blob_id string | repo_name string | path string | length_bytes int64 | score float64 | int_score int64 | text string | is_english bool |
|---|---|---|---|---|---|---|---|
3cee29d8ee596d508122d7e4d1349b6e226f148e | iampaavan/Pure_Python | /Exercise-95.py | 260 | 4.21875 | 4 | """Write a Python program to convert a byte string to a list of integers."""
def byte_str_list_integers(string):
"""return list of integers."""
my_list = list(string)
return my_list
print(f"List of integers: {byte_str_list_integers(b'Abc')}")
| true |
1ae7da4a80ff13e6e38a6f1ef9ec8c092208002d | iampaavan/Pure_Python | /Exercise-15.py | 494 | 4.3125 | 4 | from datetime import date
"""Write a Python program to calculate number of days between two dates."""
first_date = date(2014, 7, 2)
last_date = date(2014, 7, 11)
delta = last_date - first_date
print(f"First Date: {first_date}")
print(f"Second Date: {last_date}")
print(f'*****************************************************************')
print(f"Difference between the two dates are: {delta.days} days.")
print(f'*****************************************************************')
| true |
9b279f74643634d02aae4969267a3fee44e10a78 | iampaavan/Pure_Python | /Exercise-4.py | 372 | 4.3125 | 4 | from math import pi
"""Write a Python program which accepts the radius of a circle from the user and compute the area."""
def radius():
"""Calculate the radius"""
r = int(input(f"Enter the radius of the circle:"))
Area = pi * (r * r)
return f"Area of the circle is: {Area}"
c = radius()
print(c)
print(f'*******************************************')
| true |
c35a1b8ecdab55a9c6a3dcabc2bdbd241a5e93f5 | iampaavan/Pure_Python | /Exercise-28.py | 291 | 4.21875 | 4 | """Write a Python program to concatenate all elements in a list into a string and return it."""
def concatenate(my_list):
"""Return a string."""
output = ''
for i in my_list:
string = str(i)
output += string
return output
print(concatenate([1, 2, 3, 4, 5]))
| true |
d8b11d176fb47a44390c72b979370931cf9f6ce8 | mtrunkatova/Engeton-projects | /Project2.py | 2,521 | 4.25 | 4 | def welcoming():
print("WELCOME TO TIC TAC TOE")
print("GAME RULES:")
print("Each player can place one mark (or stone) per turn on the 3x3 grid")
print("The WINNER is who succeeds in placing three of their marks in a")
print("* horizontal,\n* vertical or\n* diagonal row\nLet's start the game")
def creating_board():
list1 = list(range(1,3**2+1))
list2 = [' ']*3**2
board = dict(zip(list1,list2))
return board
def print_board(board):
x = 6*'-'
print(x)
print(board[1] + '|' + board[2] + '|' + board[3])
print(x)
print(board[4] + '|' + board[5] + '|' + board[6])
print(x)
print(board[7] + '|' + board[8] + '|' + board[9])
print(x)
def win_horizontal(board):
we_have_winner = False
if board[1] == board[2] == board[3] != ' ' \
or board[4] == board[5] == board[6] != ' ' \
or board[7] == board[8] == board[9] != ' ':
we_have_winner = True
return we_have_winner
def win_vertical(board):
we_have_winner = False
if board[1] == board[4] == board[7] != ' '\
or board[2] == board[5] == board[8] != ' '\
or board[3] == board[6] == board[9] != ' ':
we_have_winner = True
return we_have_winner
def win_cross(board):
we_have_winner = False
if board[1] == board[5] == board[9] != ' '\
or board[3] == board[5] == board[7] != ' ':
we_have_winner = True
return we_have_winner
def switch_players(whose_turn):
if whose_turn == "O":
whose_turn = "X"
else:
whose_turn = "O"
return whose_turn
def no_winner(board):
if ' ' not in board.values():
return True
else:
return False
def main():
welcoming()
board = creating_board()
print_board(board)
print("========================================")
end = False
whose_turn = "O"
