blob_id string | repo_name string | path string | length_bytes int64 | score float64 | int_score int64 | text string | is_english bool |
|---|---|---|---|---|---|---|---|
87df1e1450436998657e85387bf327b296890e47 | zhangzheng888/Python | /Python 3/004A Tuple Accessors.py | 1,721 | 4.9375 | 5 | """
Data Type: Tuples
Tuple Accessors
Indexing
The index operator [] to access an item in a tuple, where the index starts from 0.
A tuple having 6 elements will have indices from 0 to 5. Trying to access an index outside of the
tuple index range(6,7,... in this example) will raise an IndexError.
The index must be an integer, float or other types can't be used. This will result in TypeError.
Likewise, nested tuples are accessed using nested indexing.
"""
# Accessing tuple elements using indexing
simple_tuple = ('s', 'i', 'm', 'p', 'l', 'e')
print(simple_tuple[0]) # 's'
print(simple_tuple[5]) # 'e'
# IndexError: list index out of range
# print(simple_tuple[6])
# Index must be an integer
# TypeError: list indices must be integers, not float
# simple_tuple[2.0]
# nested tuple
nested_tuple = ("spice", [8, 4, 6], (1, 2, 3))
# nested index
print(nested_tuple[0][3]) # 'c'
print(nested_tuple[1][1]) # 4
"""
Negative Indexing
The index of -1 refers to the last item, -2 to the second last item and so on.
"""
# Negative indexing for accessing tuple elements
negative_tuple = ('h', 'o', 'r', 's', 'e', 's')
# Output: 't'
print(negative_tuple[-1])
# Output: 'p'
print(negative_tuple[-6])
"""
Slicing
A range of items in a tuple by using the slicing operator colon :.
"""
# Accessing tuple elements using slicing
slice_tuple = ('s', 'l', 'i', 'c', 'i', 'n', 'g', 's')
# elements 2nd to 4th
# Output: ('l', 'i', 'c')
print(slice_tuple[1:4])
# elements beginning to last
# Output: ('s', 's')
print(slice_tuple[:-7])
# elements 8th to end
# Output: ('g', 's')
print(slice_tuple[6:])
# elements beginning to end
# Output: ('s', 'l', 'i', 'c', 'i', 'n', 'g', 's')
print(slice_tuple[:])
| true |
6acdf1c3b85a9d24913d1cd76ef07ee0977af257 | xiaoqing928/DataStructure | /Queue and Stack/1.2queue.py | 835 | 4.25 | 4 | '''
用一个固定数组创建一个队列
思路：start和end两个指针，start返回值，end加入值，同时判断和数组size的大小
相当于start和end在转圈
'''
class Queue:
start, end = 0, 0
arr = [[]]
size = 0
def __init__(self, l):
self.arr = self.arr * l
def push(self, num):
if self.size < len(self.arr):
self.arr[self.end % len(self.arr)] = num
self.end += 1
self.size += 1
else:
print('arr is full')
def poll(self):
if self.size > 0:
self.size -= 1
self.start += 1
return self.arr[(self.start-1) % len(self.arr)]
else:
print('arr is null')
a = Queue(2)
a.push(1)
a.push(2)
print(a.arr)
a.push(3)
print(a.poll())
a.push(3)
a.push(3)
print(a.poll()) | false |
0a9b7939cb27d11136da319770de28a1f8e25b87 | kadr19-333/GitITEA | /HomeWork_3/hw3_6.py | 827 | 4.4375 | 4 | # Если решать по тому как написано в задаче "Используя условные операторы if - elif осуществить проверку:"
number = float(input('Введите ЦЕЛОЕ число '))
text = 'Нет результата!'
if number >= 0 and number <= 5:
if number == 0:
print(0)
elif number == 1:
print(1)
elif number == 2:
print(2)
elif number == 3:
print(3)
elif number == 4:
print(4)
elif number == 5:
print(5)
else:
print('Ошибка:', text)
# Но можно упростить всё:
# number = int(input('Введите число: '))
# text = 'Нет результата!'
# if number >= 0 and number <= 5:
# print(number)
# else:
# print('Ошибка:', text) | false |
86274e0dcb9ece9bd082fb2c17f8e12a08006f93 | taking1fortheteam/pands-problem-set | /Problem3.py | 962 | 4.28125 | 4 | # Aidan Conlon - 21 March 2019
# This is the solution to Problem 3
# Write a program that prints all numbers between 1,000 and 10,000 that are divisible
# by 6 but not 12.
i = 1000 # Set i to 1000
x = 1000 # Set x to 1000
# Print to screen following comment
print("The Values divisible by 6 but not divisible by 12 between 1,000 & 10,000 are:")
while i <= 10000: # While loop - while i is less than 10001 continue with this loop
if x % 6 == 0 and x % 12 != 0: # if x is divisible by 6 (no remainder) AND not divisible by 12 (remainder)
print(x) # print the value of x to the screen
i = i + 1 # increment the value of i (for next iteration)
x = x + 1 # increment value of x (for next iteration)
quit() # Finish | true |
21354b9cf2e1b31c94b65c3acc5f59b31a6fc172 | dbbaskette/FinanceDataGen | /objects/Transaction.py | 705 | 4.125 | 4 | class Transaction(object):
"""A customer of with a checking account. Customers have the
following properties:
Attributes:
name: A string representing the customer's name.
balance: A float tracking the current balance of the customer's account.
"""
def __init__(self):
"""Return a Customer object whose name is *name* and starting
balance is *balance*."""
self.customerNumber = 0
self.streetAddress = ""
self.city=""
self.state = ""
self.zip = ""
self.latitude = 0
self.longitude = 0
self.transactionTimestamp = 0
self.amount = 0
self.id= 0
self.flagged = 0
| true |
c373d27eb30c8e8b92f8742814a760b79790229f | miguelr22/Lab-Activity-4 | /M4P2.py | 334 | 4.15625 | 4 | #This program will take a password from the user and print an appropriate message
#The 'in' keyword checks to see if a value exists somewhere in the given string.
greeting = input("Hello, possible pirate! What's the password?")
if greeting in ("Arrr!"):
print("Go away, pirate.")
else:
print("Greetings, hater of pirates!")
| true |
29bac457e5708c1da32de148ecd088b82564dde8 | richbon75/practice | /CCI_6ed/Chapter_3/ch3_3_plates.py | 1,680 | 4.25 | 4 | """Set of stacks - make new stack when current stack reaches max height."""
from collections import deque
class SetOfStacks(object):
"""Make a new stack when current stack reaches max height.
Should be transparent to user - push() and pop() should work
simply as expected."""
def __init__(self, maxheight):
"""maxheight is the maximum height of a single stack"""
self.maxheight = maxheight
self.stax = deque()
def __len__(self):
return sum([len(x) for x in self.stax])
def push(self, value):
"""Push a value onto the set of stacks"""
if not self.stax or len(self.stax[-1]) >= self.maxheight:
self.stax.append(deque())
self.stax[-1].append(value)
def pour(self, iterable):
"""Push in lots of values"""
for value in iterable:
self.push(value)
def pop(self):
"""Pop a value off the set of stacks"""
if not self.stax:
raise IndexError('stack empty')
value = self.stax[-1].pop()
self.trim()
return value
def pop_substack(self, i):
"""Pop a value out of a substack"""
value = self.stax[i].pop()
self.trim()
return value
def trim(self):
"""Check the tail for empty stacks, and remove them."""
while self.stax and len(self.stax[-1]) == 0:
self.stax.pop()
def print(self):
"""output the SetOfStacks so we can visualize it"""
for i, x in enumerate(self.stax):
print('{i} : {x}'.format(i=i, x=x))
if __name__ == "__main__":
merp = SetOfStacks(7)
merp.pour(range(0, 55))
merp.print()
| true |
1288b1bb228cb31061e023620d7e96acf8c3891a | richbon75/practice | /CCI_6ed/Chapter_5/ch5_1_insertion.py | 1,425 | 4.15625 | 4 | """Bit manipulation
You are given two 32-bit numbers, N and M, and two bit positions, i and j.
Write a method to insert M into N such that M starts at bit j and ends at bit i.
You can assume that the bits j through i have enough space to fit all of M.
That is, if M = 10011, you can assume that there are at least 5 bits between j
and i. You would not, for example, have j = 3 and i = 2, because M could not
fully fit between bit 3 and bit 2.
EXAMPLE
Input: N = 10000000000, M = 10011, i 2, j 6
Output: N = 10001001100
"""
# Looks like we're just overwriting the bits in positions j to i.
# Note: Because of the way integers work in Python (no fixed byte length)
# you can't make a one step "AND" mask of zeros where you want to clear
# values and 1s everywhere else, because the "everywhere else" space is
# effectively infinite. So I'll do it in a two-step process with a mask
# that has 1s where I want to clear values and 0s "everywhere else" (since
# that's the default)
def binary_overlay(N, M, i, j):
# make a mask
mask = 1
for _ in range(j-i):
mask = (mask << 1) | 1
mask <<= i
# shift M to the right place
M <<= i
# clear out old bits. Remember - no standard bit width in Python integers
N |= mask
N ^= mask
# overlay new bits
N |= M
return N
if __name__ == '__main__':
print('{0:b}'.format(binary_overlay(0b10000000000, 0b10011, 2, 6)))
| true |
0607f44c4bf7c629d7bf5142492d64c98f32d561 | TrollAndRoll/PythonMiniGames | /rps/play_game.py | 2,009 | 4.21875 | 4 | import random
ROCK = "rock"
PAPER = "paper"
SCISSORS = "scissors"
possible_actions = [ROCK, PAPER, SCISSORS]
def get_user_input():
return input("Enter a choice (rock, paper, scissors): ")
def get_computer_input():
return random.choice(possible_actions)
def determine_winner(user_choice, computer_choice):
"""
:param user_choice: string if user chose rock, paper, or scissors
:param computer_choice: computers string of rock, paper, or scissors
:return: 1 if user wins, 0 if tie, -1 if computer wins
"""
if user_choice == computer_choice:
print(f"Both players selected {user_choice}. It's a tie!")
return 0
elif user_choice == ROCK:
if computer_choice == SCISSORS:
print("Rock smashes scissors! You win!")
return 1
else:
print("Paper covers rock! You lose.")
return -1
elif user_choice == PAPER:
if computer_choice == ROCK:
print("Paper covers rock! You win!")
return 1
else:
print("Scissors cuts paper! You lose.")
return -1
elif user_choice == SCISSORS:
if computer_choice == PAPER:
print("Scissors cuts paper! You win!")
return 1
else:
print("Rock smashes scissors! You lose.")
return -1
def player_play_again():
return input("Do you want to play again? (y/n): ") == 'y'
def validate_input(user_choice):
return user_choice in possible_actions
def run_game():
print("Get ready to play rock, paper scissors!")
play_again = True
while play_again:
user = get_user_input()
while not validate_input(user):
print("Please choose between rock, paper, or scissors")
user = get_user_input()
computer = get_computer_input()
determine_winner(user, computer)
play_again = player_play_again()
print("Thank you for playing!")
if __name__ == '__main__':
run_game()
| true |
60b3491525e02ce518a3d7a7cb9aba21c83de340 | asya0107/candy-project | /candy_problem/main.py | 1,712 | 4.40625 | 4 | '''
DIRECTIONS
==========
1. Given the list `friend_favorites`, create
a new data structure in the function `create_new_candy_data_structure`
that describes the different kinds of candy paired with a list of friends that
like that candy.
friend_favorites = [
[ "Sally", [ "lollipop", "bubble gum", "laffy taffy" ]],
[ "Bob", [ "milky way", "licorice", "lollipop" ]],
[ "Arlene", [ "chocolate bar", "milky way", "laffy taffy" ]],
[ "Carlie", [ "nerds", "sour patch kids", "laffy taffy" ]]
]
2. In `get_friends_who_like_specific_candy()`, return friends who like lollipops.
3. In, `create_candy_set()`, return a set of all the candies from
the data structure made in `create_new_candy_data_structure()`.
4. Write tests for all of the functions in this exercise.
'''
# data = friend_favorites
def create_new_candy_data_structure(data):
# create function that describes differnet kinds of candy paired with a LIST of friends that like that candy
candy_types = []
# this makes a list of all the candy types
for friend in data:
for candy in friend[1]:
if candy not in candy_types:
candy_types.append(candy)
candy_dict = {}
friends_list = []
# make each value in candy types a key in candy dict
for candy in candy_types:
candy_dict[candy] = []
for candy in candy_types:
for friend in data:
if candy in friend[1]:
candy_dict[candy].append(friend[0])
return candy_dict
# def get_friends_who_like_specific_candy(data, candy_name):
# # return friends who like lollipops
# def create_candy_set(data):
# # return all candy sets made in the first function
| true |
57e69baedf18144687ea75d0d1ddad445f8cd0d6 | yahya-10/beginner_shoppinCart_with_python | /__main__.py | 1,802 | 4.46875 | 4 | import sys
def main_menu():
while True:
print()
print('''###SHOPPING CART
Select an option from the list below
1.Show all items in the cart
2.Add item to the cart
3.Remove item from cart
4.Check if item exists in the cart
5.How many items in the cart
6.Cleat the cart
7.Exit
''')
selection = input("Enter your option: ")
if selection == "1":
show_items()
elif selection == "2":
add_item()
elif selection == "3":
remove_item()
elif selection == "4":
check_item()
elif selection == "5":
count_of_items_in_cart()
elif selection == "6":
clear_cart()
elif selection == "7":
sys.exit()
else:
print("Wrong option!!")
shopping_cart = ["Milk", "Whole Grain Bread", "Chicken Breasts", "Bananas"]
def show_items():
for i in shopping_cart:
print("-> ", i)
def add_item():
new_item = input("Enter the new item you want to add: ")
shopping_cart.append(new_item)
print(new_item + " has been added to the cart")
def remove_item():
unwanted_item = input("Enter the item you want to remove from cart: ")
shopping_cart.remove(unwanted_item)
print(unwanted_item + " has been removed successfully!!")
def check_item():
item_to_check = input("Enter the item you're looking for: ")
for i in shopping_cart:
if i == item_to_check:
print(item_to_check + " exists in the cart!")
def count_of_items_in_cart():
print("There are ", len(shopping_cart), " item in cart")
def clear_cart():
shopping_cart.clear()
print("Shopping cart is clear!!")
main_menu()
| true |
656191fc7588e1fdc4eb6dc09a8509980d028ba7 | romaross/pythonProject | /task_1_5.py | 268 | 4.125 | 4 | import math
def triangle_square(a, b):
return 0.5 * a * b
res = input('a b (separated by space)')
a, b = map(float, res.split())
print('Hypotenuse of right-angled triangle -', math.hypot(a, b))
print('Square of right-angled triangle -', triangle_square(a, b))
| true |
71a98f3cdfe88981b6cd2ff3413cacfc613f5a8c | chnghyn/ABS | /chapter1/helloworld.py | 1,059 | 4.1875 | 4 | # This program says hello and asks for my name.
