blob_id stringlengths 40 40 | repo_name stringlengths 5 127 | path stringlengths 2 523 | length_bytes int64 22 3.06M | score float64 3.5 5.34 | int_score int64 4 5 | text stringlengths 22 3.06M |
|---|---|---|---|---|---|---|
dd7c7d610693962c38b03cad154d0bcedd243ad8 | smile0304/py_asyncio | /chapter11/thread_sync.py | 552 | 3.671875 | 4 | import threading
from threading import Lock,RLock
total = 0
lock = RLock()
def add():
global lock
global total
for i in range(1000000):
lock.acquire()
lock.acquire()
total += 1
lock.release()
lock.release()
def desc():
global lock
global total
for i in ra... |
7d22306762bb54a09f2d764a83bc8ed60ea8c706 | smile0304/py_asyncio | /chapter05/biset_test.py | 314 | 3.8125 | 4 | import bisect
#用来处理已排序的序列，用来维持已排序序列(升序)
#二分查找
inter_list = []
bisect.insort(inter_list,3)
bisect.insort(inter_list,2)
bisect.insort(inter_list,5)
bisect.insort(inter_list,1)
bisect.insort(inter_list,6)
print(bisect.bisect_left(inter_list,3))
print(inter_list) |
ae591c5ccda56d20dd75a06bbbec5ba51556efe0 | Hassiad/100-days-of-Code-Python | /20_21-Snake_Game/main.py | 1,236 | 3.796875 | 4 | # Day 20 and 21 Snake Game
from turtle import Turtle, Screen
from snake import Snake
from food import Food
from scoreboard import Scoreboard
import time
"""Screen setup"""
screen = Screen()
screen.setup(width=600,height=600)
screen.bgcolor("black")
screen.title("Timmy Snake Game")
screen.tracer(0)
"""Initial 3 segme... |
acf5a86f8be7589c70c6156e8935379ba7295a1d | akashsury/Operations | /Operations.py | 1,683 | 3.8125 | 4 | class Operations:
def __init__(self):
pass
def aboveBelow(self, values, comparison):
dict = {"above": 0, "below": 0} # created as local variable, each call it has to have zero
if not isinstance(comparison, int) or not isinstance(values, list):
return False... |
f0206ac8aa63ea9b17d861d3115932b014abcbfc | sensecollective/bioinformatics-coursera | /find_sequence.py | 524 | 3.578125 | 4 | #!/usr/bin/env python
import sys
import operator
#first argument is a single-line fasta input
seq_f = sys.argv[1]
f = open(seq_f)
#f.readline() #toss input line
seq = f.readline() #grab fasta
seq = seq.strip()
kmer = f.readline().strip() #grab kmer
f.close()
#second argument is the k-mer to find
#kmer = sys.argv[2... |
97001f35773373489d615f21fd25d2cba18a20ef | Nisarga-AC/python-for-mbas | /Part 1/strings.py | 720 | 4.0625 | 4 | # Strings are text surrounded by quotes
# Both single (' ') and double (" ") can be used
kanye_quote = ('My greatest pain in life is that I will never '
'be able to see myself perform live.')
print(kanye_quote)
# Switch to single quotes if your string uses double-quotes
hamilton_quote = 'Well, the word got around, th... |
1c1c65e73e9072be54e5428f4a9c4f95745658e9 | yyassif/My_Util_Scripts | /multi_threading.py | 1,450 | 3.578125 | 4 | import threading
import time
class MyThread(threading.Thread):
def __init__(self, threadID, name, counter):
threading.Thread.__init__(self)
self.threadID = threadID
self.name = name
self.counter = counter
def run(self):
print("Starting", self.name, "\n")
thread_lock.acquire()
print_time(s... |
ba84d228a663b9c9dce7dc8fc2916a55fca1305b | PrajaktaJadhav2/Problems-on-DS-and-Algos | /The_Celebrity_Problem.py | 1,421 | 4.03125 | 4 | # Problem Statement
# A celebrity is a person who is known to all
# but does not know anyone at a party.
# If you go to a party of N people, find if there is a celebrity in the party or not.
import numpy as np
def celebrity():
n = int(input("Enter the number of people"))
matrix = []
#Get the number of peo... |
da34092f0d87be4f33072a3e2df047652fb29cf3 | sudj/24-Exam3-201920 | /src/problem2.py | 2,658 | 4.125 | 4 | """
Exam 3, problem 2.
Authors: Vibha Alangar, Aaron Wilkin, David Mutchler, Dave Fisher,
Matt Boutell, Amanda Stouder, their colleagues and
Daniel Su. January 2019.
""" # DONE: 1. PUT YOUR NAME IN THE ABOVE LINE.
def main():
""" Calls the TEST functions in this module. """
run_tes... |
b2e763e72dd93b471294f997d2c18ab041ad665c | Evanc123/interview_prep | /gainlo/3sum.py | 1,274 | 4.375 | 4 | '''
Determine if any 3 integers in an array sum to 0.
For example, for array [4, 3, -1, 2, -2, 10], the function
should return true since 3 + (-1) + (-2) = 0. To make things simple,
each number can be used at most once.
'''
'''
1. Naive Solution is to test every 3 numbers to test if it is zero
2. is it possible to c... |
c40c9fcd27f8cf8f9ecf46cf7c576507450fb6e3 | Evanc123/interview_prep | /dynamic_programming/edit_distance.py | 2,307 | 3.96875 | 4 | # A Naive recursive Python program to fin minimum number
# operations to convert str1 to str2
def editDistance(str1, str2, m , n):
# If first string is empty, the only option is to
# insert all characters of second string into first
if m==0:
return n
# If second string is empty, the only op... |
408aa37f423d41ae5b724fbae02a5a92228ddfb0 | packetbuddha/SecretSanta | /secret_santa/secret_santa.py | 4,923 | 3.53125 | 4 | #!/usr/bin/env python
"""
Description: Plays traditional Secret Santa game, printing results to stdout,
to files or send via email.
Author: Carl Tewksbury
Last Update: 2019-11-27 - Py3-ize; remove unneeded passing of couples var;
email auth password now in file.