while not end: # pojedou maximalne 9 kol
game = int(input("Player {} Please enter your move number: ".format(whose_turn)))
if board[game] == ' ':
board[game] = whose_turn
else:
continue
print_board(board)
x = win_horizontal(board)
y = win_vertical(board)
z = win_cross(board)
w = no_winner(board)
if x or y or z:
print("Congratulations, the player {} WON!".format(whose_turn))
end = True
elif w:
break
else:
whose_turn = switch_players(whose_turn)
if __name__ == "__main__":
main() | true |
e53fe6f2b910c285b7851783b61c4d7939638b71 | manoznp/LearnPython-Challenge | /DAY3/list.py | 1,519 | 4.34375 | 4 | #list
names = ['raju', 'manoj']
length_names = len(names)
print("Length of the array name is {}".format(length_names))
names.append("saroj")
length_names = len(names)
print("Length of the array name is {}".format(length_names))
print(names)
names.insert(1, "sudeep")
print("After inserting sudeep at position second: {}".format(names))
for name in names:
print(name)
# extending the list
other_names = ['dipesh', 'pankaj', 'prmaod']
names.extend(other_names)
print(names)
# Dictionary
student = {'name':'manoz',
'time':'1.5 hrs',
'course': 'Data Science',
'fee': 25000,
'hasVeichle': False
}
# print the value of the dictionary with key
print(student['course'])
student_all_key = student.keys()
print(student_all_key)
student_all_value = student.values()
print(student_all_value)
# iterate in the dictionary
for key in student.keys():
print("{} : {}".format(key, student[key]))
# change the value of the dictionary
student['course'] = 'Advanced Java'
print(student)
# lets add the key that doesnot exist in the dictionary student
student['gender'] = "male"
print("after adding gender")
print(student)
# delete the key course
del(student['course'])
# after deleting the key course
print(student)
# convert the student to list
student = list(student)
# after converting to the list
print(student)
# check the key is exist in the dictionary or not
isCourseExists = "courses" in student
print("course exists in student : {} ". format(isCourseExists))
| true |
6db5f977cc64e90e243d62f3b882c7d77371a86f | Williano/Python-Scripts | /turtle_user_shape/turtle_user_shape.py | 2,734 | 4.1875 | 4 | import turtle
window = turtle.Screen()
window.setup()
window.title("Draw User Shape")
window.bgcolor("purple")
mat = turtle.Turtle()
mat.shape("turtle")
mat.color("black")
mat.pensize(3)
mat.speed(12)
drawing = True
while drawing:
mat.penup()
SQUARE = 1
TRIANGLE = 2
QUIT = 0
shape_choice = int(input("What do you want me to draw? (1 = square, 2 = triangle, 0 = quit):"))
if shape_choice == SQUARE:
side_length = int(input("How long do you want the sides of your square to be?"
"Please enter the number of pixels (e.g. 100):"))
mat.pendown()
x_pos = int(mat.xcor())
y_pos = int(mat.ycor())
heading = int(mat.heading())
print("My 1st corner is at: {}, {} and my heading is {}".format(x_pos, y_pos, heading))
mat.forward(side_length)
x_pos = int(mat.xcor())
y_pos = int(mat.ycor())
heading = int(mat.heading())
print("My 2nd corner is at: {}, {} and my heading is {}".format(x_pos, y_pos, heading))
mat.left(90)
mat.forward(side_length)
x_pos = int(mat.xcor())
y_pos = int(mat.ycor())
heading = int(mat.heading())
print("My 3rd corner is at: {}, {} and my heading is {}".format(x_pos, y_pos, heading))
mat.left(90)
mat.forward(side_length)
x_pos = int(mat.xcor())
y_pos = int(mat.ycor())
heading = int(mat.heading())
print("My 4th corner is at: {}, {} and my heading is {}".format(x_pos, y_pos, heading))
mat.left(90)
mat.forward(side_length)
elif shape_choice == TRIANGLE:
print("I will draw an equilateral triangle.")
side_length = int(input("How long do you want the sides of your triangle to be?"
