blob_id string | repo_name string | path string | length_bytes int64 | score float64 | int_score int64 | text string |
|---|---|---|---|---|---|---|
63bcc3398aa5a360cb9c2cb6b9a81ebf44c3a0ab | chxj1992/leetcode-exercise | /55_jump_game/_1.py | 1,008 | 3.71875 | 4 | import unittest
from typing import List
class Solution:
def canJump(self, nums: List[int]) -> bool:
"""
Timeout!
"""
cache = {}
def recursion(index: int):
if index in cache:
return cache[index]
if len(nums[index:]) == 1 or not nums[i... |
5748c6b5d7a4fd16e633d70728212ba3853109d3 | chxj1992/leetcode-exercise | /169_majority_element/_2.py | 651 | 3.65625 | 4 | import unittest
from typing import List
class Solution:
def majorityElement(self, nums: List[int]) -> int:
l = len(nums) // 2
nums.sort()
prev = nums[0]
count = 0
for i in nums:
if i == prev:
count += 1
else:
prev = i
... |
ca3f41a8b2e0ead234da83522cec6712f0f93ba3 | chxj1992/leetcode-exercise | /49_group_anagrams/_1.py | 1,539 | 3.765625 | 4 | import unittest
from typing import List
class Solution:
def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
"""
Time: O(n^2)
Space: O(1)
Timeout!
"""
res = []
while len(strs) > 0:
curr = strs.pop(0)
row = [curr]
i... |
a40a82211b52554af5b14c1674bb22a80939b619 | chxj1992/leetcode-exercise | /subject_lcof/29/_1.py | 1,156 | 3.890625 | 4 | import unittest
from typing import List
class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
if len(matrix) == 0:
return []
curr = (0, 0)
left, right, top, bottom = 0, len(matrix[0]) - 1, 0, len(matrix) - 1
res = [matrix[0][0]]
while rig... |
5e208e7700e6cfab63cf67e6e6ebe2f3a50e036b | chxj1992/leetcode-exercise | /190_reverse_bits/_1.py | 353 | 3.515625 | 4 | import unittest
class Solution:
def reverseBits(self, n: int) -> int:
b = bin(n)[2:]
b = '0' * (32 - len(b)) + b
return int(b[::-1], 2)
class Test(unittest.TestCase):
def test(self):
s = Solution()
self.assertEqual(964176192, s.reverseBits(43261596))
if __name__ ==... |
769a79c8ad14b52843a548f46897ceeff73a43d1 | chxj1992/leetcode-exercise | /subject_lcof/22/_1.py | 976 | 3.9375 | 4 | import unittest
# Definition for singly-linked list.
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
prev = head
i = 1
while head:
if i > k:
p... |
0cf283edd673daeff56b6ed2e4211c3c292665d6 | chxj1992/leetcode-exercise | /198_house_robber/_2.py | 574 | 3.78125 | 4 | import unittest
from typing import List
class Solution:
def rob(self, nums: List[int]) -> int:
dp_list = [0]
for i, x in enumerate(nums):
if i < 2:
dp_list.append(max(nums[:i + 1]))
else:
dp_list.append(max(dp_list[i], dp_list[i - 1] + x))
... |
c4619837564f4be4a25a01ed14789b1ca5ea657d | GeekTemo/Python_Algorithms | /sort/merge_sort.py | 614 | 3.859375 | 4 | __author__ = 'gongxingfa'
def merge(arr, lo, mid, hi):
for i in range(lo, hi):
aux[i] = arr[i]
i = lo
j = mid + 1
k = lo
while k <= hi: # or (i < mid or j < hi)
if i > mid:
arr[k] = aux[j]
j += 1
elif j > hi:
arr[k] = aux[i]
... |
2a079f90c7e359d61aca111a7d3d8a6cb29349a1 | axiomiety/exercism-io | /python/clock.py | 455 | 3.765625 | 4 |
class Clock(object):
def __init__(self, hours, minutes):
self.minutes = minutes % 60
self.hours = (hours + minutes // 60) % 24
def add(self, minutes):
m = (self.minutes + minutes) % 60
h = (self.hours + (self.minutes + minutes) // 60) % 24
return Clock(h,m)
def __repr__(self):
return '... |
7280cc558f163f80889c0ed161b31fb34e89f48b | Kaustav-Byte/Assignments | /Untitled.py | 413 | 3.90625 | 4 | #!/usr/bin/env python
# coding: utf-8
# In[11]:
for i in range (2000, 3000,1):
if i%7== 0 and i/5!=0:
print (i,',',end="")
else:
continue
# In[15]:
def my_function(x):
return x[::-1]
mytxt = my_function(input('Enter your First name= '))
myend =my_function(input('Enter your Second ... |
21e0efc165d96630552f72405d8c99c61dcdc252 | green-fox-academy/dzsofa | /PythonPractice/0104/linearsearch.py | 204 | 3.84375 | 4 | numbers = [4, 5, 6]
def linear_search(my_list, sought):
for number in my_list:
if number == sought:
return my_list.index(number)
return -1
print(linear_search(numbers, 6))
|
c137c714f180a04d18ee4427edd9021ffcf1fa73 | green-fox-academy/dzsofa | /PythonPractice/1211/PrintBigger.py | 393 | 4.125 | 4 | # Write a program that asks for two numbers and prints the bigger one
number1 = int(input("Give me the first number: "))
number2 = int(input("Give me the second number"))
def printbigger(number1, number2):
if number1 > number2:
print(number1)
elif number2 > number1:
print(number2)
else:
... |
445b08a82edb6952c0e4d7eff6a56d2f0491a0dd | penguinsss/interface | /requests库/请求/get请求/getBaiduRequestParams.py | 496 | 3.640625 | 4 | """
Requests库是用Python编写的，基于urllib开发的工具，能够进行HTTP协议的接口测试和调试工作，使用简单，功能强大，完全满足HTTP测试需求；
"""
import requests
# 3种方式访问百度搜索，params的值为一个字符串或字典
resp = requests.get('http://www.baidu.com/S?wd=嗯哼')
resp1 = requests.get('http://www.baidu.com/S', params="wd=杨幂")
resp2 = requests.get('http://www.baidu.com/S', params={"wd": "秋刀鱼"})
... |
915c64980a5d689f4f1333ebdc3116f59f231c74 | penguinsss/interface | /requests库/事务.py | 1,244 | 3.8125 | 4 | """
事务提交机制：
自动提交：
1 建立连接设置autocommit=True来设置自动提交
2 建立连接后，通过conn.autocommit(True)来设置
手动提交：
1 conn.rollback 进行回滚
2 conn.commit 进行提交
"""
import pymysql
conn, cursor = None, None
try:
conn = pymysql.connect('localhost', 'root', 'root', 'books') # utf8, utf-8不行
cursor = co... |
3d3e9db12ff25265fe6994218ba6052cc6f9568a | kvsaijayanthkrishna/Python_Programming | /2_character_frequency.py | 404 | 4.40625 | 4 | #2. Write a Python program to count the number of characters (character frequency) in a string.
