blob_id string | repo_name string | path string | length_bytes int64 | score float64 | int_score int64 | text string |
|---|---|---|---|---|---|---|
7ad07e51fefa4c5118192f6bab9b52c4fe3aa542 | mouyang2001/SOFTENG-284 | /assignment-1/task1.py | 748 | 3.609375 | 4 | # Author: Matthew Ouyang
# Task: find the maximum profit from a list of stock prices
# Commands to execute code:
# > python3 task1.py < canvas.in > myout1.txt
# > diff myout1.txt canvas.out1
import sys
# more efficient solution O(n)
def find_max_profit(prices, length):
max_profit = prices[1] - prices[0]
min_price... |
53d7acce0c9cbdc2b171ede0a72f9207d7a19f01 | jaychsu/algorithm | /lintcode/611_knight_shortest_path.py | 1,336 | 4.03125 | 4 | """
Definition for a point.
class Point:
def __init__(self, a=0, b=0):
self.x = a
self.y = b
"""
class Solution:
V = (
(-2, -1),
( 2, 1),
(-2, 1),
( 2, -1),
(-1, -2),
( 1, 2),
(-1, 2),
( 1, -2),
)
"""
@param: G: a... |
f6ca0ba07ae817412cd3c200e61d8cd4ca111b14 | hun-a/learn-the-python-basic | /src/03.list.py | 1,158 | 4.09375 | 4 | # https://wikidocs.net/14
# List
odd = [1, 3, 5, 7, 9]
a = []
b = [1, 2, 3]
c = ['Life', 'is', 'too', 'short']
d = [1, 2, 'Life', 'is']
e = [1, 2, ['Life', 'is']]
print(a, b, c, d, e)
# Indexing and slicing
print(b[0], b[1], b[2])
print(b[0] + b[1] + b[2])
print(b[-1])
a = [1, 2, 3, ['a', 'b', 'c']]
print(a[0])
prin... |
6c741be9668e504c84b48a0d27aba7f6fe0a1d1b | Mandeep5138/Study | /PYTHON CODES/Rolling_Hash_String_compare.py | 646 | 3.71875 | 4 | text = "Python"
pattern='th'
X=len(text)
Y=len(pattern)
d=256 #positional base
#sum for pattern
sumP=0
for i in range(Y):
sumP=sumP+(ord(pattern[i])*(d**(Y-i-1)))
print(sumP)
#sum of text
sumT=0
flag=0
for i in range(Y):
sumT=sumT+(ord(text[i])*(d**(Y-i-1)))
if sumP==sumT:
print("Patt... |
5def0d88a071d4d836e66fbbeac357caa6ae7c5c | handabaldeep/leetcode | /arrays/two_sum.py | 392 | 3.546875 | 4 | # Link - https://leetcode.com/problems/two-sum/
from typing import List
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
nums_idx_dict = {}
for i in range(len(nums)):
if target - nums[i] not in nums_idx_dict:
nums_idx_dict[nums[i]] = i
... |
6272690c68253082e2596ad3de8b375f7c9478cd | spacecase123/cracking_the_code_interview_v5_python | /18.4_number_of_2s.py | 881 | 3.953125 | 4 | '''
Write a method to count the number of 2s between0 and
'''
def num_2s_brute_force(n):
twos = 0
for i in range(2, n+1):
x = str(i)
for j in range(len(x)):
if x[j] == "2":
twos += 1
return twos
def num_2s(n):
if n < 2: return 0
twos = 0
... |
860dc28d80a0e9a631b3731afcc3a303fbaa6f3d | SATHANASELLAMUTHU/MYPROJECT | /B72.py | 171 | 3.734375 | 4 | n=input("Enter the sentence:")
c=0
for i in n:
if(i=="a" or i=="e" or i=="i" or i=="o" or i=="u"):
c=c+1
if c==0:
print("No")
else:
print("Yes")
|
ffc943eab1f639db284c2a79e3087344d1f8f076 | kagerman80/Recursion | /recursion.py | 455 | 4.125 | 4 | """Digital root is the recursive sum of all the digits in a number.
Given n, take the sum of the digits of n. If that value has more than one digit,
continue reducing in this way until a single-digit number is produced.
The input will be a non-negative integer."""
def calc_digits(x):
result = sum(list(map(int,s... |
aa60daaec83b46ef58da28f6f6219701855c8e86 | kevinlondon/leetcode | /0000-0999/026_remove_duplicates_from_array_easy.py | 463 | 3.5625 | 4 | class Solution:
def removeDuplicates(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
unique_nums = sorted(list(set(nums)))
length = len(unique_nums)
print(unique_nums)
for index in range(len(nums)):
if index < length:
... |
5201720365f8967343fbe88a073f3c3ef654c665 | jonliu6/Programming | /PythonDemo/interests.py | 659 | 3.5625 | 4 | initialAmount = 100
annualAmount = 10
interests = 0.1
asset = 0
#for x in range(0,20):
# asset = (initialAmount + annualAmount * x) * (1 + interests) ** x
#asset = (initialAmount) * (1 + interests) ** x # with a fixed initial amount, no annual deposite
#print("year", x, ", asset=" , asset)
# print(asset)
def... |
e4a009b37ed9a3c8a625664d88d02e87c5044709 | asifbux/Python-Course-ENSF-592 | /Assignment4/A04_analyze_book1.py | 2,652 | 3.78125 | 4 | """This module contains a code example related to
Think Python, 2nd Edition
by Allen Downey
http://thinkpython2.com
Copyright 2015 Allen Downey
License: http://creativecommons.org/licenses/by/4.0/
"""
from __future__ import print_function, division
from Histogram import Histogram as hg
import string
def process_f... |
e88b4192ce368b93ab88f978bc3649fd3d16b242 | psycho-pomp/CodeChef | /KS2.py | 270 | 3.703125 | 4 | # cook your dish here
def sum_digits(n): return sum(map(int, str(n)))
def findRoundNumber(order):
s = sum_digits(order) % 10
return order * 10 + (0 if s == 0 else (10 - s))
t=int(input())
for _ in range(t):
n=int(input())
print(findRoundNumber(n))
|
aa513a0a1b66b4f40a5dac8e83a14ee5767272ed | Nuttanan29445/Python-Daily | /Day028.py | 932 | 3.515625 | 4 | def readword(n):
read = ['soon','neung','song','sam','si','ha','hok','chet','paet','kao']
return read[n]
def to_Thai(n):
ans = ''
k = n
if k == 0 : ans+=readword(k)
else :
if len(str(n))>3:
p = k//1000
if p>0 :
ans+=readword(p)
ans... |
cba8fa8380a41c876bbd89f23cd134abfb40f731 | irina-baeva/algorithms-with-python | /data-structure/queue.py | 589 | 3.828125 | 4 | class Queue:
def __init__(self, head=None):
self.storage = [head]
def enqueue(self, new_element):
self.storage.append(new_element)
def peek(self):
return self.storage[0]
def dequeue(self):
return self.storage.pop(0)
# Setup
q = Queue(12)
q.enqueue(20)
q.enqueue(1... |
3abfb0c19e5398bab0b3b2618242d562ab9f6330 | IveDeletedYourPostAsItsADuplicate/codonPython | /codonPython/suppression.py | 1,381 | 3.9375 | 4 | def central_suppression_method(valuein: int, rc: str = "5", upper: int = 5000000000) -> str:
"""
Suppresses and rounds values using the central suppression method.
