blob_id string | repo_name string | path string | length_bytes int64 | score float64 | int_score int64 | text string | is_english bool |
|---|---|---|---|---|---|---|---|
fe703b6364b1835b433410da896166adaf129654 | elijahsk/cpy5python | /practical04/q05_count_letter.py | 509 | 4.1875 | 4 | # Name: q05_count_letter.py
# Author: Song Kai
# Description: Find the number of occurences of a specified letter
# Created: 20130215
# Last Modified: 20130215
# value of True as 1 and value of False as 0
def count_letter(string,char):
if len(string)==1: return int(string[0]==char)
return int(string[0]==char)+count_letter(string[1:],char)
# input the string and char
string=input("Enter a string: ")
char=input("Enter a specific char you want to serach: ")
# print
print(count_letter(string,char))
| true |
66488b56ef595a4ce86f4936262ce9c47c2030d0 | elijahsk/cpy5python | /practical03/q08_convert_milliseconds.py | 813 | 4.1875 | 4 | # Name: q08_convert_milliseconds.py
# Author: Song Kai
# Description: Convert milliseconds to hours,minutes and seconds
# Created: 20130215
# Last Modified: 20130215
# check whether the string can be converted into a number
def check(str):
if str.isdigit():
return True
else:
print("Please enter a proper number!")
return False
# the convert process
def convert_ms(n):
string=""
# calculate the number of hours
string+=str(n//3600000)
n%=3600000
# calculate the number of minutes
string+=":"+str(n//60000)
n%=60000
# calculate the number of seconds
string+=":"+str(n//1000)
return string
# input the time
n=input("Enter the time in milliseconds: ")
while not check(n):
n=input("Enter the time in milliseconds: ")
# print
print(convert_ms(int(n)))
| true |
6dce3c4965f32564385623ffb9ef1528a180a016 | siddk/hacker-rank | /miscellaneous/python_tutorials/mobile_number.py | 469 | 4.125 | 4 | """
mobile_number.py
Given an integer N, and N phone numbers, print YES or NO to tell if it is a valid number or not.
A valid number starts with either 7, 8, or 9, and is 10 digits long.
"""
n = input()
for i in range(n):
number = raw_input()
if number.isdigit():
number = int(number)
if (len(str(number)) == 10) and (int(str(number)[0]) in [7,8,9]):
print "YES"
else:
print "NO"
else:
print "NO"
| true |
b90e89de330807f02410bee5f6faa321376c0a26 | siddk/hacker-rank | /miscellaneous/python_tutorials/sets.py | 684 | 4.3125 | 4 | """
sets.py
You are given two sets of integers M and N and you have to print their symmetric difference in ascending order. The first line of input contains the value of M followed by M integers, and then N and N integers. Symmetric difference is the values that exist in M or N but not in both.
"""
m = input()
m_lis = raw_input()
n = input()
n_lis = raw_input()
m_list = m_lis.split()
m_list = list(map(int, m_list))
n_list = n_lis.split()
n_list = list(map(int, n_list))
m_set = set(m_list)
n_set = set(n_list)
intersect = m_set.intersection(n_set)
for i in intersect:
m_set.remove(i)
n_set.remove(i)
union = m_set.union(n_set)
for i in sorted(union):
print i
| true |
cbb33de2960bed126ad694a163a15aca703bc40a | askwierzynska/python_exercises | /exercise_2/main.py | 702 | 4.125 | 4 | import random
print('-----------------------------')
print('guess that number game')
print('-----------------------------')
print('')
that_number = random.randint(0, 50)
guess = -1
name = input("What's your name darling? ")
while guess != that_number:
guess_text = input('Guess number between 0 an 50: ')
guess = int(guess_text)
if guess < that_number:
print('{0}Your guess of {1} was too low. Keep trying!'.format(name, guess))
elif guess > that_number:
print('{0}Your guess of {1} was too big. Keep trying!'.format(name, guess))
else:
print('Huge congrats {0}! {1} was that number!'.format(name, guess))
print('')
print('done. thanks ;)')
| true |
1607baf5554044a4870c9402e79c1ae0867faa0e | obayomi96/Algorithms | /Python/simulteneous_equation.py | 786 | 4.15625 | 4 | # solving simultaneous equation using python
print('For equation 1')
a = int(input("Type the coefficient of x = \n"))
b = int(input("Type the coefficient of y = \n"))
c = int(input("Type the coefficient of the constant, k = \n"))
print("For equation 2")
d = int(input("Type the coefficient of x = \n"))
e = int(input("Type the coefficient of y = \n"))
f = int(input("Type the coefficient of the constant, k = \n"))
def determinant():
return((a*e) - (b*d))
x = determinant()
# print(x) to output: a*e - b*d
def cofactors():
return((e*c) + (-1 * d * f)) / determinant(), ((-1 * b * c) + (a * 5)) / determinant()
z = cofactors()
#print(z) output: (e*c) + (-1 * d * f) / determinant(), ((-1 * b * c) + (a * 5)) / determinant()
print("Answer = (x,y)")
print(cofactors())
| true |
9b6acd981eb913452c020d5bd66aac70f329fd6c | heyhenry/PracticalLearning-Python | /randomPlays.py | 1,096 | 4.15625 | 4 | import random
randomNumber = random.randint(1, 10)
randomNumber = str(randomNumber)
print('✯ Welcome to the guessing game ✯')
username = input('✯ Enter username: ')
guessesTaken = 0
print(username, 'is it? \nGreat name! \nOkay, the rules are simple.\n')
print('⚘ You have 6 attempts. \n⚘ Enter a number ranging between 1 - 10. \n⚘ Heed the warnings after each attempt.\n')
while guessesTaken < 6:
guess = input('Guess the number:')
guess = str(guess)
guessesTaken += 1
if guess < randomNumber:
print("A little higher...")
elif guess > randomNumber:
print("A little lower...")
elif guess == randomNumber:
print("\nCongratulations! You guessed it right!")
break
if guess != randomNumber:
print('\nOh no! You ran out of attempts. The correct number was', randomNumber + '.\n')
print('Taking you to the stats screen now...\n')
print('\nStats: ')
print('The actual number: ', randomNumber, '\nLast guess: ', guess, '\nAttempts taken: ', guessesTaken)
print('\nThank you. Come again ( ͡👁️ ͜ʖ ͡👁️)✊!')
| true |
de5a9fdc707edaee79994df174ca98968d07228c | ivy-liu/re-Automation | /0311_01.py | 980 | 4.34375 | 4 | #算术运算符
a=21
b=10
c=0
c=a+b
print("a+b=",c)
c=a//b
print("a//b=",c)#取整除 - 返回商的整数部分（向下取整）
c=a/b
print("a/b=",c)
c=a**b
print("a**b=",c)
c=a*b
print("a*b=",c)
#条件语句，判断
jin=90
qu=75
ts=(jin-qu)/qu*100
print('小明成绩提升百分点：%.1f' % ts+'%')
flag=False
name="xiaoming"
if name=='小明'or'xiaoming':
flag=True
print('是的，对，我是小明')
else:
print('不是')
num=5
if num==3:
print('我是3')
elif num==4:
print('我是4')
else:
print('我好困，我是5')
num=9
if num >=0 and num <=10:
print('我大于等于0小于等于10\n')
height=1.68
weight=53
bmi=weight/(height**2) #体重除以身高的平方
print('bmi=',bmi)
if bmi<18.5:
print('过轻')
elif bmi>=18.5 and bmi <=25:
print('正常')
elif bmi>25 and bmi<=28:
print('过重')
elif bmi>28 and bmi <=32:
print('肥胖')
elif bmi>32:
print('严重肥胖')
else:
print('异常')
| false |
39f360ae4828952b4da6249bacfadda4671911d2 | aryashah0907/Arya_GITSpace | /Test_Question_2.py | 454 | 4.40625 | 4 | # 3 : Write a Python program to display the first and last colors from the following list.
# Example : color_list = ["Red","Green","White" ,"Black"].
# Your list should be flexible such that it displays any color that is part of the list.
from typing import List
color_list = ["Red", "Green", "White", "Black", "Pink", "Azure", "Brown"]
newColorlist = color_list[0]
finalColorlist = color_list[len(color_list) - 1]
print(newColorlist + finalColorlist)
| true |
ac9de1f5797579203874fef062220415cc7a3c13 | cvhs-cs-2017/sem2-exam1-LeoCWang | /Function.py | 490 | 4.375 | 4 | """Define a function that will take a parameter, n, and triple it and return
the result"""
def triple(n):
n = n * 3
return (n)
print(triple(5))
"""Write a program that will prompt the user for an input value (n) and print
the result of 3n by calling the function defined above. Make sure you include
the necessary print statements and address any issues with whitespace. """
def usertriple(n):
n = 3 *n
return n
print (usertriple(float(input('Please enter a number'))))
| true |
1ce5245bebfd0e12605f9a1fac47a3fd985c080c | luffysk/coderesources | /python/1.base/11.str_format.py | 255 | 4.28125 | 4 | # 格式化输出字符串有三种方式
# 第一种
a = 'str'
b = 'str2'
print('a is ' + a + ', b is ' + b)
# 第二种
a = 's1'
b = 's2'
print('a is %s, b is %s' % (a, b))
# 第三种, 推荐此种
a = 'fstr'
b = 'fstr2'
print(f'a is {a}, b is {b}') | false |
9040c3eeb0c6b4bf7cdff293644912e879a07eee | dexterpengji/practice_python | /fishC/038_class_inherit.py | 919 | 4.15625 | 4 | # -*- coding: utf-8 -*
import random as r
class Fish:
def __init__(self):
self.x = 100
self.y = 100
def move(self):
self.x += r.randint(-5,5)
self.y += r.randint(-5,5)
print("position",self.x,self.y)
class Gold_Fish(Fish):
def __init__(self):
#Fish.__init__(self) # way 1
super().__init__() # way 2 - the better one
self.size = 20
def inflate(self):
self.size += 10
def shrink(self):
self.size -= 10
class Carp_Fish(Fish):
pass
class Salmon_Fish(Fish):
pass
class Shark_Fish(Fish):
def __init__(self):
#Fish.__init__(self) # way 1
super().__init__() # way 2 - the better one
self.hungry = True
def eat(self):
if self.hungry:
print("Hungry...")
self.hungry = False
else:
print("Full...")
class Gold_Shark_Fish(Gold_Fish, Shark_Fish): # 多重继承可能会出现不可预料的bug，尽量不要使用
pass | false |
1712892d8d87ea7ffc0532bedeb31afd088c47bc | damianserrato/Python | /TypeList.py | 693 | 4.46875 | 4 | # Write a program that takes a list and prints a message for each element in the list, based on that element's data type.
myList = ['magical unicorns',19,'hello',98.98,'world']
mySum = 0
myString = ""
for count in range(0, len(myList)):
if type(myList[count]) == int:
mySum += myList[count]
elif type(myList[count]) == str:
myString += myList[count]
myString = "String: " + myString
print myString
mySum = "Sum: " + myString
if mySum > 0 and len(myString) > 1:
print "The list you entered is of a mixed type"
elif mySum > 0 and len(myString) == 0:
print "The list you entered is of an integer type"
else:
print "The list you entered is of a string type" | true |
2d5273b86616f827bed2a4496b4fe6854acf5ac3 | damianserrato/Python | /FindCharacters.py | 388 | 4.15625 | 4 | # Write a program that takes a list of strings and a string containing a single character, and prints a new list of all the strings containing that character.
word_list = ['hello','world','my','name','is','Anna']
char = 'o'
newList = []
for count in range(0, len(word_list)):
for c in word_list[count]:
if c == "o":
newList.append(word_list[count])
print newList | true |
8a833841f94024fa0c6e5185a78d4b404716ba8c | Hikareee/Phyton-exercise | /Programming exercise 1/Area of hexagon.py | 266 | 4.3125 | 4 | import math
#Algorithm
#input the side
#calculate the area of the hexagon
#print out the area
#input side of hexagon
S = eval(input("input the side of the hexagon: "))
#area of the hexagon
area=(3*math.sqrt(3)*math.pow(S,2))/2.0
#print out the area
print (area) | true |
51150a87bda43b7f1a9c1787005001a034b73232 | rvrn2hdp/informatorio2020 | /FuncionesComp/func9.py | 1,182 | 4.21875 | 4 | '''Ejercicio 9: ¿Un string representan un entero?
En este ejercicio escribirá una función llamada es_entero que determina
si los caracteres en una cadena representan un número entero válido.
Al determinar si un string representa un número entero, debe ignorar cualquier
espacio en blanco inicial o final. Una vez que se ignora este espacio en blanco,
una cadena representa un número entero si su longitud es al menos
1 y solo contiene dígitos, o si su primer carácter es + o -
y el primer carácter va seguido de uno o más caracteres,
todos los cuales son dígitos
Escriba un programa principal que lea una cadena del usuario
e informe si representa o no un número entero.
Sugerencia: Puede encontrar los métodos lstrip,
rstrip y / o strip para cadenas útiles cuando complete este ejercicio.
'''
def es_entero(cadena):
cadena = cadena.strip(' ')
print(cadena)
if len(cadena) > 0 or (cadena[0] == '+' or cadena[0] == '-') and cadena[0:].isdigit():
return "Es un número entero."
else:
return "No es un número entero."
def principal():
micadena = input("Ingrese un numero: ")
print(es_entero(micadena))
principal() | false |
9d0d3dd9396aa1135f9d3f84d5d6ba7c76c3cffa | rvrn2hdp/informatorio2020 | /FuncionesComp/func6.py | 731 | 4.125 | 4 | '''Ejercicio 6: Centrar una cadena en la terminal
Escriba una función que tome una cadena de caracteres como primer parámetro
y el ancho de la terminal en caracteres como segundo parámetro.
Su función debe devolver una nueva cadena que consta de la cadena original
y el número correcto de espacios iniciales para que la cadena original
aparezca centrada dentro del ancho proporcionado cuando se imprima.
No agregue ningún carácter al final de la cadena.
Incluya un programa principal que use su función.'''
def centrar_cadena(cadena, ancho):
return cadena.center(ancho, ' ')
def funcion_principal():
miCadena = input("Ingrese una cadena: ")
print(centrar_cadena(miCadena, 79))
funcion_principal()
| false |
2580e93a9665e26326c9478950be89a27e1709e0 | rvrn2hdp/informatorio2020 | /Desafios1/desafiorepetitiva5.py | 1,350 | 4.125 | 4 | """Se está desarrollando un sistema de control de vehículos
desde donde se han tirado restos de basura a la vía pública.