"Please enter the number of pixels (e.g. 100): "))
mat.pendown()
x_pos = int(mat.xcor())
y_pos = int(mat.ycor())
heading = int(mat.heading())
print("My 1st corner is at: {}, {} and my heading is {}".format(x_pos, y_pos, heading))
mat.forward(side_length)
x_pos = int(mat.xcor())
y_pos = int(mat.ycor())
heading = int(mat.heading())
print("My 2nd corner is at: {}, {} and my heading is {}".format(x_pos, y_pos, heading))
mat.left(120)
mat.forward(side_length)
x_pos = int(mat.xcor())
y_pos = int(mat.ycor())
heading = int(mat.heading())
print("My 3rd corner is at: {}, {} and my heading is {}".format(x_pos, y_pos, heading))
mat.left(120)
mat.forward(side_length)
mat.left(120)
else:
drawing = False
else:
print("Thank you for playing")
| true |
532f8be7df48120a466f060ae0b80ebb6b2a7dfa | p3dr051lva/hanoi-s_tower.py | /hanoi.py | 1,293 | 4.1875 | 4 | print('Ola, este eh o jogo, torre de hanoi, o jogo tem o objetivo de voce transferir \
os numero da tore um para a torre 3, sem que numeros maiores fiquem em cima de numeros menores, voce\
so pode mover o disco do "topo", e uma por vez, a primeira torre de baixo para cima eh a t1, a segunda a\
t2, e a terceira a t3, para move-los use o seguinte metodo:t1>t2. para movimentar o disco do "topo" da t1 para o "topo" da t2, bom jogo meu paladino')
def hanoi():
t1= [3,2,1]
t2= []
t3= []
torres = [t1, t2, t3]
aux = 0
print(t1)
print(t2)
print(t3)
while len(torres[-1]) < 3:
y1, y2 = [int(i[-1]) - 1 for i in input('digite a origem>destino: ').split('>')]
while len(torres[y1]) == 0 or (len(torres[y2]) > 0 and torres[y1][-1] > torres[y2][-1]):
print('jogada invalida')
print(t1)
print(t2)
print(t3)
y1, y2 = [int(i[-1]) - 1 for i in input('digite a origem>destino: ').split('>')]
else:
torres[y2].append(torres[y1].pop())
aux += 1
print(t1)
print(t2)
print(t3)
else:
print('parabens voce completou o quebra-cabeca em %d movimentos' %(aux))
hanoi()
| false |
830802e627fe7aaed1057d4ed66ccadf532bf357 | tanpv/awesome-blockchain | /algorithm_python/move_zeros_to_end.py | 378 | 4.25 | 4 | """
Write an algorithm that takes an array and moves all of the zeros to the end,
preserving the order of the other elements.
move_zeros([false, 1, 0, 1, 2, 0, 1, 3, "a"])
returns => [false, 1, 1, 2, 1, 3, "a", 0, 0]
The time complexity of the below algorithm is O(n).
"""
input_list = [false, 1, 0, 1, 2, 0, 1, 3, "a"]
def move_zeros(self, input_list):
| true |
eda4f62ffe3d1771eb868d4325d8310348d521c7 | Jayson22341/CS362HW3 | /LeapYearProg.py | 493 | 4.21875 | 4 | def leap(n):
if n % 4 == 0:
if n % 100 == 0:
if n % 400 == 0:
print(n, 'is a leap year')
return
print(n, 'is not a leap year')
return
print(n, 'is a leap year')
return
print(n, 'is not a leap year')
return
import time
print("This program will determine whether a year you enter is a leap year or not")
value = int(input("Enter in a year number: "))
leap(value)
time.sleep(5)
| false |
048ab8c2e21afc334d5cd093b47e7633f3530520 | ravenusmc/algorithms | /HackerRank/algorithms/diagonal_diff.py | 938 | 4.1875 | 4 | # Given a square matrix, calculate the absolute difference between the sums of its diagonals.