#Sample String : google.com'
#Expected Result : {'o': 3, 'g': 2, '.': 1, 'e': 1, 'l': 1, 'm': 1, 'c': 1}
string=input("enter a string\t:")
letter=""
print("output:-")
for s in string:
if s not in letter:
... |
8afb9e8b2813f3c14161304f4786948f20a5015f | kvsaijayanthkrishna/Python_Programming | /addition of two numbers.py | 177 | 4.03125 | 4 | #. Write a program to add 2 numbers. Accept the input in a single line.
n1,n2=(input("enter 2no's\t:")).split()
n1=int(n1);n2=eval(n2)
print(n1,"+",n2,"=",n1+n2)
print()
|
2ccd3e41c313528e0fb0eb75f79d969bdcdcce16 | kvsaijayanthkrishna/Python_Programming | /sum_of_first_n_nos.py | 153 | 4.0625 | 4 | #sum of first n numbers
n=int(input("enter a number\t:"))
sum=0
for i in range(1,n+1):
sum+=i
print("sum of {0} numbers is {1}".format(n,sum))
|
f9a50021eca47f4d830454f08a46dfc80d2afd6f | kvsaijayanthkrishna/Python_Programming | /5_long_word.py | 406 | 4.28125 | 4 | #5. Write a Python function that takes a list of words and returns the length of the longest one.
string=input("enter list of words with spaces\t:")
words_list=string.split()
length=len(words_list[0])
longest=""
for word in words_list:
if len(word)>length:
length=len(word)
longest=word
prin... |
2ce0060c14988f64d14dcd84a6014994cce06384 | kvsaijayanthkrishna/Python_Programming | /7_remove_odd_index.py | 259 | 4.34375 | 4 | #7. Write a Python program to remove the characters
#which have odd index values of a given string.
string=input("enter a string\t:")
new_string=""
i=0
for s in string:
if(i%2==0):
new_string+=s
i=i+1
print("new string:-",new_string)
|
d81b3164e349450e7d9daf6c0543c85e927b037a | kvsaijayanthkrishna/Python_Programming | /triangle2.py | 129 | 3.78125 | 4 | m=int(input("enter size\t:"))
for i in range(m):
for i in range(m-i,0,-1):
print("*",end=" ")
print()
|
5df716b35bcabe42bec9830de5b6cb76aeaaaad0 | kvsaijayanthkrishna/Python_Programming | /string_without_vowels_for.py | 269 | 4.34375 | 4 | #lec-19/slide-8
#Write a program to accept a string from the user and display it vertically
#but don’t display the vowels in it using for loop.
string=input("enter a string\t:-")
for ch in string:
if ch in "aeiou":
continue
print(ch)
print()
|
0331f1354dc9537c01f4294d4d6ae40fd630fe05 | victorhook/mqtt-broker | /src/utils/security.py | 4,410 | 3.578125 | 4 | import binascii
import hashlib
import getpass
import os
import sys
CREDENTIALS_FILE = 'passwd'
BASE_DIR = os.path.join(os.path.dirname(sys.argv[0]), 'etc')
CREDENTIALS_PATH = os.path.join(BASE_DIR, CREDENTIALS_FILE)
PASSWORD_LIMIT = 4
def hash_password(password):
""" hashes a given password with sha256 and r... |
f5078fcf3bcf7f46e168969a33c6c334e17685cf | Vasia228/hello | /Сдать/BinTree.py | 5,599 | 3.703125 | 4 | class node():
def __init__(self,data,left,right,mother):
self.Data=data
self.Left=left
self.Right=right
self.Mother=mother
def inputNode(root,data):
curCheck=root
while(1):
if data>curCheck.Data:
if curCheck.Right==0:
curCheck.... |
d5beac50f1ed6a986a635201e8745aaf55ea117d | pavel-malin/pavel | /buttondot.py | 375 | 3.640625 | 4 | from tkinter import *
root = Tk()
var = IntVar()
rbutton1 = Radiobutton(root, text='1', variable=var, value=1)
rbutton2 = Radiobutton(root, text='2', variable=var, value=1)
rbutton3 = Radiobutton(root, text='3', variable=var, value=1)
rbutton1.pack()
rbutton2.pack()
rbutton3.pack()
root.mainloop()
"""
point select... |
e2fcf0917844232594be77ed02d59d72a217ea57 | MechaEdgar/PracticasPython | /hello.py | 246 | 4.09375 | 4 | print ("Hola mundo!")
hola =input("hola ¿cual es tu nombre? ")
print ("mucho gusto en conocerte " + hola)
edad = input("Cual es tu edad: ")
print ("Entonces tu nombre es: " +hola+ " Y tu edad es: "+ edad)
a= "a"
print ("Esto es una letra " + a)
|
3446ddab720a3a900616a85172dce04e87dc5621 | Schmidty88/ICP3 | /Source/File2.py | 418 | 4.3125 | 4 | #This function is used for counting vowels then
def VowelCount(Sentence):
#this set contains all the vowels we will be using if
#it sees a letter in the set it will add to the counter
Vowel = set("aeiou")
counter = 0
for letter in Sentence:
if letter in Vowel:
counter += 1
print("Vow... |
edd9b8b1f15497f1dc1c246027a5c294cf9d0566 | kevinmolina-io/GrokkingInterview | /Two_Pointer/tripletSumCloseToTarget.py | 1,834 | 4.09375 | 4 | def triplet_sum_close_to_target(arr, target_sum):
"""
HIGH LEVEL:
This is a similar approach to triplet sum, with a little twist.