Para ello la ciudad cuenta con sistemas de monitoreo de patentes
que devuelve 3 letras y un valor numérico de 5 dígitos a la Central
con el siguiente significado:
3 letras: Correspondientes a la patente.
Del valor numérico:
Los 3 primeros números corresponden a la patente
El 4 número indica
1 - Tiró basura a la vía pública
0 - No tiró basura a la vía pública
El 5 número indica
1 - Ya había sido multado el vehículo
0 - Vehículo sin multas.
Deberás informar cantidad de vehículos observados, cantidad de vehículos
que han tirado basura y porcentaje de éstos que ya habían sido multados.
"""
print("\nBienvenido\n")
tiro_basura = 0
ya_multado = 0
analizadas = 0
while True:
patente = input("\nIngrese la patente: ")
analizadas += 1
if patente [6] == '1':
tiro_basura += 1
if patente [7] == '1':
ya_multado += 1
seguir = int(input("\n1 - Siguiente patente\n2 - Terminar de analizar patentes\n"))
if seguir == 2:
break
print("\nSe analizaron", analizadas, "patentes")
print(tiro_basura, "tiraron basura")
print("y", ya_multado, "ya habían sido multados.\n") | false |
7633108013a04f6e5ef0c4287c65002bd12e70c6 | rvrn2hdp/informatorio2020 | /FuncionesComp/func10.py | 1,074 | 4.21875 | 4 | '''Ejercicio 10: Precedencia del operador
Escriba una función llamada precedencia que devuelve un número entero
que representa la precedencia de un operador matemático.
Una cadena que contiene el operador se pasará a la función como su único parámetro.
Su función debe devolver 1 para + y -, 2 para * y /, y 3 para ˆ.
Si la cadena que se pasa a la función no es uno de estos operadores,
la función debería devolver -1.
Incluya un programa principal que lea un operador del usuario
y muestre la precedencia del operador o un mensaje
de error que indique que la entrada no era un operador.
En este ejercicio, se usa ˆ para representar la exponenciación,
en lugar de la elección de Python de **, para facilitar el desarrollo de la solución.
'''
def precedencia(s):
if s == '+' or s == '-':
return 1
elif s == '*' or s == '/':
return 2
elif s == '^':
return 3
else:
return -1
def principal():
print("precedencia del operador ingresado:")
s = input()
print (precedencia(s))
principal() | false |
e74ddfcc4b04789ce051adee50598c592547d7fb | congyingTech/Basic-Algorithm | /medium/hot100/2-add-two-numbers.py | 2,341 | 4.25 | 4 | """
给你两个 非空 的链表，表示两个非负的整数。它们每位数字都是按照 逆序 的方式存储的，并且每个节点只能存储 一位 数字。
请你将两个数相加，并以相同形式返回一个表示和的链表。
你可以假设除了数字 0 之外，这两个数都不会以 0 开头。
输入：l1 = [2,4,3], l2 = [5,6,4]
输出：[7,0,8]
解释：342 + 465 = 807.
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/add-two-numbers
解题思路：
每个数位都是逆序的，所以可以从头遍历两个链表，把两个链表的值相加
"""
# Definition for singly-linked list.
class ListNode(object):
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
p1 = l1
p2 = l2
res_Node = ListNode()
head_res = res_Node
temp = 0
while p1 or p2 or temp:
if p1 and not p2:
res_val = p1.val + temp
elif not p1 and p2:
res_val = p2.val + temp
elif p1 and p2:
res_val = p1.val + p2.val + temp
elif not p1 and not p2 and temp:
res_val = temp
else:
break
temp = 0
if res_val < 10:
res = ListNode(res_val)
else:
res = ListNode(res_val % 10)
temp = res_val // 10
head_res.next = res
head_res = head_res.next
if p1 and p1.next:
p1 = p1.next
else:
p1 = None
if p2 and p2.next:
p2 = p2.next
else:
p2 = None
res = res_Node.next
return res
if __name__ == "__main__":
s = Solution()
l1 = [9, 9, 9, 9, 9, 9, 9]
l2 = [9, 9, 9, 9]
l1_node = ListNode(l1[0])
p1 = l1_node
l2_node = ListNode(l2[0])
p2 = l2_node
for i in range(1, len(l1)):
p1.next = ListNode(l1[i])
p1 = p1.next
for i in range(1, len(l2)):
p2.next = ListNode(l2[i])
p2 = p2.next
s.addTwoNumbers(l1_node, l2_node)
| false |
070b897a2f9365ae21c150490d0949a4b3d47960 | congyingTech/Basic-Algorithm | /data_structure/3.directed-graph.py | 1,586 | 4.21875 | 4 | # encoding:utf-8
"""
问题描述：有向（可能有环图）图
"""
class DirectedGraph(object):
def __init__(self,vertices):
self.vertices = vertices
self.graph = [[0]*self.vertices for i in range(self.vertices)]
def add_edges(self, src, dest):
self.graph[src][dest] = 1
def print_graph(self):
for i in range(self.vertices):
connect_vertices = []
for j in range(self.vertices):
if self.graph[i][j] == 1:
connect_vertices.append(j)
print('head:{}, connect with {}'.format(i, connect_vertices))
# 深度搜索(backtrack)find path
def print_path(self, i, temp, res):
"""
i是第i个点
"""
if len(set(temp))<len(temp):
print('闭环:{}'.format(temp))
inner_res = temp[:]
res.append(inner_res)
return
if 1 not in self.graph[i] and temp not in res :
print(temp)
inner_res = temp[:]
res.append(inner_res)
return
for j in range(self.vertices):
if self.graph[i][j] == 1:
temp.append(j)
self.print_path(j, temp, res)
temp.pop()
if __name__ == "__main__":
v_num = 4
graph = DirectedGraph(v_num)
graph.add_edges(0,1)
graph.add_edges(0,2)
graph.add_edges(1,3)
graph.add_edges(2,3)
graph.add_edges(2,0)
graph.print_graph()
temp = [0]
res = []
graph.print_path(0, temp, res)
print(res)
| false |
30698c43e3ce59495f42fb6d0bff2c5cce4a525b | congyingTech/Basic-Algorithm | /getOffer/17.merge-two-sorted-lists.py | 2,753 | 4.125 | 4 | # encoding:utf-8
"""
问题描述：合并两个上升排序的链表，使之合并后有序
解决方案：递归的方案，
非递归的方案:先比较第一个节点，节点小的那一个作为主链表，把循环遍历另一条链表，把其中的元素插入到主链表中
在主链表设置两个指针：mainHead/mainNext，次链表只有secondHead一个指针，做次链表单节点插入主链表的动作。
"""
# Definition for singly-linked list.
class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None
class Solution1(object):
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if not l1:
return l2
if not l2:
return l1
if l1.val < l2.val:
mainHead = l1
secondHead = l2
else:
mainHead = l2
secondHead = l1
mainNext = mainHead.next
mergeHead = mainHead
while secondHead:
if not mainNext:
mainHead.next = secondHead
break
# 当第二个linkedlist的node的值小于主list的时候，把其插入主list
if secondHead.val<mainNext.val:
mainHead.next = secondHead
secondHead = secondHead.next
mainHead.next.next = mainNext
mainHead = mainHead.next
else:
mainHead = mainHead.next
mainNext = mainNext.next
return mergeHead
def printLinkList(self, head):
while head:
print(head.val)
head = head.next
class Solution(object):
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if not l1:
return l2
if not l2:
return l1
if l1.val<l2.val:
l1.next = self.mergeTwoLists(l1.next, l2)
return l1
else:
l2.next = self.mergeTwoLists(l1, l2.next)
return l2
def printLinkList(self, head):
while head:
print(head.val)
head = head.next
if __name__ == "__main__":
s = Solution()
l1 = ListNode(1)
l1.next = ListNode(2)
l1.next.next = ListNode(4)
# l1.next.next.next = ListNode(7)
# l1.next.next.next.next = ListNode(9)
l2 = ListNode(1)
l2.next = ListNode(3)
l2.next.next = ListNode(4)
# l2.next.next.next = ListNode(8)
# # 递归的方法
# l = s.mergeTwoLists(l1, l2)
# s.printLinkList(l)
s1 = Solution1()
l1 = s1.mergeTwoLists(l1,l2)
s1.printLinkList(l1) | false |
0fd112e01e2545fb78af3211f1f1d12dd0159d24 | Lukhanyo17/Intro_to_python | /Week7/acronym.py | 538 | 4.15625 | 4 | def acronym():
acro = ''
ignore = input("Enter words to be ignored separated by commas:\n")
title = input("Enter a title to generate its acronym:\n")
ignore = ignore.lower()
ignore = ignore.split(', ')
title = title.lower()
titleList = title.split()
titleList.append("end")
for k in range(len(titleList)-1):
if(not titleList[k] in ignore):
word = titleList[k]
acro += word[0]
print("The acronym is: "+acro.upper())
acronym()
| true |
a51357020c8422857735e9092ed7b35fa3992060 | samvelarakelyan/ACA-Intro-to-python | /Practicals/Practical5/Modules/pretty_print.py | 667 | 4.25 | 4 |
def simple_print(x:int):
"""
The function gets an integer and just prints it.
If type of function argument isn't 'int' the function print 'Error'.
"""
if isinstance(x,int):
print("Result: %d" %x)
else:
print("Error: Invalid parametr! Function parametr should be integer")
def pro_print(x:int):
"""
The function gets an integer and just prints it.
If type of function parametr isn't 'int' the function print 'Error'.
"""
if isinstance(x,int):
print("The result of the operation is %d" %x)
else:
print("Error: Invalid parametr! Function parametr should be integer")
| true |
76e9d759bbee0884ff88b287e491ae097580f41b | imsk003/Python | /reverse_words.py | 224 | 4.25 | 4 | def reverseWords(input):
inputWords = input.split(" ")
inputWords=inputWords[-1::-1]
output = ' '.join(inputWords)
return output
if __name__ == "__main__":
input = 'hello python'
print(reverseWords(input))
| false |
6ca26100985fc23524b13578e97cde54a42f2f70 | twhay/Python-Scripts | /Dice Simulator.py | 509 | 4.21875 | 4 | # Python Exercise - Random Number Generation - Dice Generator
# Import numpy as np
import numpy as np
# Set the seed
np.random.seed(123)
# Generate and print random float
r = np.random.rand()
print(r)
# Use randint() to simulate a dice
dice = np.random.randint(1,7)
print(dice)
# Starting step
step = 50
# Finish the control construct
if dice <= 2 :
step = step - 1
elif dice <=5 :
step = step + 1
else :
step = step + np.random.randint(1,7)
# Print out dice and step
print(dice)
print(step) | true |
6065b6d6ecb45e18d532a5a3bcf765851e97d1eb | iMeyerKimera/play-db | /joins.py | 929 | 4.34375 | 4 | # -*- coding: utf-8 -*-
# joining data from multiple tables
import sqlite3
with sqlite3.connect("new.db") as connection:
c = connection.cursor()
# retrieve data
c.execute("""
SELECT population.city, population.population,
regions.region FROM population, regions
WHERE population.city = regions.city ORDER BY population.city ASC
""")
rows = c.fetchall()
for r in rows:
print "City: " + r[0]
print "Population: " + str(r[1])
print "Region: " + r[2]
print
# Take a look at the SELECT statement.
# Since we are using two tables, fields in the SELECT statement must adhere
# to the following format: table_name.column_name (i.e., population.city ).
# In addition, to eliminate duplicates, as both tables include the city name,
# we used the WHERE clause as seen above.
# finally organize the outputted results and clean up the code so it’s more compact | true |
2909e9da710ee8da8f33f9c67687f697c8db7385 | lorenanicole/abbreviated-intro-to-python | /exercises/guessing_game_two.py | 1,143 | 4.34375 | 4 | #!/usr/bin/env python
# -*- coding: utf-8 -*-
from __future__ import print_function
import random
"""
ITERATION THREE: Now that you have written a simple guessing
game, add some logic in that determines if the game is over
when either:
* user has correctly guesses the number
* user has reached 5 guesses
Let's use two functions:
* is_guess_correct
* display_user_output
STRETCH: In the case user loses, show correct number.
In the case user wins tell them how many guesses it took for
them to win. Both situations show user all guesses they made.
"""
def display_user_output(num_to_guess, guessed_num):
"""
:param: guessed_num - number user guessed
:param: num_to_guess - number the user has to guess
:returns string saying if the guess is too high, too low, or match
"""
def is_guess_correct(num_to_guess, guessed_num):
"""
:param: guessed_num - number user guessed
:param: num_to_guess - number the user has to guess
:returns boolean True if correct else False
"""
game_not_over = False
turns = 0
while game_not_over:
# TODO: ask for user input
# determine what message to show user
# determine if guess is correct | true |
b9a599544b5dbf2e0cf855cfe9b848f3fec098eb | juliakyrychuk/python-for-beginners-resources | /final-code/oop/car.py | 777 | 4.25 | 4 | class Car:
"""Represents a car object."""
def __init__(self, colour, make, model, miles=0):
"""Set initial details of car."""
self.colour = colour
self.make = make
self.model = model
self.miles = miles
def add_miles(self, miles):
"""Increase miles by given number."""
self.miles += miles
def display_miles(self):
"""print the current miles value."""
print(
f'{self.make} {self.model} ({self.colour}) '
f'has driven {self.miles} miles.'
)
astra = Car('Red', 'Vauxhall', 'Astra')
astra.display_miles()
astra.add_miles(100)
astra.display_miles()
prius = Car('Blue', 'Toyota', 'Prius', 1000)
prius.display_miles()
prius.add_miles(50)
prius.display_miles() | true |
70504fde426af3e446033709871f1d5219844539 | Farmerjoe12/PythonLoanLadder | /pythonLoanLadder/model/Loan.py | 1,234 | 4.1875 | 4 | import numpy
import math
class Loan:
""" A representation of a Loan from a financial institution.
Loans are comprised of three main parts, a principal
or the amount for which the loan is disbursed, an interest rate
because lending companies are crooked organizations which charge
you money for borrowing their money, and a term for which
the repayment period is defined.