# 이 프로그램은 인사를 하고 내 이름을 묻는 프로그램이다.
print('Hello world!')
print('What is your name? ') # ask for their name 그들의 이름을 묻는다
myName = input()
print('It is good to meet you, ' + myName) # 내게 인사한다
print('THe length of your name is:')
print(len(myName))
print('What is your age? ') # ask for their age 그들의 나이를 묻는다
myAge = input()
print('You will be ' + str(int(myAge) + 1) + ' in a year.') # 일 년 후에 몇 살이 될거라고 말한다
# print(): 전달하는 문자열을 화면에 표시한다.
# input(): 사용자가 키보드로 텍스트를 입력하고 Enter 키를 누를 때까지 기다린다.
# len(): 문자열을 전달할 수 있고, 문자열의 문자 개수를 정수 형식으로 리턴한다.
# str(): 전달되는 값을 문자열 형식으로 리턴한다.
# int(): 전달되는 값을 정수 형식으로 리턴한다.
# float(): 전달되는 값을 부동 소수점 형식으로 리턴한다.
| false |
d2dce500f9e3ec8c0347d43461a99a9439cd972d | MarinaFirefly/Python_homeworks | /5/at_lesson/square.py | 975 | 4.125 | 4 | #find squres of the simple numbers using map
list_numers = [1,2,3,45,6,8,9,12,8,8,17,29,90,77,113]
#function that calculates squares
def sqrt_num(num):
return num**2
#function that return new list consisting of simple numbers
def dividers(list_num):
new_list = []
for num in list_num:
i = 2
cnt = 1 if num == 1 else 0 #take 1 as not simple number
while i <= num/2:
if num%i == 0:
cnt+=1 #if num has dividors between 2 and num/2 cnt is equal 1
break #break if cycle find out some dividor between 2 and num/2
else: i+=1
if cnt == 0: new_list.append(num)
return new_list
print(list(map(sqrt_num, dividers(list_numers))))
#ugly method that uses 2 functions: 1 finds dividors and 2 makes a list of simple numbers (besides 1 is simple here)
def div_for_num(num):
return [i for i in range (1,num) if num%i == 0]
def div_for_list(list_num):
return [i for i in list_num if len(div_for_num(i))<2]
print(list(map(sqrt_num, div_for_list(list_numers)))) | true |
7894adf619c48a61cec7f861d0efb4212055f8fa | JunctionChao/LeetCode | /next_bigger_number.py | 1,463 | 4.15625 | 4 | #!/usr/bin/env python
# -*- coding: utf-8 -*-
# Date : 2020-11-05
# Author : Yuanbo Zhao (chaojunction@gmail.com)
"""
get the next bigger number using the same digits of a number
ex: 123 -> 132
"""
def next_bigger(number):
# 将数字转化为list
number_to_list = []
while number:
number, mod = divmod(number, 10)
number_to_list.append(mod)
number_to_list = number_to_list[::-1]
# 先找到右边比左边大的第一个位置
size = len(number_to_list)
for x in range(size-1, -1, -1):
if number_to_list[x-1] < number_to_list[x]:
break
if x > 0:
# 找第二层较大的数
for y in range(size-1, -1, -1):
if number_to_list[x-1] < number_to_list[y]:
number_to_list[x-1], number_to_list[y] = \
number_to_list[y], number_to_list[x-1]
break
# 后续的数是降序的，做置换调整
for z in range((size-x)//2):
number_to_list[x+z], number_to_list[size-z-1] = number_to_list[size-z-1], number_to_list[x+z]
# 恢复为数字
res, ex = 0, 0
while number_to_list:
res += number_to_list.pop() * 10**ex
ex += 1
return res
# x==0说明左边的数字总是比右边的大
else:
return "the bigger number is not exist"
if __name__ == '__main__':
print(next_bigger(4321))
print(next_bigger(1342))
print(next_bigger(1243))
| false |
fcaac4ea634139f796f0256c20ce361fb83609a2 | manotasce/python | /PythonApplicationPractice/PythonApplicationPractice/MontlyLoanCalc.py | 423 | 4.25 | 4 | #
costOfLoan=input("Enter the cost of loan ")
rate=input("Enter the rate of loan ")
years=input("How many years you will pay the loan? ")
#Converting values entered
costOfLoan=float(costOfLoan)
rate=float(rate)
years=int(years)
#Calculate monthly payment
monthlyPayment=costOfLoan*((rate*(1+rate)**years)/(1+rate)**(years-1))
print("Monthly Payment %.3f " % monthlyPayment)
#print(costOfLoan)
#print(rate)
#print(years) | true |
e24400335960ffcf3adfb5236a9be56152ba3659 | tandiawanedgar/EdgarTandiawan_ITP2017TugasPAKBAGUS | /Tugas 3/Tugas 3-8/3-8.py | 267 | 4.15625 | 4 | place=['japan','paris','italy','us','cleveland']
print(place)
print(sorted(place))
print(place)
print(sorted(reversed(place)))
print(place)
place == reversed(place)
print(place)
place.reverse()
print(place)
print(sorted(place))
print(sorted(reversed(place)))
| true |
cae63bb302d25fd70c99763ee56c3b919675548c | norrismei/coding-fun | /linked_list_remove_duplicates_from_sorted_ll.py | 2,255 | 4.28125 | 4 | # You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order.
# Delete nodes and return a sorted list with each distinct value in the original list.
# The given head pointer may be null indicating that the list is empty.
#
# For your reference:
#
# SinglyLinkedListNode:
# int data
# SinglyLinkedListNode next
class SinglyLinkedListNode:
def __init__(self, node_data):
self.data = node_data
self.next = None
class SinglyLinkedList:
def __init__(self):
self.head = None
self.tail = None
def insert_node(self, node_data):
node = SinglyLinkedListNode(node_data)
if not self.head:
self.head = node
else:
self.tail.next = node
self.tail = node
def print_singly_linked_list(node):
while node:
print(str(node.data))
node = node.next
def removeDuplicates(head):
# if linked list is empty, return None
if head is None:
return None
# if linked list only has one node, then there are no duplicates to remove
if head.next is None:
return head
else:
# starting with first two nodes as prev and curr, move through list, comparing current to previous
previous = head
current = head.next
while current:
# if current is equal to previous, then this is a duplicate that we have to delete
if current.data == previous.data:
# link previous's next node to current's next node so the linkedlist stays intact
previous.next = current.next
# current node updates to the next node we just linked from previous
current = current.next
# continue with loop to compare new current to previous
continue
# else current is greater than previous, keep moving through linked list
else:
# previous becomes the node we just looked at
previous = current
# current node moves on to the next node
current = current.next
return head
ll = SinglyLinkedList()
for num in [1, 2, 2, 3, 3, 5]:
ll.insert_node(num)
print_singly_linked_list(removeDuplicates(ll.head))
| true |
ce9675386e0d42018fe283a2670bbede7d4f5fc0 | 2275114213/test | /06.范型递归树的递归/98.验证二叉搜索树.py | 1,776 | 4.125 | 4 | """
给定一个二叉树，判断其是否是一个有效的二叉搜索树。
假设一个二叉搜索树具有如下特征：
节点的左子树只包含小于当前节点的数。
节点的右子树只包含大于当前节点的数。
所有左子树和右子树自身必须也是二叉搜索树。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/validate-binary-search-tree
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
"""
# Definition for a binary tree node.
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
# def isValidBST(self, root):
# """
# :type root: TreeNode
# :rtype: bool
# """
# if root == None:
# return True
# res = []
#
# def inorder(root, res):
# if root:
# inorder(root.left, res)
# res.append(root.val)
# inorder(root.right, res)
#
# inorder(root, res)
# for i in range(len(res) - 1):
# if res[i] >= res[i + 1]:
# return False
# return True
def isValidBST(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
if root == None:
return True
self.last = ""
def inorder(root):
if root:
inorder(root.left)
if root.val < self.last:
return False
self.last = root.val
inorder(root.right)
else:
return True
res = inorder(root)
return res
| false |
8b52358ff78a127df4ef26cebdcc799b89872fdb | AquaMagnet/Python-Samples | /OddorEven.py | 2,068 | 4.28125 | 4 | # -*- coding: utf-8 -*-
"""
Created on Sat Sep 1 02:18:20 2018
@author: JUSTINE
"""
"""
Ask the user for a number. Depending on whether the number is even or odd,
print out an appropriate message to the user. Hint: how does an even / odd number
react differently when divided by 2?
Extras:
If the number is a multiple of 4, print out a different message.
Ask the user for two numbers: one number to check (call it num) and one number to
divide by (check). If check divides evenly into num, tell that to the user.
If not, print a different appropriate message.
"""
def oddOrEven(number,divisor):
if number%2 == 0:
if number%4 == 0:
print("{} is an even number and a multiple of 4!".format(number))
else:
print("{} is an even number!".format(number))
else:
print("{} is an odd number!".format(number))
print()
print()
print()
print("Now checking whether the number is divided by the divisor evenly: ")
if number%divisor == 0:
print("{} is divided evenly by {}".format(number,divisor), end = '\n')
else:
print("{} is not divided evenly by {}".format(number,divisor), end = '\n')
print("This program determine whether the user input is an odd or an even number", end = '\n')
print("This program also determines whether the user's chosen divider will divide the first number evenly.", end = '\n')
#First number:
while True:
try:
usernumber = int(input("Input the number you wanted to check: "))
if usernumber < 0:
print("Input a non-negative number")
else:
break
except:
print("Input a valid non-negative number!")
#Second number:
while True:
try:
divisor = int(input("Input the divisor: "))
if usernumber < 0:
print("Input a non-negative number")
else:
break
except:
print("Input a valid non-negative number!")
oddOrEven(usernumber,divisor)
| true |
8a41ad2803241d4f5ae0f9e8f0f405ad934c3f79 | Cemal-y/Python_Fundamentals | /fundamentals/Opdracht_6.py | 753 | 4.25 | 4 | def list_manipulation(list, command, location, value=""):
if command == "remove" and location == "end":
return list.pop(-1)
elif command == "remove" and location == "beginning":
return list.pop(0)
elif command == "add" and location == "beginning":
list.insert(0, value)
return list
elif command == "add" and location == "end":
list.append(value)
return list
list1 = [1, 2, 3]
print(list_manipulation(list=list1, command="remove", location="beginning"))
print(list_manipulation(list=list1, command="remove", location="end"))
print(list_manipulation(list=list1, command="add", location="beginning", value=20))
print(list_manipulation(list=list1, command="add", location="end", value=30))
| true |
c85d1bb469df4f643420b4ddba242cb3b03e280d | isutare412/python-cookbook | /04_Iterators_and_Generators/05_iterating_in_reverse.py | 791 | 4.125 | 4 | class Countdown:
def __init__(self, start):
self._start = start
def __iter__(self):
n = self._start
while n > 0:
yield n
n -= 1
class CountdownWithR(Countdown):
def __reversed__(self):
n = 1
while n <= self._start:
yield n
n += 1
if __name__ == '__main__':
a = [1, 2, 3, 4]
for x in reversed(a):
print(x)
print()
for cnt in Countdown(3):
print(cnt)
print()
# reversed() only works if the object in question has a size that can
# be determined or the object implements a __reversed__() special method.
for cnt in reversed(list(Countdown(3))):
print(cnt)
print()
for cnt in reversed(CountdownWithR(3)):
print(cnt)
| true |
06a16a7bbd31fa758edaa0bd8c9bcb423894a3b3 | nakshatar/python2_assignment | /pg4.py | 290 | 4.125 | 4 | #string="nakshatra"
string = raw_input("enter the string:\t")
vowel=0
for i in string:
if(i=='a'or i=='e' or i=='i' or i=='o' or i=='u' or i=='A' or i=='E' or i=='I' or i=='O' or i=='U'):
vowel = vowel+1
print("\n The totel number of vowels present in the string are:\t")
print(vowel) | false |
43b3e1f7f59ace3b173bda5caaadd43e814b3aaf | nakshatar/python2_assignment | /pg5.py | 230 | 4.15625 | 4 | # TO CHECK THE ELEMENT IN THE LIST HAS WORD SOIS OR NOT
list = ["vlsi","sois","SOIS","ewt","bigdata","vir"]
var = raw_input("enter the key element to be found :\t")
if var in list:
print("SOIS IS FOUND")
else:
print("NOT FOUND") | true |
11c06ebe7f86611a8979ab60d5f2884f9bd07fa7 | cristianodeoliveira/Chapter8-1-lists | /BasicListInput.py | 414 | 4.34375 | 4 | #it keeps the numbers in a list
numlist = list()
while True:
inp = input ('Enter a number: ')
if inp == 'done': break
try:
value = float (inp)
except:
continue
numlist.append(value)
#outside the loop it calculates the sum ()
#then it shows the average by dividing the sum
#by the number of items in the list
average = sum(numlist) / len (numlist)
print (numlist)
print (average) | true |
f96478a60f94d457c7ec1202e36a1f4dc0cd0840 | mjbenitez95/connect-py | /connect-four.py | 1,673 | 4.15625 | 4 | import random
NUM_ROWS = 6
NUM_COLS = 7
PLAYER_NUM = 1
CPU_NUM = 2
class Board:
def __init__(self):
self.board = []
for _ in range(NUM_ROWS):
single_row = [0] * NUM_COLS
self.board.append(single_row)
def __str__(self):
board_display = ""
for row in self.board:
board_display += str(row)
board_display += "\n"
return board_display
def display(self):
for row in self.board:
print row
print "-" * NUM_COLS * 3
def drop(self, column, player):
if self.board[0][column] != 0:
return False
for row in range(1, NUM_ROWS):
if self.board[row][column] != 0:
self.board[row-1][column] = player
return True
elif (row == NUM_ROWS - 1):
self.board[row][column] = player
return True
def prompt_player_move():
print "Where would you like to drop a coin?"
player_input = input()
return int(player_input)
if __name__ == "__main__":
board = Board()
turn_number = 1
while True:
if turn_number % 2 == PLAYER_NUM:
move = prompt_player_move()
while board.drop(move, PLAYER_NUM) is False:
print "Invalid move. Please try again."