You want to use a two pointer approach to solve it efficiently.
Here's how it goes:
Variables:
closest_sum
global_difference
left, right... |
f968395c2a1f85a5121e66d80bb0b52a80012fc0 | kevinmolina-io/GrokkingInterview | /Sliding_Window/permutationInAString.py | 2,018 | 3.984375 | 4 | def find_permutation(str, pattern):
"""
HIGH LEVEL:
Create a hashmap that keeps track of frequency count of each letter in pattern.
You want to use a sliding window technique, and keep expanding the window until you reach
len(pattern), at that point you need to shrink the window
In bet... |
ab3de7e5c050ac72c931c3f039cf479a54350cd8 | kevinmolina-io/GrokkingInterview | /Sliding_Window/longestSub_sameLetters_replacement.py | 1,198 | 4.125 | 4 | def length_of_longest_substring(str, k):
"""
HIGH LEVEL:
This is another sliding window problem.
The trick to these problems where they ask to find the longest substring AFTER REPLACEMENT,
is to keep track of the max repeating character.
If the difference between the current substring and... |
76ad5a738f26b25f109febcc2777a489d5d41d18 | sophiaperson/ConLingo | /syntax.py | 752 | 3.5 | 4 | # Sophia Ho
# Sentence Class
# andrewID: swho
# Recitation: P
class Sentence(object):
def __init__(self, writtenSent, pronuncation, meaning):
self.writtenSent = writtenSent
self.pronunciation = pronunciation
self.meaning = meaning
def getHashables(self):
return (self.writtenSent... |
06ea4f9e9cb4c0912cae59fe0ae4ed2eb730b65e | BenjaminAage/Kattis_Problems | /Level_1.3/greetings.py | 179 | 3.75 | 4 |
greeting = input()
returnGreeting = ""
count = 0
for i in greeting:
if i == "e":
returnGreeting += "ee"
else:
returnGreeting += i
print(returnGreeting)
|
734f5cffc4e86a353170af1f5217c3bb0435ded0 | BenjaminAage/Kattis_Problems | /Level_1.4/spavanac.py | 307 | 3.796875 | 4 | time = input()
Hours, Minutes = time.split(" ")
if int(Minutes) <= 44:
if int(Hours) != 0:
Hours = int(Hours) - 1
Minutes = int(Minutes) + 15
else:
Hours = 23
Minutes = int(Minutes) + 15
else:
Minutes = int(Minutes) - 45
print(str(Hours) + " " + str(Minutes))
|
c3db0feccf0bab678a1e4aa816e26cf72f5f3903 | BenjaminAage/Kattis_Problems | /Level_1.3/qualityAdjustedLifeYear.py | 153 | 3.53125 | 4 | N = int(input())
count = 0.0
for i in range(N):
qaly = input()
num1, num2 = qaly.split(" ")
count += float(num1) * float(num2)
print(count) |
20d04fa9553f7cdb8951580c57c431daa24c0f62 | BenjaminAage/Kattis_Problems | /Level_1.4/alphabetSpam.py | 426 | 4.03125 | 4 | sentence = input()
whitespace = 0.0
lowercase = 0.0
uppercase = 0.0
symbols = 0.0
for i in sentence:
if i == "_":
whitespace += 1
elif i.islower():
lowercase += 1
elif i.isupper():
uppercase += 1
else:
symbols += 1
print(float(whitespace / len(sentence)))
print(float(... |
2adc0f98b406a943027c44cf721477151bdb25cb | BenjaminAage/Kattis_Problems | /Level_1.4/acmContestScoring.py | 674 | 3.671875 | 4 |
correct = {}
incorrect = {}
problem = ""
solved = 0
time = 0
while problem != "-1":
problem = input()
if len(problem) < 3:
break
problem = problem.split(" ")
key = problem[1]
if problem[2] == "right":
solved += 1
if key in incorrect:
value = incorrect.get(key... |
c5f748a26f2ed3b68c9d3e1fe826ce372d855de6 | tong800/ICS-32 | /project0.py | 366 | 3.53125 | 4 | user = int(input())
if user < 1000:
i = 1
x = " "
print("+-+")
print("| |")
if user > 1:
print("+-+-+")
elif user ==1:
print ("+-+")
while i != user:
print(x*i + "| |")
if i+1 != user:
print(x*i + "+-+-+")
else:
... |
8fa1324afa4f9d46559a5997f5c90b56fe947e79 | jpch89/effectivepython | /ep015_nonlocal.py | 2,203 | 4.25 | 4 | # 把 numbers 中出现在 group 里面的数字放在前面
def sort_priority(values, group):
def helper(x):
if x in group:
return (0, x)
return (1, x)
values.sort(key=helper)
numbers = [8, 3, 1, 2, 5, 4, 7, 6]
group = {2, 3, 5, 7}
sort_priority(numbers, group)
print(numbers)
"""
[2, 3, 5, 7, 1, 4, 6, 8]
"""
... |
ed7035dec39bf19299ac149805a2be8aa829ecc1 | jpch89/effectivepython | /ep028_collectionsabc.py | 616 | 3.84375 | 4 | class FrequencyList(list):
def __init__(self, members):
super().__init__(members)
def frequency(self):
counts = {}
for item in self:
counts.setdefault(item, 0)
counts[item] += 1
return counts
# 继承于 list 的类拥有 list 提供的全部标准功能
foo = FrequencyList(['a', 'b', ... |
b345d66b818e6d3816b3cad8b19b77ae135df6c1 | jpch89/effectivepython | /ep008_nolongcomprehensions.py | 1,396 | 3.71875 | 4 | # 使用两级列表推导展开矩阵
# 推导顺序是从左到右
matrix = [[1, 2, 3],
[4, 5, 6],
[7, 8, 9]]