If value is 0 then it will remain as 0.
If value is 1-7 it will be suppressed and appear as 5.
All other values will be rounded to the ... |
60a87e5fbc228c7dcc0c2c4d303d9855c95e94ec | PedroPadilhaPortella/Curso-Em-Video | /Curso de Python Mundo 2/055-estatistica.py | 662 | 3.765625 | 4 | olderName = ''
older = 0
sumAges = 0
youngWomen = 0
for pessoa in range(1, 5):
print(f"-----{pessoa} PESSOA-----")
nome = input(f"Nome da Pessoa {pessoa}: ")
idade = int(input(f"Idade da Pessoa {pessoa}: "))
sexo = input(f"Sexo da Pessoa {pessoa} [M/F]: ").strip().upper()
sumAges += idade
if(ida... |
9bf8325a657b4848a1ec8330a19c74eb0e4ecae9 | patil16nit/algorithms | /algorithm/permutationPalindrom.py | 559 | 3.59375 | 4 |
def permutationPalindrom(msg, start, end):
if start == end:
print ''.join(msg)
else:
for i in xrange(start, end+1):
msg[start], msg[i] = msg[i], msg[start]
permutationPalindrom(msg, start+1, end)
msg[start], msg[i] = msg[i], msg[start]
def main():... |
afee1c02dfa24cfa0350c399990dd1c047923107 | ww-tech/primrose | /primrose/transformers/categoricals.py | 6,727 | 3.546875 | 4 | """Module to run a basic decision tree model
Author(s):
Mike Skarlinski (michael.skarlinski@weightwatchers.com)
"""
import pandas as pd
import numpy as np
import logging
from sklearn import preprocessing
from primrose.base.transformer import AbstractTransformer
class ExplicitCategoricalTransform(AbstractTransf... |
c142babbbfdefbf54f5b53b511e7630aa3e726b4 | Mehedi-Hasan-NSL/Python | /bfs_shortest_reach.py | 2,208 | 3.90625 | 4 | # -*- coding: utf-8 -*-
"""
Created on Wed Jul 7 17:09:46 2021
@author: DELL
"""
from collections import defaultdict
# This class represents a directed graph
# using adjacency list representation
class Graph:
# Constructor
def __init__(self):
# default dictionary to store graph
... |
b47e77a5dc42c5fd191af10f58195c9a6df2286c | ZunLayNwe12/GitSample | /Variable.py | 1,946 | 4.3125 | 4 | #21.12.2019
>>> text = "This is a string"
>>> text
'This is a string'
>>> 2 + 2
4
>>> 50 - 5 * 6
20
>>> (50 - 5 * 6) / 4
5.0
>>> round(5.0)
5
>>> 5.123456
5.123456
>>> round(5.123456,2)
5.12
>>> 17 / 3 # division returns a float
5.666666666666667
>>> 17 // 3 # discards the fraction
5
>>> 17 % 3 # ... |
e67835f8263a0934afbb015f4eaf51f2479c0b98 | zaid-sayyed602/Python-Basics | /star patterns with for loop.py | 3,942 | 4 | 4 | n=int(input("Enter the number of lines you want\n:"))
print("1")
for i in range(n):
for j in range(i+1):
print("*",end="")
print()
print("\n")
print("\n")
print("2")
for i in range(n):
for j in range(n-i-1):
print(" ",end="")
for k in range(i+1):
print("*",end=... |
e6cc8873afeac49f80de87d51e221dd982dea1f9 | roisinhanlon/hello-world | /whilecounter.py | 81 | 3.578125 | 4 | counter = 0
while counter < 5:
counter +=1
print(counter, 'test loop') |
4d1455a28a83758176391f13a392f9590d2c10ab | ahmedzaabal/Python-Demos | /functions.py | 796 | 3.984375 | 4 | from datetime import datetime
#this is a functions current date and time
#custom messages
# def print_time(task_name):
# print(task_name)
# print(datetime.now())
# print()
# first_name = "Ahmed"
# print_time("task is done")
# for x in range(0, 10):
# print(x)
# print_time("Task is done")
def ge... |
ae6c737b08c1eff5d4e169eef23c7db9455d737a | LeunamBk/MSC | /regmod/pcaPython/argvValidate.py | 314 | 3.734375 | 4 | # validate user input
import sys
def validateArgv(argv):
def is_intstring(s):
try:
int(s)
return True
except ValueError:
return False
for arg in argv:
if not is_intstring(arg):
sys.exit("All arguments must be integers. Exit.")