move = prompt_player_move()
else:
# make computer move
random_move = random.choice(range(NUM_COLS))
while board.drop(random_move, CPU_NUM) is False:
random_move = random.choice(range(NUM_COLS))
board.display()
turn_number += 1
| true |
cc69c5b3f7b92bfb0bffc1cbbcab61a0f1f2a206 | amalshehu/exercism-python | /sum-of-multiples/sum_of_multiples.py | 419 | 4.34375 | 4 | # File: sum_of_multiples.py
# Purpose: Write a program that, given a number, can find the sum of all
# the multiples of particular numbers up to but not including that number.
# Programmer: Amal Shehu
# Course: Exercism
# Date: Wednesday 7th September 2016, 11:00 PM
def sum_of_multiples(limit, div_list):
return sum(set(j for i in div_list if i > 0 for j in range(i, limit, i)))
| true |
464153406ddd89c324969fd943576f6a34475728 | NicolasWarlop/AdventOfCode2017 | /Day 03/2017d3.py | 2,090 | 4.21875 | 4 | """ 2017d2.py - Advent of code 2017 - day 3"""
#Globals
INPUT = 265149
def get_direction():
"""Returns the next direction modifier"""
current_dir = 0
direction = [(1, 0), (0, 1), (-1, 0), (0, -1)]
while True:
yield direction[current_dir%len(direction)]
current_dir += 1
def get_manhattan_distance(point_a, point_b):
"""Calculate the manhattan distance between both points"""
return abs(point_b[0] - point_a[0]) + abs(point_b[1] - point_a[1])
def get_cartesian_coordinates(steps):
"""get the x,y coordinates for a given input """
current_x = 0
current_y = 0
next_dir = get_direction()
cur_dir = next(next_dir)
vector_length = 1
length_counter = 0
steps_taken = 0
for _ in range(1, steps):
current_x += cur_dir[0]
current_y += cur_dir[1]
steps_taken += 1
#are we at the tip of the vector?
if steps_taken == vector_length:
cur_dir = next(next_dir)
steps_taken = 0
length_counter += 1
#do we need to increase the vector length?
if length_counter % 2 == 0:
vector_length += 1
return (current_x, current_y)
def solve_part_1():
""" part 1 solver"""
print("Part 1: " + str(get_manhattan_distance((0, 0), get_cartesian_coordinates(INPUT))))
def solve_part_2():
""" part 2 solver"""
seen = {}
seen[(0, 0)] = 1
steps = 2
return_value = 0
while return_value < INPUT:
return_value = 0
#get the next coordinate
cur_pos = get_cartesian_coordinates(steps)
for i in range(-1, 2):
for j in range(-1, 2):
if (cur_pos[0]+i, cur_pos[1]+j) in seen:
return_value += seen[(cur_pos[0]+i, cur_pos[1]+j)]
seen[cur_pos] = return_value
steps += 1
print("Part 2: " + str(return_value))
def main():
"""main function"""
solve_part_1()
solve_part_2()
#Main execution
if __name__ == "__main__":
main()
| true |
580266fd0bd4f8d963d0bbdab7839622fb805c02 | joaosvictor/practice | /src/Initial/check_odd_even.py | 457 | 4.46875 | 4 | # check if a input number is odd or even
# int input
number = int(input('Enter a number: '))
# "%" operator check the remainder of a division
# so, 8 / 2 = 4. No remainder!
# otherwise, 8 / 3 = 2,66. It has remainder number!
if number % 2 == 0: # zero represent the "remainder"
print('{} is Even number'.format(number))
else:
print('{} is Odd number'.format(number))
# .format() will call the variable, so you put '{}' to the .format find it
| true |
3a64c1343bed86cdc050ea442a5543c74046c389 | RichardEsquivel/Recursive-Sorting | /src/recursive_sorting/cursion.py | 979 | 4.3125 | 4 | def my_recursion(n):
print(n)
if n == 3:
return
print("Before First", n)
my_recursion(n + 1)
print("Before Second:", n)
my_recursion(n + 1)
print("Before Third:", n)
my_recursion(n + 1)
print("Before Fourth:", n)
my_recursion(n + 1)
print("After Fourth:", n)
my_recursion(1)
"""
output expected is :
1
Before First 1
2
Before First 2
3
Before Second: 2
3
Before Third: 2
3
Before Fourth: 2
3
After Fourth: 2
Before Second: 1
2
Before First 2
3
Before Second: 2
3
Before Third: 2
3
Before Fourth: 2
3
After Fourth: 2
Before Third: 1
2
Before First 2
3
Before Second: 2
3
Before Third: 2
3
Before Fourth: 2
3
After Fourth: 2
Before Fourth: 1
2
Before First 2
3
Before Second: 2
3
Before Third: 2
3
Before Fourth: 2
3
After Fourth: 2
After Fourth: 1
"""
# As each value is returned back to it's caller within the recursion loop until we have the last 4th my recursion which has no original caller other than the invocation.s
| true |
661eb30c3cf306b90aee846309045354a0a77705 | sandyhsia/codeWarwww | /string中的重复字符.py | 1,136 | 4.15625 | 4 | ## My solution
def is_isogram(string):
#your code here
if type(string) != str:
raise TypeError('Argument should be a string')
elif string == "":
return True
else:
string = string.lower()
repeat = 0
for char in string:
if string.count(char) > 1:
repeat += 1
if repeat > 0:
return False
else:
return True
#Best Practice
def isogram(n):
if not isinstance(n, str):
return False
elif len(n) < 1:
return True
n = n.lower()
if len(n) == len(set(n)): ## we check if the length of the input is equal to the length of the set(n).
return True
else:
return False
# The function set() converts a collection or a sequence or an iterator object into a set.
# For example: set('lists') returns {'s', 't', 'l', 'i'}, as you can see the letter 's' which appears twice in 'lists', does not appear in the set.
# This is useful to check if the length of the set is equal to the length of the input, if there is a letter which appears twice in the input the condition is False.
| true |
923be6b25ef4609f1c45c4a21692d684100fdcd8 | VanessaVanG/word_count | /word_count.py | 836 | 4.21875 | 4 | '''Alright, this one might be a bit challenging but you've been doing great so far, so I'm sure you can manage it.
I need you to make a function named word_count. It should accept a single argument which will be a string. The function needs to return a dictionary. The keys in the dictionary will be each of the words in the string, lowercased. The values will be how many times that particular word appears in the string.
Check the comments below for an example.'''
# E.g. word_count("I do not like it Sam I Am") gets back a dictionary like:
# {'i': 2, 'do': 1, 'it': 1, 'sam': 1, 'like': 1, 'not': 1, 'am': 1}
# Lowercase the string to make it easier.
def word_count(phrase):
words = phrase.lower().split()
counter = {}
for word in words:
counter.update({word: words.count(word)})
return counter | true |
b52661f383b25062efd0cd2776d7366d6bd87027 | sbstnzcr/project-euler | /problem_002.py | 804 | 4.1875 | 4 | # Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
# 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
# By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
def fibonacci(n: int) -> int:
"""Find the nth fibonacci number."""
sequence = [1, 2]
for i in range(2, n+1):
sequence.append(sequence[i-1] + sequence[i-2])
return sequence[n]
if __name__ == '__main__':
even_terms = []
i = 0
while True:
element = fibonacci(i)
i += 1
if element > 4_000_000:
break
elif not element % 2:
even_terms.append(element)
result = sum(even_terms)
print(result)
| true |
286396e606464f35516d7cdc96207e109f069bbe | Immaannn2222/holbertonschool-higher_level_programming | /0x07-python-test_driven_development/0-add_integer.py | 348 | 4.21875 | 4 | #!/usr/bin/python3
""" the add module"""
def add_integer(a, b=98):
""" adds two integers a and b"""
if not (isinstance(a, int) or isinstance(a, float)):
raise TypeError("a must be an integer")
if not (isinstance(b, int) or isinstance(b, float)):
raise TypeError("b must be an integer")
return (int(a) + int(b))
| true |
62313b3d2063e28e3d1b6d6ec78b637bf435980a | philkrause/python-basics | /lists.py | 564 | 4.15625 | 4 | # A List is a collection which is ordered and changeable. Allows duplicate members.
#Create a list
numbers = [1,2,3,4,5]
fruits = ['Apples', 'Oranges', 'Grapes', 'Pears']
#Use a constructor
# numbers2 = list((1,2,3,4,5))
print(fruits[1])
print(len(fruits))
#Append to list
fruits.append('Mangos')
#Remove from list
fruits.remove('Grapes')
#Insert into position
fruits.insert(2, 'Strawberries')
#Remove with pop
fruits.pop(2)
fruits.reverse()
fruits.sort() #alphabetical sort
fruits.reverse()
#change value
fruits[0] = 'Blueberries '
print(fruits) | true |
07576aad1ffe85389a340b42ac930fa3baac2106 | tcollins1984/CollatzFun | /LICENSE.md/pwrsof2.py | 1,621 | 4.53125 | 5 | #The collatz algorithm ends when you hit a powewr of 2. All subsequent steps will
#be to divide by 2 until you reach one.
#The only way to initally "hit" a power of 2 is by taking an oddd number X and performing 3X + 1.
#Let's look at the first 100 powers of 2 and see if they will ever be that power of 2 which ends
#the algorigthm when by being reached from an odd number
#We'll create a dataframe in pandas to show the power of 2 that ends the algorithm and the odd number which
#will land us there. We'll also have a brief text description saying whether or not that particular power of 2 is a terminal number
import pandas as pd
max_power = 100
pwr2 = []
for i in range(0,max_power+1):
pwr2.append(2**i)
#Let's all each element of pwrd Ni. Since it must be reached by running an odd number X through 3X+1,
#we can find out if this is so by inverting the function (Ni - 1)/3 = X
#Since X must be a whole number, we know that Ni is reached through the application
# of the 3X + 1 algorith when (Ni - 1)%3 = 0
cc = []
for n in range(0,len(pwr2)):
if (pwr2[n]-1)%3 == 0:
cc.append([pwr2[n],(pwr2[n]-1)/3,'will converge form here'])
else:
cc.append([pwr2[n], 'na','cannot be reached by 3n + 1'])
number = []
odd = []
result = []
for i in range(0,len(cc)):
number.append(cc[i][0])
odd.append(cc[i][1])
result.append(cc[i][2])
powerTwo = pd.DataFrame({"2 to the Index":number, "Corresponding Odd if Any":odd,"Result":result})
print(powerTwo)
#use the following code to print to csv file in the directory of your choice
#powerTwo.to_csv('CollatzTerminalNumbers.csv')
| true |
1a17b195885eaf95ddf3310203c68669691c5f4f | srinivasreddy/euler | /src/35.py | 1,149 | 4.125 | 4 | import math
"""
function to generate infinite prime numbers!!!
"""
def gen_prime():
yield 2
x=3
while True:
if x>1000000:
break
elif(is_prime(x) and should_not(x)):
yield x
x=x+2
def is_prime(x):
if x==1:
return False
for i in range(3,int(math.sqrt(x))+1,2):
if(x%i==0):
return False
return True
"""
the idea is -if it cotains any even number ,then it will not be a prime after rotations..
and remove zero as it's presence will make 103 as 31 and 13 then we will loss one number in the digit.
"""
def should_not(n):
cache =str(n)
return (not "0" in cache) and (not "2" in cache) and (not "4" in cache) and (not "6" in cache) and (not "8" in cache)
if __name__ =="__main__":
total=0
for i in gen_prime():
flag=True
original=i
if len(str(i))==1:
total=total+1
else:
for j in range(len(str(i))):
if(not is_prime(i)):
flag=False
i=int(str(i)[1:]+str(i)[:1])
if flag:
total=total+1
print total
| true |
e71c8cfdaabf3d4eb9e04dc145afb6a33047228e | harrydadson/Python-DataStructures-Algo | /03-Searching-and-Sorting/05-insertion-sort.py | 832 | 4.375 | 4 | # Implementation of Insertion Sort Algorithm
# Insertion look at the next index values and pick and put
# them at their right sport in ascending order
# Sorts a sequence in ascending order using Insertion
def insertionSort(theSeq):
n = len(theSeq)
# Starts with the first item as the only sorted entry
for i in range(1, n):
# Save the value to be positioned
value = theSeq[i]
# find the position where the value fits in the ordered part of the list
pos = i
while pos > 0 and value < theSeq[pos - 1]:
# shift the items to the right during search
theSeq[pos] = theSeq[pos - 1]
pos -= 1
# Put the saved value into the open slot
theSeq[pos] = value
return theSeq
print(insertionSort([8,3,22,15,5,13,9]))
| true |
39692630b6a7f7d305d0aef4fb52ee298c882d60 | mzqshine/test | /day1/test1.py | 1,444 | 4.3125 | 4 | # '''
# 编写时间：2021-06-27
# 编写者：马志强
# 功能描述：第一天
# '''
# # #注释
# # #打印helloworld
# # print('helloworld')
# #
# # # 变量 --简单断点
# # his_name = '小明'
# # print(his_name)
# # print(his1name)
#
# #数据类型
# m=123
# print(type(m))
# m = 0.123
# n = 'sad'
# print(type(m))
# print(type(n))
# print(type(True))
# b = (1,2)
# v = {1,2}
# c = {'s':1,'b':2}
# print(type(b))
# print(type(v))
# print(type(c))
# 格式化输出 ctrl+alt+l 可以对代码进行格式调整
his_name = '小飞'
age = 18
weight = 50.12
num = 1
print('我的名字是：%s' % his_name) # 字符串
print('我的年龄是：%d' % age) # 整形
print('我的体重是：%f' % weight) # 浮点，默认小数点后6位
print('我的学号是：%05d' % num) # 必须0开始，5代表总长度
print('我的年龄是：%.2f' % weight) # 。2代表小数点位数
print('我的名字是：%s,我的年龄是：%d,我的体重是：%f' % (his_name, age, weight)) # 整合
print('我的名字是：%s,我的年龄是：%d,我的明天体重是：%.3f' % (his_name, age, weight - 1)) # 打印加减函数
print(f'我的名字是：{his_name},我的年龄是：{age},我的明天体重是：{weight}') # 3.7版本上有f参数使用
print('shjdkahdjka\nshd\tadjk\n\tsdhka') # 转义换行符 \n换行 \t非行开头空格1个字符 \t行开头是tab 4个字符
print('结束', end='-------') # 结束标识
| false |
6108d7674abd9d8930f8779ba130db204164b69d | jamalie/Reverse3 | /reverse3.py | 549 | 4.3125 | 4 | #Given an array of ints length 3, return a new array with the elements in reverse order, so {1, 2, 3} becomes {3, 2, 1}.