flat = [x for row in matrix for x in row]
print(flat)
print('-' * 50)
# 对矩阵每个元素求平方，组成新矩阵
squared = [[x ** 2 for x in row] for row in matrix]
print(squared)
print('-' * 50)
# 假如是三维列表，列表推导要拆成几行才好看
# 不推荐这种写法，因为没有比嵌套循环更清晰方便
my_lists... |
e8013f373923e2f9dbb39515196b7279b9587fc2 | yyoonca/Python | /hackerrank/BuiltIn/G.py | 1,335 | 3.796875 | 4 | '''
@author: simon.park
version 1
'''
N=int(input())
students = []
lowest = 100
for i in range(N):
lst = []
x = input()
y = float(input())
lowest = min(lowest,y)
lst.append(x)
lst.append(y)
students.append(lst)
# It will creat another list which remove the students having the lowest grade... |
0893c94c67b71eb46f6781df5e33b28f3747310d | sidvanvliet/py-fibonacci-sequence | /main.py | 315 | 3.671875 | 4 | times = int(input('Enter the sequence amount (e.x.: 19): '))
print("Writing out " + str(times) + " Fibonacci sequences:\n- - -")
previous_y = 1
previous_z = 2
for time in range(times):
seq_answer = (previous_y + previous_z)
previous_y = previous_z
previous_z = seq_answer
print(str(seq_answer))
|
9f32fb1e545ff04eed882ce994985a1415a01ef7 | myusf01/100-days-of-code | /R1D2/edx_interst.py | 1,510 | 3.8125 | 4 | """
author= Muhammed Yusuf
bln: stands for balance
aIR: for Annual Interest Rate
mPr: for Monthly Payment Rate
"""
### Version 0.1
def interest(bln, aIr, mPr):
def clc(bln, aIr, mPr):
for i in range(12):
monthlyInterstRate = aIr / 12.0
minMonthlyPayment = mPr * bln
mo... |
6f51919565828d00d6fdb14ae8bf292a07949522 | mallison/herdcats | /herdcats/players.py | 3,690 | 3.734375 | 4 | """Players (owners and cats)."""
from . import tube
def create(number):
"""Returns list of cats and owners positioned at random stations."""
owner_and_cats = []
for i in xrange(number):
owner_and_cats.append(_create())
return owner_and_cats
def move(owners_and_cats, turn):
"""Returns lis... |
188e698e8700061f8661b234de6312ce1fd39ab3 | msinghnanhre/Python-Coding-Challenges | /hackerRank/timeConversion.py | 281 | 3.875 | 4 | timeString = "07:05:45PM"
def timeConversion(s):
if s[-2:] == "AM" and s[0:2] == "12":
return("00" + s[2:-2])
elif s[-2:] == "PM" and s[0:2] == "12" or s[-2:] == "AM":
return(str(s[:-2]))
elif s[-2:] == "PM":
return(str(int(s[:2])+12)+s[2:-2])
|
e9db3680392dcf98ef17c701c46f6365e87ddccf | robertperimov/amis_python71 | /km71/Perimov_Robert/mylabs/3/task1.py | 198 | 4.125 | 4 | print("this program will add together 3 numbers")
x = int(input("enter first number "))
y = int(input("enter second number "))
z = int(input("enter third number "))
answer = x + y + z
print(answer)
|
a82225eb7ccb800305ecbe88ab91fed75b17ca5a | ldc84/python-ex | /basic/20.logic.py | 532 | 3.640625 | 4 | # def return_false():
# print('함수 return_false')
# return False
# def return_true():
# print('함수 return_true')
# return True
# print('테스트1')
# a = return_false()
# b = return_true()
# if a and b:
# print(True)
# else:
# print(False)
# print('테스트2')
# if return_false() and return_true(): # 단락평가
# print... |
cf7a0debe3c0675cdfd2c0ac62f247a53be5f05b | ldc84/python-ex | /basic/16.tuple_packing.py | 200 | 3.671875 | 4 | a, b = 1, 2
print(a, b)
c = (3,4)
print(c)
d, e = c
print(d, e)
f = d, e
print(f)
x = 5
y = 10
print(x, y)
x, y = y, x
print(x, y)
def tuple_func():
return 1, 2
q, w = tuple_func()
print(q, w) |
edf10b96c29c5d23ab8b295612994e72fd709136 | noraibraheem/Hangman_project | /hangman.py | 2,657 | 3.90625 | 4 | import random
def hangman():
word = random.choice([
"pugger", "littlepugger", "tiger", "superman", "thor", "pokemon",
"avengers", "savewater", "earth", "annable"
])
turns = 10
while turns > 0:
print("guess the word:", "-" * len(word))
guess = input()
if guess =... |
256e512e252e02cb2d4648c8056477a9b35e968d | teamneem/projecteuler | /euler2.py | 489 | 4 | 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 term... |
a3f797e5bd75d6c2c556a4b6dea813ec773740a8 | droy78/Leetcode | /63.UniquePaths2.py | 2,953 | 3.703125 | 4 | class Solution(object):
#You can reach any cell either from the cell above it and to its left. However, if the cell has
#a value 1 that is a blocked cell. Similar to problem no. 62-Unique paths, we create another grid
#called table where each cell will hold the no. of ways of reaching it. If the obstablegri... |
c894619a7072dbe1f8234d572fae0f14c8c4f347 | sauravsingh243/IISc-LAP-Coding-Assignment | /solution.py | 1,343 | 3.890625 | 4 | import random
class Solution:
def __init__(self):
pass
def MontyHall(self, N):
ans = [0.0, 0.0, 0.0, 0]