|
3254a02f11d8420fd64cc069f47a879d4108775c | sky-bot/Interview_Preparation | /LeetCode/Binary_tree_paths.py | 825 | 3.734375 | 4 | class Solution:
def binaryTreePaths(self, root: TreeNode) -> List[str]:
all_paths = list()
if root.left is None and root.right is None:
return [str(root.val)]
self.inorder(root.left, str(root.val), all_paths)
self.inorder(root.right, str(root.va... |
a4b9949e0c7e9e01bcd704589b6edcbdab540c0b | tzkcnm/test | /python_lecture_20161028/context_manager.py | 809 | 3.78125 | 4 | # -*- coding:utf-8 -*-
# class SkipMe(Exception):
# pass
SkipMe = type('SkipMe', (Exception,), {})
class ContextManager(object):
def __enter__(self):
print '=== START ==='
def __exit__(self, exc_type, exc_val, exc_tb):
if exc_type is SkipMe:
return True
print '=== END ... |
8db2f626046f6cbcc3037e5e2fc513b9e8cf88d1 | lujin123/algorithms | /leetcode/python/remove_linked_list_elements.py | 1,315 | 3.875 | 4 | # Created by lujin at 3/3/
#
# 203. Remove Linked List Elements
#
# Description:
#
# Remove all elements from a linked list of integers that have value val.
#
# Example
# Given: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6
# Return: 1 --> 2 --> 3 --> 4 --> 5
#
class ListNode(object):
def __init__(self, x):
... |
6bfaf0fd0dc86d3d408d03319f2e2751e385706b | mfkiwl/Sanajeh | /src/call_graph.py | 7,927 | 3.53125 | 4 | # -*- coding: utf-8 -*-
# Define Python call graph nodes' data structure
from typing import Set
# Block tree Node
class CallGraphNode:
node_count = 0
def __init__(self, node_name):
self.name: str = node_name # node name
self.declared_variables: Set[VariableNode] = set() # varia... |
f8515820f299607c0b8ec6a4f5988e175fe9f5f5 | Subhajit-Roy-18/Mensuration | /Mensuration.py | 4,831 | 4.34375 | 4 | print("This is a program which will help you in Mensuration.")
print("NOTE: While you Input the Measurements, DO NOT TYPE ITS UNIT. \n")
def main():
print("On which Topic do you want help in ?")
print("For Perimeter, Type 1.")
print("For Area, Type 2.")
print("For Volume, Type 3.")
Type... |
51eed8ad2f705c8e9d872a2ff9cb6245f0b5fb44 | mpkuchera/PHY200_homeworks | /Homework5/intensity.py | 1,262 | 3.75 | 4 | """
name: intensity.py
Compute and visualize diffraction of waves upon encountering a straight edge.
Problem 5.11 from Newman's Computational Physics.
author: Darcy Lewis
date created: 15/3/2021
"""
# Reminder: you can create other helper functions as long as the function
# behaviors below remain the same
#... |
c5e51a72df7a385de192d1721d28d11385249961 | iakonk/MyHomeRepos | /python/examples/leetcode/easy/find_smallest_contigius_subarr.py | 609 | 3.59375 | 4 | def find_max_freq(arr):
window_start, max_freq = 0, 0
numbers_freq = {}
for window_end in range(len(arr)):
right_num = arr[window_end]
numbers_freq.setdefault(right_num, 0)
numbers_freq[right_num] += 1
while len(numbers_freq) > 1:
left_num = arr[window_start]
... |
89f7b5212b15c7540d786400e44793ced58c2cfa | ncrubin/NielsenNNBookExercise | /network.py | 5,664 | 4.40625 | 4 | ''''
Feed forward neural network from Michael Nielsen's book on deep learning
http://neuralnetworksanddeeplearning.com/
Network is set up as number of layers and a list of sizes. The weights and
biases for network are selected at random. The cost function is quadratic.
This program demonstrates a simple neural n... |
c59d570c507ebb1c3c67ddd04bb3fb9a4d8629c1 | mohinishmd/Speech-Recognition-Python | /wav_transcribe.py | 821 | 3.8125 | 4 | import speech_recognition as sr
# obtain path to "english.wav" in the same folder as this script
from os import path
WAV_FILE = path.join(path.dirname(path.realpath(__file__)), "microphone-results.wav")
# use "english.wav" as the audio source
r = sr.Recognizer()
with sr.WavFile(WAV_FILE) as source:
audio = r.reco... |
c1db1a2125e0c87fc0b10d8d3c7ee41e07cad3ed | hegde10122/python_training | /numpy/numpy6.py | 2,700 | 4.4375 | 4 | '''
INDEXING AND SLICING of ndarrays
'''
import numpy as np
arr = np.arange(10)
print(arr)
print(arr[5:8]) #index 8 is not included ---only 3 elemnts at index 5,6,7
arr[5:8] = 19
print(arr)
'''
If you assign a scalar value to a slice, as in arr[5:8] = 19, the value is propagated (or broadcasted henceforth)
to the ... |
a3c448ce545883a8b5cd713ac47ca825c40c2a1d | swordwielder/python3 | /printdiamond.py | 390 | 3.921875 | 4 | n = int(input())
print(n*'#')
for i in range(1, (2*n+1)//2 + 1):
for j in range((2*n+1)//2 - i):
print(" ", end = "")
for k in range((i*2)-1):
print("#", end = "")
print()
for i in range((2*n+1)//2 + 1, 2*n + 1):
for j in range(i - (2*n+1)//2 -1):
print(" ", end = "")
for k ... |
a04466fc4c154a59dd983fe4a56befe8b530be68 | huwanping001/Python | /chapter4/demo9.py | 225 | 3.8125 | 4 | # 学校：四川轻化工大学
# 学院：自信学院
# 学生：胡万平
# 开发时间：2021/9/18 15:00
age = int(input('请输入你的年龄：'))
if age:
print(age)
else:
print('年龄为：', age) |
c9296d6322695116decd3cc66604a16fae3f3a30 | Slendercoder/Triqui | /my_app/funciones_logica.py | 13,988 | 3.5625 | 4 | import re
conectivos = ['O', 'Y', '>', '=']
Nfilas = 3
Ncolumnas = 3
Nnumeros = 3
Nturnos = 2
class Tree(object):
def __init__(self, label, left, right):
self.left = left
self.right = right
self.label = label
def inorder(A):
#Convierte un Tree en una cadena de símbolos
#Input: A, ... |
c2da35c69dbc9990c64b9d228fc5a68059486bca | jiinmoon/Algorithms_Review | /Archives/Leet_Code/Old-Reviews/Current-Reviews/Top-Review-01/0179-Largest-Number.py | 421 | 3.671875 | 4 | # 179 Largest Number