"""
# stdlib
imp... |
0664bf4c09687ee98fb8f81d11c6691ddaa67642 | CrazyCoder4Carrot/lintcode | /python/Reverse Linked List II.py | 1,155 | 3.890625 | 4 | """
Definition of ListNode
class ListNode(object):
def __init__(self, val, next=None):
self.val = val
self.next = next
"""
class Solution:
"""
@param head: The head of linked list
@param m: start position
@param n: end position
"""
def reverseBetween(self, head, m, n):... |
fa7f6f44b5a764f35e32120ecb1dba1595c16dd6 | CrazyCoder4Carrot/lintcode | /python/Subsets.py | 499 | 3.75 | 4 | class Solution:
"""
@param S: The set of numbers.
@return: A list of lists. See example.
"""
def subsets(self, S):
# write your code here
res = []
S.sort()
self.search(S, [], 0, res)
return res
def search(self, nums, temp, index, res):
if index == ... |
152113e43cceee7807ab807267ca54fb2a1d1c19 | CrazyCoder4Carrot/lintcode | /python/Search a 2D Matrix.py | 352 | 3.765625 | 4 | class Solution:
"""
@param matrix, a list of lists of integers
@param target, an integer
@return a boolean, indicate whether matrix contains target
"""
def searchMatrix(self, matrix, target):
# write your code here
for row in matrix:
if target in row:
... |
96df8f1ee63a40fde99d66d6218987cf119a12e6 | CrazyCoder4Carrot/lintcode | /python/Swap Two Nodes in Linked List.py | 1,319 | 3.953125 | 4 | # Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
# @param {ListNode} head, a ListNode
# @oaram {int} v1 an integer
# @param {int} v2 an integer
# @return {ListNode} a new head of singly-linked list
... |
5ebd4c91bd2fd1b34fbfdcff3198aef77ceb8612 | CrazyCoder4Carrot/lintcode | /python/Insert Node in a Binary Search Tree.py | 1,676 | 4.25 | 4 | """
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
"""
Revursive version
"""
class Solution:
"""
@param root: The root of the binary search tree.
@param node: insert this node into the binary search tree.
@return... |
d14d74df34371a880c57f2580e2f27cc9a35aee6 | CrazyCoder4Carrot/lintcode | /python/Insertion Sort List.py | 1,696 | 3.921875 | 4 | """
932ms version
insert node in the same list
"""
class Solution:
"""
@param head: The first node of linked list.
@return: The head of linked list.
"""
def insertionSortList(self, head):
# write your code here
cur = head
index = head
tempmax = head.val
while... |
1a5645ba8a8b640312045416a97fcabd7ff14e6e | CrazyCoder4Carrot/lintcode | /python/Rotate List.py | 918 | 3.96875 | 4 | # Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
# @param head: the list
# @param k: rotate to the right k places
# @return: the list after rotation
def rotateRight(self, head, k):
# write your ... |
0046fd9fa359ebfaa81b7fb40ecf2c5f6d278273 | sfagnon/stephane_fagnon_test | /QuestionA/QaMethods.py | 1,120 | 4.21875 | 4 | #!/usr/bin/python
# -*- coding: utf-8 -*-
#Verify if the two positions provided form a line
def isLineValid(line_X1,line_X2):
answer = True
if(line_X1 == line_X2):
answer = False
print("Points coordinates must be different from each other to form a line")
return answer
#Verify i... |
7a50c599822eba23ea3de2e98fe95a7fdc7c8152 | PanQnshT/Projekt_X | /ship.py | 1,149 | 3.53125 | 4 | import pygame
from pygame.math import Vector2
class Ship(object):
def __init__(self, game):
self.game = game
size = self.game.screen.get_size()
self.speed = 1.5
self.opor = 0.85
self.gravity = 0.5
self.pos = Vector2(size[0]/2,(size[1]/2)-(size[1]/4))
self.v... |
f65179cd3fa2f47d95a538cded047baaf18c4d1c | ares5221/python-Learning | /day04/MaxtRotation.py | 546 | 3.578125 | 4 | #!/usr/bin/env python
# -*- coding: utf-8 -*-
"""
__mtime__ = '2018/6/7'
"""
array=[[col for col in range(4)] for row in range(4)] #初始化一个4*4数组
for row in array: #旋转前先看看数组长啥样
print(row)
print('-----开始旋转矩阵----')
for i, row in enumerate(array):
for index in range(i, len(row)):
tmp = array[index][i] # get... |
e5a2135657bb8bf74063da10c85701b820cf386e | mc-cabe/CompareLists | /compareLists.py | 192 | 3.59375 | 4 | arr1 = []
arr1 = raw_input("please enter list1:")
arr2 = []
arr2 = raw_input("please enter list2:")
if arr1 == arr2:
print "arrays are the same"
else:
print "arrays are not the same"
|
f15c286da0a0de84375e3f71b88b3a2711e74180 | umemotsu/deimscripts | /generateVoteData.py | 1,724 | 3.578125 | 4 | #coding:utf-8
# generateVoteData
import csv, sys
# read participan data
def readParticipants(filename):
partList = {}
with open(filename,'rb') as csvfile:
reader = csv.reader(csvfile,delimiter=',',quotechar='"',quoting=csv.QUOTE_MINIMAL)
for row in reader:
person = {}
... |