# Write your code here. Please do not change the return statement.
# Update ans[0], ans[1], ans[2] and ans[3] as mentioned in the question.
numberOfDoors = 3
for i in range(0,N):
... |
c23321e9ee8c8541b0d4dc662288167806a2aeae | GeuberLucas/linguagem_python | /python/classes/Alunos.py | 1,809 | 3.890625 | 4 | class Alunos():
nome=None
prova1=None
prova2=None
trabalho=None
matricula=None
nota_max_prova1=None
nota_max_prova2=None
peso_trabalho=None
situacao=None
nota_max_trab=None
nota=None
nota_max=None
def __init__(self,nota_max_prova1,nota_max_prova2,nota_max_trab,peso_t... |
49680d296d916bb5ca7cf7dfc188b5e564c76c5b | GeuberLucas/linguagem_python | /python/herança/exec1_her/Aluno.py | 558 | 3.765625 | 4 | from Pessoa import Pessoa
class Aluno (Pessoa):
matricula: int = None
curso = None
def setmat(self):
self.matricula = int(input('insira a matricula: '))
def getmat(self):
return self.matricula
def setcurso(self):
self.curso=input('insira o curso do aluno: ')
def get... |
33c991eada4d781694798864dbd5c287eb751c1e | AvaniVerma/Python-codes | /profanity_editor.py | 692 | 3.71875 | 4 | import urllib
def read_text():
#Give the absolute path of the file you want to check for profanity as input for open
input_file = open("C:\Users\Avani\Desktop\Python udacity\Input_for_profanity_editor.txt")
input_text = input_file.read()
#print(input_text)
input_file.close()
profanity_check(inp... |
4768933db52051acade495c26dd96e9e185dc458 | ajlongcoy21/StoreInventory | /product.py | 3,580 | 3.59375 | 4 | import datetime
from peewee import *
# Define Product
db = SqliteDatabase('inventory.db')
# Define Product
class Product(Model):
product_id = PrimaryKeyField()
product_name = CharField(max_length=255, unique=True)
product_qty = IntegerField(default=0)
product_price = IntegerField(default=0)
dat... |
75ac11613e8843d21d34d5ab3f1388780051d032 | LegendKrazy/Learning | /Project Euler/pe6.py | 772 | 3.625 | 4 | # Project_Euler_Problem_6
# The sum of the squares of the first ten natural numbers is,
# 12 + 22 + ... + 102 = 385
# The square of the sum of the first ten natural numbers is,
# (1 + 2 + ... + 10)2 = 552 = 3025
# Hence the difference between the sum of the squares of the first ten natural numbers and
# the squar... |
ada7d3aa25694537e9dcf3ca68d28333697e99b7 | archanasheshadri/Python-coding-practice | /queueclass.py | 1,481 | 4.21875 | 4 | class Queue(object):
"""A Queue is a set of integers
The value is represented by a list of ints, self.vals.
Each int in the set occurs in self.vals exactly once."""
def __init__(self):
"""Create an empty set of integers"""
self.vals = []
def insert(self, e):
"""Assumes e is... |
98615a7a5289609427243b88b0bd1821f020530c | archanasheshadri/Python-coding-practice | /alphabetsub.py | 2,176 | 4.3125 | 4 | #author Archana
'''
Write a program that prints the longest substring of s in which the letters occur in alphabetical order. For example, if s = 'azcbobobegghakl',
then your program should print
Longest substring in alphabetical order is: beggh
In the case of ties, print the first substring. For example, if s = 'abcbc... |
565f18341cb930b461515aec81a526098d13f341 | archanasheshadri/Python-coding-practice | /squareroot.py | 539 | 3.8125 | 4 | x = 23
epsilon = 0.01
step = 0.1
guess = 0.0
while abs(guess**2-x) >= epsilon:
if guess <= x:
guess += step
else:
break
if abs(guess**2 - x) >= epsilon:
print 'failed'
else:
print 'succeeded: ' + str(guess)
#Second approach--- runs into infinite loop
#x = 25
#epsilon = 0.01
#step =... |
e16c4918c005be136424c46af4ba19978b2c5aac | konrei/lecture2 | /dictionaries.py | 89 | 3.546875 | 4 | ages = {"Samet": 18, "Eren": 21}
ages["Ömer"] = 21
ages["Samet"] += 1
print(ages)
|
f0477b3e615e9ae843a7c89e72f8689c27b0e672 | mkrotos/Algorithms | /algorithms/extra/dijkstra_algorithm.py | 2,490 | 3.578125 | 4 | from .graph import *
class Road:
def __init__(self, weight: float, parents: dict = None, finish=None):
self._weight = weight
self._parents = parents
self._path = self.find_path(parents, finish)
@staticmethod
def find_path(parents, finish):
path = []
node = finish
... |
4eb1c05afc2e11b94ad1ac5c2609d0cf82912ae5 | mkrotos/Algorithms | /test/test_quicksort.py | 546 | 3.65625 | 4 | from unittest import TestCase