#
# We create a custom comparator that compares by string concatenation.
from functools import cmp_to_key
class Solution:
def largestNumber(self, nums):
def larger(a, b):
return -1 if a + b > b + a else 1
nums = map(str, nums)
nums.sort(key=cmp_to_key(larg... |
9e1dd814bbc4e447b48cd212ac51444085b650e6 | Bublinz/Python-Expert | /Exercises/student_fidel/triangle.py | 319 | 4.0625 | 4 | # write a program that will calculate the area of a traingle
print('calculate the area of a triangle')
def rate(length):
b = int(input('Input the Base: '))
h = int(input('Input the Height: '))
# area = 0.5 * b * h
area = 0.5 * b * h
print('The area of the triangle is ' + str(area))
rate('length')
|
333e740a6f337c4bbc1d107ad2aef8146905098f | Bansi-Parmar/Python_Basic_Programs | /python_program-062.py | 275 | 3.671875 | 4 | ## 62. 1
## 2 2
## 3 3 3
## 4 4 4 4
## 5 5 5 5 5
print('\n\t patterns')
print('\t...........\n')
n = int(input("Enter No of Rows :- "))
i=0
for r in range(1,n+1):
print((('{} '.format(r)*r) .ljust(n*2)))
i=i+1
|
44b93eaafdb68529be2bae98d6b00d67317c3b21 | vitaedrinker/python | /JohnEmailScrape.py | 882 | 3.890625 | 4 | # read a webpage
import urllib2
response = urllib2.urlopen('http://python.org/')
html = response.read()
import re #Regex
#Author: John Nicholson
#Scrapes Webpage and pulls all lines with an email to a .txt
def WebToTxt():
print ("Would you like to use a different url other than https://pastebin.com/n9phcmx4... |
f3c3bc512463b91bf81ec035b760b41ee4b89eb3 | walxc1218/git | /圆的周长面积.py | 1,152 | 4 | 4 | #打印圆的周长和面积
r = 3
p = 3.14
l = 2*p*r
s = p*r**2
print("圆的周长是：",l,"厘米")
print("圆的面积是：",s,"平方厘米")
在交互模式下查看当前作用域的所有变量
help("＿＿main＿＿")
del 语句
作用:
用于删除变量，同时解除与对象的关联，如果可能则释放对象
语法：
del 变量名１，
is／is not
小整数对象池：
cpython中，－５至２５６的数永远存在于小整数
id(x)函数
作用：
返回一个
练习：
1.在终端输出图形：
*
***
*****
*******
2.中国古代的称是１６两一斤，请问古代的２１６两... |
c2af09a3372d309a5c0e2a0349ea2a9ea3d1acba | L200184092/Praktikum-AlgoStruk | /Praktikum 8/5.py | 957 | 3.78125 | 4 | class PriorityQueue(object):
def __init__(self):
self.qlist = []
def isEmpty(self):
return len(self) == 0
def __len__(self):
return len(self.qlist)
def enqueue(self, data, priority):
entry = PriorityQEntry(data, priority)
self.qlist.append(entry)
def... |
efcc91ae4fbbacb2e3d66a967c3dde6a51fd1b7b | senioryta/Python-OOP | /12_Super()/main.py | 840 | 3.671875 | 4 | # superclass
class Estudent:
# private class attribute
__total = 0
# constructor __init__()
def __init__(self,name,id,classroom):
self.__name = name
self.__id = id
self.__class = classroom
Estudent.__total += 1
# method
def showinfo(self):
print("{} \n\t id : {} \n\t class : {} \n".format(self.__name,... |
bbac10925ea7bebbd6ddbdcd6cb460dc401acc40 | VinayakBagaria/Python-Programming | /New folder/Highest Prime Factor.py | 435 | 3.734375 | 4 | #Largest Prime Factor
def prime(a):
for i in range(2,a):
if a%i==0:
return -1
return 1
def calc_prime(n):
p=2
while p*p<=n:
if (n%p==0 and prime(p)!=-1):
n=n//p
else:
p=p+1
print(n)
global take
test=int(input())
list=[No... |
c4e7ea0c065b5f60ef259afb67cb5503857ee448 | bpull/fall16 | /graphics/hw/4/circle.py | 1,299 | 3.578125 | 4 | #!/usr/bin/python3