Fields:
name: A descriptive name for a loan
interest_rate: The interest rate given by a lender (%5.34)
principal: The amount of money that is lent
term: the repayment period (in years)
"""
def __init__(self, name, interest_rate, principal):
self._name = name
self._interest_rate = interest_rate
self._principal = principal
def get_interest_rate(self):
return self._interest_rate
def get_principal(self):
return self._principal
def get_name(self):
return self._name
def to_string(self):
result = "Loan Name: {}\n".format(self._name)
result += "Principal: {}\n".format(self._principal)
result += "Interest Rate: %{}".format(self._interest_rate)
return result | true |
0a0558e50b0bd8f9f41348596aae5c06ac66c7e7 | LIZETHVERA/python_crash | /chapter_7_input_while/parrot.py | 788 | 4.3125 | 4 | message = input ("Tell me something, and i will repetar it back to you: ")
print (message)
name = input ("please enter your name: ")
print ("hello, "+ name + "!")
prompt = "If you tell us who you are, we can personalize the messages you see"
prompt += "\nWhat is your first name?"
name = input (prompt)
print ("\nHello," + name + "!")
prompt = "Tell me something, and i will repeat the message to you: "
prompt += "\nEnter 'quit' to end the program. ?"
message = ""
while message != "quit":
message = input (prompt)
print (message)
prompt = "Tell me something, and i will repeat the message to you: "
prompt += "\nEnter 'quit' to end the program. ?"
message = ""
while message != "quit":
message = input (prompt)
if message != "quit":
print (message)
| true |
f25f1e4333a6c8a4ef1f22eb18a430ad3be862ab | olsgaard/adventofcode2019 | /day01_solve_puzzle1.py | 801 | 4.5 | 4 | """
Fuel required to launch a given module is based on its mass. Specifically, to find the fuel required for a module, take its mass, divide by three, round down, and subtract 2.
For example:
For a mass of 12, divide by 3 and round down to get 4, then subtract 2 to get 2.
For a mass of 14, dividing by 3 and rounding down still yields 4, so the fuel required is also 2.
For a mass of 1969, the fuel required is 654.
For a mass of 100756, the fuel required is 33583.
"""
def calculate_fuel(mass):
return int(mass / 3) - 2
assert calculate_fuel(12) == 2
assert calculate_fuel(14) == 2
assert calculate_fuel(1969) == 654
assert calculate_fuel(100756) == 33583
with open("input.txt01") as f:
lines = f.readlines()
result = sum([calculate_fuel(int(mass)) for mass in lines])
print(result)
| true |
a00d0f8ef6e6f6ba51524c3c0309ebe863ab9581 | osmandi/programarcadegames | /Capítulo 4: Adivianzas con números aleatorios y bucles/ejemplos_de_while.py | 1,633 | 4.21875 | 4 | """
# Sample Python/Pygame Programs
# Simpson College Computer Science
# http://programarcadegames.com/
# http://simpson.edu/computer-science/
"""
# Podemos emplear un bucle while allí donde, también, podríamos usar un bucle for:
i = 0
while i < 10:
print(i)
i = i + 1
# Es lo mismo que:
for i in range(10):
print(i)
# Es posible simplificar el código:
# i=i+1
# Con lo siguiente:
# i += 1
# Esto lo podemos hacer también con la resta y la multiplicación.
i = 0
while i < 10:
print(i)
i += 1
# ¿Qué imprimiremos aquí?
i = 1
while i <= 2 ** 32:
print(i)
i *= 2
# Una tarea muy habitual, es iterar hasta que el usuario realiza una petición de salida.
salir = "n"
while salir == "n":
salir = input ("¿Quieres salir? ")
# Existen diversas formas para salir de un bucle. Un operador booleano que dispare el evento es una
# forma de conseguirlo.
hecho = False
while not hecho:
salir = input ("¿Quieres salir? ")
if salir == "s":
hecho = True;
ataca = input ("¿El elfo atacó al dragón? ")
if ataca == "s":
print("Mala elección, estás muerto.")
hecho = True;
valor = 0
incremento = 0.5
while valor < 0.999:
valor += incremento
incremento *= 0.5
print(valor)
# -- Problemas habituales con los bucles while --
# El programador quiere hacer una cuenta atrás empezando en 10.
# ¿Qué es lo que está mal, y cómo lo podemos arreglar?
i = 10
while i == 0:
print (i)
i -= 1
# ¿Qué es lo que está mal en este bucle que intenta contar hasta 10?
# ¿Qué sucederá cuando lo ejecutemos?
i = 1
while i < 10:
print (i) | false |
77da7fdefd5a5ddc68ce5652094f4ad1b627d3a3 | stefantoncu01/Pizza-project | /pizza_project.py | 2,981 | 4.125 | 4 | class Pizza:
"""
Creates a pizza with the attributes: name, size, ingredients
"""
def __init__(self, name, size):
self.name = name
self.size = size
self.ingredients = None
@property
def price(self):
"""
Calculates the price based on size and ingredients
"""
price_per_ingredient = 3
size_price = {
"small": 1,
"medium": 1.2,
"large": 1.5
}
return size_price[self.size] * len(self.ingredients) * price_per_ingredient
class VeganPizza(Pizza):
"""
Creates a vegan pizza by inheriting attributes from Pizza class
"""
def __init__(self, name, size):
super().__init__(name, size)
self.ingredients = ['tomato_sauce', 'pepper', 'olives']
class CarnivoraPizza(Pizza):
"""
Creates a carnivora pizza by inheriting attributes from Pizza class
"""
def __init__(self, name, size):
super().__init__(name, size)
self.ingredients = ['tomato_sauce', 'cheese', 'chicken', 'parmesan', 'spinach']
class PepperoniPizza(Pizza):
"""
Creates a pepperoni pizza by inheriting attributes from Pizza class
"""
def __init__(self, name, size):
super().__init__(name, size)
self.ingredients = ['tomato_sauce', 'cheese', 'salami', 'habanero_pepper']
class HawaiianPizza(Pizza):
"""
Creates a hawaiian pizza by inheriting attributes from Pizza class
"""
def __init__(self, name, size):
super().__init__(name, size)
self.ingredients = ['tomato_sauce', 'cheese', 'pineapple', 'coocked_ham', 'onion']
class Client:
"""
Creates a client with the attributes name, address, has_card(bool)
"""
def __init__(self, name, address, has_card=True):
self.name = name
self.address = address
self.has_card = has_card
class Order:
"""
Creates an order based on the Client class and products (a list of Pizza objects)
"""
def __init__(self, client, products):
self.client = client
self.products = products
@property
def total_price(self):
"""
Calculates the total price of the order based on products attributes. If the client has_card a 10% discount should be applied.
"""
price = sum([product.price for product in self.products])
if self.client.has_card:
return price * 0.9
return price
@property
def invoice(self):
"""
Table formatted string containing all products associated with this order, their prices, the total price, and client information
"""
result ='\n'.join([f'{product.name} - {product.price}' for product in self.products])
result += f'\nThe total price is {self.total_price}! \nThe delivery will be in {self.client.address}!'
return result
| true |
e378f29d1bd2dbf43f88f0a0d2333f811150be2f | scvetojevic1402/CodeFights | /CommonCharCount.py | 660 | 4.15625 | 4 | #Given two strings, find the number of common characters between them.
#Example
#For s1 = "aabcc" and s2 = "adcaa", the output should be
#commonCharacterCount(s1, s2) = 3.
#Strings have 3 common characters - 2 "a"s and 1 "c".
def commonCharacterCount(s1, s2):
num=0
s1_matches=[]
s2_matches=[]
for i in range(0,len(s1)):
if i not in s1_matches:
for j in range(0,len(s2)):
if j not in s2_matches:
if s1[i]==s2[j]:
num+=1
s1_matches.append(i)
s2_matches.append(j)
break
return num
| true |
c2d0e8489e7783edf1fc6a5548825a77da605e57 | dayanandtekale/Python_Basic_Programs | /basics.py | 1,303 | 4.125 | 4 | #if-else statements:
#score=int(input("Enter your score"))
#if score >=50:
# print("You have passed your exams")
# print("Congratulations")
#if score <50:
# print("Sorry,You have failed Exam")
#elif statements:
#score=109
#if score >155 or score<0:
# print("Your score is invalid")
#elif score>=50:
# print("You have passed your exam")
# print("Congratulations!")
#else:
# print("Sorry,You have failed Exam")
#while loop:
#count= 5
#while count <=10:
# print(count)
# count= count + 1
#number = int(input("Enter a number: "))
#count = 1
#while count >= 10:
# product = number * count
# print(number, "x" ,count, "=", product)
# count = count + 1
#for loop in python:
#text = "Python"
#for character in text:
# print(character)
#languages = ["English","French","German"]
#for language in languages:
# print(language)
#count=1
#while count <= 5:
# print(count)
# count = count + 1
#using for loop in 1 sentence we get same op:
#for count in range(1,6):
# print(count)
#for table of any number in (1-11)
#number = int(input("Enter an integer: "))
#for count in range(1,11):
# product = number * count
# print(number, "X", count, "=", product)
| true |
49063cf5cbae4fc79e96e59d6dfd07178f16b211 | cxdy/CSCI111-Group4 | /project2/final.py | 597 | 4.40625 | 4 | # Find the distance between two points
# Class: CSCI 111 - Intro to Computer Science
# Group: Project Group 4
import math
# Ask for the 2 xy coordinates
x1 = float(input("Point #1 x-coord: "))
y1 = float(input("Point #1 y-coord: "))
x2 = float(input("Point #2 x-coord: "))
y2 = float(input("Point #2 y-coord: "))
# Square the difference between the (x2 - x1) vars and (y2 - y1) vars
xMath = (x2 - x1) ** 2
yMath = (y2 - y1) ** 2
# Find the square root of the sum of the xMath and yMath vars
squareRoot = math.sqrt(xMath + yMath)
print(f"The distance between the two points is: {squareRoot}") | false |
613a833ae062123c4f5a81af2e957fb28fea74cd | cxdy/CSCI111-Group4 | /project1/GroupProect1 BH.py | 1,054 | 4.40625 | 4 | firstname1 = input("Enter a first name: ")
college = input("Enter the name of a college: ")
business = input("Enter the name of a business: ")
job = input("Enter a job: ")
city = input("Enter the name of a city: ")
restaurant = input("Enter the name of a restaurant: ")
activity1 = input("Enter an activity: ")
activity2 = input("Enter another activity: ")
animal = input("Enter a type of animal: ")
firstname2 = input("Enter a different first name: ")
store = input("Enter the name of a store: ")
# Display the paragraph with all the inputs in the console
print(f"There once lived a person named {firstname1} who recently graduated from {college}. After accepting a job with {business} as a(n) {job}, {firstname1} decided to first take a vacation to {city}. Upon arriving in {city}, {firstname1} planned to visit {restaurant} before enjoying an evening of {activity1} and {activity2}. The next day, {firstname1} adopted a(n) {animal} named {firstname2} found in front of an abandoned {store}. {firstname1} and {firstname2} lived happily ever after .")
| true |
10e396f5019fd198a588d543c6746a642867496c | DSR1505/Python-Programming-Basic | /04. If statements/4.05.py | 309 | 4.15625 | 4 | """ Generate a random number between 1 and 10. Ask the user to guess the number and print a
message based on whether they get it right or not """
from random import randint
x = randint(1,10)
y = eval(input('Enter a number between 1 and 10: '))
if(x == y):
print('You get it right')
else:
print('Try again') | true |
4c590e8a0ff2058228a40e521d4dc31b1978ee9e | DSR1505/Python-Programming-Basic | /02. For loops/2.14.py | 276 | 4.34375 | 4 | """ Use for loops to print a diamond. Allow the user to specify how high the
diamond should be. """
num = eval(input('Enter the height: '))
j = (num//2)
for i in range(1,num+1,2):
print(' '*j,'*'*i)
j = j - 1
j = 1
for i in range(num-2,0,-2):
print(' '*j,'*'*i)
j = j + 1 | true |
1ddc261cf174c109583fd0ead1f537673d29090a | athirarajan23/luminarpython | /regular expression/validation rules/rules with eg.py | 2,250 | 4.15625 | 4 | #rules used for pattern matching
# #1. x='[abc]' either a,b or c
#eg:
# import re
# x="[abc]"
# matcher=re.finditer(x,"abt cq5kz")
# for match in matcher:
# print(match.start())
# print(match.group())
#2. x='[^abc]' except abc
#eg:
# import re
# x="[^abc]"
# matcher=re.finditer(x,"abt cq5kz")
# for match in matcher:
# print(match.start())
# print(match.group())
#3. x='[a-z]' a to z ^ cap means that is not included
#eg
# import re
# x="[a-z]"
# matcher=re.finditer(x,"abt cq5kz")
# for match in matcher:
# print(match.start())
# print(match.group())
#eg with ^
# import re
# x="[^a-z]"
# matcher=re.finditer(x,"abt cq5kz")
# for match in matcher:
# print(match.start())
# print(match.group())
#4. x='[A-Z]' A TO Z
# import re
# x="[A-Z]"
# matcher=re.finditer(x,"abt SC5kZ")
# for match in matcher:
# print(match.start())
# print(match.group())
#5.X="[a-zA-Z]" BOTH LOWER AND UPPERCASE ARE CHECKED
import re
x="[a-zA-Z]"
matcher=re.finditer(x,"abtABIkz")
for match in matcher:
print(match.start())
print(match.group())
#6. X="[0-9]"
# import re
# x="[0-9]"
# matcher=re.finditer(x,"ab1z7")
# for match in matcher:
# print(match.start())
# print(match.group())
#7.x="[a-zA-Z0-9]"
# import re
# x="[a-zA-Z0-9]"
# matcher=re.finditer(x,"ab72ABIkz")
# for match in matcher:
# print(match.start())
# print(match.group())
#8.x='\s' check space
# import re
# x="\s"
# matcher=re.finditer(x,"ab tAB Ikz")
# for match in matcher:
# print(match.start())
# print(match.group())
#9.x='\d' check the digits
# import re
# x="\d"
# matcher=re.finditer(x,"ab7tAB12kz")
# for match in matcher:
# print(match.start())
# print(match.group())
#9. x='\D' except digits
# import re
# x="\D"
# matcher=re.finditer(x,"ab001tAB5236Ikz")
# for match in matcher:
# print(match.start())
# print(match.group())
#10. x='\w' all words except special characters
# import re
# x="\w"
# matcher=re.finditer(x,"ab %tAB @Ikz")
# for match in matcher:
# print(match.start())
# print(match.group())
#11.x='\W' for special characters
# import re
# x="\W"
# matcher=re.finditer(x,"ab!!tAB@Ikz")
# for match in matcher:
# print(match.start())
# print(match.group())
| false |
af8f6a301de6b7bcf12dce96898944ec99928b6d | vickylee745/Learn-Python-3-the-hard-way | /ex30.py | 925 | 4.46875 | 4 |
people = 30
cars = 40
trucks = 15
if cars > people:
print("We should take the cars.")
elif cars < people:
print("We should not take the cars.")
else:
print("We can't decide.")
if trucks > cars:
print("That's too many trucks.")
elif trucks < cars:
print("Maybe we could take the trucks.")
else:
print("We still can't decide.")
if people > trucks:
print("Alright, let's just take the trucks.")
else:
print("Fine, let's stay home then.")