06176687b50a2f17535536bf922dc6254b5c348d | artmukr/Homework_4 | /Homework_4.2.py | 167 | 3.515625 | 4 | link = input('Enter your link :')
link = link.replace('https://', "")
link=link.replace('www.', '')
value_of_start = link.rfind('/')
print(link[:value_of_start])
|
70d3fb39df6fd9c1e6f48569d0057d9c11c32dec | apotato369550/breadth-first-search-python | /main3.py | 4,142 | 3.796875 | 4 | # i think i get this
# x and y
# test it with another maze configuration
def create_maze():
return [
"O..#.",
".#.#.",
"....#",
"#...#",
"X.###"
]
def create_maze_2():
return [
"...O##X",
".###...",
".####..",
"...#...",
"..###... |
89df09062d2ccaefd9526ee559ee67bf9a837a39 | AmrinderKaur1/turtle-crossing-start | /car_manager.py | 2,061 | 3.640625 | 4 | from turtle import Turtle
import random
COLORS = ["red", "orange", "yellow", "green", "blue", "purple"]
STARTING_MOVE_DISTANCE = 5
MOVE_INCREMENT = 10
X = [-300, 300]
Y_AXES_LIST = [[120, 220], [20, 120], [-120, -20], [-220, -120]]
class CarManager:
def __init__(self):
self.all_cars = []
self.ca... |
e606335002d2e02b40b59538d2dc59aba70eed6f | ganjingcatherine/leetcode_python | /sort_by_leetcode/math/hard/780 reaching point.py | 1,487 | 3.59375 | 4 |
#!/usr/bin/python
# -*- coding: utf-8 -*-
"""
一开始的思路是逆向推，目的坐标(tx, ty)肯定是由(tx-ty, ty)或(tx, ty-tx)，但是坐标限定不为负数，因此比较tx、ty的大小可以唯一确定上一步的步骤
类似搜索的题，从终点往起点看，比较容易猜到正解。
这个题的关键点是，如果 ty > tx，那么必然是从 (tx, ty-tx)到达(tx, ty)这个状态的。
想明白上面，还需要知道一个优化的思路：取模不要一次一次相减
题目大意：
从点(x, y)出发经过一次移动可以到达(x + y, y)或者(x, x + y)
给定点(sx, sy)与(tx, ty)，判断(... |
23167d9292e97e858d70c4836fe3f6cb4f09fc5f | getoverlove/Majiang | /eye/32-不放回摸球直到摸到a白b黑球的比例.py | 1,309 | 3.609375 | 4 | from fractions import Fraction
from sympy import *
f_dict = {}
def f(m, n, a, b):
"""
m个白球，n个黑球，想要摸到a个白球，b个黑球
用递推式的方式计算准确结果，结果使用分数表示
"""
assert a >= 0 and b >= 0 and m >= 0 and n >= 0
assert a <= m and b <= n
param = (m, n, a, b)
if param in f_dict:
return f_dict[param]
if... |
c99ba21ebb47076b3fcefefc8d29e9ad6bcf50e3 | getoverlove/Majiang | /eye/40-多目标不放回摸球每个目标与总目标的关系.py | 2,083 | 3.796875 | 4 | """
当前局面到胡牌局面的差
D={d1,d2,d3......}
问题等价于：从一个袋子里面摸球，期望摸几次才能使pattern符合D中的任意一个。
例如2个白球3个黑球，摸到1个白球或者1个黑球时停止，此问题的答案会小于f(2,3,1,0)和f(2,3,0,1).显然此问题的答案会变成1.
麻将问题应该是一个多目标概率问题。
称此问题为：多目标摸球问题
"""
from fractions import Fraction
from copy import deepcopy
noput_dict = {}
def noput(m, n, a, b):
"""
m个白球，n个黑球，想要摸到a个白球，b个... |
d83e883c6595e1e5eb646f96a9f12635b81839c0 | shtakai/cd-python | /0502-am/assignment/scoresandgrades.py | 789 | 4.03125 | 4 | def scores_and_grades():
scores = []
for value in range(10):
score = raw_input("Input Score {}:".format(value + 1))
scores.append(score)
print "Scores and Grades"
for score in scores:
grade = ""
score = int(score)
if score >= 60 and score < 70:
# print... |
753165690888d3364f9a6bb9de23a4e24b148f0f | shtakai/cd-python | /oop/assignment/linkedlist/linkedlist.py | 2,725 | 4.0625 | 4 |
class Linkedlist(object):
def __init__(self, value):
self.value = value
self.prev_node = None
self.nxt_node = None
print 'initialized node', self
def insert_last(self, last_node):
if last_node:
last_node.nxt_node = self
self.prev_node = last... |
0d7c64b809aaf2495587c01e21f76792733ed566 | shtakai/cd-python | /0502-am/list.py | 356 | 3.65625 | 4 | ninjas = ['Rozen', 'KB', 'Oliver']
my_list = ['4' ,['list', 'in', 'a', 'list'], 987]
empty_list = []
print ninjas
print my_list
print empty_list
fruits = ['apple', 'banana', 'orange', 'strawberry']
vegetables = ['lettuce', 'cucumber', 'carrots']
fruits_and_vegetables = fruits + vegetables
print fruits_and_vegetables... |
efbe44be5d1ab0fd9f75f373ed7c7bf1ea2cdda7 | tt1998215/pythoncode1 | /begin end.py | 712 | 3.671875 | 4 | # import functools
# def log(func):
# # @functools.wraps(func)
# def wrapper(*args,**kw):
# print("begin call %s():"%func.__name__)
# res=func(*args,**kw)
# print("the result of the program is {}".format(res))
# print("end call %s():"%func.__name__)
# #resturn res
# r... |
cdaee67ad742ae35588e82fd5682391750e4f583 | tt1998215/pythoncode1 | /Mydatetime.py | 515 | 3.515625 | 4 | from datetime import datetime,timezone,timedelta
# print(datetime.now())
# utc_dt=datetime.utcnow().replace(tzinfo=timezone.utc)
# print(utc_dt)
# bj_dt=utc_dt.astimezone(timezone(timedelta(hours=8)))
# print(bj_dt)
import re
def to_timestmap(dt_str,tz_str):
input_dt=datetime.strptime(dt_str)
time_zone_num=re.m... |
22ca752af618837890c074dfe20776c44f40fece | matteblack9/infoRetrieval | /searchEngine5.py | 6,636 | 3.984375 | 4 |
"""
This file contains an example search engine that will search the inverted
index that we build as part of our assignments in units 3 and 5.