#reverse3([1, 2, 3]) -> [3, 2, 1]
#reverse3([5, 11, 9]) -> [9, 11, 5]
#reverse3([7, 0, 0]) -> [0, 0, 7]
def reverse3(nums):
for i in range(len(nums)-1):
if i <= (len(nums)-1-i):
x = nums[i]
nums[i] = nums[len(nums)-1-i]
nums[len(nums)-1-i] = x
else:
return nums
return nums
print reverse3([1, 2, 3])
print reverse3([5, 11, 9])
print reverse3([7, 0, 0])
| true |
2e09fea6272e4c0cc1a1c28b046d1025cc228722 | estela-ramirez/PythonCourses | /Homework 3 #1.py | 1,611 | 4.1875 | 4 | '''
ICS 31 Homework 3 Problem 1
Author: UCI_ID: 18108714 Name: Estela Ramirez Ramirez
'''
def main():
hint_list = ["I am round.", "I can be seen during the day.",
"I am very hot.", "I am brihgt.", "I am in the sky."]
introduction()
is_Guess_correct("answer")
ask_question(hint_list, "answer" )
def is_Guess_correct(answer:str)->str: # asks the question # does not give hints, but if the guess is right/wrong it works
hint_number = 5
for hints in range(hint_number):
user_answer = input("Now make a guess: ")
if user_answer== "sun":
print("Good job you guessed correctly!")
break
elif user_answer == "the sun":
print("Good job you guessed correctly!")
break
else:
print("Nope, Guess again! ")
def introduction():
print("Let's play a guessing game! ")
print("I've picked something, and I will give you five hints. ")
print("You get to make a guess after each hint. ")
print("If you get it right, I will let you know otherwise ")
print("You will keep guessing until I run out of hints. ")
print("Let's get started ")
def ask_question(hint_list:list, answer:str): # gives hints, asks for guess
hint_number = 5
for x in hint_list:
print()
print("Here is your hint: ","\n", x)
user_answer = input("Now make a guess: ")
user_answer = user_answer.lower()
hint_number -=1
if hint_number < 1:
print("\n","Ohh, I must not have given good hints. I was the sun. ")
main()
| true |
a26dfcfb34a20b6340fff0ce648100025ca8a352 | estela-ramirez/PythonCourses | /LAB 3 Q1.py | 1,340 | 4.125 | 4 | '''
ICS 31 Lab 3 Question 1
Driver: UCI_ID: 18108714 Name: Estela Ramirez Ramirez
Navigator: UCI_ID: 69329009 Name: Karina Regalo
'''
def main():
i = get_n_of_people()
ns, ag = getsAgesAndNames(i)
maxb, index = maxAge(ag)
minb, index2 = minAge(ag)
s = sumAge(ag)
printOldestYoungestAverage(index,index2,s,i,ns,ag)
def get_n_of_people():
return int(input("Number of people: "))
def getsAgesAndNames(number_of_people)->list:
a =[]
b =[]
for x in range(1, number_of_people+1):
name = input("What is Person #{}'s name: ". format(x))
a.append(name)
age = int(input("What is {}'s age? ". format(name)))
b.append(age)
return a, b
def maxAge(b:list)->int:
maxb = 0
for x in b:
if x >= maxb:
maxb = x
index = (b.index(maxb))
return maxb, index
def minAge(b:list)->int:
minb = 100000000000
for x in b:
if x <= minb:
minb = x
index = (b.index(minb))
return minb, index
def sumAge(b:list)->int:
for x in (b):
sumb = sum(b)
def printOldestYoungestAverage(ma,mia,ta,np,ns,ag):
print( ns[mia], "is the youngest and is", ag[mia], "years old")
print( ns[ma], "is the oldest and is ", ag[ma], "years old")
print("Average Age is:",(sum(ag)/len(ag)))
main()
| false |
2bebea84337e7ca8e2146a7cf59dd982694175a5 | estela-ramirez/PythonCourses | /Pre Lab 4 P 3.py | 2,207 | 4.125 | 4 | #This function checks if the data in num is an integer.
def isInt(num: str) -> bool:
try:
int(num)
return True
except ValueError:
return False
#IN OPERATOR
L = [1,2,3]
val = 1
if val in L:
print( val , "is in the list")
else:
print(val , "is not in the list")
S = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
sub_Str = "JK"
if sub_Str in S:
print( sub_Str , "is in the list")
else:
print(sub_Str , "is not in the list")
#Also note that the empty string is considered in every string
empty = ""
if empty in S:
print("S has the empty_string")
#String INDEXING
S = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
first_15 = S[0:15] #Contains the first 15 letters, or the letters in indexes 0 - 14
last_15 = S[len(S) - 15 : len(S)] #Specify the end index + 1
last_15_alt = S[len(S) - 15 : ] #Say start here and grab all
print(first_15)
print(len(first_15))
print()
print(last_15)
print(len(last_15))
print()
print(last_15_alt)
print(len(last_15_alt))
#INDEXING BACKWARDS
# 0 1 2 3 4
# H E L L O
# -5 -4 -3 -2 -1
first_15 = S[ -len(S) : (-len(S) + 15)]
last_15 = S[-15: len(S)] #Contains the last 15 letters, or the letters in indexes -15 to -1
last_15_alt = S[-15 : ] #Say start here and grab all
print(first_15)
print(len(first_15))
print()
print(last_15)
print(len(last_15))
print()
print(last_15_alt)
print(len(last_15_alt))
#FIND
str1 = "My Example, Sample, Text, String";
str2 = "exam";
#This will return -1, as find is case sensitive
print(str1.find(str2))
#This will return 3, as that is the index of the substring
print(str1.lower().find(str2.lower()))
#This will return -1, as it will start searching after the substring
print(str1.lower().find(str2.lower(), 10))
#This will return 1, as find goes from the front of the string
print(str1.find(str2))
#R.FIND searches backwards and returns the highest rather than lowest index
#This will return 13 as rfind goes from the back of the string
print(str1.rfind(str2))
#This will return 1, as their is no later occurrence before index 10
print(str1.rfind(str2, 0, 10))
#This will return 13, as their is no later occurrence after index 10
print(str1.rfind(str2, 10))
| true |
72dd09f5af0d18b42e2584ced6503e8a18958ed1 | alee1412/activities | /Lesson_3_Python/Activity_5.py | 1,973 | 4.21875 | 4 | prices = ["24", "13", "16000", "1400"]
price_nums = [int(price) for price in prices]
print(prices)
print(price_nums)
print("**************************************")
dog = "poodle"
letters = [letter for letter in dog]
print(letters)
print(f"We iterate over a string into a list: {letters}")
print("**************************************")
capital_letters = [letter.upper() for letter in letters]
capital_letters = []
for letter in letters:
capital_letters.append(letter.upper())
print(capital_letters)
print("**************************************")
#First way
no_o = [letter for letter in letters if letter != 'o']
print(no_o)
#Second way
no_o = []
for letter in letters:
if letter != 'o':
no_o.append(letter)
print(no_o)
print("**************************************")
june_temperature = [72,65,59,87]
july_temperature = [87,85,92,79]
august_temperature = [88,77,66,100]
temperature = [june_temperature, july_temperature, august_temperature]
print("**************************************")
#short hand
lowest_summer_temperature = [min(temps) for temps in temperature]
maximum_summer_temperature = [max(temps) for temps in temperature]
print(sum(maximum_summer_temperature)/len(maximum_summer_temperature))
print(lowest_summer_temperature[0])
print(lowest_summer_temperature[1])
print(lowest_summer_temperature[2])
print("=" * 30)
print("**************************************")
#long hand
lowest_summer_temperature = []
for temps in temperature:
lowest_summer_temperature.append(min(temps))
print(lowest_summer_temperature[0])
print(lowest_summer_temperature[1])
print(lowest_summer_temperature[2])
print(sum(lowest_summer_temperature)/len(lowest_summer_temperature))
print("**************************************")
def name(parameter):
return "Hello " + parameter
print(name("Albert"))
def average(data1,data2):
return (sum(data1)/len(data1)) + (sum(data2)/len(data2))
print("=" * 40)
print(average([1,2,3,4,5],[2,3,4,5,6])) | false |
b4215a396e6e080c19154eeacf2a91ad1a80ca85 | stge4code/CodePractice | /stepic.org/Python_Programming/3.7-4.py | 739 | 4.15625 | 4 | # put your python code here
class Turtle:
def __init__(self, x=0, y=0):
self.x = x
self.y = y
def movement(self, s):
(cmd, steps) = (s.split())
self.CMDs[cmd](self, int(steps))
def nord(self, steps):
self.y += steps
def east(self, steps):
self.x += steps
def west(self, steps):
self.x -= steps
def south(self, steps):
self.y -= steps
def getcoords(self):
return str(self.x) + ' ' + str(self.y)
CMDs = {
"север": nord,
"юг": south,
"восток": east,
"запад": west
}
turtle = Turtle()
n = int(input())
for i in range(n):
turtle.movement(input())
print(turtle.getcoords()) | false |
fb077e5b1a676165bbe69a90211b076cf1d35903 | VenkateshSatagopan/Data-Structures | /Other_problems/array_binary_sum.py | 1,140 | 4.25 | 4 | def add_arr(arr_1,arr_2):
'''
Function to add 2 binary numbers array and return the result
'''
arr_3=[0]*(max(len(arr_1),len(arr_2))+1)
k=len(arr_3)-1
i=len(arr_1)-1
j=len(arr_2)-1
carry=0
while(i>=0 and j >= 0):
arr_3[k] = (arr_1[i]+arr_2[j] + carry )%2
carry=(arr_1[i]+arr_2[j]+ carry)//2
k-=1
j-=1
i-=1
while(i>=0):
arr_3[k]=(arr_1[i]+carry)%2
carry=(arr_1[i]+carry)//2
i-=1
k-=1
while(j>=0):
arr_3[k]=(arr_2[j]+carry)%2
carry=(arr_2[j]+carry)//2
j-=1
k-=1
if carry:
arr_3[k]=carry
return arr_3
if __name__ == '__main__':
# Test case 1
arr1=[1,1,1,1,1,0]
arr2=[1,1,1,1,1,1]
assert [1,1,1,1,1,0,1] == add_arr(arr1,arr2)
# Test case 2
arr1=[1,0,1]
arr2=[1,1,1,1]
assert [1,0,1,0,0] == add_arr(arr1,arr2)
# Test case 3
arr1=[1,0,0,0,1,0]
arr2=[1,1]
assert [0,1,0,0,1,0,1] == add_arr(arr1,arr2)
# Test case 4
arr1=[1,1,1,1,1,1]
arr2=[1,1,1,1,1,1]
assert [1,1,1,1,1,1,0] == add_arr(arr1,arr2)
| false |
53e46e1fc8a9445e723c34bdbdbfe6a3d308a368 | jos-h/Python_Exercises | /List_Container/merge_unsortedlist_insertion.py | 638 | 4.15625 | 4 | #!/usr/bin/python3
def sort_list(list_3):
key = 0
i = 1
j = 0
while i < len(list_3):
j = i - 1
key = list_3[i]
while j >= 0 and key < list_3[j]:
list_3[j + 1] = list_3[j]
j -= 1
list_3[j + 1] = key
i += 1
return list_3
def main():
list_1 = eval(input("accept the first list"))
list_2 = eval(input("accept the second list"))
list_3 = []
list_3.extend(list_1)
list_3.extend(list_2)
print("Merged list is {}".format(list_3))
print("sorted list is {}".format(sort_list(list_3)))
if __name__ == '__main__':
main() | false |
ab2d94b7c694176eedcd4aa2e9878a94c0cef134 | jos-h/Python_Exercises | /substring_string.py | 1,366 | 4.125 | 4 | """
Given two strings s1, s2, write a function that returns true if s2 is a special substring s1.
A special substring s2 is such that the s1 contains s2
where each character in s2 appears in sequence in s1,
but there can be any characters in s1 in between the sequence.
Example:
isSpecialSubstring('abcdefg', 'abc') => true
isSpecialSubstring('abcdefg', 'acg') => true
isSpecialSubstring('abcdefg', 'acb') => false;
The first two are abc and acg appears in 'abcdefg' in that order, although there might be multiple chars between the next character in s2.
The last one is false, because in 'acb' the character 'b' appears before the character 'c' in 'abcdefg'
"""
def check_dest(dest):
des_len = len(dest) - 1
i = 0
j = 0
while i < des_len and j <= des_len:
j = i + 1
if dest[i] > dest[j]:
return False
i += 1
def is_special_substring(org, dest):
dest_len = len(dest)
flag = 0
val = check_dest(dest)
if val == False:
return False
else:
for j in dest:
for i in org:
if j == i:
flag += 1
break
if flag == 3:
return True
def main():
val = is_special_substring('abcdefg', 'bad')
if val:
print("True")
else:
print("False")
if __name__ == '__main__':
main()
| true |
940bdcee82e02161b8002b6fc031a4d47ae51101 | jos-h/Python_Exercises | /List_Container/Union_list.py | 1,447 | 4.21875 | 4 | #!/usr/bin/python3
def menu():
print("1:Union of Lists\n"
"2:Intersection of Lists\n"
"3:Difference in Lists\n"
"4:Exit")
choice = eval(input("Accept the choice"))
return choice
def union_lists(list_1, list_2):
list_3 = []
list_3.extend(list_1)
for x in list_2:
if x not in list_3:
list_3.append(x)
return list_3
def intersection_list(list_1, list_2):
list_3 = []
for x in list_1:
for y in list_2:
if x == y:
list_3.append(x)
return list_3
def difference_lists(list_1, list_2):
list_3 = []
j = 0;
list_3.extend(list_1)
while j < list_3.__len__():
i = 0
while i < list_2.__len__():
if list_3[j] == list_2[i]:
list_3.remove(j)
i += 1
j += 1
return list_3
def main():
list_1 = eval(input("Accept 1st list"))
list_2 = eval(input("Accept 2nd list"))
while True:
choice = menu()
if choice == 1:
print("Union of list is {}".format(union_lists(list_1,list_2)))
elif choice == 2:
print("Intersection of list is {}".format(intersection_list(list_1, list_2)))
elif choice == 3:
print("Difference between lists is {}".format(difference_lists(list_1, list_2)))
else:
print("bye")
return
if __name__ == "__main__":
main() | false |
d4706392e28c352dc53d2d54580be308a8f53acd | jiten-kmar/python-code | /largest_number_list.py | 342 | 4.4375 | 4 | #Python Program to Find the Largest Number in a List
list1=[]
total_num=int(input("Enter how many numbers needed for the list: "))
for n in range(0, total_num):
number=input("Enter the number for the list ")
list1.append(number)
new_list=sorted(list1, key=int, reverse=True)
print(new_list)
print ("Largest number is ", new_list[0])
| true |
12a9d741f8120cfb240b36572b1ab357926ea033 | kiranprabakaran/rock_paper_scissors | /main.py | 1,283 | 4.15625 | 4 | import random
while True:
print("rock, paper, scissors?")