''' or above decision process can be written:'''
def what_to_take(people, cars, trucks):
if trucks >= people and trucks < cars:
print ("Alright take trucks")
elif trucks < people and cars >= people:
print ("not enough trucks, take cars")
else:
print ("Can't fit everyone")
people = input("number of people: ")
cars = input("number of cars: ")
trucks = input("number of trucks: ")
what_to_take(people, cars, trucks)
| true |
8684285a1e580d4249b60a838d4eec6bc85222db | kkashii/Python-NRS568 | /Class 2/Challenge2.2.py | 391 | 4.125 | 4 | # List Overlap
list_a = ['dog', 'cat', 'rabbit', 'hamster', 'gerbil']
list_b = ['dog', 'hamster', 'snake']
def overlap(list_a, list_b):
list_c=[value for value in list_a if value in list_b]
return list_c
print(overlap(list_a, list_b))
for x in list_a:
if x not in list_b:
print(x)
for y in list_b:
if y not in list_a:
print (y)
| false |
33dffd72dbc71e9bf6d0669e3570557c5102418d | Chyi341152/pyConPaper | /Concurrency/codeSample/Part4_Thread_Synchronuzation_Primitives/sema_signal.py | 1,236 | 4.65625 | 5 | #!/usr/bin/env python3
# -*- coding:utf-8 -*-
# sema_signal.py
#
# An example of using a semaphore for signaling between threads
import threading
import time
done = threading.Semaphore(0) # Resource control.
item = None
def producer():
global item
print("I'm the producer and I produce data.")
print("Producer is going to sleep.")
time.sleep(5)
item = "Hello"
print("Producer is alive. Signaling the consumer.")
done.release() # Increments the count and signals waiting threads
def consumer():
print("I'm a consumer and I want for date.")
print("Consumer is waiting.")
done.acquire() # Waits for the count is 0, otherwise decrements the count and continues
print("Consumer got", item)
t1 = threading.Thread(target=producer)
t2 = threading.Thread(target=consumer)
t1.start()
t2.start()
"""
Semaphore Uses:
1. Resource control
You can limit the number of threads performing certain operations.For example, performing database queries making network connections
2. Signaling
Semaphores can be used to send "signals" between threads. For example, having one thread wake up another thread
"""
| true |
3df062d10ddff464e32ed814ddd2012dc97d326d | DataKind-DUB/PII_Scanner | /namelist.py | 1,231 | 4.25 | 4 | #!/usr/bin/python3 -i
"""This file contain functions to transform a list of names text file to a set
"""
def txt_to_set(filename):
"""(str) -> set()
Convert a text file containing a list of names into a set"""
res = set()
with open(filename,'r') as f:
for line in f:
l = line.strip().split()
for item in l:
try: int(item[0])
except ValueError: res.add(item)
return res
def cleaned_txt_to_set(filename):
"""(str) -> set()
Convert a cleaned text file containing a list of names into a set
a clean text is formatted as a name each line
Same effect as txt_to_set but quicker"""
res = set()
with open(filename,'r') as f:
for line in f:
res.add(line.strip())
return res
def clean_txt(filename):
"""(str) -> None
Rewrite a file with a list of name under the format name, pass line, name"""
tmp = []
with open(filename,'r') as f:
for line in f:
l = line.split()
for item in l:
try: int(item[0])
except ValueError: tmp.append(item)
with open(filename,'w') as f:
for item in tmp: f.write(item+"\n")
if __name__=="__main__":
print(txt_to_set("Irish Name Dict with Number.txt"))
print(cleaned_txt_to_set("Irish Name Dict with Number.txt"))
| true |
061f34d0ae2055ecaa8c26aaa5ef2d43dcb473cf | tzpBingo/github-trending | /codespace/python/tmp/strings.py | 661 | 4.15625 | 4 | """
字符串常用操作
Version: 0.1
Author: 骆昊
Date: 2018-02-27
"""
str1 = 'hello, world!'
print('字符串的长度是:', len(str1))
print('单词首字母大写: ', str1.title())
print('字符串变大写: ', str1.upper())
# str1 = str1.upper()
print('字符串是不是大写: ', str1.isupper())
print('字符串是不是以hello开头: ', str1.startswith('hello'))
print('字符串是不是以hello结尾: ', str1.endswith('hello'))
print('字符串是不是以感叹号开头: ', str1.startswith('!'))
print('字符串是不是一感叹号结尾: ', str1.endswith('!'))
str2 = '- \u9a86\u660a'
str3 = str1.title() + ' ' + str2.lower()
print(str3)
| false |
307b9a3b8ac5d9d65ad3c9a834cabab1b51c6c62 | tzpBingo/github-trending | /codespace/python/tmp/example16.py | 1,731 | 4.25 | 4 | """
魔术方法
如果要把自定义对象放到set或者用作dict的键
那么必须要重写__hash__和__eq__两个魔术方法
前者用来计算对象的哈希码，后者用来判断两个对象是否相同
哈希码不同的对象一定是不同的对象，但哈希码相同未必是相同的对象（哈希码冲撞）
所以在哈希码相同的时候还要通过__eq__来判定对象是否相同
"""
class Student():
__slots__ = ('stuid', 'name', 'gender')
def __init__(self, stuid, name):
self.stuid = stuid
self.name = name
def __hash__(self):
return hash(self.stuid) + hash(self.name)
def __eq__(self, other):
return self.stuid == other.stuid and \
self.name == other.name
def __str__(self):
return f'{self.stuid}: {self.name}'
def __repr__(self):
return self.__str__()
class School():
def __init__(self, name):
self.name = name
self.students = {}
def __setitem__(self, key, student):
self.students[key] = student
def __getitem__(self, key):
return self.students[key]
def main():
# students = set()
# students.add(Student(1001, '王大锤'))
# students.add(Student(1001, '王大锤'))
# students.add(Student(1001, '白元芳'))
# print(len(students))
# print(students)
stu = Student(1234, '骆昊')
stu.gender = 'Male'
# stu.birth = '1980-11-28'
print(stu.name, stu.birth)
school = School('千锋教育')
school[1001] = Student(1001, '王大锤')
school[1002] = Student(1002, '白元芳')
school[1003] = Student(1003, '白洁')
print(school[1002])
print(school[1003])
if __name__ == '__main__':
main()
| false |
9a049b0cd58d5c48f564591f9edb5ff6b19e1ae9 | Ishita46/PRO-C97-NUMBER-GUESSING-GAME | /gg.py | 551 | 4.25 | 4 | import random
print("Number Guessing Game")
Number = random.randint(1,9)
chances = 0
print("Guess a number between 1-9")
while chances < 5:
guess = int(input("Enter your guess"))
if guess == Number:
print("Congratulations you won!")
break
elif guess < Number:
print("You loose, try to choose a higher number.",guess)
else:
print("Your guess was too high, guess a low number than that.",guess)
chances += 1
if not chances < 5:
print("You loose! The actual number is: ",Number) | true |
36a5612ef360eeb868da742bc9f29f535cb3fb84 | saparia-data/data_structure | /geeksforgeeks/maths/2_celcius_to_fahrenheit.py | 852 | 4.5 | 4 | """
Given a temperature in celsius C. You need to convert the given temperature to Fahrenheit.
Input Format:
The first line of input contains T, denoting number of testcases. Each testcase contains single integer C denoting the temperature in celsius.
Output Format:
For each testcase, in a new line, output the temperature in fahrenheit.
Your Task:
This is a function problem. You only need to complete the function CtoF that takes C as parameter and returns temperature in fahrenheit( in double).
The flooring and printing is automatically done by the driver code.
Constraints:
1 <= T <= 100
1 <= C <= 104
Example:
Input:
2
32
50
Output:
89
122
Explanation:
Testcase 1: For 32 degree C, the temperature in Fahrenheit = 89.
def cToF(C):
hints:
1) �F = �C x 9/5 + 32
"""
coding="utf8"
def cToF(C):
return (C * 9/5) + 32
print(cToF(32)) | true |
795bc9954ddcd5e52ad3477bf197c5bedd9a9650 | saparia-data/data_structure | /geeksforgeeks/linked_list/segregate_even_and_odd_nodes_in_linked_list_difficullt.py | 2,808 | 4.15625 | 4 | '''
Given a Linked List of integers, write a function to modify the linked list such that all even numbers appear before all the odd numbers in the modified linked list.
Also, keep the order of even and odd numbers same.
https://www.geeksforgeeks.org/segregate-even-and-odd-elements-in-a-linked-list/
'''
class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
def __init__(self):
self.head = None
def push(self, data):
new_node = Node(data)
new_node.next = self.head
self.head = new_node
def printList(self):
temp = self.head
while(temp):
print(temp.data)
temp = temp.next
def segregateEvenOdd(self):
# Starting node of list having
# even values.
evenStart = None
# Ending node of even values list.
evenEnd = None
# Starting node of odd values list.
oddStart = None
# Ending node of odd values list.
oddEnd = None
# Node to traverse the list.
currNode = self.head
while(currNode != None):
val = currNode.data
# If current value is even, add
# it to even values list.
if(val % 2 == 0):
if(evenStart == None):
evenStart = currNode
evenEnd = evenStart
else:
evenEnd.next = currNode
evenEnd = evenEnd.next
# If current value is odd, add
# it to odd values list.
else:
if(oddStart == None):
oddStart = currNode
oddEnd = oddStart
else:
oddEnd.next = currNode
oddEnd = oddEnd.next
# Move head pointer one step in
# forward direction
currNode = currNode.next
# If either odd list or even list is empty,
# no change is required as all elements
# are either even or odd.
if(oddStart == None or evenStart == None):
return
# Add odd list after even list.
evenEnd.next = oddStart
oddEnd.next = None
# Modify head pointer to
# starting of even list.
self.head = evenStart
llist = LinkedList()
llist.push(11)
llist.push(10)
llist.push(9)
llist.push(6)
llist.push(4)
llist.push(1)
llist.push(0)
llist.printList()
llist.segregateEvenOdd()
print("Linked list after re-arranging")
llist.printList() | true |
17d7ca663bf8697b79bd7824a49e561839818cf3 | saparia-data/data_structure | /pepcoding/dynamic_programming/4_climb_stairs_with_minimum_moves.py | 1,375 | 4.15625 | 4 | '''
1. You are given a number n, representing the number of stairs in a staircase.
2. You are on the 0th step and are required to climb to the top.
3. You are given n numbers, where ith element's value represents - till how far from the step you
could jump to in a single move. You can of-course fewer number of steps in the move.
4. You are required to print the number of minimum moves in which you can reach the top of
staircase.
Note -> If there is no path through the staircase print null.
Sample Input:
10
3
3
0
2
1
2
4
2
0
0
Sample Output:
4
https://www.youtube.com/watch?v=d42uDPBOXSw&list=PL-Jc9J83PIiG8fE6rj9F5a6uyQ5WPdqKy&index=4
https://www.youtube.com/watch?v=Zobz9BXpwYE&list=PL-Jc9J83PIiG8fE6rj9F5a6uyQ5WPdqKy&index=5
'''
import sys
def climbStairsWithMinMoves(arr):
dp = [None] * (len(arr) + 1)
dp[n] = 0
for i in range(n-1, -1, -1):
minn = sys.maxsize
j = 1
while(j <= arr[i] and i+j <= len(dp)):
if(dp[i + j] != None):
minn = min(minn, dp[i + j])
j += 1
if(minn != sys.maxsize):
dp[i] = minn + 1
return dp[0]
if __name__ == "__main__":
arr = [3,3,0,2,1,2,4,2,0,0]
n = 10
print(climbStairsWithMinMoves(arr)) | true |
fea46e295115747dfcc38e182eb865603b501e74 | saparia-data/data_structure | /pepcoding/recursion/1_tower_of_hanoi.py | 961 | 4.15625 | 4 | '''
1. There are 3 towers. Tower 1 has n disks, where n is a positive number. Tower 2 and 3 are empty.
2. The disks are increasingly placed in terms of size such that the smallest disk is on top and largest disk is at bottom.
3. You are required to
3.1. Print the instructions to move the disks.
3.2. from tower 1 to tower 2 using tower 3
3.3. following the rules
3.3.1 move 1 disk at a time.
3.3.2 never place a smaller disk under a larger disk.
3.3.3 you can only move a disk at the top.
sample input:
3
10
11
12
sample output:
1[10 -> 11]
2[10 -> 12]
1[11 -> 12]
3[10 -> 11]
1[12 -> 10]
2[12 -> 11]
1[10 -> 11]
https://www.youtube.com/watch?v=QDBrZFROuA0&list=PL-Jc9J83PIiFxaBahjslhBD1LiJAV7nKs&index=12
'''
def toh(n, t1, t2, t3):
if(n == 0):
return
toh(n-1, t1, t3, t2)
print(str(n) + "[" + str(t1) + " -> " + str(t2) + "]")
toh(n-1, t3, t2, t1)
toh(3, 10, 11, 12) | true |
60937f08311c74e6773fa09eaec056097efb9496 | saparia-data/data_structure | /pepcoding/generic_tree/17_is_generic_tree_symmetric.py | 1,776 | 4.1875 | 4 | '''
The function is expected to check if the tree is symmetric, if so return true otherwise return false.
For knowing symmetricity think of face and hand. Face is symmetric while palm is not.
Note: Symmetric trees are mirror image of itself.