"""
import sys,os,re
import math
import sqlite3
import time
# use simple dictionnary data structures in Python to maintain lists with hash keys
docs = {}
resultslist... |
b24483771fe532d0641a9025558c07fab2559f75 | kushaswani/Insight_DE_Challenge | /src/consumer_complaints.py | 1,995 | 3.6875 | 4 | import csv
import sys
# from read_csv import read_csv
# importing read_and_transform_csv function which lets you apply custom functions to rows and get the transformed dataframe as dict of arrays
from read_csv import read_and_transform_csv
# importing the custom function which we need to apply to get intended results... |
c3625fd73193cbcbc96060b39e37b3b9a9104828 | Bn-com/myProj_octv | /myset_scripts/ttteeemmm.py | 97 | 3.515625 | 4 | abc = "ABCcdedDD"
print(abc.lower())
print(abc.swapcase())
print(abc.title())
print(abc.upper())
|
ad847ca987ab258b897d0d436a537863b9aeefd5 | yangzhaonan18/detect_traffic_light | /kmeans.py | 2,705 | 3.640625 | 4 | # -*- coding=utf-8 -*-
# K-means 图像分割 连续图像分割 （充分利用红绿灯的特点）
import cv2
import matplotlib.pyplot as plt
import numpy as np
def seg_kmeans_color(img, k=2):
# 变换一下图像通道bgr->rgb，否则很别扭啊
h, w, d = img.shape # (595, 1148, 3) # 0.47 聚类的时间和图像的面积大小成正比，即与图像的点数量成正比。
b, g, r = cv2.split(img)
img = cv2.merge([r, g... |
84b65a981960d7895747edd6e9090df0f34c71e7 | hoylemd/practice | /pattern_matching/pattern_matching.py | 837 | 3.890625 | 4 | import argparse
import operator
# set up the args parser
parser = argparse.ArgumentParser(description='This is a script to find all positions of a substring in a string')
parser.add_argument("file", help="The file to operate on")
args = parser.parse_args()
# open the file and read it
f = open(args.file)
input_lines =... |
78d4b7482374fd8a57c06b76b7474b0a373e1d3d | prakashdivyy/advent-of-code-2019 | /01/run.py | 566 | 3.546875 | 4 | import math
total_part_1 = 0
total_part_2 = 0
filename = 'input.txt'
with open(filename,) as file_object:
for line in file_object:
# Part 1
x_part_1 = int(line,10)
res_part_1 = x_part_1 / 3
total_part_1 += math.floor(res_part_1) - 2
# Part 2
x_part_2 = int(line,10)... |
1916ef020df5cb454e7887d2d4bb051d308887e3 | sammysun0711/data_structures_fundamental | /week1/tree_height/tree_height.py | 2,341 | 4.15625 | 4 | # python3
import sys
import threading
# final solution
""" Compute height of a given tree
Height of a (rooted) tree is the maximum depth of a node,
or the maximum distance from a leaf to the root.
"""
class TreeHeight:
def __init__(self, nodes):
self.num = len(nodes)
self.parent = nodes
... |
0723f91d492b9e11767bc6b5a019977ed0f3eb73 | kbtony/my-teleprompt | /src/teleprompter/run.py | 4,560 | 3.84375 | 4 | import csv
import sys
from datetime import datetime
from datetime import timedelta
from datetime import timezone
USAGE = f"usage: python {sys.argv[0]} file_name.csv datetime"
class CommandLine:
def __init__(self, argv):
# More than two cmd arguments
if len(argv) > 3:
print("error: at m... |
61dd1f00de80a02605fdab226828683769b6bf96 | araceli24/Ejercicios-Python | /Modificar.py | 667 | 3.5 | 4 | lista = [29, -5, -12, 17, 5, 24, 5, 12, 23, 16, 12, 5, -12, 17]
def modificar(l):
l = list(set(l)) # Borrar los elementos duplicados (recrea la lista a partir de un nuevo diccionario)
l.sort(reverse=True) # Ordenar la lista de mayor a menor
for i,n in enumerate(l): # Eliminar todos los números impa... |
95ef4825c2a198c3ebabcdb5a26e4328ab4bb4ed | xudalin0609/leet-code | /dynamic/generateTrees.py | 887 | 4.09375 | 4 | """
Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1 ... n.
Example:
Input: 3
Output:
[
[1,null,3,2],
[3,2,null,1],
[3,1,null,null,2],
[2,1,3],
[1,null,2,null,3]
]
Explanation:
The above output corresponds to the 5 unique BST's shown below:
... |
83176af41a32baa87bf1ce06b002e1d5c82865f4 | xudalin0609/leet-code | /BinarySearch/findPeek.py | 1,335 | 3.75 | 4 | #LintCode 75
"""
75. 寻找峰值
中文English
你给出一个整数数组(size为n)，其具有以下特点：
相邻位置的数字是不同的
A[0] < A[1] 并且 A[n - 2] > A[n - 1]
假定P是峰值的位置则满足A[P] > A[P-1]且A[P] > A[P+1]，返回数组中任意一个峰值的位置。
样例
样例 1:
输入: [1, 2, 1, 3, 4, 5, 7, 6]
输出: 1 or 6
解释:
返回峰顶元素的下标
样例 2:
输入: [1,2,3,4,1]
输出: 3
挑战
Time complexity O(logN)
注意事项
数组保证至少存在一个峰
如果... |
740a31e63edbde67c2add9e33219f5616c5af689 | xudalin0609/leet-code | /BFS/sequenceReconstruction.py | 2,434 | 3.828125 | 4 | """
605. 序列重构
中文English
判断是否序列 org 能唯一地由 seqs重构得出. org是一个由从1到n的正整数排列而成的序列，1<=n<=10^4 。
重构表示组合成seqs的一个最短的父序列 (意思是，一个最短的序列使得所有 seqs里的序列都是它的子序列).