from algorithms.quicksort import quicksort
class Test(TestCase):
def test_single_element_list(self):
# given & when
actual = quicksort([1])
# then
self.assertEqual([1], actual)
def test_sort_list(self):
# given & when
actual = qui... |
1bcc8610e5fb62d1ecee7097b9a6720e46568a18 | JaeZheng/jianzhi_offer | /07.py | 1,199 | 3.90625 | 4 | """
题目描述:
大家都知道斐波那契数列，现在要求输入一个整数n，请你输出斐波那契数列的第n项（从0开始，第0项为0）。
n<=39
解决思路:
递归,但是代入公式复杂度是2的n次方，会超时。所以改用数组存储或者是两个临时变量存储，时间换空间
"""
# -*- coding:utf-8 -*-
# # 递归公式，会超时
# class Solution:
# def Fibonacci(self, n):
# if n == 0:
# return 0
# if n == 1 or n == 2:
# return 1
# ... |
98e40f016721fed39d1e2518b51fd8aa5472d99c | JaeZheng/jianzhi_offer | /25.py | 1,324 | 3.59375 | 4 | """
题目描述
输入一个复杂链表（每个节点中有节点值，以及两个指针，一个指向下一个节点，另一个特殊指针指向任意一个节点），
返回结果为复制后复杂链表的head。（注意，输出结果中请不要返回参数中的节点引用，否则判题程序会直接返回空）
解决思路：
用一个另外的字典来储存对应关系
"""
# -*- coding:utf-8 -*-
class RandomListNode:
def __init__(self, x):
self.label = x
self.next = None
self.random = None
class Solution:
# 返回 Ra... |
0f253e5c5f347c7bdffef7ffb44801947f58b268 | JaeZheng/jianzhi_offer | /62.py | 768 | 3.75 | 4 | """
题目描述
给定一棵二叉搜索树，请找出其中的第k小的结点。例如， （5，3，7，2，4，6，8） 中，按结点数值大小顺序第三小结点的值为4。
"""
# -*- coding:utf-8 -*-
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
# 返回对应节点TreeNode
def KthNode(self, pRoot, k):
list = self.inorderTra... |
487812a5162d74d0c45558084ee907922b0af91d | JaeZheng/jianzhi_offer | /17.py | 948 | 3.671875 | 4 | """
题目描述:
输入两棵二叉树A，B，判断B是不是A的子结构。（ps：我们约定空树不是任意一个树的子结构）
解决思路:
写一个函数判断A树是否包含B树，再递归调用
"""
# -*- coding:utf-8 -*-
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
def HasSubtree(self, pRoot1, pRoot2):
if pRoot1 == None or pRoot... |
d4aea444e400224109d23fb804a0a40a6829e1ae | JaeZheng/jianzhi_offer | /38.py | 1,393 | 3.9375 | 4 | """
题目描述:
输入一棵二叉树，求该树的深度。从根结点到叶结点依次经过的结点（含根、叶结点）形成树的一条路径，最长路径的长度为树的深度。
解决思路：
第一种：递归。即求孩子节点的最大深度，要么是左孩子，要么是右孩子，那么我们只需要对传入的孩子节点递归调用即可。
第二种：BFS。层次遍历，每经过一层就深度加一。
"""
# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# 第一种
# c... |
2c018b879e226a3e3dcd068ebc132b7f21978ede | JaeZheng/jianzhi_offer | /11.py | 943 | 3.796875 | 4 | """
题目描述:
输入一个整数，输出该数二进制表示中1的个数。其中负数用补码表示。
解决思路:
第一种：n&(n-1),把一个整数减去1，再和原整数做与运算,会把该整数最右边一个1变成0.
那么一个整数的二进制有多少个1，就可以进行多少次这样的操作。
第二种：转为字符串之后计数
第三种：用1和n的每一位按位与
"""
# -*- coding:utf-8 -*-
# 第一种
# class Solution:
# def NumberOf1(self, n):
# # python需要把负数统一转为补码
# n = n & 0xffffffff
# count = 0
#... |
07dd3c78d5cb418ea7e09268e54331259e033fd5 | JaeZheng/jianzhi_offer | /36.py | 1,513 | 3.84375 | 4 | """
题目描述:
输入两个链表，找出它们的第一个公共结点。
解决思路:
第一种: 把两个链表的结点都入栈，然后同时出栈（即从后向前遍历），出现相同时加入结果数组，结果数组出栈第一个元素即为第一个公共结点。
第二种：两个链表不知谁长谁短，但是相加起来，到最后的公共结点部分长度是相同的，所以遍历时出现null就跳到另一个链表。
"""
# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
# 第一种
# class Solution:
# d... |
1a5db3ec15b799f081375487c9923c7e2aca79e3 | avinashnarasimha/Python_Training_Trishana | /forDemo2.py | 160 | 4.15625 | 4 | #program to print only odd or even numbers
for data in range(0,11,2):
print(data)
#to print only odd numbers
for v in range(1,11,2):
print(v)
|
0c777c1dc256a7f99b24274a16f3c09e9821cbd9 | avinashnarasimha/Python_Training_Trishana | /inputDemo2.py | 257 | 4.125 | 4 | #python code to read numbers
x =float(input("enter a number")) #reading in string
y = float(input("enter a nnumber"))
total=int(x)+int(y) #converting float into int.
print("sum of %f and %f is %d " %(x,y,int(total)))
print("sum of"x,y," is", total)
|
92d64e2db89df1297cc37d2500b2abfe847a7852 | nicolasdonato/reverse-snake.github.io | /test_flipper.py | 3,779 | 3.734375 | 4 | # Takes a test (defined below) and builds the other 7 symmetrical tests
# To automate rotation, the test needs to fit within a 10x10. The program checks for this, and can shift the snake to match.
snake = [[7, 2], [7, 3], [7, 4], [7, 5], [7, 6], [6, 6], [5, 6], [4, 6], [3, 6], [3, 5], [3, 4], [3, 3], [3, 2], [3, 1], [... |
93569b03705db06fa90d626300205d3006540000 | JoosikHan/Effective-Python-temp | /files/BetterWay21_ForceKeywordArgument.py | 4,047 | 3.578125 | 4 | # 83쪽. 키워드 전용 인수로 명료성을 강요하자.
# 2016/10/27 작성.
#######################################################################################
"""
키워드로 인수를 넘기는 방법은 파이썬 함수의 강력한 기능이다.
키워드 인수의 유연성 덕에 코드의 쓰임새를 명료하게 정의할 수 있다.
예를 들어 어떤 숫자를 다른 숫자로 나눈다고 해보자. 어떤 에러를 발생시키지만 예외를 조심해야 한다.