choice = input()
choice = choice.lower()
print("my choice is ", choice)
choices = ['rock', 'paper', 'scissors']
computer_choice = choices[random.randint(0, len(choices)-1)]
print("computer choice is ", computer_choice)
if choice in choices:
if choice == 'rock':
if computer_choice == 'rock':
print("It is a Draw")
elif computer_choice == 'paper':
print("You Lost :(")
elif computer_choice == 'scissors':
print("Winner, Winner!!!!")
if choice == 'paper':
if computer_choice == 'rock':
print("Winner, Winner!!!!")
elif computer_choice == 'paper':
print("It is a Draw")
elif computer_choice == 'scissors':
print("You Lost :(")
if choice == 'scissors':
if computer_choice == 'rock':
print("You Lost :(")
elif computer_choice == 'paper':
print("Winner, Winner!!!!")
elif computer_choice == 'scissors':
print("It is a Draw")
else:
print("Your answer doesn't make sense. Try again")
print()
| false |
0a1126a739b46f19a8c2ec082ad880e08e7c46db | ashleynguci/pythonproj | /point.py | 902 | 4.21875 | 4 | # -*- coding: utf-8 -*-
"""
class that implements 2D point
"""
class Point:
""" class that knows x and y coordinates """
def __init__(self, _x=0, _y=0):
self.__x = _x
self.__y = _y
@property
def x(self):
return self.__x
@property
def y(self):
return self.__y
def xy(self, _x, _y):
""" sets new values to point's coordinates """
self.__x = _x
self.__y = _y
#objects methods
# def length(self):
# """ Point distance from origo """
# return pow((self.__x**2 + self.__y**2), 0.5)
def length(self, _point=None):
""" Point distance from origo if _point is None else
distance from given point
"""
if _point is None:
_point = Point()
return pow(((self.__x - _point.x)**2 + (self.__y - _point.y)**2), 0.5)
| false |
cb208030a3d35d607ded29d8121d3a16ebf2b390 | Nadjamac/Mod1_Blue | /Aula8 Exerc5.py | 321 | 4.125 | 4 |
num = int(input("Digite um numero :\n"))
contador = 0
for x in range (1,num+1):
if num % x == 0:
print("{} é divisivel : {} ".format(num, num/x))
contador += 1
if contador == 2:
print("é um numero primo ")
else:
print("não é um numero primo ")
| false |
6372b4c622b02fe034aa8aec03d3dabb1545670f | mohapatralovleen4/Python-task-0 | /3rd program.py | 238 | 4.34375 | 4 | #program to count the total number of digits in a given number
n = int(input("Enter a number to calculate the number of digits\n"))
c = 0
while(n>0):
c = (c+1)
n = (n//10)
print("The number of digits in the given number are",c)
| true |
18e2f774f328af2d0036a75a291a2e860fb24d4d | TIBTHINK/damien | /lesson4/lesson4.3/beer.py | 1,106 | 4.28125 | 4 | from os import system, name
# Creating a function for clearing the screen
# (You will thank me later)
def clear():
if name == 'nt':
_ = system('cls')
else:
_ = system('clear')
beer_count = 0 # Sets the counter to 0 as a base line
try:
print("all systems go")
clear()
print("how many beers do you order?") # Asking the user for a number
question = int(input("[1-13]: ")) # Saving there answer as a integer
while beer_count < 13: # Checking every cycle if beer_count = 13, if not it will continue the loop until it does
beer_count = beer_count + question # Adds your answer to the beer counter plus what its current at
print("beer count: " + str(beer_count)) # Tells the user how many they have ordered
question = int(input("[1-13]: ")) # Reasks the questions to start the loop
else: # If you have ordered 13 the program will exit
exit("you have had enough to drink")
except KeyboardInterrupt:
exit("good bye") | true |
8df245c5c4de0f7655fffcb29496f491630b6980 | kz/compsci-homework | /1. AS Level/5. Encryption Exercise/Task 1.py | 2,046 | 4.75 | 5 | # Task 1
# Author: Kelvin Zhang
# Date Created: 2016-03-22
# This program takes in a word and makes it uppercase (e.g., "computing" -> "COMPUTING". It then takes in a keyword.
# An encrypted message is generated by adding the alphabet value of the message to the value of the keyword.
# If the word is longer than the keyword, the keyword repeats (e.g., "GCSE" -> "GCSEGCSEG" for "COMPUTING").
def convert_character_to_alphabet_value(char):
return ord(char) - ASCII_A + 1 # Incremented by one as the alphabet does not use zero-based numbering
def convert_alphabet_value_to_character(val):
return chr(ASCII_A + val - 1)
# This function takes a position of a message (starting from position zero [zero-based numbering])
# and returns the equivalent keyword letter for this position.
# E.g., if message is "COMPUTING", keyword is "GCSE" and we are taking the position five ("T"),
# the equivalent keyword letter would be "GCSE"[5 % 4] = "GCSE"[1] = "C"
def get_keyword_letter_for_position(keyword, pos):
return keyword[pos % len(keyword)]
# ASCII value for the letter A
ASCII_A = 65
# Prompt for input
rawMessage = input("Enter your message: ")
keyword = input("Enter your keyword: ")
# Initialise empty string for the encrypted message
encryptedMessage = ""
# Loop through the raw message, generating the encrypted message
for i in range(0, len(rawMessage)):
# Get the letter of the keyword as an alphabet value
keywordLetter = get_keyword_letter_for_position(keyword, i)
keywordValue = convert_character_to_alphabet_value(keywordLetter)
# Get the alphabet value for the letter of the message
messageValue = convert_character_to_alphabet_value(rawMessage[i])
# Obtain the encrypted letter
encryptedValue = (keywordValue + messageValue) % 26 # Modulo 26 is used to bring overflow values back to the start
encryptedLetter = convert_alphabet_value_to_character(encryptedValue)
# Append the letter to the message
encryptedMessage += encryptedLetter
# Output to user
print(encryptedMessage) | true |
113e7df40ee698ec6788c9ad34cce936811ea356 | Kyle5150/CSE111 | /CSE111/TireVolume.py | 1,068 | 4.21875 | 4 | print()
with open("/Users/kylejohnson/Documents/Python/CSE111/volumes.txt", "at") as volume_file:
import math
w = int(float(input("Enter the width of the tire in mm (ex 205): ")))
a = int(float(input("Enter the aspect ratio of the tire (ex 60): ")))
d = int(float(input("Enter the diameter of the wheel in inches (ex 15): ")))
v = (math.pi * w ** 2 * a * (w * a + 2540 * d)) / 10000000
liters = v / 1000
print()
print(f"The approximate volume is {v:.1f} milliliters\nor\n{liters:.1f} liters")
print()
from datetime import datetime
current_date_time = datetime.now()
buy_tires = input("Would you like to buy tires (yes/no)? ")
if buy_tires == "yes":
phone_number = input("Please enter your phone number ((555)555-5555): ")
print(f"{current_date_time}, {w}, {a}, {d}, {v:.1f}, {phone_number}", file = volume_file)
print("Thank you for your time.")
else:
print(f"{current_date_time}, {w}, {a}, {d}, {v:.1f}", file = volume_file)
print("Thank you for your time.")
print() | true |
4b32a03bacf3057df50147de4d873a06178943a6 | JohnnyChingas/Algorithms | /LinkedList.py | 2,161 | 4.15625 | 4 | # Some linked list operations
# T. Melano
class Node:
def __init__(self, data = None):
# Node initiator
# @param data is any object contained in the node
self.data = data
self.next = None
def insert(self, data):
# Add nodes to linked list
# @param data for new node added to linked list
if self.next == None:
self.next = Node(data)
else:
self.next.insert(data)
def traverse(self):
# Traverse and print linked list
print self.data
if self.next != None:
self.next.traverse()
def reverse(self):
# Reverse linked list
last = None
head = self
while head != None:
next = head.next
head.next = last
last = head
head = next
return last
def delete(self,data):
if self.data == data:
return self.next
else:
head = self
n = self
flag = 0
while n.next != None:
if n.next.data == data:
n.next = n.next.next
n = n.next
flag = 1
else:
n = n.next
if not flag:
print "Element not found"
return head
def remove_duplicates(linked_list):
if isinstance(linked_list,Node):
prev = linked_list
htable = {}
n = prev.next
htable[prev.data] = None
while n != None:
if n.data in htable:
prev.next = n.next
n = n.next
else:
htable[n.data] = None
prev = n
n = n.next
#Write function that reverses a linked list using recursion
def main():
list = [1,2,3,4,5]
head = Node(0)
for elem in list:
head.insert(elem)
print "Print list"
head.traverse()
print "Reversing list"
head = head.reverse()
head.traverse()
print "Reversing list"
head = head.reverse()
head.traverse()
if __name__ == "__main__":
main()
| true |
7ab53e3a31bde9b675212fdf1f0f48b490edbbbc | Anory/Python_projeck | /list_test/sample_list.py | 741 | 4.25 | 4 | # 列表的操作
name_list = ["张三", "李四", "王五", "赵六", "孙斌", "王五"]
wangwu = name_list[2] # 正向取值
print(wangwu)
wangwu = name_list[-3] # 负向取值
print(wangwu)
# 范围取值：切片(左闭右开原则：包含左边的对应数据， 不包含右边的对应数据)
name_list1 = name_list[1:3] # 从1开始到3（不包含3）
print(name_list1)
# 获取列表值的位置（index函数用于获取列表指定元素的索引值，注意：只会返回元素第一次出现的索引值）
wangwu_index = name_list.index("王五")
print(wangwu_index)
# 获取指定元素的索引值
for i in name_list:
print(i)
name_index = name_list.index(i)
if i == "王五":
print(name_index)
| false |
a0d691d843234cad7a9f151c109e966679ceff79 | RFishwick/CO_SCI_124_Python | /AsnCh10/10_1.py | 625 | 4.125 | 4 | import pet
def main():
# Local variables
pet_name = ""
pet_type = ""
pet_age = 0
# Get pet data.
pet_name = input('Enter the name of the pet: ')
pet_type = input('Enter the type of animal: ')
pet_age = int(input('Enter the age of the pet: '))
# Create an instance of Pet.
mypet = pet.Pet(pet_name, pet_type, pet_age)
# Display the data that was entered.
print('Here is the data that you entered: ')
print('Pet name: ', mypet.get_name())
print('Animal type: ', mypet.get_animal_type())
print('Age of pet: ', mypet.get_age())
# Call the main function.
main()
| true |
45e60c606aad958ee86045f71a11a9c0545a6c44 | RFishwick/CO_SCI_124_Python | /AsnCh9/ex9_5.py | 1,682 | 4.46875 | 4 | # Write a program that reads the contents of a text file.
# The program should create a dictionary in which the keys are the individual words
# found in the file and the values are the number of times each word appears.
#
# For example, if the word “the” appears 128 times, the dictionary would contain an element with 'the'
# as the key and 128 as the value. The program should either display the frequency of each word
# or create a second file containing a list of each word and its frequency.
#
def main():
input_name = 'text.txt' # Assign file name
# Open the input file and read the text.
input_file = open(input_name, 'r') # Open the file for read access
text = input_file.read() # Read the file into a string variable
words = text.split() # Split text into a list of words
# Extra steps to strip off special characters
new_text = ''
for word in words:
word = word.strip(',')
word = word.strip('.')
new_text = new_text + word + ' '
# Assign individual stripped words to words list and split again
words = new_text.split()
# Create a set containing the unique words.
unique_words = set(words)
# Create a word_count dictionary from unique words and initialize counts to zero
word_count = {}
for word in unique_words:
word_count[word] = 0
# Increment word_count dictionary value for each word in words
for word in words:
word_count[word] += 1
# Print the results
print('\n*** Word Frequency ***\n')
for key in word_count:
print(key, word_count[key])
# Close the file.
input_file.close()
# Call the main function.
main()
| true |
bdb0a0cacb35c26c02b82acda79b42f6389530cb | Nan-Do/DailyCoding | /Daily_113.py | 1,543 | 4.15625 | 4 | # Cost O(n) (we traverse the string two times)
# Traverse the string putting each word in a stack and then
# extract words from the stack that will return the words in
# a reversed order.
def reverse_words(s):
words = []
word = ''
for x in s:
if x == ' ':
words.insert(0, word)
word = ''
continue
word += x
words.insert(0, word)
return ' '.join(words)
def reverse_from(s, init, end):
while init < end:
temp = s[init]
s[init] = s[end]
s[end] = temp
init += 1
end -= 1
# Cost O(n) We traverse the string three times, one to detect the words
# another one to reverse each word and the last time to reverse the whole
# string
# The algorithm works as follows traverse the string reversing each word # and then return the reverse of the string (if we reverse each word and # then we reverse the whole string we obtain the desired answer without
# having to care about where to put the spaces on the reversed string
# which can complicate the exercise quite a lot
def reverse_words_mutable(s):
last_init = 0
for pos, x in enumerate(s):
if x == ' ':
reverse_from(s, last_init, pos - 1)
last_init = pos + 1
# Reverse last word
reverse_from(s, last_init, len(s) - 1)
reverse_from(s, 0, len(s) - 1)
print(reverse_words("hello world here"))
s = [
'h', 'e', 'l', 'l', 'o', ' ', 'w', 'o', 'r', 'l', 'd', ' ', 't', 'h', 'e',
'r', 'e'
]
reverse_words_mutable(s)
print(s)
| true |
79591707fec63b979b5e6e6d34cdee4ba97dce67 | iriasyk/python_language | /students/km63/Ryasyk_Ihor/homework_5.py | 2,635 | 4.1875 | 4 | #task1--------------------------------------------------------------------
'''
Даны четыре действительных числа: x1, y1, x2, y2.
Напишите функцию distance(x1, y1, x2, y2), вычисляющая расстояние между точкой (x1,y1) и (x2,y2).
Считайте четыре действительных числа и выведите результат работы этой функции.
'''
from math import sqrt
def distance(x1,y1,x2,y2):
return sqrt((x1-x2)**2+(y1-y2)**2)
x1=float(input())
y1=float(input())
x2=float(input())
y2=float(input())
print(distance(x1,y1,x2,y2))
#--------------------------------------------------------------------------
#task2---------------------------------------------------------------------
'''
Дано действительное положительное число a и целоe число n.
Вычислите a^n. Решение оформите в виде функции power(a, n).
'''
def power(a, n):
res=1
if n>0:
for i in range(n):
res*=a
elif n<0:
for i in range(-1*n):
res/=a
return res
print(power(float(input()), int(input())))
#--------------------------------------------------------------------------
#task3---------------------------------------------------------------------
'''
Дано действительное положительное число a и целое неотрицательное число n.
Вычислите an не используя циклы, возведение в степень через ** и функцию math.pow(), а используя рекуррентное соотношение a^n=a*a^(n-1).