判断是否有且仅有一个能从 seqs重构出来的序列，并且这个序列是org。
样例
例1:
输入:org = [1,2,3], seqs = [[1,2],[1,3]]
输出: false
解释:
[1,2,3] 并不是唯一可以被重构出的序列，还可以重构出 [1,3,2]
例2:
输入: org = [1,2,3], seqs = [[1,2]]... |
8178aa5e0cc682bec6b3935fa7eadf2c1c9bb3fa | xudalin0609/leet-code | /Greedy/canCompleteCircuit.py | 1,549 | 3.5625 | 4 | # LintCode 187
"""
187. 加油站
中文English
在一条环路上有 N 个加油站，其中第 i 个加油站有汽油gas[i]，并且从第_i_个加油站前往第_i_+1个加油站需要消耗汽油cost[i]。
你有一辆油箱容量无限大的汽车，现在要从某一个加油站出发绕环路一周，一开始油箱为空。
求可环绕环路一周时出发的加油站的编号，若不存在环绕一周的方案，则返回-1。
样例
样例 1:
输入:gas[i]=[1,1,3,1],cost[i]=[2,2,1,1]
输出:2
样例 2:
输入:gas[i]=[1,1,3,1],cost[i]=[2,2,10,1]
输出:-1
挑战
O(n)时间和O(1)额外空间
... |
ae456f697158e53e2a6a806a6344757247fd1e23 | xudalin0609/leet-code | /Greedy/canPlaceFlowers.py | 1,176 | 4 | 4 | # 605
"""
假设你有一个很长的花坛，一部分地块种植了花，另一部分却没有。可是，花卉不能种植在相邻的地块上，它们会争夺水源，两者都会死去。
给定一个花坛（表示为一个数组包含0和1，其中0表示没种植花，1表示种植了花），和一个数 n 。能否在不打破种植规则的情况下种入 n 朵花？能则返回True，不能则返回False。
示例 1:
输入: flowerbed = [1,0,0,0,1], n = 1
输出: True
示例 2:
输入: flowerbed = [1,0,0,0,1], n = 2
输出: False
注意:
数组内已种好的花不会违反种植规则。
输入的数组长度范围为 [1, 20000]。
n 是非负整... |
e3b9ee8b7a645084e7430185c65db3e59d097718 | xudalin0609/leet-code | /DFS/wordSearchII.py | 2,763 | 3.84375 | 4 | # LintCode 132
"""
132. 单词搜索 II
中文English
给出一个由小写字母组成的矩阵和一个字典。找出所有同时在字典和矩阵中出现的单词。
一个单词可以从矩阵中的任意位置开始，可以向左/右/上/下四个相邻方向移动。一个字母在一个单词中只能被使用一次。且字典中不存在重复单词
样例
样例 1:
输入：["doaf","agai","dcan"]，["dog","dad","dgdg","can","again"]
输出：["again","can","dad","dog"]
解释：
d o a f
a g a i
d c a n
矩阵中查找，返回 ["again","can","dad","dog... |
014009ae4f7e48fe843dd5dc02104ffbf887fb52 | xudalin0609/leet-code | /BFS/numIslands.py | 2,072 | 3.8125 | 4 | # LintCode 433
"""
433. 岛屿的个数
中文English
给一个 01 矩阵，求不同的岛屿的个数。
0 代表海，1 代表岛，如果两个 1 相邻，那么这两个 1 属于同一个岛。我们只考虑上下左右为相邻。
样例
样例 1：
输入：
[
[1,1,0,0,0],
[0,1,0,0,1],
[0,0,0,1,1],
[0,0,0,0,0],
[0,0,0,0,1]
]
输出：
3
样例 2：
输入：
[
[1,1]
]
输出：
1
"""
import collections
class Solution:
"""
@param grid: a boolean 2D ... |
310be24ce75fad3d656397284eaf5fbbc9f70784 | xudalin0609/leet-code | /DivideAndConquer/flatten.py | 1,974 | 4.03125 | 4 | # LintCode 453
"""
453. 将二叉树拆成链表
中文English
将一棵二叉树按照前序遍历拆解成为一个 假链表。所谓的假链表是说，用二叉树的 right 指针，来表示链表中的 next 指针。
样例
样例 1：
输入：{1,2,5,3,4,#,6}
输出：{1,#,2,#,3,#,4,#,5,#,6}
解释：
1
/ \
2 5
/ \ \
3 4 6
1
\
2
\
3
\
4
\
5
\
6
样例 2：
输入：{1}
输出：{1}
解释：
"""
"""
Defi... |
5e0b1671e4a00faa1c0481ea461ffaa089c30d5e | xudalin0609/leet-code | /DivideAndConquer/findKthLargest.py | 1,560 | 4 | 4 | """
Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.
Example 1:
Input: [3,2,1,5,6,4] and k = 2
Output: 5
Example 2:
Input: [3,2,3,1,2,4,5,5,6] and k = 4
Output: 4
Note:
You may assume k is always valid, 1 ≤ k ≤... |
6cca45c1943ad0fa138ae0e5dcf331aa54aba5b1 | xudalin0609/leet-code | /BFS/searchNode.py | 2,303 | 4.09375 | 4 | """
618. 搜索图中节点
中文English
给定一张 无向图，一个 节点 以及一个 目标值，返回距离这个节点最近 且 值为目标值的节点，如果不能找到则返回 NULL。
在给出的参数中, 我们用 map 来存节点的值
样例
例1:
输入:
{1,2,3,4#2,1,3#3,1,2#4,1,5#5,4}
[3,4,5,50,50]
1
50
输出:
4
解释:
2------3 5
\ | |
\ | |
\ | |
\ | |
1 --4
Give a node 1, target is 50
there a hash named values which ... |
b4c38e0f4f0446c2bb67bdb844f51ce3263732c7 | xudalin0609/leet-code | /DivideAndConquer/closestValue.py | 2,799 | 3.9375 | 4 | # LintCode 900
"""
900. 二叉搜索树中最接近的值
中文English
给一棵非空二叉搜索树以及一个target值，找到在BST中最接近给定值的节点值
样例
样例1
输入: root = {5,4,9,2,#,8,10} and target = 6.124780
输出: 5
解释：
二叉树 {5,4,9,2,#,8,10}，表示如下的树结构：
5
/ \
4 9
/ / \
2 8 10
样例2
输入: root = {3,2,4,1} and target = 4.142857
输出: 4
解释：
二叉树 {3,2,4,1}，表示... |
d49a28a9653203145baf58365b4c7e4861e6b01b | ryan-aday/adayR | /20_cw/app.py | 1,675 | 3.65625 | 4 | from functools import reduce
from collections import Counter
import itertools
import operator
def most_common(L):
# get an iterable of (item, iterable) pairs
SL = sorted((x, i) for i, x in enumerate(L))
# print 'SL:', SL
groups = itertools.groupby(SL, key=operator.itemgetter(0))
# auxiliary function to get... |
4cc92252d90ac9677998ec3669b3bd7a35bb8d10 | alxthedesigner/cloud_application_example | /word_number_app.py | 1,368 | 3.84375 | 4 | from flask import Flask
from flask import request
application = Flask(__name__) #App Instance created
@application.route("/") #When URL ends in '/', execute method
def home():
return WORDHTML
WORDHTML = """
<html><body>
<h2>Word and Number Game</h2>
<form action="/game">
E... |
c5ac3685195f48a4da04cd71e2ed8202c375ac28 | Teghfo/python_workhosp_zemestan98_mapsa | /J6/my_pickle.py | 1,021 | 3.71875 | 4 | import pickle
class Book:
def read_data(self):
self.title = input('Enter Book tile: ')
if not self.title:
return
self.author = input('Enter Author name: ')
self.page_count = input('Enter page count: ')
self.borrow_history = []
def add_borrow_history(self, u... |
853c4b8d92dd31f672cdc1f15cb0deedc852c961 | desperadoespoir/TheoriesDesGraphes-LangagesNaturels | /tp2/automate.py | 1,655 | 3.921875 | 4 | class Automate(object):
"""Une class automate qui permet de lire des solutions
on peut y ajouter une regle
on peut aussi resoudre avec la fonction solve et un mot de passe
"""
END = -1
NOT_FOUND = 0
START = ""
def __init__(self, name):
self.name = name
self.regles... |
1e26f26d4d308a9a9ac1c8f9975b459b945ce315 | Htrams/Leetcode | /binary_search_tree_iterator.py | 1,101 | 4.03125 | 4 | # My Rating = 2
# https://leetcode.com/explore/challenge/card/december-leetcoding-challenge/570/week-2-december-8th-december-14th/3560/
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right... |
b07d3973639dd0b42409027ff1ed2e31aace33cf | Htrams/Leetcode | /matrix_diagonal_sum.py | 621 | 3.9375 | 4 | # My Rating = 1