때로는 무한대값을 반환하거나, 0을 반환하고 싶을 수도 있다.
"""
... |
72a54bca1901187812764ea78385184ea49c2cdc | IdeaLaboratory/MachineLearning | /DataAnalytics/numpyPandas.py | 883 | 3.71875 | 4 | import numpy as np
import pandas as pd
import os
def header(msg):
print('-' * 50)
print('[ ' + msg + ' ]')
# 1. load hard-coded data into a dataframe
header("1. load hard-coded data into a df")
df = pd.DataFrame(
[['Jan',58,42,74,22,2.95],
['Feb',61,45,78,26,3.02],
['Mar',65,48,84,25,2.34],
['Apr',67,50,92,28,1... |
03be83d26008d2383aae9605b4c4d974d5c2acda | Chuphay/hadoop | /temp/invertThis.py | 320 | 3.546875 | 4 | #! /usr/bin/env python
import sys
import re
myDict ={}
numbers = "\d+"
for line in sys.stdin:
num,text = line
words = text.split()
for word in words:
try:
myDict[word].append(num)
except KeyError:
myDict[word] = [num]
for key in myDict:
print key, myDict[key]
|
71bbe30c6f5c82448da4e0da711819f30a3df535 | GanatheVyshnavi/Vyshnavi | /7th.py | 74 | 3.84375 | 4 | s=float(input('enter value of s'))
a=4*s
print("perimeter of square is",a) |
8af4889a4941a1e589d4d2e787a92b220be77d04 | GanatheVyshnavi/Vyshnavi | /assignment 9.py | 1,902 | 4.125 | 4 | # program to get smallest and largest numbers from the list
n=int(input('Enter number of items in the list: '))
l=[]
for i in range(0,n):
e=float(input('Enter a number: '))
l.append(e)
print("The smallest number in the list is",min(l))
print("The largest number in the list is",max(l))
# program to multiply a... |
9e3a5e36ab30a6068ebbb8b580a932031a0a4aa9 | maryamkarimi/map-reducer-python | /Part3/reducer2.py | 533 | 3.5625 | 4 | #!/usr/bin/env python
"""reducer.py"""
import sys
current_count = 0
# input comes from STDIN
for line in sys.stdin:
# remove leading and trailing whitespace
line = line.strip()
# convert line (currently a string) to int
try:
count = int(line)
except ValueError:
# count was not a ... |
d99020c3001a5db3fda8e17699e489b2158fea52 | Aug-G/my-awesome-python | /1.Algorithm/Fibonacci.py | 458 | 4.09375 | 4 | #coding:utf-8
# Fibonacci 数列的Python实现
#lambda 实现
fib = lambda n: 1 if n < 2 else fib(n - 1) + fib(n - 2)
# decorator 经典实现
def memo(func):
cache = {}
def wrap(*args):
if args not in cache:
cache[args] = func(*args)
return cache[args]
return wrap
def fib2(i):
if i < 2:
... |
6776dbe4ba898e4ee5954c2ee812f9c1c4154c6e | saijayadeep1998/Python-math-Module | /math.isnan() Method.py | 313 | 3.515625 | 4 | # Import math Library
import math
# Check whether some values are NaN
print (math.isnan (56))
print (math.isnan (-45.34))
print (math.isnan (+45.34))
print (math.isnan (math.inf))
print (math.isnan (float("nan")))
print (math.isnan (float("inf")))
print (math.isnan (float("-inf")))
print (math.isnan (math.nan))
|
a50ed11c72d651c6bf9a22d96146e6a55c33877c | saijayadeep1998/Python-math-Module | /math.pow() Method.py | 105 | 3.703125 | 4 | # Import math Library
import math
# Return the value of 9 raised to the power of 3
print(math.pow(9, 3))
|
9041fb5afcad7bdb9296029eeb6df521c22c89ab | choi5798/GIT_Practice | /aiya.py | 224 | 3.765625 | 4 | def fact(n):
if n == 0:
return 1
else:
return n * fact(n-1)
n = int(input("n 팩토리얼의 값을 구해줍니다. n을 입력하세요 : "))
print("값은 : " + str(fact(n)))
print('Hello World!') |
b2d384d4f4035f3bcfc9475eb3054341f2379dce | kusumachan/prak_ASD_C | /MODUL-1/MODUL1-L200170078.py | 5,192 | 3.90625 | 4 | """(NO 1)"""
def cetakSiku(x):
i=1
while i <= x:
print("*"*i)
i+=1
cetakSiku(5)
"""(NO 2)"""
def gambarlahPersegiEmpat(x):
l=x[1]
p=x[0]
jarak=l-4
i=1
while i <=l:
if i==1:
print("@"*l)
elif i==l:
print("@"*l)
... |
0ef13352aea24b617fa14bad27f171be2a4bbf6d | eabasir/algo | /utils/linked_list.py | 948 | 3.546875 | 4 |
class SinglyLinkedList(object):
def __init__(self, value, next=None):
self.value = value
self.next = next
@staticmethod
def from_list(vals: list):
head = SinglyLinkedList(vals[0])
current = head
for i in vals[1:]:
current.next = SinglyLinkedList(i)
... |
23e295bd415047317766b5ec865cbde4401e2efc | Ypman/fakeisos | /json_handler.py | 578 | 3.796875 | 4 | import json
def get_dict_from_json(json_file):
"""
Opens given json file and parse it as dictionary
:param json_file: filename of *.json in json folder
:return: json as dictionary
"""
with open("json/{}.json".format(json_file)) as file:
data = json.load(file)
file.close()
r... |
8e73593897d9ec80d263dc8c46069e556afd14ab | maymashd/webdev2019 | /week10/informatics/3)циклы/цикл while/B.py | 85 | 3.609375 | 4 | n=int(input())
b=2
while (b<=n):
if (n%b==0):
print(b)
break |
9f8341c763859fafeff033afa40423aab58c26f3 | maymashd/webdev2019 | /week10/Hackerrank/13)String split and join.py | 95 | 3.734375 | 4 | s=input()