# For example, the square matrix arr is shown below:
# URL https://www.hackerrank.com/challenges/diagonal-difference/problem?h_r=next-challenge&h_v=zen
# Rank: 1,670,172
arr = [
[1, 2, 3],
[4, 5, 6],
[9, 8, 9]
]
# arr = [
# [1, 2, 3, 4],
# [4, 5, 6, 7],
# [9, 8, 9, 6],
# [4, 5, 6, 7],
# ]
def diagonalDifference(arr):
left_to_right_sum = 0
count_left = 0
while count_left < len(arr):
left_to_right_sum = arr[count_left][count_left] + left_to_right_sum
count_left += 1
right_to_left_sum = 0
count_up = 0
count_right = len(arr) - 1
while count_right >= 0:
right_to_left_sum = arr[count_up][count_right] + right_to_left_sum
count_up += 1
count_right -= 1
absolute_difference = abs(left_to_right_sum - right_to_left_sum)
return absolute_difference
absolute_difference = diagonalDifference(arr)
print(absolute_difference)
| true |
c4bbc4e04d3ab14ecb92baad254d46959fc4329e | Sandeep0001/PythonTraining | /PythonPractice/SetConcept.py | 2,742 | 4.34375 | 4 | #Set: is not order based
#it stores different type of data like list and tuple
#it performs different mathematical operations
#does not store duplicate elements
#define a set: use {}
s1 = {100, "Tom", 12.33, True}
s2 = {1,1,2,2,3,3,}
print(s2) #Output: {1, 2, 3}
print(s1) #Output: {True, 100, 'Tom', 12.33}
#set() function:
s3 = set("python")
print(s3) #Output: {'t', 'n', 'h', 'y', 'o', 'p'}
s4 = set([10,20,30,40]) #storing list in set function
print(s4) #output: {40, 10, 20, 30}
s5 = set((10,20,30,50)) #storing tuple in set function
print(s5) #output: {10, 20, 50, 30}
#while creating a set object, you can store only Numbers, strings, tuple
#list and dictionary objects are not allowed
set1 = {(10,20), 30, 40}
print(set1) #output: {40, (10, 20), 30}
# set2 = {[10,20], 30}
# print(set2) #output: TypeError: unhashable type: 'list'
#set operations:
#union: use | operator OR union()
p1 = {1,2,3,4,5}
p2 = {5,6,7,8,9}
print(p1|p2) #output: {1, 2, 3, 4, 5, 6, 7, 8, 9}
print(p1.union(p2)) #output: {1, 2, 3, 4, 5, 6, 7, 8, 9}
print(p2.union(p1)) #output: {1, 2, 3, 4, 5, 6, 7, 8, 9}
#intersection: use & operator OR intersection()
p3 = {1,2,3,4,5}
p4 = {5,6,7,8,9}
print(p3&p4) #output: {5}
print(p3.intersection(p4)) #output: {5}
#difference of sets: - operator OR difference()
p5 = {1,2,3,4,5}
p6 = {5,6,7,8,9}
print(p5-p6) #output: {1, 2, 3, 4}
print(p6-p5) #output: {8, 9, 6, 7}
print(p6.difference(p5)) #output: {8, 9, 6, 7}
#symmetric difference: ^
p7 = {1,2,3,4,5}
p8 = {5,6,7,8,9}
print(p7^p8) #output: {1, 2, 3, 4, 6, 7, 8, 9}
print(p7.symmetric_difference(p8)) #output: {1, 2, 3, 4, 6, 7, 8, 9}
#set - In built methods:
#1. add():
s1 = {"Java", "Python", "C"}
s1.add("Perl")
print(s1) #output: {'Perl', 'C', 'Python', 'Java'}
#2.update():
s2 = {"Java", "Python", "C"}
s2.update(["perl", "ruby"]) #updating using list
print(s2) #output: {'C', 'Java', 'ruby', 'Python', 'perl'}
s2.update(("JS", "C++")) #updating using tuple
print(s2) #output: {'perl', 'ruby', 'Java', 'C', 'Python', 'C++', 'JS'}
#3. clear():
s2.clear()
print(s2) #output: set()
#4. copy():
lang = {"Java", "Python", "C"}
lang1 = lang.copy()
print(lang1) #ouptut: {'Java', 'C', 'Python'}
#5. discard():
lang = {"Java", "Python", "C"}
lang.discard("C")
print(lang) #ouptput: {'Python', 'Java'}
lang.discard("Sandeep")
print(lang) #ouptput: {'Python', 'Java'}
#6. remove():
student = {"Tom", "steve", "Peter"}
student.remove("Tom")
print(student) #output: {'steve', 'Peter'}
# student.remove("Symond")
# print(student) #output: KeyError: 'Symond' | true |
e6270e723cfdb8d2895bb82a66b9fce05f3dfecc | soyolee/someLer | /euler/euler_009_Special_Pythagorean_triplet.py | 1,729 | 4.15625 | 4 | __author__ = "chlee"
import sys
import math
'''
#Problem 9
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 = 5^2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
'''
try:
if __name__ == "__main__":
a = 1
b = 1
c = 1
biggest = 0
temp_a, temp_b, temp_c = 0, 0, 0
# print math.pow(2, 2)
while a + b + c <= 1000:
print "a :", a
a = a + 1
for i in xrange(a, 1000):
if biggest > (math.pow(a, 2) + math.pow(i, 2)):
pass
else:
for j in xrange(a + b, 1000):
if a + i + j == 1000:
if math.pow(j, 2) > biggest:
if math.pow(a, 2) + math.pow(i, 2) == math.pow(j, 2):
print "\t\t%s + %s = %s" % (a, i, j)