# https://leetcode.com/contest/biweekly-contest-34/problems/matrix-diagonal-sum/
# Given a square matrix mat, return the sum of the matrix diagonals.
# Only include the sum of all the elements on the primary diagonal and all the elements on the secondary diagonal that are not part of the primary diagon... |
f02f36b8bab69dcaef3240ae84600a3092f720ae | Htrams/Leetcode | /flipping_an_image.py | 854 | 3.875 | 4 | # My Rating = 1
# https://leetcode.com/explore/challenge/card/november-leetcoding-challenge/565/week-2-november-8th-november-14th/3526/
# Given a binary matrix A, we want to flip the image horizontally, then invert it, and return the resulting image.
# To flip an image horizontally means that each row of the image is ... |
6ab534dbbc70e049b062fd0f93c43f578531e8fa | Htrams/Leetcode | /psuedo-palindromic_paths_in_a_binary_tree.py | 1,489 | 3.859375 | 4 | # My Rating = 1
# https://leetcode.com/explore/challenge/card/december-leetcoding-challenge/573/week-5-december-29th-december-31st/3585/
# Given a binary tree where node values are digits from 1 to 9. A path in the binary tree is said to be pseudo-palindromic
# if at least one permutation of the node values in the pa... |
d7b224fc94c116eb466349f05f2657f0cd0fc27a | Htrams/Leetcode | /reverse_string.py | 308 | 3.5 | 4 | # My Rating = 1
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
length=len(s)
for i in range(length//2):
temp=s[i]
s[i]=s[length-i-1]
s[length-i-1]=temp |
7bb08a8b4b5963ff4fdf46f7a62e30959e74aac0 | Htrams/Leetcode | /swapping_nodes_in_a_linked_list.py | 1,676 | 3.71875 | 4 | # My Rating = 2
# https://leetcode.com/contest/weekly-contest-223/problems/swapping-nodes-in-a-linked-list/
# The following code interchanges the nodes, instead of the values in the node.
# You are given the head of a linked list, and an integer k.
# Return the head of the linked list after swapping the values
# of t... |
c14defaa309c4ab8868ae7c756bd89aeeb5bf678 | grsind19/Pythondeveloper | /2. Learning Python/11. Fileaccess.py | 256 | 3.53125 | 4 | def main():
# f = open('textfile.txt','w+')
# for i in range(10):
# f.write("this is line "+str(i) + "\r\n")
g = open('textfile.txt','r')
if g.mode == 'r':
contents = g.read()
print(contents)
if __name__ == "__main__":
main() |
3ef54f4dcf0751f1bc2d4be3bc387910c2e0330d | grsind19/Pythondeveloper | /1. Programmingfoundationsandalsogithms/10. Quicksort.py | 840 | 3.734375 | 4 | items=[6, 20, 8, 19, 56, 23, 87, 41, 49, 53]
def quickSort(dataset, first, last):
if first < last:
pivotrIdx = partition(dataset, first, last)
quickSort(dataset, first, pivotrIdx-1)
quickSort(dataset, pivotrIdx+1, last)
def partition(dataset, first, last):
pivotValue = dataset[first]
lower = first... |
fb2e544ffcccaacad953cfcba321bf757af6ecef | grsind19/Pythondeveloper | /5. PythonstandardLib/8. Type.py | 167 | 3.5625 | 4 | # type
# Instance
class Car:
pass
class Truck(Car):
pass
c = Car()
t = Truck()
print(type(c))
print(type(t))
print(isinstance(c,Car))
print(isinstance(t,Car)) |
683673bcb322193b98900c5dd93ec07b14f5eb8a | tron135/python_practise | /even odd.py | 137 | 4.21875 | 4 | number = int(input('Enter any number: '))
if(number%2 == 0):
print('The Number is even.')
else:
print('The Number is odd.')
|
c5d87c6d8317f501cb862fb0a8bc30b16868e759 | tron135/python_practise | /unlucky number.py | 202 | 3.875 | 4 | for i in range(1,21):
if i == 4 or i == 13:
s = str(i) + " is unlucky !!!!"
print(s)
elif i % 2 == 0:
t = str(i) + " is even !!"
print(t)
else:
u = str(i) + " is odd !!"
print(u) |
adc39c22e430ce632ebcb644797ef2addd9f8553 | hyunlove12/20191005 | /dlearn/digits/__init__.py | 1,242 | 3.671875 | 4 | import matplotlib.pyplot as plt
from digits.hand_writing import HandWriting
if __name__ == '__main__':
def print_menu():
test = ['asdfsadf', 'asdgsad', 'asdf']
print(test)
#print(test[:3, 1])
t = [2 == 2]
print(t)
print('0, EXIT')
print('1, 손글씨 인식')
pr... |
7a75cd6d39e5329b1226ddf851478f341cc163bc | m-hollow/deeper_dungeons | /old_files (archive)/main.py | 45,747 | 3.671875 | 4 | import time
from random import randint
from random import choice
import copy
# define game objects
class GameSettings():
"""Define game settings"""
def __init__(self):
self.grid_size = 5
self.starting_gold = 10
self.difficulty = 1
def user_settings_change(self):
msg = 'Choose the size of the dungeon gri... |
810cdf77367004f7764c0de8e6af34d7253011fd | m-hollow/deeper_dungeons | /old_files (archive)/determine_event.py | 1,074 | 3.53125 | 4 | from random import randint
from random import choice
import time
def determine_event(current_room):
"""look at current room type and determine event that results"""
if current_room == 'Empty':
pass
if current_room == 'Monster':
monster_event(settings, player)
if current_room == 'Treasure':
treasure_event(pl... |
3f40d22f79ba711e9ff154e109338d58804f2256 | m-hollow/deeper_dungeons | /old_files (archive)/dungeon_more_old/ui_functions.py | 3,901 | 3.765625 | 4 | import time
from random import randint, choice
import copy
# define game UI functions.
def you_are_dead(player, text=''):
if text:
print(text)
print('YOU', end='', flush=True)
time.sleep(0.5)
print('\tARE', end='', flush=True)
time.sleep(0.5)
print('\tDEAD.', end='', flush=True)
print('So passes {} the {} ... |
b1080291935da258c1711e9980f70ab10dbe28cc | m-hollow/deeper_dungeons | /old_files (archive)/encounter_work.py | 1,846 | 3.90625 | 4 | from random import randint
from random import choice
import time
def user_input():
msg = '> '
choice = input(msg)
return choice
def press_enter():
msg = '...'