a=""
for i in s:
if i==' ':
a+="-"
else:
a+=i
print(a) |
dac64dccf4958e8ca42761ff68811aa62309eb70 | maymashd/webdev2019 | /week10/CodeingBat/logic-1/in1to10.py | 173 | 3.75 | 4 | def f(a,ok):
if a>=1 and a<=10:
return True
elif ok:
return True
else:
return False
a=int(input())
b=bool(input())
print(f(a,b)) |
2bd1c90d6dfef947eb1d1018855642cef83f38ef | maymashd/webdev2019 | /week10/informatics/4)массивы/A.py | 154 | 3.625 | 4 | list1=[]
n=int(input())
for i in range(0,n):
a=int(input())
list1.append(a)
for i in range(0,n):
if (i%2==0):
print(list1[i])
|
6937044db9946d9195790d0d5a78505ff86718ce | maymashd/webdev2019 | /week10/CodeingBat/String-1/make_tags.py | 125 | 3.5 | 4 | def make_tag(tag,words):
return '<'+tag+'>'+words+"</"+tag+'>'
tag=input()
words=input()
print(make_tag(tag,words))
|
b3938edda72cc1ec752a7f9f7195963821e2a2fa | maymashd/webdev2019 | /week10/informatics/5)функции/B.py | 99 | 3.6875 | 4 | def power1(a,n):
return pow(a,n)
a1=int(input())
n1=int(input())
print(power1(a1,n1))
|
49bfb275f013e755054e63238bbf604e4d506389 | jul-star/Stepik_BasicUse | /03/src/ex_3_3_08.py | 518 | 3.5625 | 4 | import sys
import re
def zz3(_str):
"""
Выведите строки, содержащие две буквы "z", между которыми ровно три символа.
:param _str:
:return: True/False
"""
# pattern = r"(z.{3}z)"
# return len(re.findall(pattern, _str)) > 0
pattern = r"z.{3}z"
return re.search(pattern, _str) is not ... |
0f989ade4b30dcd79223488e82bab1500b723b39 | ameykasbe/algorithms | /1. searching_algorithms/1. linear_search.py | 250 | 3.765625 | 4 | def linear_search(arr, key):
for i in range(len(arr)):
if arr[i] == key:
return i
return -1
if __name__ == "__main__":
arr = [55, -78, 88, 1, -50]
print(linear_search(arr, 88))
print(linear_search(arr, 100))
|
7d303f61d3ed5e7c9bdf558a848d9c2abc4b3c4e | ttknight2020/py | /GaussNaive.py | 989 | 3.9375 | 4 | def GaussNaive(A, b):
'''
GaussNaive: naive Gauss elimination
x = GaussNaive(A, b): Gauss elimination without pivoting
input:
A = coefficient matrix
b = right hand side vector
output:
x = soultion vector
'''
import numpy as np
m, n = A.shape
if m != n:... |
d564ef3c7a178cc19f45f9ae545bb1b8f0466577 | karakumm/puzzle | /puzzle.py | 3,086 | 3.9375 | 4 | '''
Playing board for logic puzzle
'''
def check_column(board: list, column: int) -> bool:
'''
Checks if the column is valid. Returns True if yes, and False if not.
>>> check_column([\
"**** ****",\
"***1 ****",\
"** 3****",\
"* 4 1****",\
" 9 5 ",\
" 6 83 *",\
"3 1 **",\
" 8 2***",\
" 2 ****"... |
31531a1fd7549bb2e999d070bd43f9209119c6b2 | lbain/exercism-python | /rna-transcription/dna.py | 276 | 3.578125 | 4 | def convert(char):
transcription = {'G': 'C',
'C': 'G',
'T': 'A',
'A': 'U'}
return transcription[char]
def to_rna(test_string):
transcribed = map(convert, test_string)
return ''.join(transcribed)
|
1b51258b43ffb71800ceddbed5b6ebcd99632e49 | WilbertHo/leetcode | /easy/count_and_say/py/countandsay.py | 566 | 3.78125 | 4 | import re
import sys
class Solution(object):
# @return a string
def countAndSay(self, n):
num = '1'
for i in range(1, n):
say = ''
while num:
digit = num[0]
digit_count = len(re.search(r'^{d}+'.format(d=digit), num).group(0))
... |
277454aa808b0394d9ce372c006c1f4e955cda0f | troutstick/magic_words | /get_magic.py | 1,854 | 3.625 | 4 | import create_magic
import random
seen_words_file = 'seen_magic.txt'
def reset():
try:
f = open(seen_words_file, 'w')
f.close()
except FileNotFoundError:
f = open(seen_words_file, 'x')
f.close()
print("List of seen words has been reset!")
def get_magic(ignore_seen=False):... |
2feecefe98cf4882672213d2713742a0db19c52b | bartoszmaleta/dojos_katas_exercise_bank | /katas_codewars/7 kyu/alphabetical_addition.py | 1,258 | 4.125 | 4 | # Your task is to add up letters to one letter.
# The function will be given a variable amount of arguments, each one being a letter to add.
# Notes:
# Letters will always be lowercase.
# Letters can overflow (see second to last example of the description)
# If no letters are given, the function should return 'z'
# E... |
15ee729c38b581fa56f254b1111a4c76336e63fe | yanivr78/MyProjects | /Python/Nested_lists.py | 980 | 4.28125 | 4 | #/usr/bin/python3
fruits = ["Streberries", "Nectarines", "Apples", "Grapes", "Peaches", "Cherries", "Pears"]
vegetables = ["Spinach", "Kale", "Tomatos", "Celery", "Potatoes"]
dirty_dosen = [fruits, vegetables]
print(dirty_dosen)
print(dirty_dosen[1][1])
print(dirty_dosen[0][1])
# Nested List Game
row1 = ["⬜️","⬜️",... |
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