Sample Input:
20
10 20 50 -1 60 -1 -1 30 70 -1 80 -1 90 -1 -1 40 100 -1 110 -1 -1 -1
Sample Output:
true
https://www.youtube.com/watch?v=ewEAjK83ZVM&list=PL-Jc9J83PIiEmjuIVDrwR9h5i9TT2CEU_&index=39
https://www.youtube.com/watch?v=gn2ApElF2i0&list=PL-Jc9J83PIiEmjuIVDrwR9h5i9TT2CEU_&index=40
'''
class Node:
def __init__(self):
self.data = None
self.children = []
def create_generic_tree(arr):
root = Node()
st = [] # create empty stack
for i in range(len(arr)):
if(arr[i] == -1):
st.pop()
else:
t = Node()
t.data = arr[i]
if(len(st) > 0):
st[-1].children.append(t)
else:
root = t
st.append(t)
return root
def areMirror(root1, root2):
if(len(root1.children) != len(root2.children)):
return False
for i in range(len(root1.children)):
j = len(root1.children) - 1 - i
c1 = root1.children[i]
c2 = root2.children[j]
if(areMirror(c1, c2) == False):
return False
return True
def isSymmetric(root):
return areMirror(root, root) # symmetric trees are mirror image of itself
if __name__ == "__main__":
arr = [10, 20, 50, -1, 60, -1, -1, 30, 70, -1, 80, -1, 90, -1, -1, 40, 100, -1, 110, -1, -1, -1]
root = create_generic_tree(arr)
print(root.data)
print(isSymmetric(root)) | true |
008b19ff7efc7fc9be540903c086a792aed6c0d2 | saparia-data/data_structure | /geeksforgeeks/maths/7_prime_or_not.py | 1,297 | 4.34375 | 4 | '''
For a given number N check if it is prime or not. A prime number is a number which is only divisible by 1 and itself.
Input:
First line contains an integer, the number of test cases 'T'. T testcases follow. Each test case should contain a positive integer N.
Output:
For each testcase, in a new line, print "Yes" if it is a prime number else print "No".
Your Task:
This is a function problem. You just need to complete the function isPrime that takes N as parameter and returns True if N is prime else returns false.
The printing is done automatically by the driver code.
Constraints:
1 <= T <= 100
1 <= N <= 103
Example:
Input:
2
5
4
Output:
Yes
No
Explanation:
Testcase 1: 5 is the prime number as it is divisible only by 1 and 5.
Hints:
Count the number of factors. If the factors of a number is 1 and itself, then the number is prime.
You can check this optimally by iterating from 2 to sqrt(n) as the factors from 2 to sqrt(n) have multiples from sqrt(n)+1 to n.
So, by iterating till sqrt(n) only, you can check if a number is prime.
def isPrime(N):
'''
import math
def isPrime(N):
if(N == 1):
return True
else:
for i in range(2, int(math.sqrt(N)) + 1):
if (N % i == 0):
return False
return True
print(isPrime(9)) | true |
3fb0fefa594130bd865c342d60d0f7ff34127f7f | saparia-data/data_structure | /geeksforgeeks/linked_list/4_Insert_in_Middle_of_Linked_List.py | 1,343 | 4.15625 | 4 | '''
Given a linked list of size N and a key. The task is to insert the key in the middle of the linked list.
Input:
The first line of input contains the number of test cases T. For each test case,
the first line contains the length of linked list N
and the next line contains N elements to be inserted into the linked list and the last line contains the element to be inserted to the middle.
Output:
For each test case, there will be a single line of output containing the element of the modified linked list.
User Task:
The task is to complete the function insertInMiddle() which takes head reference and element to be inserted as the arguments.
The printing is done automatically by the driver code.
Expected Time Complexity : O(N)
Expected Auxilliary Space : O(1)
Constraints:
1 <= T <= 100
1 <= N <= 104
Example:
Input:
2
3
1 2 4
3
4
10 20 40 50
30
Output:
1 2 3 4
10 20 30 40 50
Explanation:
Testcase 1: The new element is inserted after the current middle element in the linked list.
Testcase 2: The new element is inserted after the current middle element in the linked list and Hence, the output is 10 20 30 40 50.
'''
def insertInMid(head,new_node):
slow = head
fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
new_node.next = slow.next
slow.next = new_node | true |
296acd0cb1655e3fd6d57e80b9bd72529ffb4ecc | saparia-data/data_structure | /geeksforgeeks/matrix/8_Boundary_traversal_matrix.py | 2,384 | 4.34375 | 4 | '''
You are given a matrix A of dimensions n1 x m1.
The task is to perform boundary traversal on the matrix in clockwise manner.
Input:
The first line of input contains T denoting the number of testcases. T testcases follow.
Each testcase two lines of input. The first line contains dimensions of the matrix A, n1 and m1.
The second line contains n1*m1 elements separated by spaces.
Output:
For each testcase, in a new line, print the boundary traversal of the matrix A.
Your Task:
This is a function problem. You only need to complete the function boundaryTraversal()
that takes n1, m1 and matrix as parameters and prints the boundary traversal.
The newline is added automatically by the driver code.
Constraints:
1 <= T <= 100
1 <= n1, m1<= 30
0 <= arri <= 100
Examples:
Input:
4
4 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
3 4
12 11 10 9 8 7 6 5 4 3 2 1
1 4
1 2 3 4
4 1
1 2 3 4
Output:
1 2 3 4 8 12 16 15 14 13 9 5
12 11 10 9 5 1 2 3 4 8
1 2 3 4
1 2 3 4
Explanation:
Testcase1: The matrix is:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
The boundary traversal is 1 2 3 4 8 12 16 15 14 13 9 5
Testcase 2: Boundary Traversal will be 12 11 10 9 5 1 2 3 4 8.
Testcase 3: Boundary Traversal will be 1 2 3 4.
Testcase 4: Boundary Traversal will be 1 2 3 4.
hints:
Traverse the top boundary first then the right boundary then the bottom boundary
then the left boundary and print the elements simultaneously.
'''
def BoundaryTraversal(a, n, m):
"""
a: matrix
n: no. of rows
m: no. of columns
"""
if(min(n,m) == 1):
for row in a:
print(*row, end = " ")
return
row_idx, col_idx = 0, 0
#print first row
while(col_idx < m):
print(a[row_idx][col_idx], end = " ")
col_idx += 1
#print last column
col_idx = m - 1
row_idx += 1
while(row_idx < n):
print(a[row_idx][col_idx], end = " ")
row_idx += 1
#print last row
row_idx = n - 1
col_idx -= 1
while(col_idx > -1):
print(a[row_idx][col_idx], end = " ")
col_idx -= 1
#print first column
row_idx -= 1
col_idx = 0
while(row_idx > 0):
print(a[row_idx][col_idx], end = " ")
row_idx -= 1
a = [
[1,2,3],
[4,5,6],
[7,8,9]
]
n, m = 3, 3
BoundaryTraversal(a, n, m) | true |
789288e7c8987df76436b463f7e8fa6f2d9b1a8a | saparia-data/data_structure | /geeksforgeeks/tree/6_height_of_binary_tree.py | 698 | 4.125 | 4 | '''
Hint:
1. If tree is empty then return 0
2. Else
(a) Get the max depth of left subtree recursively i.e.,
call maxDepth( tree->left-subtree)
(a) Get the max depth of right subtree recursively i.e.,
call maxDepth( tree->right-subtree)
(c) Get the max of max depths of left and right
subtrees and add 1 to it for the current node.
max_depth = max(max dept of left subtree,
max depth of right subtree)
+ 1
(d) Return max_depth
'''
def height(root):
if(root is None):
return 0
return (1 + max(height(root.left), height(root.right)))
| true |
770ac8812cac09b77d0990edacc7fed78c612480 | saparia-data/data_structure | /geeksforgeeks/maths/1_absolute_value_solved.py | 1,062 | 4.375 | 4 | '''
You are given an interger I. You need to print the absolute value of the interger I.
Input Format:
The first line of input contains T, denoting number of testcases. Each testcase contains single integer I which may be positive or negative.
Output Format:
For each testcase, in a new line, output the absolute value.
Your Task:
This is function problem. You only need to complete the function absolute that takes integer I as parameter and returns the absolute value of I.
All other things are taken care of by the driver code.
Constraints:
1 <= T <= 100
-106 <= I <= 106
Example:
Input:
2
-32
45
Output:
32
45
Explanation:
Testcase 1: Since -32 is negative, we prints its positive equavalent, i.e., 32
Testcase 1: Since 45 is positive, we prints its value as it is, i.e., 45
Hints:
1. Absolute value of an integer means value >=0. So if a number is negative you can find its absolute value by multiplying it by -1.
For positive numbers no need to multiply as it's already positive.
'''
def printAbs(n):
return abs(n)
print(printAbs(-32)) | true |
35030f181df15a7fb1ad550279aadc98a6baf954 | saparia-data/data_structure | /geeksforgeeks/sorting/11_Counting_Sort.py | 1,265 | 4.25 | 4 | '''
Given a string S consisting of lowercase latin letters, arrange all its letters in lexographical order using Counting Sort.
Input:
The first line of the input contains T denoting number of testcases.Then T test cases follow. Each testcase contains positive integer N denoting the length of string.The last line of input contains the string S.
Output:
For each testcase, in a new line, output the sorted string.
Your Task:
This is a function problem. You only need to complete the function countSort() that takes char array as parameter.
The printing is done by driver code.
Constraints:
1 <= T <= 105
1 <= N <= 105
Example:
Input:
2
5
edsab
13
geeksforgeeks
Output:
abdes
eeeefggkkorss
Explanation:
Testcase 1: In lexicographical order , string will be abdes.
Testcase 2: In lexicographical order , string will be eeeefggkkorss.
hints:
1)
Store the count of elements. And use this count to sort the elements accordingly.
'''
def countingSort(s,n):
freq=[0 for i in range(256)]
print(freq)
for char in s:
freq[ord(char)] += 1
print(freq)
for i in range(256):
for j in range(freq[i]):
print(chr(i),end="")
s = "geeksforgeeks"
n = len(s)
countingSort(s, n) | true |
d476ee8753b489e9a160984ca1a71e49eed86f2d | saparia-data/data_structure | /geeksforgeeks/array/3_majority_in_array_solved.py | 2,439 | 4.40625 | 4 | '''
We hope you are familiar with using counter variables. Counting allows us to find how may times a certain element appears in an array or list.
You are given an array arr[] of size N. You are also given two elements x and y. Now, you need to tell which element (x or y) appears most in the array.
In other words, print the element, x or y, that has highest frequency in the array. If both elements have the same frequency,
then just print the smaller element.
Input Format:
The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase contains 3 lines of input.
The first line contains size of array denoted by n. The second line contains the elements of the array separated by spaces.
The third line contains two integers x and y separated by a space.
Output Format:
For each testcase, in a newline, print the element with highest occurrence in the array. If occurrences are same, then print the smaller element.
Your Task:
Since, this is a function problem. You don't need to take any input, as it is already accomplished by the driver code.
You just need to complete the function majorityWins() that takes array, n, x, y as parameters and return the element with highest
Constraints:
1 <= T <= 100
1 <= n <= 103
0 <= arri , x , y <= 108
Examples:
Input:
2
11
1 1 2 2 3 3 4 4 4 4 5
4 5
8
1 2 3 4 5 6 7 8
1 7
Output:
4
1
Explanation:
Testcase 1: n=11; elements = {1,1,2,2,3,3,4,4,4,4,5}; x=4; y=5
x frequency in arr is = 4 times
y frequency in arr is = 1 times
x has higher frequency, so we print 4.
Testcase 2: n=8; elements = {1,2,3,4,5,6,7,8}; x=1; y=7
x frequency in arr is 1 times
y frequency in arr is 1 times
both have same frequency, so we look for the smaller element.
x=1 is smaller than y=7, so print 1.
'''
def majorityWins(arr, n, x, y):
c_x = 0
c_y = 0
for i in range(n):
if(arr[i] == x):
c_x = c_x + 1
elif(arr[i] == y):
c_y = c_y + 1
else:
continue
#print(c_x)
#print(c_y)
if(c_x > c_y):
return x
elif(c_x < c_y):
return y
elif(c_x == c_y):
return min(x,y)
arr = [854,190,562,906,2,625,678,667,779,755,699,842,970,81,253,155,456,830,575,640,962,44,124,209,841,571,211,213,904,993,130,996,866,938,553,114,3,79,18,823,176,75,740,83,391,223,289,561,927,351,744]
n = len(arr)
x = 10
y = 40
print(majorityWins(arr,n,x,y))
| true |
8f2e60355dfe700cfde7b30b569d431ce959b7e5 | saparia-data/data_structure | /geeksforgeeks/strings/6_check_if_string_is_rotated_by_two_places.py | 1,986 | 4.21875 | 4 | '''
Given two strings a and b. The task is to find if the string 'b' can be obtained by rotating another string 'a' by exactly 2 places.
Input:
The first line of input contains an integer T denoting the number of test cases.
Then T test cases follow. In the next two lines are two string a and b respectively.
Output:
For each test case in a new line print 1 if the string 'a' can be obtained by rotating string 'b' by two places else print 0.
User Task:
The task is to complete the function isRotated() which checks if given strings can be formed by rotations.
The function returns true if string 1 can be obtained by rotating string 2 by two places, else it returns false.
Expected Time Complexity: O(N).
Expected Space Complexity: O(N).
Challenge: Try doing it in O(1) space complexity.
Constraints:
1 <= T <= 50
1 <= length of a, b < 100
Example:
Input:
2
amazon
azonam
geeksforgeeks
geeksgeeksfor
Output:
1
0
Explanation:
Testcase 1: amazon can be rotated anti-clockwise by two places, which will make it as azonam.
Testcase 2: If we rotate geeksforgeeks by two place in any direction , we won't get geeksgeeksfor.
hints:
Step1: There can be only two cases:
a) Clockwise rotated
b) Anti-clockwise rotated
Step 2: If clockwise rotated that means elements are shifted in right. So, check if a substring[2.... len-1] of string2
when concatenated with substring[0,1] of string2 is equal to string1. Then, return true.
Step 3: Else, check if it is rotated anti-clockwise that means elements are shifted to left.
So, check if concatenation of substring[len-2, len-1] with substring[0....len-3] makes it equals to string1. Then return true.
Step 4: Else, return false.