input(msg)
def run_attempt():
escape_num = 10 # this will be determined by mosnster difficulty in actual game
print('Press enter to roll your Run ... |
566276fb5aa5e56e446404520131bff5b19a6a91 | arumsey93/py-class | /pizzajoin.py | 624 | 3.640625 | 4 | class Pizza:
def __init__(self):
self.size = 0
self.style = ""
self.new_topping = []
def add_topping(self, topping):
self.new_topping.append(topping)
def print_order(self):
space = " and "
new_string = f"I would like a {self.size}-inch, {self.style} pizza"
... |
e442af8261e751871962eaedc86d8afc61156284 | irina-baeva/algorithms-with-python | /algorithms/big-o-complexity/constant_linear-complexity.py | 507 | 3.96875 | 4 | # Constant complexity example
# O(1) - the same time no matter how much input
def sum(a, b, c):
return a + b + c
print(sum(3,2,1))
# takes the same time as:
print(sum(1000,300,111))
# Linear complexity example
# O(n) - grows with the input 1:1
foodArray = ['pizza', 'burger', 'sushi', 'curry']
def findFoodInThe... |
7cafa9fc6b348af6d2f11ad8771c5cca59b1bea7 | irina-baeva/algorithms-with-python | /data-structure/stack.py | 1,702 | 4.15625 | 4 | '''Imlementing stack based on linked list'''
class Element(object):
def __init__(self, value):
self.value = value
self.next = None
class LinkedList(object):
def __init__(self, head=None):
self.head = head
def append(self, new_element):
current = self.head
... |
12e37d46b3b3b1dbcac643020f6faa1c28202d70 | amitesh1201/Python_30Days | /day1_ex03.py | 193 | 4.28125 | 4 | # Calculate and print the area of a circle with a radius of 5 units. You can be as accurate as you like with the value of pi.
pi = 3.14
radius = 5
area = pi * radius ** 2
print("Area: ", area)
|
c022ad739dbd4634b682b04f4036a18014b6283f | amitesh1201/Python_30Days | /OldPrgms/Day4_Ex2.py | 437 | 3.921875 | 4 | # 2. Use the input function to gather information about another movie. You need a title, director’s name, release year, and budget.
movies = [
("The Room", "Tommy Wiseau", "2003", "$6,000,000")
]
title = input("Title: ")
director = input("Director: ")
year = input("Year of release: ")
budget = input("Budget: ")
# m... |
358227b3759aa81b96ea6b1e4be87e066be878ee | amitesh1201/Python_30Days | /day2_ex01.py | 184 | 4.21875 | 4 | # Ask the user for their name and age, assign theses values to two variables, and then print them.
name = input("Enter your name: ")
age = input("Enter your age: ")
print(name, age)
|
01ece2001beaf028b52be310a8f1df24858f4e59 | amitesh1201/Python_30Days | /OldPrgms/Day3_prc05.py | 579 | 4.375 | 4 | # Basic String Processing : four important options: lower, upper, capitalize, and title.
# In order to use these methods, we just need to use the dot notation again, just like with format.
print("Hello, World!".lower()) # "hello, world!"
print("Hello, World!".upper()) # "HELLO, WORLD!"
print("... |
36ab5e0899e94be5d9ad720ace9341e4866a541c | amitesh1201/Python_30Days | /OldPrgms/Day4_Ex6.py | 374 | 3.890625 | 4 | # 6) Print both movies in the movies collection.
movies = [
("The Room", "Tommy Wiseau", "2003", "$6,000,000")
]
title = input("Title: ")
director = input("Director: ")
year = input("Year of release: ")
budget = input("Budget: ")
new_movie = title, director, year, budget
print(f"{new_movie[0]} ({new_movie[2]})")
m... |
f138eb288de47bb1d2262157d93e1802f0e1252e | bolyb/CS361 | /down_up.py | 744 | 3.765625 | 4 | class therapist:
name = ""
rank = ""
__init__(self, set_name):
name = set_name
rank = 0
#Discription:
#the votetally function is the crux of the task, the rest is used to showcase what the functionality is supposed to do.