Решение оформите в виде функции power(a, n).
'''
def power(a, n):
if n == 0:
return 1
else:
return a * power(a, n - 1)
print(power(float(input()), int(input())))
#--------------------------------------------------------------------------
#task4----------------------------------------------------------------------
'''
Напишите функцию fib(n), которая по данному целому неотрицательному n возвращает n-e число Фибоначчи.
В этой задаче нельзя использовать циклы — используйте рекурсию.
'''
def fib(n):
if n == 1 or n == 2:
return 1
else:
return fib(n - 1) + fib(n - 2)
n = int(input())
print(fib(n))
#--------------------------------------------------------------------------
| false |
e23596e870f260c2df15a847431b678c5eba3bf1 | Ran-oops/python | /3/06描述符/07.py | 1,061 | 4.3125 | 4 | #声明类
#自定义类来继承系统的整形类
class myint(int):
#魔术方法 触发时机：当前对象+另外一个数值的时候触发
def __add__(self,other):
#print('add被触发')
return int(self) - int(other)
#魔术方法 触发时机：当前对象-另外一个数值的时候触发
def __sub__(self,other):
#print('sub备触发')
return int(self) + int(other)
#魔术方法 触发时机：当前对象*另外一个数值的时候触发
def __mul__(self,other):
#print('mul备触发')
return int(self) / int(other)
#魔术方法 触发时机：当前对象/另外一个数值的时候触发
def __truediv__(self,other):
#print('__truediv__ 备触发')
return int(self) * int(other)
#实例化对象
one = myint(10)#相当于int（10）
#加法操作
#result = one + 20
#print(result)
#减法操作
#result = one - 2
#print(result)
#乘法操作
#result = one * 5
#print(result)
#除法操作
#result = one / 4
#print(result) | false |
3771e08cb17bb5323d042ee530a04d37b6a927e4 | Ran-oops/python | /1python基础/07Python中的nolocal关键词的使用，lambda表达式，递归，字符串操作/09.py | 973 | 4.125 | 4 | '''
# + 字符串连接运算
str1 = '你问我爱你有多深'
str2 = '大概10cm左右！'
result = str1 + '：' + str2
print(result)
# * 字符串相乘（复制）
str1 = '吕欣'
result = str1 * 10
print(result)
#[] 索引操作
# 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
str1 = '下班的时候我宝贝骑摩托车来接你！'
# ....-4-3-2-1
print(str1[3])
print(str1[-3])
'''
#取片操作
str1 = '冬天脱毛衣会有很多静电，噼里啪啦，夏天为什么没有涅？因为夏天傻子才穿毛衣！'
#从指定位置截取到最后
print(str1[17:])
#从开头截取到指定位置之前（不包含结束索引位置）
print(str1[:17])
#从开始索引截取到指定的索引之前（不包含结束索引位置）
print(str1[12:16])
#获取整个字符串
print(str1[:])
#根据指定的间隔获取字符
print(str1[::3])
#指定和截取字符串的范围同时指定间隔数值
print(str1[17:30:2])
| false |
7ad7c535a5a6e60090ebe38b333a1956ba9df1f2 | Ran-oops/python | /2/02列表 元组 字典/04.py | 669 | 4.4375 | 4 | #元组的操作
#1.创建空元组
#tuple1 = ()
tuple1 = tuple()
print(type(tuple1))
print(tuple1)
#2.创建具有一个元素的元组
tuple1 = ('不知妻美刘强东',)#tuple1 = '不知妻美刘强东',
print(type(tuple1))
print(tuple1)
#3.创建多个元素的元素
tuple1 = ('马云','马化腾','王健林','丁磊')#tuple1 = '马云','马化腾','王健林','丁磊'
print(type(tuple1))
print(tuple1)
#元组的基本操作（除了访问啥也干不了）
#使用索引访问元素
tuple1 = ('王思聪','牛广林','牛少鸟')
print(tuple1[2])
#删除整个元组
tuple1 = ('王思聪','牛广林','牛少鸟')
del tuple1
print(tuple1)
| false |
15786735b73566332996aeef2d6cc3133bfd6ea2 | Ran-oops/python | /2/02列表 元组 字典/05.py | 837 | 4.375 | 4 | #元组的序列操作
#元组相加
tuple1 = ('肉夹馍','土豆丝卷饼','山东烧饼','煎饼','煎饼果子')
tuple2 = ('抄手','燃面','砂锅面','砂锅土豆粉')
result = tuple1 + tuple2
print(result)
#模拟修改元组
tuple1 = ('肉夹馍','土豆丝卷饼','山东烧饼','煎饼','煎饼果子')
result = tuple1[0:3] + ('糁',) + tuple1[3:]
print(result)
#元组相乘
tuple1 = ('雍建凯','李艳玲')
result = tuple1 * 5
print(result)
#分片操作
tuple1 = ('张飞','岳飞','双飞','王菲','飞飞','咖妃','尿嫔')
print(tuple1[2:5])
print(tuple1[2:])
print(tuple1[:5])
print(tuple1[:])
print(tuple1[::2])
#成员检测
tuple1 = ('张建宇','邢天宇','田宇','王雷雨','王晓雨')
result = '田宇' in tuple1
print(result)
result = '吕欣' not in tuple1
print(result)
| false |
dcfecea3b168804ef867b088677fce2187e6ec0a | Ran-oops/python | /1python基础/01 Python基础/04/repeat.py | 2,557 | 4.34375 | 4 | #书写代码
'''
print('我们对于女朋友提出的要求向来是不太注重，草草了事！')
print('我们对于女朋友提出的要求向来是不太注重，草草了事！')
print('我们对于女朋友提出的要求向来是不太注重，草草了事！')
print('我们对于女朋友提出的要求向来是不太注重，草草了事！')
print('我们对于女朋友提出的要求向来是不太注重，草草了事！')
print('我们对于女朋友提出的要求向来是不太注重，草草了事！')
print('我们对于女朋友提出的要求向来是不太注重，草草了事！')
print('我们对于女朋友提出的要求向来是不太注重，草草了事！')
print('我们对于女朋友提出的要求向来是不太注重，草草了事！')
print('我们对于女朋友提出的要求向来是不太注重，草草了事！')
'''
'''
#声明一个变量,作为计数变量使用
num = 0
while num<100:
#要循环的内容
print('我们对于女朋友提出的要求向来是不太注重，草草了事！')
#变量变化
num += 1
'''
'''
执行过程：
1.声明了变量num 并且赋值为0
2.进入while循环，判断num<100是否为真->0<100 真
3.执行循环代码组，输出了打印内容
4.将mum +1 变为了1
5.再次进入while循环，判断num<100 是否为真-> 1<100 真
6.执行循环代码组，输出了打印内容
7.将num +1 变为2
.....
再次进入while循环，判断num<100是否为真 -> 99<100 真
执行循环代码组，输出了打印内容
将num +1 变为100
再次进入while循环，判断num<100 是否为真 -> 100<100 假
循环结束
'''
'''
#使用while循环实现1-100的和
#声明计数变量
num = 1
#声明一个累加数值的变量
total = 0
while num<=100:
#将当前的num值添加到total当中
total += num #total = total + num
#改变num的值
num += 1
#顺序结构区域
print(total)
'''
'''
#格式2
num = 100
while num<50:
print('悲催的职业，炮兵连的炊事兵')
num += 1
else:
print('条件为假')
'''
#判断循环的第一次是否就是假值
#计数变量
num = 200
#是否开始循环的标志变量
flag = False
while num<100:
#进入循环时设定标志为True 表示来过循环区域执行
flag = True
print('悲催的职业，炮兵连的炊事兵')
num+=1
else:
#判断是否第一次就是假值
if flag == False:
print('起始值为假，循环未能执行')
else:
print('起始值为真，进入过循环')
| false |
ec41d9fc44f34b45f90aa8e63e77b37891603005 | GeekyShacklebolt/PYM_exercises | /day_1/variable_and_datatypes.py | 1,628 | 4.25 | 4 | #!/usr/bin/python3
a = 21
b = 34
c = a + b
print(c)
print('India')
print("India")
print('India\'s')
print("\n")
# reading input from the keyboard
num = int(input("Enter a number: "))
print(num)
if num < 10:
print("this is less than 10")
else:
print("this is bigger than 10")
print("\n")
# INTEREST PROBLEM
amt = float(input("Enter the amount: "))
interest = float(input("Enter the rate of interest: "))
period = float(input("Enter the period: "))
year = 1
tot = 0
while year <= period:
amt = amt + (interest*amt)
print("After %d year, your amount will be %f" % (year,amt))
year = year + 1
print("\n")
# AVERAGE OF N NUMBERS
N = int(input("Enter how many numbers would you like to enter: "))
ctr = 1
SUM = 0
average = 0
while ctr <= N:
num = int(input("Enter the no. {} : " .format(ctr)))
SUM = SUM + num
ctr = ctr + 1
average = float(SUM)/N
print("N is %d, SUM of \'N\' numbers is %d, Average is %f" % (N,SUM,average))
print("\n")
# TEMPERATURE CONVERSION
celcius = 0
while celcius <= 350:
farhenheit = 1.8 * celcius + 32
print("Celcius = %d and Farhenheit = %f" % (celcius,farhenheit))
celcius = celcius + 25
print("\n")
# MULTIPLE ASSIGNMENT
a, b = 50, 10
print(a, '+', b)
# swapping
a, b = b, a
print("%d + %d" % (a,b))
print("\n")
# TUPLE PACKING AND UNPACKING
tuple = ("shiva","Btech","Python")
name, course, language = tuple
print(name, course, language)
print("\n")
# FORMATTING STRINGS
msg = "{0} is studying in {1} and learning {2}".format(name,course,language)
print(msg)
msg2 = f"{name} is studying in {course} an learning {language}"
print(msg2)
| true |
cd96511ea7d7a311542ea00ad6b0c27921ed1ca0 | Youth-Avenue-2021/All_Programming | /PYTHON/LIST/double space identifier.py | 392 | 4.28125 | 4 | #This programs returns the index value where double space occure else it returns -1
a = "This program enables you to find if the string conatins double space!"
#a = "This program enables you to find if the string conatins double space!"
double_space = a.find(" ")
print(double_space)
#now the further code replaces double space (if any) with single space
a = a.replace(" "," ")
print(a) | true |
8d89fbba4e0f120c138ed4bc009cc502976c045b | carlossarante/python100days | /EventsPlayGround/turtle_race.py | 1,838 | 4.21875 | 4 | import turtle
import random
from turtle import Turtle, Screen
def get_turtle_race(screen_width, screen_height, turtle_colors):
starting_point_x = - screen_width / 2 + 20
gap = screen_height / len(turtle_colors)
turtle_list = []
turtle_y_positions = [-(screen_height / ) + gap * color_index for color_index in range(0, len(turtle_colors))]
for turtle_index in range(0, len(turtle_colors)):
turtle_created = Turtle(shape="turtle")
turtle_created.penup()
turtle_created.color(turtle_colors[turtle_index])
turtle_created.goto((starting_point_x, turtle_y_positions[turtle_index]))
turtle_list.append(turtle_created)
return turtle_list
def start_race(screen_width, screen_height):
screen = Screen()
is_race_on = False
screen.setup(width=screen_width, height=screen_height)
user_bet = screen.textinput(title="Make your first bet", prompt="Which turtle will win the race? Enter a color: ")
turtle_colors = ['red', 'pink', 'blue', 'black', 'yellow', 'gray']
turtle_list = get_turtle_race(screen_width=screen_width, screen_height=screen_height, turtle_colors=turtle_colors)
if user_bet:
is_race_on = True
while is_race_on:
# turtle size is 40
for turtle_created in turtle_list:
if turtle_created.xcor() > screen_width / 2 - (40 / 2):
is_race_on = False
winner_color = turtle.pencolor()
if user_bet.lower() == winner_color:
print(f'You\'ve won! turtle {turtle_created.pencolor()} is the winner!')
else:
print(f'You\'ve lost! turtle {turtle_created.pencolor()} is the winner!')
random_distance = random.randint(0, 10)
turtle_created.forward(random_distance)
screen.exitonclick() | true |
ba4278fba6ee31001612f952843c93d39a28b66e | warmer90/ttas | /面向对象/bike.py | 1,466 | 4.28125 | 4 | """
一个自行车类（Bicycle），run（骑行），调用显示骑行里程km（骑行里程传入的数字）
写一个EBicycle 继承 Bicycle，添加电量valume 属性通过参数传入，有两个方法
1. fill_charge(vol)，用来充电。vol 为电量
2. run（km） 方法骑行， 10km消耗一度电，电量耗尽调用Bicycle的run方法骑行
"""
class Bicycle:
# 定义一个run方法，需要传入km参数
def run(self,km):
print(f"自行车骑行的里程数:{km}")
class EBicycle(Bicycle):
#初始化
def __init__(self,volume):
#实例变量
self.valume = volume
def get_valume(self):
print("当前电量:",self.valume)
def fill_charge(self,vol):
print("充电电量",vol)
def run(self,km):
#电量支持的最大里程数
e_miles = self.valume*10
## 假如电瓶有10度电，我们支持的里程 10*10=100
miles = km - e_miles
if miles<=0:
print("电瓶车骑了：",km)
else:
#因为子类中有个run，把父类的run覆盖掉了
#self.run
#应该传入参数是除了电瓶车之外的里程数
print("电瓶车骑了",e_miles)
super().run(miles)
# bicycle= Bicycle()
# bicycle.run(100)
#类在初始化时候，给init中定义的参数传参
# bike = EBicycle(100)
# bike.get_valume()
bike = EBicycle(16)
bike.run(250)
| false |
c38bd931c248c1b4f1b4ce75d79f12195c8ddd62 | martinbolger/hein_scraping | /code/modules/data_manipulation_functions.py | 1,779 | 4.34375 | 4 | #!/usr/bin/env python
import pandas as pd
#Removes commas from all values in a row of a dataframe
def remove_commas(df1):
for col in df1.columns:
df1[col] = df1[col].str.replace(',', '')
return df1
#This function checks to see if a file with the papers from a professor has already been created
#If it has, their name is skipped (their data is not rescraped)
def check_files(fm_name, last_name, current_files):
done = False
for cur_file in current_files:
if fm_name.lower() in cur_file.lower() and last_name.lower() in cur_file.lower():
done = True
break
return done
# This function converts a list of strings to a comma
# separated string.
def list_to_comma_separated_string(list_of_strings):
list_str = ', '.join(list_of_strings)
list_str = list_str.replace('\'', '')
list_str.replace('\"', '')
return list_str
# This function is used to concatenate strings
# in a panda dataframe. It can handle cases when
# one of the strings is missing. It is used in an
# apply statement.
def concat_function(x, y):
if not pd.isna(x) and not pd.isna(y):
return x + ", " + y
elif not pd.isna(x):
return x
elif not pd.isna(y):
return y
else:
return x
# This function removes substrings from one list of
# comma separated strings from another list. It is used
# to remove error names from the alt names lists.
def remove_err_names(names, err_names):
if names:
names_list = names.split(", ")
else:
names_list = []
if err_names:
err_names_list = err_names.split(", ")
else:
err_names_list = []
names_list = [x for x in names_list if x not in err_names_list]
return list_to_comma_separated_string(names_list) | true |
2bf68d11dad29d6d5e4bdac9df597f79e6f6ffce | dragan-nikolic/python_ex | /operators.py | 659 | 4.125 | 4 | """
Created on 2012-02-12
@author: dnikolic
"""
def use_logical_operators(os_name='WINDOWS', os_version='5.1'):
if os_name == 'WINDOWS' and os_version == '5.1':
print 'OS is WinXP!'
else:
print 'OS is not XP'
if ((os_name == 'WINDOWS') and
(os_version == '5.1')):
print 'Told you, WinXP!'
else:
print 'Told you, not XP!'
def use_arithmetic_operators():
x = 5
print 'x=%d' % x
x += 9
print 'x=%d' % x
x /= 2
print 'x=%d' % x
if __name__ == '__main__':
use_logical_operators()
use_logical_operators('LINUX')
#use_arithmetic_operators()
pass
| false |
21f0da384a2b5a8e4960ff3384fa07f7eda8e272 | dragan-nikolic/python_ex | /tools/bulk_rename.py | 1,674 | 4.25 | 4 | """
The utility for renaming all the files in a directory which name contains
PATTERN_IN_ORIGINAL_NAME text.