'''
def isRotated(s,p):
n=len(p)
if(n<3):
return p==s
anticlock_str=p[2:]+p[0:2]
clockwise_str=p[-2]+p[-1]+p[:n-2]
if(s==anticlock_str or s==clockwise_str):
return True
return False
s = "amazon"
p = "onamaz"
print(isRotated(s, p)) | true |
b6bf8024d4250adafec321751f4317591391a1c9 | saparia-data/data_structure | /geeksforgeeks/tree/26_Foldable_Binary_Tree.py | 1,007 | 4.4375 | 4 | '''
Given a binary tree, find out if the tree can be folded or not.
-A tree can be folded if left and right subtrees of the tree are structure wise mirror image of each other.
-An empty tree is considered as foldable.
Consider the below tree: It is foldable
10
/ \
7 15
\ /
9 11
'''
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
def foldableUtil(root1, root2):
if(root1 is None and root2 is None):
return True
if(root1 is None or root2 is None):
return False
return foldableUtil(root1.left, root2.right) and foldableUtil(root1.right, root2.left)
def foldable(root):
if(root is None):
return True
result = foldableUtil(root.left, root.right)
return result
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.right = Node(5)
root.right.left = Node(4)
print(foldable(root)) | true |
b05134767f55b443af849edfa92df6ae5937491b | saparia-data/data_structure | /geeksforgeeks/matrix/11_Reversing_the _columns_Matrix.py | 1,919 | 4.40625 | 4 | '''
You are given a matrix A of dimensions n1 x m1.
The task is to reverse the columns(first column exchanged with last column and so on).
Input:
The first line of input contains T denoting the number of testcases. T testcases follow.
Each testcase two lines of input. The first line contains dimensions of the matrix A, n1 and m1.
The second line contains n1*m1 elements
separated by spaces.
Output:
For each testcase, in a new line, print the resultant matrix.
Your Task:
This is a function problem. You only need to complete the function reverseCol()
that takes n1, m1, and matrix as parameter and modifies the matrix.
The driver code automatically appends a new line.
Constraints:
1 <= T <= 100
1 <= n1, m1 <= 30
0 <= arri <= 100
Examples:
Input:
2
4 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
2 3
4 3 2 1 5 6
Output:
3 2 1 7 6 5 11 10 9 15 14 13
2 3 4 6 5 1
Explanation:
Testcase 1: Original array is as follows:
1 2 3
5 6 7
9 10 11
13 14 15
Array after exchanging columns:
3 2 1
7 6 5
11 10 9
15 14 13
Testcase 2: Original matrix is as follows:
4 3 2
1 5 6
After reversing the column of matrix
2 3 4
6 5 1
hints:
Take two pointers low and high, low at 0 and high at n-1.
Start swapping elements at two pointers and increasing the low and decreasing the high one.
Do the above for all the rows.
Keep in mind, when number of rows is odd, no need to take care of middle column.
'''
#arr1 is matrix
#n1 is rows
#m1 is cols
def reverseCol(n1, m1, arr1):
for i in range(n1):
for j in range(m1//2):
t = arr1[i][j]
arr1[i][j] = arr1[i][m1 - 1 - j]
arr1[i][m1 - 1 - j] = t
for i in range(n1):
for j in range(m1):
print(arr1[i][j], end = " ")
print("\n")
arr1 = [[8, 9, 7, 6],
[4, 7, 6, 5],
[3, 2, 1, 8],
[9, 9, 7, 7]]
n1, m1 = 4, 4
reverseCol(n1, m1, arr1) | true |
e76d09cc97d9d6d8677ae32ba52425ec5fa34a0f | cginiel/si507 | /lecture/week3/2020.01.21.py | 2,638 | 4.46875 | 4 | # # import datetime
# # date_now = datetime.datetime.now()
# # print(date_now.year)
# # print(date_now.month)
# # print(date_now.day)
# # print(type(date_now))
# # class is a type of thing, object is a particular instance of that class!!!!!
# # BEGIN CLASS DEFINITION
# """
# class Dog:
# def __init__(self, nm, br):
# self.name = nm
# self.breed = br
# """
# # END CLASS DEFINITION
# # every time we call "Dog" the program is going to
# # set up the function written above (which is how
# # we "index" or call out the specific names below)
# d1 = Dog('Fido', 'German Shepherd')
# d2 = Dog('Rufus', 'Lhasa Apso')
# print (d1.name, 'is a', d1.breed)
# print (d2.name, 'is a', d2.breed)
# # "self" essentially equals d1 because self corresponds
# # to the object in question
# # in that sense, "self," in the case of d2
# # equals d2
# # "all class methods will have to have self as the first
# # parameter"
# class Dog:
# large_dogs = ['German Shepherd', 'Golden Retriever',
# 'Rottweiler', 'Collie',
# 'Mastiff', 'Great Dane']
# small_dogs = ['Lhasa Apso', 'Yorkshire Terrier',
# 'Beagle', 'Dachshund', 'Shih Tzu']
# def __init__(self, nm, br):
# self.name = nm
# self.breed = br
# def speak(self):
# if self.breed in Dog.large_dogs:
# print('woof')
# elif self.breed in Dog.small_dogs:
# print('yip')
# else:
# print('rrrrr')
# d1 = Dog('Fido', 'German Shepherd')
# d2 = Dog('Rufus', 'Lhasa Apso')
# d3 = Dog('Fred', 'Mutt')
# d1.speak()
# d2.speak()
# d3.speak()
class Dog:
large_dogs = ['German Shepherd', 'Golden Retriever',
'Rottweiler', 'Collie',
'Mastiff', 'Great Dane']
small_dogs = ['Lhasa Apso', 'Yorkshire Terrier',
'Beagle', 'Dachshund', 'Shih Tzu']
def __init__(self, nm, br, a):
self.name = nm
self.breed = br
self.age = a
def speak(self):
if self.breed in Dog.large_dogs:
print('woof')
elif self.breed in Dog.small_dogs:
print('yip')
else:
print('rrrrr')
def print_dog_years(d):
'''prints a dog's name and age in dog years
dog years = actual years * 7
Parameters
----------
dog : Dog
The dog to print
Returns
-------
none
'''
dog_years = d.age * 7
print(f"{d.name} is a {d.breed} who is {dog_years} years old in dog years")
pass #TODO: implement
kennel = [
Dog('Fido', 'German Shepherd', 4),
Dog('Rufus', 'Lhasa Apso', 7),
Dog('Fred', 'Mutt', 11)
]
for dog in kennel:
print_dog_years(dog) | true |
8288e5fd26c497ab9b285289e0218d5ab2cba302 | jpchato/pdx_code | /programming_101/unit_3/exercise_1.py | 1,026 | 4.15625 | 4 | import math
# 1.1
tri_side_1 = 1
tri_side_2 = 2
hypotenuse = (tri_side_1*tri_side_1 + tri_side_2*tri_side_2)
def triangle_perimeter(tri_side_1, tri_side_2, hypotenuse):
print(tri_side_1 + tri_side_2 + hypotenuse)
return tri_side_1 + tri_side_2 + hypotenuse
# 1.2
def triangle_area(tri_side_2, tri_side_1):
print((tri_side_1*tri_side_2)/2)
return (tri_side_1*tri_side_2)/2
#1.3
radius = 5
pi = math.pi
print(pi)
def circle_circumference(radius):
print(int(radius*math.pi*2))
return int(radius*math.pi*2)
# 1.4
def sphere_volume(radius):
volume = (math.pi*radius*radius*radius*4)/3
print (volume)
return volume
# 1.5
radius_2 = 3
def annulus_area(radius, radius_2):
area = int(math.pi*(radius*radius-radius_2*radius_2))
print (area)
return area
if __name__ == "__main__":
triangle_perimeter(tri_side_1, tri_side_2, hypotenuse)
triangle_area(tri_side_2, tri_side_1)
circle_circumference(radius)
sphere_volume(radius)
annulus_area(radius, radius_2)
| false |
1d07ee37723cc22548908a0a75b02facb6664100 | alexbenko/pythonPractice | /classes/magic.py | 787 | 4.125 | 4 | #how to use built in python methods like print on custom objects
class Car():
speed = 0
def __init__(this,color,make,model):
this.color = color
this.make = make
this.model = model
def __str__(this):
return f'Your {this.make},{this.model} is {this.color} and is currently going {this.speed} mph'
def accelerate(this,toGoTo):
print(f'Accelerating car to {toGoTo},currently going {this.speed} mph...')
while this.speed < toGoTo:
this.speed += 1
print(f'{this.speed} mph...')
print(f'Accelerated to {toGoTo} !')
volt = Car(color='White',make='Chevorlet',model='Volt')
print(volt)
#=>Your Chevorlet,Volt is White and is currently going 0 mph
volt.accelerate(100)
print(volt)
#=>Your Chevorlet,Volt is White and is currently going 100 mph
| true |
b1ec77cac2953516ced165066ea926e67f236c14 | xenron/sandbox-github-clone | /qiwsir/algorithm/delete_space.py | 508 | 4.375 | 4 | #! /usr/bin/env python
#coding:utf-8
#删除一个字符串中连续超过一次的空格。
def del_space(string):
split_string = string.split(" ") #以空格为分割，生成list，list中如果含有空格，则该空格是连续空格中的后一个
string_list = [i for i in string if i!=""]
result_string = " ".join(string_list)
return result_string
if __name__=="__main__":
one_str = "Hello, I am Qiwsir."
string = del_space(one_str)
print one_str
print string
| false |
b821bd68661a42ae46870bd01178d181656149e6 | xenron/sandbox-github-clone | /qiwsir/algorithm/divide.py | 2,005 | 4.125 | 4 | #! /usr/bin/env python
#coding:utf-8
def divide(numerator, denominator, detect_repetition=True, digit_limit=None):
# 如果是无限小数，必须输入限制的返回小数位数:digit_limit
# digit_limit = 5，表示小数位数5位，注意这里的小数位数是截取，不是四舍五入.
if not detect_repetition and digit_limit == None:
return None
decimal_found = False
v = numerator // denominator
numerator = 10 * (numerator - v * denominator)
answer = str(v)
if numerator == 0:
return answer
answer += '.'
# Maintain a list of all the intermediate numerators
# and the length of the output at the point where that
# numerator was encountered. If you encounter the same
# numerator again, then the decimal repeats itself from
# the last index that numerator was encountered at.
states = {}
while numerator > 0 and (digit_limit == None or digit_limit > 0):
if detect_repetition:
prev_state = states.get(numerator, None)
if prev_state != None:
start_repeat_index = prev_state
non_repeating = answer[:start_repeat_index]
repeating = answer[start_repeat_index:]
return non_repeating + '[' + repeating + ']'
states[numerator] = len(answer)
v = numerator // denominator
answer += str(v)
numerator -= v * denominator
numerator *= 10
if digit_limit != None:
digit_limit -= 1
if numerator > 0:
return answer + '...'
return answer
if __name__=="__main__":
print "5divide2",
print divide(5,2)
print "10divide3",
print divide(10,3)
print divide(10,3,5)
print "15divide7"
print divide(15,7)
print divide(15,7,True,3)
| true |
465074ef23a66b649ac36eb6a259c9ace2732cb3 | YYYYMao/LeetCode | /374. Guess Number Higher or Lower/374.py | 1,424 | 4.15625 | 4 | 374. Guess Number Higher or Lower
We are playing the Guess Game. The game is as follows:
I pick a number from 1 to n. You have to guess which number I picked.
Every time you guess wrong, I ll tell you whether the number is higher or lower.
You call a pre-defined API guess(int num) which returns 3 possible results (-1, 1, or 0):
-1 : My number is lower
1 : My number is higher
0 : Congrats! You got it!
Example:
n = 10, I pick 6.
Return 6.
# The guess API is already defined for you.
# @param num, your guess
# @return -1 if my number is lower, 1 if my number is higher, otherwise return 0
# def guess(num):
#65ms
class Solution(object):
def guessNumber(self, n):
"""
:type n: int
:rtype: int
"""
L , R = 1 , n
M = (L+R)>>1
m = guess(M)
while m != 0 :
# print L , R ,M ,m
if m == -1:
R = M-1
elif m == 1 :
L = M+1
M = (R+L)>>1
m = guess(M)
return M
#46ms
class Solution(object):
def guessNumber(self, n):
"""
:type n: int
:rtype: int
"""
L , R = 1 , n
while 1 :
M = (R+L)>>1
m = guess(M)
if m == -1:
R = M-1
elif m == 1 :
L = M+1
elif m == 0:
return M | true |
b7cfe79cf0003bdb6d174d1471fbb672b0700b7f | codeAligned/interview_challenges | /sort_and_search/binary_search.py | 442 | 4.25 | 4 | def binary_search(iterable, target):
"""Determine if target value is in sorted iterable containing numbers"""
sorted_iterable = sorted(iterable)
low = 0
high = len(sorted_iterable) - 1
while low <= high:
midpoint = (high + low) // 2
if sorted_iterable[midpoint] == target:
return True
elif sorted_iterable[midpoint] > target:
high = midpoint - 1
elif sorted_iterable[midpoint] < target:
low = midpoint + 1
return False
| true |
c92ba78a30f78a134befad3de91f7572a201b1ef | szk0139/python_tutorial_FALL21 | /mysciW4/readdata.py | 1,018 | 4.1875 | 4 |
def read_data(columns, types = {}, filename= "data/wxobs20170821.txt"):
"""
Read data from CU Boulder Weather Stattion data file
Parameters:
colums: A dictonary of column names mapping to column indices
types: A dictonary of column names mapping to the types to which to convert each column of data
filename: A string path pointing to the CU Boulder Wather Station data file
"""
#Initialize my data variable
data = {}
for column in columns:
data[column] = []
with open(filename, "r") as datafile:
# read first three line (header)
for _ in range(3):
#print(_)
datafile.readline()
# Read and parse the rest of the file
for line in datafile:
split_line = line.split()
for column in columns:
i = columns[column]
t = types.get(column, str)
value = t(split_line[i])
data[column].append(value)
return data
| true |
9b34c6cc778d0a45f819a5667308a2e4f5f7f1f8 | Stepancherro/Algorithm | /stack.py | 1,879 | 4.1875 | 4 | # -*- encoding: utf-8 -*-
# Stack() creates a new stack that is empty.
# It needs no parameters and returns an empty stack.
# push(item) adds a new item to the top of the stack.
# It needs the item and returns nothing.
# pop() removes the top item from the stack.
# It needs no parameters and returns the item. The stack is modified.
# peek() returns the top item from the stack but does not remove it.
# It needs no parameters. The stack is not modified.
# isEmpty() tests to see whether the stack is empty.
# It needs no parameters and returns a boolean value.
class Stack():
def __init__(self, size=10):
"""
Initialize python List with size of 10 or user given input.