# Our required libraries!
import math
from OpenGL.GL import *
from OpenGL.GLU import *
from OpenGL.GLUT import *
#drange definition taken from http://stackoverflow.com/questions/477486/python-decimal-range-step-value
def drange(start, stop, step):
r = start
while r < stop:
yield r
... |
a0b687706fd97a4a69699b2aac6bf4b0365a912a | JSYoo5B/TIL | /PS/BOJ/11723/11723.py | 857 | 3.65625 | 4 | #!/usr/bin/env python3
from sys import stdin
input = stdin.readline
if __name__ == '__main__':
op_cnt = int(input())
sets = set()
for _ in range(op_cnt):
oper = list(input().split())
operation, operand = oper[0], 0
if len(oper) == 2:
operand = int(oper[1])
... |
82d35b1f595648d45986f0d94fe041baf8538e5b | MartinGildea/tutorial2_primeFactor_competition | /martingildea_tutorial2_competition.py | 981 | 3.875 | 4 | # primeFactors.py
# import test code
from lab2Test import speedTestFun
# --------------------------------
# primeFactors()
# input: A positive integer n (e.g. 19162234)
# output: A list of the prime factors of n with multiplicity (e.g. [2, 7, 7, 13, 13, 13, 89])
# method: Divide n by all possible integers in increasin... |
f28640f6e7c1f8ab046edd4b0ee495fc24716bac | HongsenHe/algo2018 | /036_valid_sudoku.py | 3,468 | 3.71875 | 4 | # 07282021
class Solution:
def isValidSudoku(self, board: List[List[str]]) -> bool:
rows = len(board)
cols = len(board[0])
# verify each row
for i in range(rows):
visited = set()
for j in range(cols):
if not self.valid(board, i, j, vis... |
854cc0ef95d03cb7b1f3d149ad445ef8a2b154ca | dnuffer/dpcode | /determine_if_one_string_is_a_permutation_of_another/python/start.py | 608 | 3.984375 | 4 | """
>>> are_permutations("abc", "cab")
True
>>> are_permutations("abc", "aab")
False
>>> are_permutations("abcd", "abc")
False
>>> are_permutations("", "")
True
>>> are_permutations("a", "a")
True
>>> are_permutations("aaa", "aaa")
True
>>> are_permutations("aaa", "aaaa")
False
>>> are_permutations("aab", "abb")
False
... |
65ea12ccd51b4bc5dd85300bf7a2ce962a20f704 | AndyAtari/PyPong | /pong.py | 2,806 | 3.8125 | 4 | import turtle
# window screen
window = turtle.Screen()
window.title("Pong by AndyAtari")
window.bgcolor("black")
window.setup(width=800, height=600)
window.tracer(0)
# player 1
paddle_one = turtle.Turtle()
paddle_one.speed(0)
paddle_one.shape("square")
paddle_one.shapesize(stretch_wid=5, stretch_len=1)
paddle_one.col... |
92821beaaca67c43ac128590d27098ea30b250ef | Cyntha-K/bioinfo-lecture-2021-07 | /0706/020.py | 53 | 3.5 | 4 | a = input('word1')
b = input('word2')
print(a + b)
|
b1a3712100763fe1b397d0bbe7624566e417b345 | choco9966/Algorithm-Master | /programmers/level2/code/조이스틱.py | 1,746 | 3.71875 | 4 | # ord를 이용해서 알파벳의 위치를 출력하고 순차적으로 진행하는게 빠른지, 반대로 진행하는게 빠른 지 계산
def cal_change_alphabet(solution):
diff = min(abs(ord(solution)-ord('A')), abs(26 - (ord(solution)-ord('A'))))
return diff
def change_alphabet(text, index):
text_ = list(text)
text_[index] = 'A'
text = ''.join(text_)
return text
def... |
a6629d987219ccf10ba4f10674bd7e890108c71b | 111110100/my-leetcode-codewars-hackerrank-submissions | /hackerrank/plusminus.py | 1,660 | 4 | 4 | '''
Given an array of integers, calculate the ratios of its elements that are positive, negative, and zero. Print the decimal value of each fraction on a new line with places after the decimal.
Note: This challenge introduces precision problems. The test cases are scaled to six decimal places, though answers with abs... |
b044ac8e0900ea8f23ccdf6b01f9f6c2b56640c7 | 15929134544/wangwang | /leetcode/14 最长公共前缀.py | 1,274 | 3.515625 | 4 | class Solution(object):
def longestCommonPrefix(self, strs):
"""
:type strs: List[str]
:rtype: str
"""
# # 思路一：
# # 空字符串
# if not strs:
# return ''
# for index in range(len(strs[0])):
# # 把第一个字符串的每个字母依次和后面每个字符串的字母做比较
#... |
1a8556a871fa5a328ee8a0c600da163cb3d8277b | onitonitonito/k_mooc_reboot | /tortoise_hare.py | 1,376 | 4 | 4 | """
# finde duplicated number
given an array nums containing n+1 integers where each inter is between
1 & n (inclusve)
prove that at least one duplicate num must exist. Assue that there is only
duplicate num but it could be repeated more than once.
find the duplicate one.
Example 1:
- Input : [1,2,4,2,2]
- Output: ... |
a5494c7da9b4739273c40e84b05daeb06c8e3ae3 | smitgabani/algorithms-python3 | /sorting/selection-sort.py | 323 | 3.71875 | 4 | def selsort(A):
r = len(A)
for x in range(r):
minimum = x
for y in range(x + 1, r):
if A[y] < A[minimum]:
minimum = y
(A[x], A[minimum]) = (A[minimum],A[x])
mylist = input('Enter the list values to be stored: ').split()
selsort(mylist)
print(mylist... |
5f3b816c0dad6a85f6f79ed2f49a928aff9dc750 | RianMarlon/Python-Geek-University | /secao7_colecoes/exercicios1/questao23.py | 740 | 4.125 | 4 | """
23) Ler dois conjuntos de números reais, armazenando-os em vetores
e calcular o produto escalar entre eles. Os conjuntos têm 5 elementos
cada. Imprimir os dois conjuntos e o produto escalar, sendo que o produto
escalar é dado por: x1 * y1 + x2 * y2 + ... + xn * yn
"""
lista1 = []
lista2 = []
for i in range(5):
... |
2d68a23199e80a85907d665fd0af9f248e37e9a4 | Sa4ras/amis_python71 | /km71/Likhachov_Artemii/3/Task7.py | 437 | 4 | 4 | print('This program calculate the count of hours and minutes after 00:00')
while True:
X = input('Input minutes: ')
try:
if int(X) >= 0:
h = int(X)//60
m = int(X)%60
print(h, 'hours', m, 'minutes')
break
else:
print('Try again... |
8a394b2e6973a1610215e78c524527c5e066bf60 | ElcimarSilva/InterfaceGraficaPy | /main.py | 620 | 4.09375 | 4 | from tkinter import *
def infos_no_print():
texto = "Esta é minha mensagem no método"
texto_do_metodo["text"] = texto
janela = Tk()
janela.title("Minha janela criada em Python")
# janela.geometry("400x400")
texto_orientacao = Label(janela, text="Clique aqui para ver infos do método")
texto_orientacao.grid(... |
04e01661aa2d540993e644027e1791859de1016e | JaberKhanjk/LeetCode | /Array Problems/Sort Array By Parity II.py | 547 | 3.5 | 4 | class Solution(object):
def sortArrayByParityII(self, A):
n = len(A)
odd = []
even = []
for each in A:
if each % 2 == 0:
even.append(each)
else:
odd.append(each)
final = [0]*n
... |
f04fc11d9f06c6deddca4bfc1d92e8f52739bd69 | code-mic/Sololearn-Python-Core | /Fibonacci.py | 163 | 4.15625 | 4 | num = int(input())
def fibonacci(n):
if n <= 1:
return n
else:
return fibonacci(n-1) + fibonacci(n-2)
for number in range(num):
print(fibonacci(number))
|
14b6b7cfc759cdcb0ce848f67e576ebcdfe27070 | DipendraDLS/Python_OOP | /14.Thread_&_Multithreading/13.Single_Tasking_Using_Thread.py | 572 | 3.671875 | 4 | '''
- When multiple tasks are executed by a thread one by one, then it is called single tasking.