#This function takes in a user vote and a therapist to be voted on and manipulates the ... |
6b17d9aa0fb29c58334ed4cf199d972a62706301 | Atabuzzaman/AI-code-with-Python | /waterJug.py | 432 | 3.671875 | 4 | def pour(jug1, jug2):
print(jug1,'\t', jug2)
if jug2 is 2:
return
elif jug2 is 4:
pour(0, jug1)
elif jug1 != 0 and jug2 is 0:
pour(0, jug1)
elif jug1 is 2:
pour(jug1, 0)
elif jug1 < 3:
pour(3, jug2)
elif jug1 < (4 - jug2):
pour(0, ... |
980610302335d57bf25946c94f4bd2c7cc0eb97f | skidmarker/bugatti | /hash/42578.py | 264 | 3.515625 | 4 | def solution(clothes):
answer = 1
c_dict = {}
for c, type in clothes:
if type in c_dict:
c_dict[type] += 1
else:
c_dict[type] = 1
for cnt in c_dict.values():
answer *= cnt + 1
return answer - 1
|
2417d550b5488eaec6800336beda0cbd8748a747 | nataliaruzhytska/Password-generator | /src/app/api_pass/utils.py | 2,597 | 3.59375 | 4 | import random
import string
from flask import jsonify
def random_chars(n=12, uppercase=False, symbol=False, digit=False):
"""
The function generates a random password using user-defined parameters
and evaluates its difficulty
:param n: str
:param uppercase: bool
:param symbol: bool
:param... |
f6c183fb7de25707619a330224cdfa841a8b44be | maryliateixeira/pyModulo01 | /desafio005.py | 229 | 4.15625 | 4 | # Faça um programa que leia um número inteiro e mostre na tela o seu sucessor e antecessor
n= int(input('Digite um número:'))
print('Analisando o número {}, o seu antecessor é {} e o sucessor {}' .format(n, (n-1), (n+1)))
|
13307a93bfccb71f40106daf37b2851a994ff070 | maryliateixeira/pyModulo01 | /desafio011.py | 534 | 4.09375 | 4 | #faça um programa que leia a altura e a largura de uma parede em metros, calcule sua area e a quantidade de tinta
# necessária para pintá-la, sabendo que cada litro de tinha, pinta uma area de 2 metros quadrados
larg= float(input('Digite a largura da parede em metros:'))
alt = float(input('Digite a altura da parede em ... |
47ba0dd3e658857ff57767aa6b70bba17abce29e | ssiddoway/IT567 | /port.py | 5,137 | 3.609375 | 4 | #!/usr/bin/python
import sys, argparse, socket
def main():
parser = argparse.ArgumentParser()
parser.add_argument("-i","--input", help="Input file")
parser.add_argument("-t", help="TCP port sccaner range, i.e. -t 1-1024")
parser.add_argument("-u", help="UDP port scanner range, i.e. -u 1-1024")
par... |
067bba31d0f51150264914637f39224fe6d0cecd | renol767/python_Dasar | /DQLab/Kelas Persiapan/chartnamakelurahan.py | 589 | 3.640625 | 4 | import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
table = pd.read_csv("https://storage.googleapis.com/dqlab-dataset/penduduk_gender_head.csv")
table.head()
x_label = table['NAMA KELURAHAN']
plt.bar(x=np.arange(len(x_label)), height=table['LAKI-LAKI WNI'])
#Memberikan nilai axis dari data CSV
plt.x... |
f52d016e2684b8d6cfceb2366bd27331b4ddd71b | renol767/python_Dasar | /Latihan Python Dasar/35.3 Latihan Mengecek Suatu Password.py | 360 | 3.625 | 4 | def userx():
user = input("Masukan Username : ")
if user == 'admin':
print("Lanjut Gan, Masukin Passwordnya")
pw = input("Masukan Password : ")
if pw == 'admin':
print("Login Sukses !")
else:
print("Password salah!")
else:
print("Salah Coy User... |
2d7bfd605e1105eb79716bb0b93357760887bd4a | renol767/python_Dasar | /Latihan Python Dasar/Tutorialsebelumpackage/11 Return and Lambda.py | 489 | 3.6875 | 4 | # return
a = 8
def kuadrat (a):
total = a**2
print("Kuadrat dari",a,"adalah",total)
return total
a = kuadrat(a)
print (a)
print (50*'-')
# return multiple argument
def tambah (a,b):
total = a+b
print(a, "+" , b , '=', total)
return total
def kali (a,b):
total = a*b
print(a, "x" , b , "="... |
047ca459460989c8c06d76cf16c341cee0e838a4 | renol767/python_Dasar | /Latihan Python Dasar/35.4 Latihan Deret Fibonaci.py | 154 | 3.953125 | 4 | suku = int(input('Masukan Suku Ke Berapa : '))
a = 0
b = 1
c = 3
print(a)
print(b)
while c <= suku:
c = a+b
a = b
b = c
print(c)
c+=1
|
3e3618fc47c05008bf92518dcd953d350c265918 | avsingh999/rl-tictactoe | /tictactoe.py | 10,944 | 4.0625 | 4 | """
Reference implementation of the Tic-Tac-Toe value function learning agent described in Chapter 1 of
"Reinforcement Learning: An Introduction" by Sutton and Barto. The agent contains a lookup table that
maps states to values, where initial values are 1 for a win, 0 for a draw or loss, and 0.5 otherwise.
At every mov... |
a3fb45777567b8e29bb0b31fb734bcd1dc8579ca | mikeyhu/project-euler-python | /test/test_Problem1.py | 374 | 3.8125 | 4 | import unittest
from main import Problem1
class TestProblem1(unittest.TestCase):
def test_sum_of_natural_numbers_below_10_that_are_multiples_of_3_or_5(self):
p1 = Problem1.Problem1()
self.assertEqual(23,p1.solve(10))
def test_sum_of_natural_numbers_below_1000_that_are_multiples_of_3_or_5(self):
p1 = Problem1... |
f003866db45687ef25b32772941b73413ae552cf | mikeyhu/project-euler-python | /main/Problem5.py | 447 | 3.5 | 4 | import itertools
class Problem5(object):
def solve(self,number):
uniqueProducts = self.products(number)
for x in itertools.count(1):
valid = True
for i in uniqueProducts:
if not x % i == 0:
valid = False
if valid:
return x
def products(self,number):
result = []
for i in range(number,... |
8d19c27a48eef36d5924f38be89fee9a362d95bc | igorbustinza/theegg_ai | /tarea23/tarea23.py | 7,606 | 3.703125 | 4 | ##Tarea 23
import random
##funciones
##función que cifra y descifra el texto
def cifra_descifra(cifraTextoDescifra,listaCartasFuncion,operacion):
#creamos un diccionario con las letras y su correpondiente valor
valoresLetras = {"A":1,"B":2,"C":3,"D":4,"E":5,"F":6,"G":7,"H":8,"I":9,"J":10,"K":11,"L":12,"M":13,"N":14... |
b9151241f8234f5c1f038c733a5d0ff46da376d3 | louishuynh/patterns | /observer/observer3.py | 2,260 | 4.15625 | 4 | """
Source: https://www.youtube.com/watch?v=87MNuBgeg34
We can have observable that can notify one group of subscribers for one kind of situation.
Notify a different group of subscribers for different kind of situation.
We can have the same subscribers in both groups.
We call these situations events (different kinds o... |
bb1e695e901d9f80fd97ac45ad64891229f2d79c | timmichanga13/python | /fundamentals/fundamentals/demos/data_types.py | 1,411 | 4.03125 | 4 | # many of the same types; booleans, values, strings
# python interpreter is an environment where the computer can immediately implement code
# use for testing/remembering if something works
# booleans start with capital T and F
#values are split between integers and floats
# floats are imperfect representations
# s... |
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