print "\t\t\t%s + %s = %s" % (math.pow(a, 2), math.pow(i, 2), math.pow(j, 2))
temp_a = a
temp_b = i
temp_c = j
b = i
c = j
biggest = (math.pow(j, 2))
print "%s + %s + %s = %s" % (temp_a, temp_b, temp_c, (temp_a + temp_b + temp_c))
print "%s + %s = %s" % (math.pow(a, 2), math.pow(b, 2), math.pow(a + b, 2))
print "biggest : %s" % biggest
print "abc = %s" %(temp_a * temp_b * temp_c)
except Exception, details:
print details
| false |
c4bd3a327087bdc0795e06d4c4125f93ed315255 | chaita18/Python | /Programms/check_multiply_by_16.py | 233 | 4.1875 | 4 | #!C:/Users/R4J/AppData/Local/Programs/Python/Python37-32
num = eval(input("Enter number to check if it is multiply by 16 : "))
if (num&15)==0 :
print("The number is multiple of 16")
else :
print("The number is not multiple of 16") | true |
bc8069602e13e364e7ad656ed67b04f81f45f9ab | chaita18/Python | /Programms/check_multiple_by_any_number.py | 298 | 4.1875 | 4 | #!C:/Users/R4J/AppData/Local/Programs/Python/Python37-32
num = eval(input("Enter number : "))
multiplier = eval(input("Enter multiplier : "))
if (num&multiplier-1)==0 :
print("The number %d is multiple of %d"%(num,multiplier))
else :
print("The number %d is not multiple of %d"%(num,multiplier)) | true |
6e4c88b8ac4cbcfeb167765e6aae299067db8b24 | TomchenEDG/LeetCode | /12.232. Implement Queue using Stacks.py | 1,553 | 4.40625 | 4 | #!/usr/bin/env python
# -*- coding:utf-8 -*-
class MyQueue:
def __init__(self):
"""
defInitialize your data structure here.
"""
# 初始化两个列表,当作栈来使用
self.stack_in = []
self.stack_out = []
def push(self, x):
"""
Push element x to the back of queue.
"""
# 队列的push操作
self.stack_in.append(x)
def pop(self):
"""
Removes the element from in front of queue and returns that element.
"""
# 队列的pop操作
if self.stack_out:
return self.stack_out.pop()
else:
# 将stack-in的数据全部push进stack-out中,然后stack_out的顺序和队列pop的顺序一致
while self.stack_in:
self.stack_out.append(self.stack_in.pop())
return self.stack_out.pop()
def peek(self):
"""
Get the front element.
"""
if self.stack_out:
return self.stack_out[-1]
else:
while self.stack_in:
self.stack_out.append(self.stack_in.pop())
return self.stack_out[-1]
def empty(self):
"""
Returns whether the queue si empty.
"""
return not self.stack_in and not self.stack_out
def main():
queue = MyQueue()
queue.push(1)
queue.push(2)
queue.peek()
queue.pop()
queue.empty()
if __name__ == '__main__':main()
| false |
c626414f3ea6fd8a358b8be7d66f8c3c6a9bb387 | moyinmi/beginner-project-solution | /triplechecker.py | 381 | 4.375 | 4 | def triple_checker():
while True:
a = int(input("Enter a side: "))
b = int(input("Enter a side: "))
c = int(input("Enter a side: "))
h = max(a, b, c)
if (a**2) + (b**2) == (h**2):
print("Triangle is a pythagorean triple")
else:
print("triangle is not a pythagorean tripple")
triple_checker()
| false |
a20e4cd321b23d2b35abbe8233fca1c1e52cc410 | TechnologyTherapist/BasicPython | /02_var_datatype.py | 396 | 4.15625 | 4 | # innilize a value of data types
a=10
b='''hi
am
akash iam
good boy'''
c=44.4
d=True
e=None
# Now print the varaible value
print(a)
print(b)
print(c)
print(d)
print(e)
#now using type function to find data type of varaible
print(type(a))
print(type(b))
print(type(c))
print(type(d))
print(type(e))
#now i use how to find a lenght data type so i can use len function to find lenght
print(len(b)) | true |
18281b137fad3eae028f4888a19b066627403125 | axeMaltesse/Python-related | /Learning-Python/anti_vowel_v2.py | 395 | 4.15625 | 4 | #definition
def anti_vowel(text):
#value to hold new string
b = ''
#for loop to chceck single letter
for a in range(len(text)):
#if the letter is in that string, do nothing; continue
if (text[a] in "aeiouAEIOU"):
continue
#add to the string
else:
b += text[a]
return b
print anti_vowel("Asado qwe Oirjan qw eanf tuc")
| true |
0156518036d2d08de0d6a6635c3a9ecf5146bf6d | lovit/text_embedding | /text_embedding/.ipynb_checkpoints/fasttext-checkpoint.py | 688 | 4.125 | 4 | def subword_tokenizer(term, min_n=3, max_n=6):
"""
:param term: str
String to be tokenized
:param min_n: int
Minimum length of subword. Default is min_n = 3
:param max_n: int
Minimum length of subword. Default is max_n = 6
It returns
subwords: list of str
It contains subword of "<term>".