New name will be NAME_PREFIX + PATTERN_IN_NEW_NAME + NAMME_SUFFIX.
Examples:
old.E23.name.mp4 -> new_E23_name.m4v
"""
import os
import sys
import re
# --- Modify these variables to customize the renaming ---
PATTERN_IN_ORIGINAL_NAME = '\.E\d{1,3}\.'
PATTERN_IN_NEW_NAME = '\d{1,3}' # PATTERN_IN_NEW_NAME pattern is subpattern of the PATERN_IN_ORIGINAL_NAME
NAME_PREFIX = 'new_'
NAME_SUFFIX = '_name.m4v'
# --------------------------------------------------------
try:
dir_name = sys.argv[1]
except Exception:
dir_name = "."
def find_pattern_in_string(pattern, string):
'''Returns string that satisfies pattern if found
Otherwise returns None
'''
res = None
try:
res = re.compile(pattern).search(string).group()
except Exception:
res = None
return res
def rename_file(old_name, new_name):
os.rename(old_name, new_name)
for root, dirs, files in os.walk(dir_name):
for filename in files:
filepath = os.path.abspath(os.path.join(root, filename))
print(filepath)
filepath_parts = os.path.split(filepath)
filename = filepath_parts[1]
print(filename)
print(type(filename))
result = find_pattern_in_string(PATTERN_IN_ORIGINAL_NAME, filename)
print('result1: ', result)
print(type(result))
result = find_pattern_in_string(PATTERN_IN_NEW_NAME, result)
print('result2: ', result)
if result:
rename_file(filepath, os.path.join(filepath_parts[0], NAME_PREFIX + result + NAME_SUFFIX))
| true |
8c974518142abb05afd1becb891d60e530b0ba1c | kemoelamorim/Entra21_Resolucao | /Resolucao_Aula007/Ex001.py | 436 | 4.21875 | 4 | """
--- Exercício 1 - Funções - 1
--- Escreva uma função que imprima um cabeçalho
--- O cabeçalho deve ser escrito usando a multiplicação de carácter
--- O cabeçalho deev conter o nome de uma empresa, que será uma variável
--- Realize a chamada da função na ultima linha do seu programa
"""
empresa = input('Digite o nome da empesa: ')
def cabecalho(empresa):
print(f"{empresa:*^30}".upper() )
cabecalho(empresa) | false |
ac2af637e4aebb77a5b3637d37f446b2f5aad852 | gdmhw/automateTheBoringStuffWithPython | /collatz_sequence.py | 413 | 4.15625 | 4 | def collatz(number):
if number % 2 == 0:
res = number // 2
print(res)
return res
elif number % 2 == 1:
res = 3 * number + 1
print(res)
return res
try:
userNum = int(input('Enter a number: '))
while userNum != 1:
userNum = collatz(int(userNum))
except ValueError:
print('Value Error - please enter an integer value')
| false |
856b03934440680775d97774fb653a1669c4bff5 | hhhoang/100DaysOfCode_HH | /Day41.py | 908 | 4.125 | 4 | """
Create a GUI to convert user input of ounce into gram with tkinker
"""
from tkinter import *
window = Tk()
window.title("Ounce to Gram Converter")
#window.minsize(width=500, height=300)
window.config(padx=20, pady=20)
# Input
input = Entry(width=10)
input.grid(column=1, row=1)
input.insert(END, string="0")
print(input.get())
#Label
ounces = Label(text="Ounces", font=("Arial", 10))
ounces.grid(column=2, row=1)
#my_label.config(padx=50, pady=50)
equal = Label(text="is equal to", font=("Arial", 10))
equal.grid(column=0, row=2)
result = Label(text="0", font=("Arial", 10))
result.grid(column=1, row=2)
gram = Label(text="Gram", font=("Arial", 10))
gram.grid(column=2, row=2)
# Button
def button_clicked():
ounce = float(input.get())
gram = round(ounce*28.3495231)
result.config(text=f"{gram}")
button = Button(text="Calculate", command=button_clicked)
button.grid(column=1, row=3)
| true |
c60621068444c1b118cd05cfe1f69a380186ecc9 | hhhoang/100DaysOfCode_HH | /Day21.py | 856 | 4.25 | 4 | # https://www.codewars.com/kata/after-midnight/train/python?
"""
Write a function that takes a negative or positive integer,
which represents the number of minutes before (-) or after (+) Sunday midnight,
and returns the current day of the week and the current time in 24hr format ('hh:mm') as a string.
"""
def day_and_time(mins):
dayspos = ("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday")
print(mins)
minutes, hours_final, days_final = get_min_hour_day(mins)
return "{} {}:{}".format(dayspos[days_final], str(hours_final).zfill(2),str(minutes).zfill(2))
def get_min_hour_day(mins, clock_wise=True):
hours, minutes = divmod(mins, 60)
days, hours_final = divmod(hours, 24)
days_final = days%7
print(days_final, days, hours_final, hours, minutes)
return(minutes, hours_final, days_final)
| true |
dc52ecd5042e14b08b5f913fd15f63e26cc1c5a9 | CarolFLima/hackerrank-solutions | /miscellaneous/primality.py | 268 | 4.15625 | 4 |
def primality(n):
if n == 2:
return "Prime"
if n < 2 or n % 2 == 0:
return "Not prime"
i = 3
while (i*i <= n):
if n % i == 0:
return "Not prime"
i += 2
return "Prime"
n = 5
print(primality(n))
| false |
823ac6f88f56e861802a399a7adad4fa655694b2 | tford-dev/BMI-calculator | /BMI 2.py | 2,765 | 4.125 | 4 | class bmi_metric:
def __init__(self, kgs = 1, m = 1):
self.kgs = kgs
self.m = m
def bmi_met(self):
return kgs / pow(m, 2)
class bmi_imperial:
def __init__(self, lbs = 1, ins = 1):
self.lbs = lbs
self.ins = ins
def bmi_imp(self):
return 703 * lbs / pow(ins, 2)
while True:
print("Welcome Body Mass Index Powered By @tfordfit")
print("Enter 1 to use METRIC measurements.")
print("Enter 2 to use IMPERIAL measurements.")
print("Enter 3 to calculate height in meters.")
print("Enter 4 to calculate height in inches.")
print("Enter 5 to exit.")
choice = input("Enter choice (1/2/3/4/5): ")
if choice == '1':
kgs = float(input("Enter your weight in kgs: "))
m = float(input("Enter your height in m: "))
met = bmi_metric(kgs, m)
print(f"Your BMI is {met.bmi_met()}.")
if met.bmi_met() < 19:
print("Based on your BMI, you are considered underweight.")
elif met.bmi_met() > 25:
print("Based on your BMI, you are considered overweight, but note that the BMI scale does not take into account muscle mass.")
elif met.bmi_met() >= 30:
print("Based on your BMI, you are considered obese, but note that the BMI scale does not take into account muscle mass.")
else:
print("You are in the healthy range!")
elif choice == '2':
lbs = float(input("Enter your weight in pounds(lbs): "))
ins = float(input("Enter your height in inches: "))
imp = bmi_imperial(lbs, ins)
print(f"Your BMI is {imp.bmi_imp()}.")
if imp.bmi_imp() < 19:
print("Based on your BMI, you are considered underweight.")
elif imp.bmi_imp() > 25:
print("Based on your BMI, you are considered overweight, but note that the BMI scale does not take into account muscle mass.")
elif imp.bmi_imp() >= 30:
print("Based on your BMI, you are considered obese, but note that the BMI scale does not take into account muscle mass.")
else:
print("You are in the healthy range!")
elif choice == '3':
cm = float(input("Enter your height in centimeters: "))
m = (cm * .01)
print(f"Your height in meters is {m}.")
elif choice == '4':
ft = int(input("Enter your height in feet(you will enter how many inches next): "))
ins = float(input("Enter the amount of inches you are in ADDITION to your height in feet: "))
inches_total = (ft * 12) + ins
print(f"Your height in inches is {inches_total}.")
elif choice == '5':
print("Peace be upon you.")
break
| false |
799ed2b3388d2bdf28ae143d6ced43a3d7ccdceb | ParkerCS/ch18-19-exceptions-and-recursions-bernhardtjj | /recursion_lab.py | 1,978 | 4.5625 | 5 | """
Using the turtle library, create a fractal pattern.
You may use heading/forward/backward or goto and fill commands to draw
your fractal. Ideas for your fractal pattern might include
examples from the chapter. You can find many fractal examples online,
but please make your fractal unique. Experiment with the variables
to change the appearance and behavior.
Give your fractal a depth of at least 5. Ensure the fractal is contained on the screen (at whatever size you set it).
Have fun.
(35pts)
"""
import turtle
colors = ["red", "orange", "yellow", "green", "blue", "purple", "magenta", "cyan", "black"]
my_turtle = turtle.Turtle()
my_turtle.showturtle()
my_screen = turtle.Screen()
my_screen.bgcolor('white')
# Draw Here
my_turtle.color("black")
def draw(x, y, heading, dist, depth):
my_turtle.up()
my_turtle.goto(x, y)
my_turtle.fillcolor(colors[depth % len(colors)])
my_turtle.down()
my_turtle.begin_fill()
for i in range(4):
my_turtle.setheading(heading - 90 * i)
my_turtle.forward(dist)
my_turtle.end_fill()
new_dist = dist / 1.618033 # phi
my_turtle.backward(dist - new_dist)
new_y = my_turtle.ycor()
new_x = my_turtle.xcor()
if depth > 0:
draw(new_x, new_y, heading - 90, new_dist, depth - 1)
draw(100, -250, 180, 500, 10)
my_screen.exitonclick()
# ____
# / ___|__ _ _ __ _ _ ___ _ _ __ _ _ _ ___ ___ ___
# | | / _` | '_ \ | | | |/ _ \| | | | / _` | | | |/ _ \/ __/ __|
# | |__| (_| | | | | | |_| | (_) | |_| | | (_| | |_| | __/\__ \__ \
# \____\__,_|_| |_| \__, |\___/ \__,_| \__, |\__,_|\___||___/___/
# |___/ |___/
# _ _ _ _ _ _ ___
# __ _| |__ __ _| |_ | |_| |__ (_)___ (_)___ |__ \
# \ \ /\ / / '_ \ / _` | __| | __| '_ \| / __| | / __| / /
# \ V V /| | | | (_| | |_ | |_| | | | \__ \ | \__ \ |_|
# \_/\_/ |_| |_|\__,_|\__| \__|_| |_|_|___/ |_|___/ (_)
| true |
44d0d809ed25f029d3e71cccd9737b36e74d26fe | listenviolet/leetcode | /206-Reverse-Linked-List.py | 1,452 | 4.3125 | 4 | # Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def reverseList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if head is None or head.next is None: return head
cur = head
p = head.next
n = p.next
while(n is not None):
p.next = cur
cur = p
p = n
n = p.next
p.next = cur
head.next = None
head = p
return head
# Description
# Reverse a singly linked list.
# Example:
# Input: 1->2->3->4->5->NULL
# Output: 5->4->3->2->1->NULL
# Follow up:
# A linked list can be reversed either iteratively or recursively. Could you implement both?
# Solution:
# O -> O -> O -> O -> O -> None
# |
# head
# | | |
# cur p n
# O<-> O O -> O -> O -> None
# | | |
# cur p n
# O<-> O <- O O -> O -> None
# | | |
# cur p n
# O<-> O <- O <- O O -> None
# | | |
# cur p n
# None <- O <- O <- O <- O <- O
# | |
# cur p
# |
# head
# Beats: 99.86%
# Runtime: 40ms
# easy
| false |
bb2b78245e2fb1b50f686c5733903fd0e2131f20 | lincolen/Python-Crash-Course | /7 input and while/pizza_topings.py | 271 | 4.15625 | 4 | messege = "\n what would you like me to add to your pizza?: "
messege += "\n when youve listed everything enter quit, to stop \n"
topping = ""
while topping!= "quit":
topping = input(messege)
if topping != "quit":
print("\n I will add "+topping+" to your pizza")
| true |
5be60dddfa3603fa9a63f55553249b77b0a24292 | DriveMyScream/Python | /07_Loops/Problem_No4.py | 314 | 4.25 | 4 | num = int(input("Enter a Number: "))
if(num>1):
for i in range(2, num):
if(num % 2 == 0):
isPrime = False
break
else:
isPrime = True
else:
isPrime = False
if(isPrime):
print("The Given Number is prime")
else:
print("The Given Number is Not Prime")
| true |
d45ea6047176d37a44821d3bfe2459c4ee878ee2 | DriveMyScream/Python | /02_Data Type and Variable/02_Arithmetic_Operators.py | 389 | 4.15625 | 4 | a = 10
b = 20
addition = a + b
substraction = a - b
multiplication = a * b
divide = a / b
module = a % b
print("The Addition Of Two Numbers is:",addition)
print("The Substraction of Two Numbers is:",substraction)
print("The Multiplication Of Two Numbers is:", multiplication)
print("The Division Of Two Numbers is: ", divide)
print("The Remiander of Two Operator is:", module)
| true |
235cefdfbc3438805deec087fc35e1a351afe923 | dsahney/choose_sorting_algorithm | /max_and_min_values_of_list.py | 1,017 | 4.5625 | 5 | # This function will take a list, L, and determine its maximum and minimum values.
import bubble_sort
import insertion_sort
import selection_sort
def max_min_bubble(L):
''' (list) -> tuple
Return a tuple that uses the bubble sort algorithms, to determine the min and max value of a list, respectively.