'''
from threading import Thread
from time import sleep
class MyExam(Thread):
def solve_questions(self):
self.q1()
self.q2()
self.q3()
def q1(self):
print('Q1 Solved')
s... |
05f070ee2296e17882060d7350a55d9a2a50ba9f | hsnylmzz/Python-Learning | /pyt/set4.py | 241 | 3.96875 | 4 | def kume_dondur(str_kume: str):
if str_kume.startswith("{") and str_kume.endswith("}"):
kume_elemanlari = str_kume[1:len(str_kume) -1]
v = kume_elemanlari.split(", ")
print(v)
kume_dondur("{1, 2, 3, 17}")
|
2b269158d03bb71c784dadfb2c877b56e9208618 | ruchamahabal/TXP_Tech_Events | /codemapper/q1.py | 740 | 3.75 | 4 | def prime_list(l):
result=[]
flag = 0
for i in l:
for j in range(2,i):
if i%j!=0:
flag = 1
else:
flag = 0
break
if flag == 1 or i==2 :
if i not in result:
result.append(i)
return result
... |
c76c43f407f7df9ed3497049e720eda7ec797f47 | poojithayadavalli/String | /IMEI validation.py | 1,759 | 4.25 | 4 | """
International Mobile Equipment Identity (IMEI) is a number, usually unique, to identify mobile phones, as well as some satellite phones.
The IMEI (15 decimal digits: 14 digits plus a check digit) includes information on the origin, model, and serial number of the device.
The task is to validate IMEI in following ... |
16272970aa5433ed7481251935efb29d3e76b981 | sandeepkumar8713/pythonapps | /02_string/16_possible_space_in_string.py | 1,848 | 4.34375 | 4 | # https://www.geeksforgeeks.org/print-possible-inpStrs-can-made-placing-spaces/
# Question : Given a inpStr you need to print all possible strings that can be made by placing spaces
# (zero or one) in between them.
#
# Input: str[] = "ABC"
# Output: ABC
# AB C
# A BC
# A B C
#
# Question Type :... |
ade1dda398282ab6e380793e113a0186ccdd69a0 | eucalypto/bitesofpy | /47/password.py | 1,356 | 3.953125 | 4 | import string
PUNCTUATION_CHARS = list(string.punctuation)
used_passwords = set('PassWord@1 PyBit$s9'.split())
def validate_password(password):
if len(password) < 6:
return False
if len(password) > 12:
return False
# This is an alternative for characters that need only one appearance.
... |
f4948d57dcfb16839e16488fedd316b5ab8b4732 | j-d-0630/python | /a0323_binary_tree_doi.py | 3,424 | 3.984375 | 4 | """
Binary Tree
根付き二分木の表現
id left rightの情報が付与されたときに
節点の情報をその番号が小さい順に出力する
node id, parent , depth, heigth
Preorder Tree Walk 先行順巡回
根節点, 左部分木, 右部分木の順番
Inorder Tree Walk 中間順巡回
左部分木, 根節点, 右部分木の順番
Postorder Tree Walk 後行順巡回
左部分木, 右部分木, 根節点の順番
"""
import dataclasses
@dataclasses.dataclass
class ... |
490bfd37ebfa51bcf4eaf3019f0fdf6c542e5127 | titoeb/python-desing-patterns | /python_design_patterns/bridge/ex.py | 1,235 | 3.734375 | 4 | from abc import ABC
class Shape:
def __init__(self, renderer):
self.name = None
self.renderer = renderer
class Triangle(Shape):
def __init__(self, renderer):
super().__init__(renderer)
self.name = "Triangle"
def __str__(self):
return self.renderer.render_triangle... |
ccdf74f0435d63c7efe74177bcc8f34a4c7eae1b | anandshudda/dat-chapter-2 | /numpy_practice.py | 2,022 | 3.875 | 4 | import pandas as pd
import numpy as np
# 1. Use arange to create an array of the numbers from 0 to 100.
a = np.arange(101)
print "Array of the numbers from 0 to 100: ", '\n', a, '\n'