Python List type is a dynamic array, so we have to restrict its
dynamic nature to make it work like a static array.
"""
self.size = size
self.stack = []
self.top = 0
def push(self, value):
if self.isFull():
raise IndexError("stack is full")
else:
self.stack.append(value)
self.top += 1
def pop(self):
if self.isEmpty():
raise IndexError("stack is empty")
else:
value = self.stack[self.top - 1]
self.top -= 1
self.stack.pop()
return value
def peek(self):
if self.isEmpty():
raise IndexError("stack is empty")
return self.stack[self.top - 1]
def isFull(self):
return self.top == self.size
def isEmpty(self):
return self.top == 0
def showStack(self):
print(self.stack)
if __name__ == '__main__':
stack = Stack(7)
stack.push(15)
stack.push(6)
stack.push(2)
stack.push(9)
stack.showStack()
last_element = stack.pop()
print(last_element)
stack.showStack()
print(stack.peek())
| true |
5b7ad9bbd58b59d7e7319bad5d970b5dbba1a529 | moha0825/Personal-Projects | /Resistance.py | 1,018 | 4.375 | 4 | # This code is designed to have multiple items inputted, such as the length, radius, and
# viscosity, and then while using an equation, returns the resistance.
import math
def poiseuille(length, radius, viscosity):
Resistance = (int(8)*viscosity*length)/(math.pi*radius**(4))
return Resistance
def main():
length = float(input("Please enter the length: "))
radius = float(input("Please enter the radius: "))
viscosity = float(input("Please enter the viscosity: "))
if length <= 0:
print("Failed due to input error. Please make sure your inputs are all positive. Exiting program.")
elif radius <= 0:
print("Failed due to input error. Please make sure your inputs are all positive. Exiting program.")
elif viscosity <= 0:
print("Failed due to input error. Please make sure your inputs are all positive. Exiting program.")
else:
print("The resistance is: ", poiseuille(length, radius, viscosity))
if __name__ == "__main__":
main()
| true |
bad4bb04bb518a8b45e3c0e75dabde665ef5d942 | jojadev/simpsons-test | /main.py | 383 | 4.15625 | 4 | name = input("Who is your favourite Simpson household member? ").capitalize()
if name == 'Homer':
print("You the man Homer!")
elif name == 'Marge':
print("Way to go mom!")
elif name == 'Lisa':
print("Eww, Lisa!")
elif name == 'Maggie':
print("What's up you cool baby?")
else:
print("I'm Bart Simpson, who the hell are you?")
#https://repl.it/@jojadev/simpsons-test
| false |
d0a53a4ca42e65bb86f1e4453bdfe747ea8227ee | AmineNeifer/holbertonschool-interview | /0x19-making_change/0-making_change.py | 578 | 4.25 | 4 | #!/usr/bin/python3
""" Contains makeChange function"""
def makeChange(coins, total):
"""
Returns: fewest number of coins needed to meet total
If total is 0 or less, return 0
If total cannot be met by any number of coins you have, return -1
"""
if not coins or coins is None:
return -1
if total <= 0:
return 0
change = 0
coins = sorted(coins)[::-1]
for coin in coins:
while coin <= total:
total -= coin
change += 1
if (total == 0):
return change
return -1
| true |
3b2c0d8d806dcc4b73a80bf0564a77f77b906b67 | laurenhesterman/novTryPy | /TryPy/trypy.py | 1,832 | 4.5625 | 5 | #STRINGS
# to capitalize each first letter in the word use .capitalize() method, with the string before .capitalize
characters = "rick"
print characters.capitalize()
#returns Rick
#returns a copy of the whole string in uppercase
print characters.upper()
#returns RICK
#.lower() returns a copy of the string converted completely to lowercase
print characters.lower()
#returns rick
#count returns the number of occurances of the substr in original string. Uses (str[start:end]) as paramaters.
#defaults start and end to 0
string ="today is my birthday"
strg= "day"
print string.count("day")
#returns 2
#.find() returns lowest index at which substring is found. (str[start:end]) start and end default to entire string
#-1 on failure
print string.find(strg)
#returns 2, as "day" begins on 2nd index of the string
#.index() is similar to .find(), but returns ValueError when not found
print string.index("day")
#returns 2
""".split() reutrns a list of words of the string. (str[, sep[, maxsplit]]) are parameters. By default
they separate with whitespace, but 'sep' can replace a string as the separateor.
maxsplit defaults to 0, but that number specifies how many splits will occur at most"""
string ="today is my birthday"
print string.split("y")
#.join() concentrates a list or words to string at intervening occurances of sep (default whitespace)
string2 ="today is my birthday"
print string2.join("now")
# returns: ntoday is my birthdayotoday is my birthdayw (wraps each character of "now around" occurances of string2)
#replaces occurances of first argument within string with second argument
print string.replace("day", "month")
#returns tomonth is my birthmonth
#.format() formats string with replacement holders
string3 = "The sum of 1 + 2 = {}"
print string3.format(1+2)
#returns The sum of 1 + 2 = 3
#NUMBERS
| true |
017ccb38921399323ccb3c169a50063b3057a118 | Alex-Reitz/Python_SB | /02_weekday_name/weekday_name.py | 566 | 4.25 | 4 | def weekday_name(day_of_week):
"""Return name of weekday.
>>> weekday_name(1)
'Sunday'
>>> weekday_name(7)
'Saturday'
For days not between 1 and 7, return None
>>> weekday_name(9)
>>> weekday_name(0)
"""
i = 0
weekdays = ["Sunday", "Monday", "Tuesday",
"Wednesday", "Thursday", "Friday", "Saturday"]
while i < len(weekdays):
if i + 1 == day_of_week:
print(weekdays[i])
i = i + 1
weekday_name(1)
weekday_name(2)
weekday_name(3)
weekday_name(4)
| true |
5c39c42c7787b07e51728b67855c4f58fa98fd0f | leios/OIST.CSC | /hello_world/python/hello_world.py | 788 | 4.15625 | 4 | #-------------hello_world.py---------------------------------------------------#
#
# In most traditional coding courses, they start with a simple program that
# outputs "Hello World!" to the terminal. Luckily, in python... this is pretty
# simple. In fact, it's only one line (the line that follows this long comment).#
# You should be able to type in:
#
# python hello_world.py
#
# and receive the following output in your terminal:
#
# Hello World!
#
# So let's get to it!
#------------------------------------------------------------------------------#
print("Hello World!")
# In python, the "print" command sends whatever is in the parentheses to your
# terminal. That's precisely what we are doing. Sending "Hello World!" to the
# terminal. Let me know if you have any trouble!
| true |
53c935628a3bcee64d664a7304417cc509031e12 | LittltZhao/code_git | /342_Power_of_Four.py | 202 | 4.125 | 4 | # -*- coding:utf-8 -*-
#判断一个数是否为4的幂数
def isPowerOfFour(num):
return num>0 and (num&(num-1))==0 and (num-1)%3==0#(num&(num-1))==0判断是否为2的幂
print isPowerOfFour(16)
| false |
ec4030fa14128f11e3354b789203588dddf333ff | flashlightli/math_question | /leetcode_question/mid_question/29_Divide_Two_Integers.py | 2,103 | 4.125 | 4 | """
给定两个整数，被除数 dividend 和除数 divisor。将两数相除，要求不使用乘法、除法和 mod 运算符(取余)。
返回被除数 dividend 除以除数 divisor 得到的商。
整数除法的结果应当截去（truncate）其小数部分，例如：truncate(8.345) = 8 以及 truncate(-2.7335) = -2
示例 1:
输入: dividend = 10, divisor = 3
输出: 3
解释: 10/3 = truncate(3.33333..) = truncate(3) = 3
示例 2:
输入: dividend = 7, divisor = -3
输出: -2
解释: 7/-3 = truncate(-2.33333..) = -2
提示：
被除数和除数均为 32 位有符号整数。
除数不为 0。
假设我们的环境只能存储 32 位有符号整数，其数值范围是 [−231, 231 − 1]。本题中，如果除法结果溢出，则返回 231 − 1。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/divide-two-integers
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
"""
class Solution:
def divide(self, dividend: int, divisor: int) -> int:
# 位运算
positive = False
if dividend == 0:
return 0
if (dividend > 0 and dividend > 0) or (dividend < 0 and dividend < 0):
positive = True
scaling = 1 << 31
result = self.divide_1(dividend, divisor)
if positive == 1:
if result > scaling:
result = scaling
return -result
if result >= scaling:
result = (scaling) - 1
return result
def divide_1(self, dividend, divisor):
# 除数向左平移
count = 0
while dividend >= divisor:
divisor_i = divisor
new_count = 1
while (divisor_i << 1) < dividend:
divisor_i = divisor_i << 1
new_count = new_count << 1 # count 记录了倍数， 即divisor*count < dividend < divisor*count*2
dividend = dividend - divisor_i
count = count + new_count
return count
test = Solution()
print(test.divide(
10, 3
))
| false |
cbabb745d29b004c700fa4edfa9c16c1d4424d5a | flashlightli/math_question | /leetcode_question/mid_question/114_Flatten_Binary_Tree_to_Linked_List.py | 1,404 | 4.25 | 4 | """
给定一个二叉树，原地将它展开为一个单链表。
例如，给定二叉树
1
/ \
2 5
/ \ \
3 4 6
将其展开为：
1
\
2
\
3
\
4
\
5
\
6
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/flatten-binary-tree-to-linked-list
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
"""
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution:
def flatten(self, root: TreeNode) -> None:
"""
Do not return anything, modify root in-place instead.
"""
# 把左子树插入到根节点和右子树之间
return self.to_right(node=root)
def to_right(self, node):
if not node:
return
self.to_right(node.left)
self.to_right(node.right)
if node.left:
pre = node.left
while pre.right:
pre = pre.right
pre.right = node.right
node.right = node.left
node.left = None
root = TreeNode(1)
# root.left = TreeNode(2)
# root.left.left = TreeNode(3)
root.right = TreeNode(2)
# root.right = TreeNode(5)
root.right.right = TreeNode(3)
test = Solution()
print(test.flatten(
root
)) | false |
457ac15fcea68d766279c72296236295daa866d1 | Hanlen520/Leetcode-4 | /src/114. 二叉树展开为链表.py | 1,746 | 4.1875 | 4 | """
给定一个二叉树，原地将它展开为链表。
"""
# Definition for a binary tree node.
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
# 其实就是先序遍历，下面采用迭代的方法
class Solution:
def flatten(self, root: TreeNode) -> None:
"""
Do not return anything, modify root in-place instead.
"""
head = TreeNode(0)
stack = [root]
while stack:
curr = stack.pop()
if curr:
stack.append(curr.right)
stack.append(curr.left)
curr.left = None
curr.right = None
head.right = curr
head = curr
# 这个是用后序遍历使用的递归方法
# 后序遍历由于可以先处理左子树和右子树，根节点在最后处理。所以先找到左子树的最右节点，然后再把右子树插到左子树的最右节点后，最后将
# 根节点的右子树变为当前的左子树，依次递归。
class Solution:
def flatten(self, root: TreeNode) -> None:
"""
Do not return anything, modify root in-place instead.
"""
def helper(root1):
if not root1:
return
else:
helper(root1.left)
helper(root1.right)
# 找到左子树的最右节点
curr = root1.left
if curr:
while curr:
pre = curr
curr = curr.right
pre.right = root1.right
root1.right = root1.left
root1.left = None
helper(root)
| false |
4508866832b2578a43788e47e00c3e9781bb6b44 | cesarmarroquin/blackjack | /blackjack_outline.py | 2,089 | 4.1875 | 4 | """
I need to create a blackjack Game. In blackjack, the player plays against the dealer. A player can win in three
different ways. The three ways are, the player gets 21 on his first two cards, the dealer gets a score higher than 21,
and if the player's score is higher than the dealer without going over 21. The dealer wins if the player goes over 21,
or if the dealer's hand is greater than the player's without going over 21. In the beginning, the player is
dealt two cards. There is also one rule which states that if the dealer's hand is greater than or equal to 17, then he
can not hit anymore. There is a point system that must be followed in the game of blackjack, all cards have a point
value based off the numeric indication on the front. The card's that don't follow this rule are ace's which could be 1
or 11, and face cards which all equal to 10. It is also important to know what a deck of cards has; a deck of cards ha
s 52 cards, 13 unique cards, and 4 suits of those 13 unique cards.
Game(player1, player2=)
The Game
Responsibilities:
* ask players if they want to hit or stand.
*
Collaborators:
* Collected into a Deck.
* Collected into a Hand for each player and a Hand for the dealer.
def play_game
def deal_two_cards_to_player
def switch_players
Player
The player
Responsibilities:
* Has to hit or stand
Collaborators:
* game takes the hit or stand response
def hit
def stand
money
Dealer
The dealer
Responsibilities:
* Has to hit, has to stand at 17 or greater
Collaborators:
* game takes the hit or stand response
def hit
def stand
money
shoe
Hand
cards_in_hand
Card
A playing card.
Responsibilities:
* Has a rank and a suit.
* Has a point value. Aces point values depend on the Hand.
Collaborators:
* Collected into a Deck.
* Collected into a Hand for each player and a Hand for the dealer.
suit
kind
Deck
card_amount
suits
kinds
if __name__ == '__main__':
"""
| true |
3883f41652fa43d96be453f5d5434aead2cf3ead | mohit131/python | /project_euler/14__Longest_Collatz_sequence.py | 1,095 | 4.21875 | 4 | '''The following iterative sequence is defined for the set of positive integers:
n → n/2 (n is even)
n → 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been
proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
Which starting number, under one million, produces the longest chain?