'''
sorted_list = bubble_sort(L)
max_value = L[-1]
min_value = L[0]
return (min_value, max_value)
def max_min_insertion(L):
''' (list) -> tuple
Return a tuple that uses the insertion sort algorithms, to determine the min and max value of a list, respectively.
'''
sorted_list = insertion_sort(L)
max_value = L[-1]
min_value = L[0]
return (min_value, max_value)
def max_min_selection(L):
''' (list) -> tuple
Return a tuple that uses the selection sort algorithms, to determine the min and max value of a list, respectively.
'''
sorted_list = selection_sort(L)
max_value = L[-1]
min_value = L[0]
return (min_value, max_value)
| true |
c1f228c7db44937757a7694374ccfef77911d0bd | acosme/project-euler | /38-pandigital_multiples.py | 1,489 | 4.25 | 4 | '''Take the number 192 and multiply it by each of 1, 2, and 3:
192 x 1 = 192
192 x 2 = 384
192 x 3 = 576
By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)
The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).
What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?'''
#time: 1.5 hour
def number_is_pandigital(concated_pandigital):
no_repeat = True
for c in concated_pandigital:
if concated_pandigital.count(c) > 1:
no_repeat = False
return ('0' not in concated_pandigital) and (len(concated_pandigital) == 9) and no_repeat
big_pandigital = 0
for multiplicand in range(0,9999):
product = 0
multiplier = 0
procced = True
concated_pandigital = ""
print multiplicand
while procced:
multiplier += 1
product = multiplicand * multiplier
concated_pandigital += str(product)
if len(concated_pandigital) >= 9:
procced = False
if number_is_pandigital(concated_pandigital):
if int(concated_pandigital) > big_pandigital:
big_pandigital = int(concated_pandigital)
print big_pandigital
print "big_pandigital => %d" % big_pandigital
| true |
59ea937883e8bf59218e31a4c5469a312a8ef0e9 | ltakuno/impacta | /POO/tratamento_erros/ex02.py | 290 | 4.1875 | 4 | from math import sqrt
try:
num = float(input('Digite um numero:'))
resultado = sqrt(num)
print ('A raiz quadrada de %.1f: %.1f' % (num, resultado) )
except ValueError:
print('Não é possível tirar a raiz quadrada!')
except Exception:
print('Erro desconhecido!!')
| false |
25448a22ca20feff72b567f5a01274e7a5618c9f | marco-zangari/math | /data-statistics/variance_standard_dev.py | 1,037 | 4.1875 | 4 | """Calculate the variance and standard deviation of a list of numbers."""
def calculate_mean(nums):
"""Calculate the meand of a given list of numbers."""
mean = sum(nums) / len(nums)
return mean
def find_differences(nums):
"""Find differences between each number in given list and mean."""
mean = calculate_mean(nums)
diff = []
for num in nums:
diff.append(num - mean)
return diff
def calculate_variance(nums):
"""Calculate variance of given list of numbers."""
diff = find_differences(nums)
squared_diff = []
for d in diff:
squared_diff.append(d**2)
sum_squared_diff = sum(squared_diff)
variance = sum_squared_diff / len(nums)
return variance
if __name__ == '__main__':
donations = [100, 60, 70, 900, 100, 200, 500, 500, 503, 600, 1000, 1200]
variance = calculate_variance(donations)
print(f'The variance in the list of numbers is {variance}.')
std = variance**0.5
print(f'The standard deviation of the list of numbers is {std}.')
| true |
573a8e549e804a81786320b263b1a49e402d9840 | fpem123/ConcurrencyProgramming | /3주차/customThread.py | 761 | 4.28125 | 4 | '''
- 20년 가을학기 분산병렬 프로그래밍
- 3장 스레드 라이프
- 사용자 정의 스래드 클래스
- 이호섭
'''
from threading import Thread
# Thread를 상속받아 만들어진 사용자 정의 스래드 클래스
class myWorkerThread(Thread):
i = 0
# 생성자
def __init__(self, i):
self.i = i
print("Hello World")
Thread.__init__(self) # 필수 호출
# run() is start()
def run(self):
print("Thread {} is now running".format(i))
myThread = []
for i in range(10):
myThread.append(myWorkerThread(i))
print("Create my Thread Object")
myThread[i].start()
print("Started my thread")
for i in range(10):
myThread[i].join()
print("My Thread finished") | false |
0fb97fc7411daac2e6550c6cc5dbe0755f40a51a | normkh21/PythonBasics | /dictionaries_loops.py | 2,116 | 4.25 | 4 | # looping dictionaries and more
student1 = {'name': 'Hamza', 'gpa': 3.8, 'lastName': 'Hamrakulov'}
student2 = {'name': 'Alexa', 'gpa': 3.9, 'lastName': 'Moseyeva'}
# looping with keys
for key in student1:
print(' key is:', key)
print()
for key in sorted(student1.keys()):
print('key is:', key)
# for info in student1.keys():
for key in student1:
print('value is:', student1[key])
for value in student1.values():
print('value is', value)
print()
# Looping the values
for dkey, dvalue in student1.items():
print('key is', dkey)
print('value is ', dvalue)
print("Nesting dictionaries in LIST ")
class_2020 = [student1, student2]
print(class_2020)
for student in class_2020:
print('Name of the student:', student['name'])
print('GPA of student:', student['gpa'])
print('Last name of student:', student['lastName'])
print("------------------------------------")
print("****** Nesting dictionaries in Dictionaries")
dclass_2020 = {'student1': student1, 'student2': student2}
print(dclass_2020)
for key, value in dclass_2020.items():
print('Key of the element: ', key)
print('Value of the element: ', value)
print('Name of the student: ', value['name'])
print('GPA of the student: ', value['gpa'])
print('Last Name of the student: ', value['lastName'])
print("----------------")
#
#
# print()
# print('Excercise 6-5')
# rivers = {'nile': 'egypt:', 'hudson' : 'usa:', 'volga': 'russia:', 'mississippi' : 'usa', 'thames': 'uk'}
#
# # The KEY runs through Value
# for river, country in rivers.items():
# # if country == 'usa' or cuntry == 'uk'
# if country in ['usa', 'uk']:
# print(f"The {river.title} runs through {country.upper()}")
# else:
# print(f"The {river.title} runs through {country.title()}")
#
# print('Rivers are: ')
# for river in rivers.keys():
# print('\t', river.title)
#
# print('Countires are: ')
# for country in sorted(rivers.values(), reverse=True):
# if country in ['usa', 'uk']:
# print('\t', country.title())
# else:
# print('\t', country.upper(), end=" | ")
#
#
| false |
aa72af9d0f2ec0b36cb7c5ef1707e8feea6f3c68 | tonyechen/python-codes | /argsParameter/main.py | 374 | 4.15625 | 4 | # *args = parameter that will pack all arguments into a tuple
# useful so that a function can accept a varying amount of arguments
def add(*stuff): # doesn't have to be args, can be named anything else
sum = 0
print(stuff)
stuff = list(stuff)
print(stuff)
stuff[0] = 0
for i in stuff:
sum += i
return sum
print(add(1,2,3)) | true |
4f3b5169cfe8767c3623305056ec1756a3e466d2 | tonyechen/python-codes | /stringMethods/main.py | 731 | 4.40625 | 4 | # string methods
name = "hello world"
# len() = length of string
print(len(name))
# str.find() - find index of the first appearance of string
print(name.find("Bro"))
# str.capitalize() - only the first letter is capitalized
print(name.capitalize())
# str.upper() + str.Lower() - all cap or all lower-case
print(name.upper())
print(name.lower())
# str.isdigit() - true or false depending on if the string is a digit
print(name.isdigit())
# str.isalpha() - are these all alphabetical characters
print(name.isalpha())
# str.count("String") - count the number of appearance
print(name.count("o"))
# str.replace("String", "String") - replace string
print(name.replace("o", "x"))
print(name * 3) # print a string multiple times
| true |
502c2f54445201c08db1e01808465dbf77fbffe3 | tonyechen/python-codes | /sort/main.py | 1,195 | 4.4375 | 4 | # sort() method = used with lists
# sort() function = used with iterables
students = list()
students = ["Squidward", "Sandy", "Patrick", "Spongebob", "Mr.Krabs"]
students.sort()
#students.sort(reverse = True) # reverse
for i in students:
print(i)
#----------------------------------------------------------------------
print()
students1 = ("Squidward", "Sandy", "Patrick", "Spongebob", "Mr.Krabs")
sorted_students = sorted(students, reverse = True)
for i in sorted_students:
print(i)
print()
#----------------------------------------------------------------------
students2 = [("Squidward", "F", 60),
("Sandy", "A", 33),
("Patrick", "D", 36),
("Spongebob", "B", 20),
("Mr.Krabs", "C", 78)]
grade = lambda grades:grades[1]
students2.sort(key=grade)
print(students2)
print()
#----------------------------------------------------------------------
students3 = (("Squidward", "F", 60),
("Sandy", "A", 33),
("Patrick", "D", 36),
("Spongebob", "B", 20),
("Mr.Krabs", "C", 78))
age1 = lambda ages:ages[2]
sorted_students3 = sorted(students3, key=age1)
print(sorted_students3)
| true |
4733c4cecc951c4d1fef7e1ae575617c9713e5ca | tonyechen/python-codes | /stringFormats/main.py | 1,106 | 4.15625 | 4 | # str.format() = oprtional method that gives users more control when displaying output
animal = "Cow"
item = "moon"
print("the " + animal + " jumped over the " + item)
print("The {} jumped over the {}".format(item, animal))
print("The {1} jumped over the {0}".format(item, animal)) # positional argument
print("The {animal} jumped over the {item}".format(item="moon", animal="cow")) # keyword argument
text = "The {} jumped over the {}"
print(text.format(animal, item))
name = "Tony"
print("Hello, my name is {:10}. Nice to meet you".format(name))
print("Hello, my name is {:<10}. Nice to meet you".format(name))
print("Hello, my name is {:>10}. Nice to meet you".format(name))
print("Hello, my name is {name:^10}. Nice to meet you".format(name="Tony"))
number = 1000
print("The number pi is {:.3f}".format(number))
print("The number pi is {:,}".format(number))
print("The number pi is {:b}".format(number)) #binary
print("The number pi is {:o}".format(number)) #Octave
print("The number pi is {:x}".format(number)) #Hex
print("The number pi is {:e}".format(number)) #scientific notation
| true |
bd61f2c7afbc677422126a6a912f90896804658b | YRTr/python_basics | /-2basic programs-/5KilometersToMiles.py | 212 | 4.15625 | 4 | K = input("Enter the number of kilometers")
try:
kilometers = float(K)
except:
print("Expected values")
Mile = (1.6)*(kilometers)
print("For a kilometers of %g -- The miles are : %g" %(kilometers, Mile))
| true |
b713e9cf1c446bc4f496ba05eed66eb1f2da95ff | dougscohen/cs-module-project-algorithms | /product_of_all_other_numbers/product_of_all_other_numbers.py | 1,655 | 4.625 | 5 |
def multiplyList(myList) :
if len(myList) == 0:
return 0
# Multiply elements one by one
else:
result = 1
for x in myList:
result = result * x
return result
def product_of_all_other_numbers(arr):
"""
Returns a list of integers consisting of the product of all numbers in the
array _except_ the number at that index.
Parameters:
arr (list): a list of integers
Returns:
arr (list): a list of integers
"""
# Loop through each item in the list
# using index as the split, separate into left and right arrays at index
# create helper function to multiply all items in a list with each other
multi = []
for i in range(1, len(arr) + 1):
if i == 1 or i == (len(arr)):
left = arr[:(i - 1)]
right = arr[i:]
left = multiplyList(left)
right = multiplyList(right)
both = left + right
multi.append(both)
else:
left = arr[:(i - 1)]
right = arr[i:]
left = multiplyList(left)
right = multiplyList(right)
both = left * right
multi.append(both)
return multi
if __name__ == '__main__':
# Use the main function to test your implementation
# arr = [1, 2, 3, 4, 5]
arr = [2, 6, 9, 8, 2, 2, 9, 10, 7, 4, 7, 1, 9, 5, 9, 1, 8, 1, 8, 6, 2, 6, 4, 8, 9, 5, 4, 9, 10, 3, 9, 1, 9, 2, 6, 8, 5, 5, 4, 7, 7, 5, 8, 1, 6, 5, 1, 7, 7, 8]
print(f"Output of product_of_all_other_numbers: {product_of_all_other_numbers(arr)}")
| true |
5c081dcbd81d1ce507a960c25465cba40045b499 | nmwalsh/project-euler-solutions | /p1.py | 781 | 4.53125 | 5 | """
Problem 1
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
"""
# Concepts used:
# set, range, union
import math
# question-specific parameters
min_val = 0
max_val = 1000
multiple_one = 3
multiple_two = 5
def sum_of_multiples(min_val, max_val, multiple_one, multiple_two):
"""
Function that returns sum of union set of all multiples between min_val and max_val for two given multiples
"""
set_one = set(range(min_val, max_val, multiple_one))
set_two = set(range(min_val, max_val, multiple_two))
union_sum = sum(set_one | set_two)
print(union_sum)
return union_sum
sum_of_multiples(min_val, max_val, multiple_one, multiple_two) | true |
72f1472ecac5a8783383fcbf62a3bbadb78f7ead | mohowzeh/Python_basics | /Python_basic_calculator.py | 924 | 4.15625 | 4 | '''
freeCodeCamp - Python and Django
BUILDING A BASIC CALCULATOR
See YouTube: https://youtu.be/jBzwzrDvZ18?t=9458
'''
# try .. except .. statement
# simple calculator
# exit by typing q as operator
while 1==1:
try:
num1 = int(input('Enter first number: '))
num2 = int(input('Enter second number: '))
except ValueError:
print('Please enter numbers.')
else:
op = input('Enter operator (+, -, *, /) or type q to quit: ')
if op == '+':
print(num1, num2, op, '=>', num1+num2)
elif op == '-':
print(num1, num2, op, '=>', num1-num2)
elif op == '*':
print(num1, num2, op, '=>', num1*num2)
elif op == '/':
print(num1, num2, op, '=>', num1/num2)
elif op == 'q':
break
else:
print('Invalid operator.')
finally:
print('=============================')
| false |
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