# 2. Create two arrays [1, 3, 5] and [2, 4, 6] and add them.
x = [1, 3, 5]
y = [2, 4, 6]
z = np.add(x,y)
print "Adding [1, 3, 5] and [... |
f1576b68a38ea069cae3dbf808f9e4a37863e00c | LisandroGuerra/zombies1 | /Lista3_desafio/questao05.py | 70 | 3.59375 | 4 | num = input("Entre com número inteiro positivo: ")
print (num[::-1])
|
29e9a3e5f31badc37a0c7d00dd0a7c3dcad510b2 | StylishSdp/ProgrammingTask | /Task2/栈/有效的括号.py | 800 | 4 | 4 | def isValid(s):
if not s:
return True
if len(s) % 2 != 0: #通过一些明显的条件去掉一些False的例子
return False
if s[0] == ')' or s[0] == ']' or s[0] == '}':
return False
if s[-1] == '(' or s[-1] == '[' or s[-1] == '{':
return False
stack = []
for i in range(len(s)):
if... |
fdcae381bf9aef293f664de508f1fe575bc250ce | simranmahindrakar/DAA-things | /gh - Copy.py | 570 | 3.625 | 4 | ##def mystery(l):
## if len(l) < 2:
#### return (l)
## else:
## return ([l[-1]] + mystery(l[1:-1]) + [l[0]])
runs = {"Test":{"Dhawan":[190,14,35,119],"Kohli":[3,103,13,42],"Pujara":[153,15,133,8]},"ODI":{"Dhawan":[37],"Kohli":[63]}}
#runs["ODI"]["Pujara"].extend([44])
#runs["ODI"]["Pujara"].a... |
4311e7c569bfb53fc8a127f9d56629f4a6671e35 | Jack-Sh/04_Math_Quiz | /08_greater_lesser.py | 1,363 | 4.15625 | 4 | import random
def choice_checker(question, valid_list, error):
valid = False
while not valid:
# Ask user for choice (and put in lowercase)
response = input(question).lower()
# iterates through list and if response is an item
# in the list (or the first letter of an item), the... |
ecbd30291b872f1099f59a37326443259464e816 | EmsDz/EmsDominoGame | /Clases/ClassHandPlay.py | 3,472 | 3.640625 | 4 |
#
class handPlay(object): # partida
"""docstring for handPlay"""
def __init__(self, players):
self.handPlayNumber = 0 # present hand play, is integer
self.players = players # players in this round, is a list
self.firstsTurn = '' # who play first
self.openGameToken = '' #... |
fff5cd69fc413a7bbd29ad8b5c76d9aff0b237b6 | Zabdielsoto/Curso-Profesional-Python | /7.- Unidad/Ejercicio2.py | 624 | 4.25 | 4 | '''Ejercicio 2: Clases
Modifique el método del ejercicio
anterior para convertirlo en una propiedad (@property)
'''
class Punto:
def __init__(self, x, y):
self.x = x
self.y = y
@property
def cuadrante(self):
if self.x > 0 and self.y > 0:
print("Primer cuadrante")
elif self.x < 0 and sel... |
850796d4461986e2e29f34a882786e899a7bad8a | brucemin/repertory | /exercise05/3.py | 776 | 3.75 | 4 | Number = int(input('请输入一个三位数：'))
#输出每一位上的数
a = int(Number/100)
b = int((Number-100*a)//10)
c = int(Number%10)
#判断a，b，c大小,并按从大到小排列,再从小到大排列
if a>b and a>c and b>c:
print(str(a) + str(b) + str(c))
print(str(c) + str(b) + str(a))
elif a>b and a>c and c>b:
print(str(a) +str(c) +str(b))
print(str(b) + str(c) ... |
f212ee7e13223eb60571b62499e56ceeef2d9bbd | Helton-Rubens/Python-3 | /exercicios/exercicio037.py | 780 | 3.9375 | 4 | num = int(input('Digite um número inteiro: '))
print('+='*10)
print('''\nEscolha uma das bases para conversão:
[\033[1;34m1\033[m] Converter para \033[4;35mBINÁRIO\033[m
[\033[1;34m2\033[m] Converter para \033[4;35mOCTAL\033[m
[\033[1;34m3\033[m] Converter para \033[4;35mHEXADECIMAL\033[m\n''')
opc = int(input('Sua op... |
ef91a079fb73758417e844ab90c3575104c746cd | ludwigwittgenstein2/Algorithms-Python | /copyThird.py | 988 | 3.828125 | 4 | #!/bin/Python
"""
Logging in Python
"""
import logging
def firstLog():
logging.basicConfig(level=logging.DEBUG)
logger = logging.getLogger(__name__)
logger.info('Start reading database')
#read database here
records = {'John':55, 'tom':66}
logger.debug('Records:%s', reco... |
32de8a183bb0b8f673d185544e8d83a2b9757e1e | jproddy/rosalind | /bioinformatics_stronghold/seto.py | 1,263 | 3.84375 | 4 | '''
Introduction to Set Operations
http://rosalind.info/problems/seto/
Given: A positive integer n (n≤20,000) and two subsets A and B of {1,2,…,n}.
Return: Six sets: A∪B, A∩B, A−B, B−A, Ac, and Bc (where set complements are taken with respect to {1,2,…,n}).
'''
filename = 'rosalind_seto.txt'
def union(A, B):
return... |
fd4c8a3f8de296ae8df949979ffd8c3602b97369 | wong2/PyOthello | /src/network.py | 1,292 | 3.75 | 4 | '''
network.py
This module defines class Server and Client. They are used in network games.
'''
import socket, sys, os
class Server:
def __init__(self):
self.socket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
port = 5543
self.socket.bind(('', port))
self.socke... |
6340c85c3136ea4615724301b87e5252fc67077e | klisostom/Python | /Projetos/decida_por_mim.py | 938 | 3.59375 | 4 | import random
"""
Objetivo: Crie um script que responda qualquer pergunta que for feita a ele. Recomendo ter
uma base de possíveis respostas (10-20 ou mais). Ex: Será que devo sair de casa hoje? Seu
script reponde: “Sim, vai lá!”
"""
conjunto_respostas = [
'Sim, vai lá!',
'Não, é estranho!',
'Talvez dê c... |
3bd64d83524d44d8229de0fc8be77de7f4c75201 | vitthalpadwal/Python_Program | /hackerrank/python_sort.py | 1,765 | 4.375 | 4 | """
You are given a spreadsheet that contains a list of athletes and their details (such as age, height, weight and so on). You are required to sort the data based on the th attribute and print the final resulting table. Follow the example given below for better understanding.
image
Note that is indexed from to , ... |
1c4f2dd4d21fbe5251870e40ad305a55e5f760b8 | jugu/projecteuler | /20160112_euler003.py | 497 | 3.765625 | 4 | '''The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of a given number N?'''
''' number of test cases '''
import math
Ts = raw_input('')
T = int(Ts)
caseN = list()
for i in range(T):
caseN.append(long(raw_input('')))
for N in caseN:
x = 2
while N%x == 0:
N = int(N... |
efcc50989912ee5f4dc8e8569ce6be68e68d4010 | Syyan7979/Kattis-Solutions-python- | /Python Kattis/Kattis124(drmmessages).py | 479 | 3.75 | 4 | import string
alphabet = string.ascii_uppercase
def message(someString):
counter = 0
out = ""
for i in someString:
counter += alphabet.index(i)
for i in someString:
out += alphabet[(alphabet.index(i) + counter)%26]
return out
def conversion(A, B):
out = ""
for i in range(len(A)):
out += alphabet[(alphabet.... |
5e113a23f68b740462a4d98cfbac6b6c233271aa | tchaitanya2288/pyproject01 | /dictionaries methods/set_update.py | 306 | 3.875 | 4 | ##6 Set Union with Mutation = a_set.update()
# Syntax: a_set.update(an_iter)
# Example:
s = set([1, 2, 3, 4, 5])
s.update(set([5, 6, 7]))
print(s)
#Output : set([1, 2, 3, 4, 5, 6, 7])
'''
Mutates a_set to be the union of set a_set and the set of
elements in iterable an_iter. Returns None.