NOTE: Once the chain starts the terms are allowed to go above one million.'''
def even(x):
return(x/2)
def odd(x):
return(3*x + 1)
def Collatz(a):
b=[a]
while (a != 1):
if(a%2==0):
b.append(even(a))
a=even(a)
else:
b.append(odd(a))
a=odd(a)
return b
c=[]
lsize=[]
'''
for i in range(1,10000):
c.append((Collatz(i)))
lsize.append((len(c[i-1])))
#lsize.append()
#print(c)
print(max(lsize))
'''
irange=range(1,1000000)
c=list(map(Collatz,irange))
lsize1=list(map(len,c))
print(max(lsize1)) | true |
c7404efed6d6384bcf49df15f0dc508c01a7fef5 | mohit131/python | /project_euler/9__Special_Pythagorean_triplet.py | 490 | 4.125 | 4 | '''A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.'''
for m in range(1,5000):
for n in range(1,500):
if m>n:
a=m*m - n*n
b= 2 * m * n
c=m*m + n*n
sum_ab=a*a + b*b
sum_c=c*c
if sum_ab == sum_c:
if a+b+c==1000:
print(a,b,c,a+b+c,a*b*c)
else:
break | false |
5a97036b637ec09c393d8fae0178db26b3db1603 | adidrmwan/basic-python | /leapYear.py | 503 | 4.1875 | 4 | def is_leap(year):
leap = False
leap_year = year % 4
leap_year_divided_100 = year % 100
leap_year_divided_400 = year % 400
if leap_year == 0:
leap = True
if leap_year_divided_100 == 0:
leap = False
if leap_year_divided_400 == 0:
leap = True
else:
leap = False
else:
leap = True
else:
leap = False
return leap
year = int(input())
print(is_leap(year)) | false |
df01dc415a7750095dcd99b64690469a7c41b983 | Chittadeepan/P342 | /lab5_q2.py | 1,328 | 4.1875 | 4 | #importing everything from math module
from math import *
#importing all functions from library
from library import *
#poly_f(x) function
def poly_f(x,coeff,n):
sum=0
for i in range(1,n+1):
sum=sum+coeff[i-1]*x**(n-i)
return sum
#main program
def main():
#initialising final absolute error(epsilon), initial guess of root(root) and number of terms in polynomial(n)
epsilon=10**(-6)
root=0.9
n=5
#initialising coeff list and appending coeff elements corresponding to their variables with decreasing power
coeff=[1,-3,-7,27,-18]
#solving P(x)=x^4-3x^3-7x^2+27x-18
print('Solving x^(4)-3x^(3)-7x^(2)+27x-18 by Laguerre method and Synthetic Division method:')
#calling Laguerre function and Synthetic Division function and displaying the roots obtained in a loop
for index in range(n,1,-1):
root=Laguerre(root,epsilon,coeff,index,poly_f)
if index>0:
coeff=Synthetic_Division(root,coeff)
print('One of the roots obtained:',round(root,6))
main()
'''
#Output
Solving x^(4)-3x^(3)-7x^(2)+27x-18 by Laguerre method and Synthetic Division method:
One of the roots obtained: 1.0
One of the roots obtained: 2.0
One of the roots obtained: 3.0
One of the roots obtained: -3.0
'''
| true |
e537efac5d4877eb9b8516a4b02cee19f7985323 | PratheekH/Python_sample | /comp.py | 201 | 4.28125 | 4 | name=input("Enter your name ")
size=len(name)
if size<3:
print("Name should be more than 3")
elif size>50:
print("Name should not be more than 50")
else:
print("Name looks fine") | true |
134a72d4bd4ca94b8e3a0d2d7c5a4f6b83bf9480 | cjrzs/MyLeetCode | /重要的算法模板/排序算法模板/冒泡排序.py | 667 | 4.28125 | 4 | """
coding: utf8
@time: 2020/12/8 17:18
@author: cjr
@file: 冒泡排序.py
"""
# 冒泡排序使用当前元素与下一个元素做比较，把符合条件的后移一位或者不动，
# 这样每次都会把最大或者最小的数放在最后一个位置。
# 重复这种排序思路。就可以从后到前的排序好数组。
def bubble_sort(nums):
n = len(nums)
for i in range(n - 1):
for j in range(n - i - 1):
if nums[j] > nums[j + 1]:
nums[j], nums[j + 1] = nums[j + 1], nums[j]
return nums
if __name__ == '__main__':
nums_ = [1, 3, 8, 4, 2, 1, 5, 10]
print(bubble_sort(nums_))
| false |
3f9f28e504e80d7a04fb13f7750d12c873a063a0 | xdyxiang/pythonlearn | /python_base/digui.py | 574 | 4.15625 | 4 | # 遍历一个盘符下的所有文件（包括子文件夹、文件）
import os
def getFile(path):
try:
filelist = os.listdir(path) # 得到该文件夹下的所有文件
for file in filelist:
file = os.path.join(path, file) # 将文件名和路径结合起来
if os.path.isdir(file):
getFile(file) # 在这里如果判断一个文件是文件夹，那么就会再次调用自己
else:
print(file)
except:
print('出错，跳过')
getFile(r'E:/')
| false |
2a8ea1f95001015fba07fc2130024b5c6a8375c7 | aashishah/LocalHackDay-Challenges | /EncryptPassword.py | 702 | 4.28125 | 4 | #Using Vignere Cipher to encrpt password of a user using a key that is private to the user.
def encrypt(password, key):
n = len(password)
#Generate key
if len(key) > n:
key = key[0:n]
else:
key = list(key)
for i in range(n - len(key)):
key.append(key[i % len(key)])
key = "" . join(key)
#Generate encrypted password:
cipher = []
for i in range(n):
letter = (ord(password[i]) + ord(key[i])) % 26
letter += ord('A')
cipher.append(chr(letter))
print("Encrypted password: " + "".join(cipher))
password = input("Enter password: ")
key = input("Enter your private key: ")
encrypt(password, key)
| true |
15c76b453894e5eb09fcaa195a0a8a66351b1048 | Karlo5o/rosalind_problems | /RNA.py | 700 | 4.1875 | 4 | """
An RNA string is a string formed from the alphabet containing 'A', 'C', 'G', and 'U'.
Given a DNA string t corresponding to a coding strand, its transcribed RNA string u is formed by replacing all occurrences of 'T' in t with 'U' in u.
Given: A DNA string t having length at most 1000 nt.
Return: The transcribed RNA string of t.
"""
def dna_transcriber(input_file):
with open('output.txt','w') as output, open(input_file) as input:
dna_string = input.readline()
rna_string = dna_string.replace("T", "U")
#Convert int to string
output.write(rna_string) #Write transcribed string in to file
dna_transcriber('input.txt') #input file name is 'input.txt'
| true |
d6c9481fd67e920623b9c3ac06f98d0051db1c95 | SaulAlekss/clienteservidor | /listas.py | 1,328 | 4.5625 | 5 | def main():
# Una lista es una estructura de datos en python
# # La ventaja aceptan datos de tipos distintos
# # Creamos una lista
lista = [1,23.01, False, "hola lista", "A",[-1,-5, "hola", 0.0], -12,"A"]
# Lista Vacia
listaVacia = []
# Accesando a elementos de la lista
for elemento in lista:
print("El elemento de la lista es: ", elemento)
for i in listaVacia:
print("El elemento de la lista es: ", elemento)
#Imprimir un elemento de la lista
print("Elemento en la posicion 3: ", lista[3])
print("Elemento en la posicion 3: ", lista[-1])
print(lista[-2])
print(lista[5])
# Leer elemento en la posicion 2 de la lista interna
print(lista[5][2])
# Metodos de las listas
"""lista.append("Agregar una cadena......")
for elemento in lista:
print("El elemento de la lista es: ", elemento)
"""
# count regresa el numero de veces que se repite un elememto en la lista
print("Elemento se repite : " , lista.count("A"))
# index() imprime el indice de un elemento en la lista
print("La posicion de False es: ", lista.index(False))
# Eliminar elementos de la lista : remove()
lista.remove("hola lista")
for x in lista:
print("El elemento de la lista es: ", x)
if __name__=="__main__":
main() | false |
d516d08d9c1af9bce8e845f69552a45377b6696b | marcusiq/w_python | /greeting.py | 1,570 | 4.4375 | 4 |
"""
Generally. There is no method overloading in python.
Overloading in other languages occurs when two methods have the same name, but different numbers of arguments.
In python, if you give more than one definition using the same name,
the last one entered rules, and any previous definitions are ignored.
"""
# Here an unsuspecting student attempts to overload a method named "greeting" with several definitions.
def greeting():
print "Hi! Are you new here?\n"
# The second takes one argument, and can be used to greet a person whose name you know.
def greeting(name):
print "Yes! It's my first day! My name is %s.\n" % name
# The third takes two arguments, and can be used to name both yourself and the addressee.
def greeting(name1, name2):
print "Nice to meet you %s! My name is %s.\n" % (name1, name2)
# The fourth takes one argument, conceived to be an integer.
def greeting(n):
print "Can I borrow $%d for coffee?\n" % n
# The fifth takes one argument, conceived to be a boolean.
def greeting(b): # A boolean.
if b:
print "Sure! Let's have coffee together.\n"
else:
print "Sorry! I'm flat broke!\n"
# Hmmm, that did not work! There is no overloading in Python!
try:
greeting()
except Exception as e:
print e.message
try:
greeting("Mork")
except Exception as e:
print e.message
try:
greeting("Mork", "Mindy")
except Exception as e:
print e.message
try:
greeting(10)
except Exception as e:
print e.message
try:
greeting(True)
except Exception as e:
print e.message
| true |
0db7eaa487bf54eb30f885b7baf11d93dfd6eabd | adamelliott1982/1359_Python_Benavides | /lab_04/1359-lab_04/drop_grade.py | 1,549 | 4.34375 | 4 | # Program: drop_grade.py
# Programmer: Adam Elliott
# Date: 02/26/2021
# Description: lab 4 - lists and for statements
########################################################
# create list of 5 grades and initialize
grades = [100, 80, 70, 60, 90]
# print report name – DROP LOWEST GRADE PROGRAM
print('DROP LOWEST GRADE PROGRAM:\n')
# show original grades using for in Loop
print('Grades:')
for grade in grades:
print(grade)
# use sum function on grades to calculate the total
total = sum(grades)
# calculate the average by dividing total by len(grades)
average = total/len(grades)
# print number of grades using len function
print(f'Number of grades: {len(grades)}')
# print average formatted to 2 decimal places
print(f'Average: {average:.2f}')
# find the lowest grade using min function and print lowest grade
lowest_grade = min(grades)
print(f'Lowest grade: {lowest_grade}')
#remove lowest grade
grades.remove(lowest_grade)
# print LOWEST GRADE DROPPED
print('\nLOWEST GRADE DROPPED:\n')
# show grades after dropping lowest grade using for in Loop
print('Grades:')
for grade in grades:
print(grade)
# use sum function to calculate the total after dropping lowest grade
total = sum(grades)
# compute new average
average = total/len(grades)
# print new number of grades
print(f'Number of grades: {len(grades)}')
# print new average
print(f'Average: {average:.2f}')
# find new lowest grade and print
lowest_grade = min(grades)
print(f'Lowest grade: {lowest_grade}')
| true |
8c2a6954dbcdb20dba141975559470e8b04ad921 | BenjaminLivingstone/fundamentos-python | /Otros/ejerciciopython1.py | 1,116 | 4.1875 | 4 | # GRUPO 2:
# Crea una funcion que dado una palabra diga si es palindroma o no.
def palindroma(palabra):
if ("hola"[::-1]==palabra):
print("La palabra",palabra,"es palindroma")
else:
print("La palabra",palabra,"NO es palindroma")
palabra="hola";
palindroma(palabra)
palabra="python"
print(palabra[2:])
# palabra[inicio:final:paso]
print("hola"[::-1])
# - Crea una función que tome una lista y devuelva el primer y el último valor de la lista. Si la longitud de la lista es menor que 2, haga que devuelva False.
def devuelvevalor(lista):
if (len(lista)<2):
return False
print(lista[0])
print(lista[len(lista)-1])
lista=[1,2,3,4,5,7]
devuelvevalor(lista)
# # # - Crea una función que tome una lista y devuelva un diccionario con su mínimo, máximo, promedio y suma.
# def devuelvedic(listadic):
# minimo=min(listadic)
# maximo=max(listadic)
# promedio=sum(listadic)/len(listadic)
# suma=sum(listadic)
# dic={"Mínimo":minimo,"Máximo":maximo,"Promedio":promedio,"Suma":suma}
# print(dic)
# listadic=[1,21,3,44,-15,6]
# devuelvedic(listadic) | false |
f371184ca3169f6fffe54319969f8ed2cf8ef2d6 | DishT/Python100- | /ex2.py | 297 | 4.1875 | 4 | def factorial(num):
number = 1
for i in range(int(num)):
number = (i+1)*number
print (number)
# 8! = 8 * 7!....... , 0! = 1
def factorial_2(num):
if num == 0 :
return 1
else:
return num * factorial_2(num-1)
def factorial_3(num):
num = input()
factorial(num)
print (factorial_2(num)) | false |
37ac60fa04dbaed484d77e43317c6069a26ad777 | marc-p-greenfield/tutor_sessions | /payment_system.py | 602 | 4.21875 | 4 | number_of_employees = int(input("How many employees do you want to enter:"))
total_payment = 0
for i in range(number_of_employees):
name = input('Please enter name: ')
hours = float(input('How many hours did you work this week? '))
rate = float(input('What is your hourly rate?'))
payment = 0
overtime = 0
if hours > 40:
overtime = hours - 40
hours = 40
payment = (hours * rate) + (overtime * rate * 1.5)
total_payment += payment
print(name)
print("${:,.2f}".format(payment))
print (total_payment)
print (total_payment/number_of_employees)
| true |
a9f5a386203fe02efc9f289408744aef6e326c57 | jinseoo/DataSciPy | /src/파이썬코드(py)/Ch08/code_8_6.py | 392 | 4.125 | 4 | #
# 따라하며 배우는 파이썬과 데이터과학(생능출판사 2020)
# 8.6 집합의 항목에 접근하는 연산, 207쪽
#
numbers = {2, 1, 3}
if 1 in numbers: # 1이라는 항목이 numbers 집합에 있는가 검사
print("집합 안에 1이 있습니다.")
numbers = {2, 1, 3}
for x in numbers:
print(x, end=" ")
for x in sorted(numbers):
print(x, end=" ") | false |
f26e9e6bc1103c6c72fbe0fc72ed2435b62281ad | KishoreMayank/CodingChallenges | /Cracking the Coding Interview/Arrays and Strings/StringRotation.py | 349 | 4.1875 | 4 | '''
String Rotation:
Check if s2 is a rotation of s1 using only one call to isSubstring
'''
def is_substring(string, sub):
return string.find(sub) != -1
def string_rotation(s1, s2):
if len(s1) == len(s2) and len(s1) != 0:
return is_substring(s1 + s1, s2) # adds the two strings together and calls is substring
return False | true |
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