py | 529 | 4.46875 | 4 | # draw color-filled square in turtle
import turtle
# creating turtle pen
t = turtle.Turtle()
# taking input for the side of the square
s = int(input("Enter the length of the side of the square: "))
# taking the input for the color
col = input("Enter the color name or hex value of color(# RRGGBB): ")
# set the fillcolor
t.fillcolor(col)
# start the filling color
t.begin_fill()
# drawing the square of side s
for _ in range(4):
t.forward(s)
t.right(90)
# ending the filling of the color
t.end_fill()
| true |
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