'''
'' |
ac8c7716c9ed9da4842bb1afdacb7cbaa4f532f2 | youinmelin/practice2020 | /recurrence_03_maze.py | 3,215 | 3.71875 | 4 | from random import choice
import sys
sys.setrecursionlimit(3000) # set the maximum depth as 3000
# 0: wall, 1:road, 2:terminal, 3:wrong road, 4:passed road
class Maze():
def __init__(self,maze_list):
self.maze_list = maze_list
self.currentx = 1
self.currenty = 1
self.stepx = 0
... |
b30707789ef8429c866ac0190ed2c0edcbe2725f | zuohd/python-excise | /python_project/turple.py | 127 | 3.890625 | 4 | tuple1 = (1,)
print(tuple1)
tuple2 = (2, 3, 4, 5, 6)
print(tuple2[0])
print(tuple2[-1])
# tuple can not change :tuple2[0]=1
|
53cbd077e5f6f5da71d972e87223357fa8600dd0 | C-CCM-TC1028-111-2113/homework-1-JennyAleContreras | /assignments/22DigitosPares/src/exercise.py | 529 | 3.890625 | 4 | def main():
#escribe tu código abajo de esta línea
number = input('Dame un número: ')
pair = int
for i in range(0, 10):
pair = 0
if int(number[0])%2 == 0:
pair = pair + 1
if int(number[1])%2 == 0:
pair = pair + 1
if int(number[2])%2 == 0:
... |
072d3165615dcb15084e2f7dbc56ce6af3470317 | hn-on-fire/Python | /Basic Projects/Arithmetic Formatter/arithmatic_formater.py | 1,790 | 3.640625 | 4 | def arithmetic_arranger(questions, answers=False):
if len(questions) > 5:
return 'Error: Too many problems.'
print1, print2, print3, print4 = "", "", "", ""
for inpNums in questions:
nums = inpNums.split()
if nums[1] != '+' and nums[1] != '-':
return 'Error: Operator must... |
be6333704328b9c7c85bc00ebd34f5080b078800 | augustkx/defer-defaults | /defer_defaults/deferral.py | 798 | 3.5 | 4 | import inspect
from functools import wraps
deferred = object()
# An implementation of the deferrable args wrapper suggested by Jonathon Fine
def deferrable_args(func):
"""
Use this to decorate a function that you want to accept "defer" arguments, which will be overwritten by defaults.
:param func: A fun... |
30043b764335efa70034fe0d312166c8dcd3c3b8 | kmurphy/coderdojo | /07-Monkey_Wars/code/Monkey_Wars.py | 3,728 | 3.90625 | 4 | import turtle
import math
import random
screen = turtle.Screen()
screen.setup (width=600, height=600, startx=0, starty=0)
screen.setworldcoordinates(-300,-10,300,500)
WINDOW_WIDTH = 25
FLOOR_HEIGHT = 40
bob = turtle.Turtle() # draws the city
red = turtle.Turtle() # red player
blue = turtle.Turtle() # blue... |
15162c4c356ea7eb2439412d745c6c515903e684 | afrayz29/python-lessons | /exercises/11-Your-First-If/app.py | 249 | 3.890625 | 4 |
total = int(input('How much money do you have in your pocket\n'))
# YOUR CODE HERE
if total>100:
print("give me your money!")
elif total>50:
print("buy me some coffee your cheap!")
elif total<50:
print("you are a poor guy, go away!")
|
71501d9ebfa06c33e9ea9c6343be65a5e1785578 | Kostimo/IT2SCHOOL | /PythonLessons spring 2021/Week4/task3.py | 162 | 3.625 | 4 | A = [1,2,3,4,5,6,7,8,9,10]
m = int(input("M: "))
print(A)
def shift(lst, sh):
x = len(lst) - (sh%len(lst))
return lst[x:] + lst[:x]
print(shift(A, m))
|
cb9ef8e6c0a1e59a7b4b8ebcce88a6206de1c031 | Ravitejagupta/Python | /lc-1315.py | 1,207 | 3.53125 | 4 | # Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
from queue import deque
class Solution:
def sumEvenGrandparent(self, root: TreeNode) -> int:
visited = set()
if not root:
retu... |
f8985f583de1e83d9efe1ceb36142456edb9fac8 | safat/pratice | /src/codeforces/A281.py | 198 | 4 | 4 | def capitalize(input):
first_char = input[0]
if 'a' <= first_char <= 'z':
first_char = str(chr(ord(first_char) - 32))
return first_char + input[1:]
print(capitalize(input()))
|
f8d64dbaef27baf1b79034d0009de8a7745568b9 | deepakag5/Data-Structure-Algorithm-Analysis | /leetcode/binarytree_second_minimum.py | 2,128 | 3.953125 | 4 | # Time Complexity: O(N)+O(N) =O(N) , N is the total number of nodes in the given tree.
# We visit each node exactly once during dfs and then scan through visited
# Space Complexity: O(N)+O(N) = O(N), for info stored on stack and visited
def findSecondMin(root):
if root is None:
return None
# this is ... |
d2dd1371d54bde7afad3259405e38f00b6894005 | w22116972/leetcode_Ender | /520. Detect Capital/520.py | 869 | 3.640625 | 4 | class Solution(object):
def detectCapitalUse(self, word):
case1 = True
case2 = True
case3 = True
lower_case = 'abcdefghijklmnopqrstuvwxyz'
upper_case = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
if word[0] in upper_case:
case2 = False
for i in range(1, len(w... |
9a7fc625754d5f87356d1c95d0440f1a5fed5247 | nurarenke/study | /anagrams_diff_permutations.py | 1,308 | 3.90625 | 4 | '''Cracking the Coding interview, problem 1.3:
Given two strings, write a method to decide if one is a permutation of the other.
Whitespace and punctuation are significant and it is case sensitive.
>>> sorted_anagram("nura", "arun")
True
>>> count_anagram("nura", "arun")
True
>>> sorted_anagram("Burger ", "Burger"... |
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