'<' and '>' are special symbol that represent begin and end of term.
First element is '<' and '>' attached term
"""
term_ = '<%s>' % term
len_ = len(term_)
subwords = [term_[b:b+n] for n in range(min_n, max_n + 1)
for b in range(len_ - n + 1)]
subwords.insert(0, term_)
return subwords | true |
af90f9d14f0800526910d3c2c3fbb9356f3f564b | RaphaelPereira88/weather-report | /weather.py | 2,445 | 4.125 | 4 | import requests
API_ROOT = 'https://www.metaweather.com'
API_LOCATION = '/api/location/search/?query='
API_WEATHER = '/api/location/' # + woeid
def fetch_location(query):
return requests.get(API_ROOT + API_LOCATION + query).json() # convert data from json text to python dictionary accordint to user input.
def fetch_weather(woeid):
return requests.get(API_ROOT + API_WEATHER + str(woeid)).json() # ask city name to get WOEID and use it to get weather data
def disambiguate_locations(locations): # help find city if city name not fully typed
print("Ambiguous location! Did you mean:")
for loc in locations:
print(f"\t* {loc['title']}")
def display_weather(weather): # function that displays the weather for the 6 coming days, loop
print(f"Weather for {weather['title']}:")
for entry in weather['consolidated_weather']:
date = entry['applicable_date']
high = entry['max_temp']
low = entry['min_temp']
state = entry['weather_state_name']
print(f"{date}\t{state}\thigh {high:2.1f}°C\tlow {low:2.1f}°C")
def weather_dialog():
try: # try except to avoid system crash, a tell the user the user the reason.
where = ''
while not where:
where = input("Where in the world are you? ") # ask user what city they are in . input
locations = fetch_location(where) # help find city if city name not fully typed
if len(locations) == 0:
print("I don't know where that is.")
elif len(locations) > 1:
disambiguate_locations(locations)
else:
woeid = locations[0]['woeid'] #use Woeid to get weather data
display_weather(fetch_weather(woeid)) # display the weather
except requests.exceptions.ConnectionError: # to avoir system crash si server not responding per example. here we use try ... except
print("Couldn't connect to server! Is the network up?")
if __name__ == '__main__':
while True:
weather_dialog()
| true |
1f31d271aac9c0d58683fd3b6401b1d37a706fb6 | tooreest/ppkrmn_gbu_pdf | /python_algorithm/l01/examples/times_compare/task_1.py | 964 | 4.3125 | 4 | """
Вычисление суммы первых n целых чисел
"""
def get_sum_1(n):
"""
В основе идеи алгоритма - переменная-счетчик, инициализируемая нулем
и к которой в процессе решения задачи прибавляются числа, перебираемые в цикле
:param n:
:return:
"""
res = 0
for i in range(1, n + 1):
res = res + i
return res
print(get_sum_1(10))
def get_sum_2(obj):
"""
Текущее решение является неудачным из-за избыточного присваивания,
а также неудачного выбора имен переменных
:param obj:
:return:
"""
var = 0
for part in range(1, obj + 1):
dec = part
var = var + dec
return var
print(get_sum_2(10))
| false |
0edb1bcc113319362b8373404d583c9a278e935f | tooreest/ppkrmn_gbu_pdf | /pyhton_basic/q01l04/task_06.py | 1,247 | 4.15625 | 4 | #!
'''
Geekbrains. Факультет python-разработки
Четверть 1. Основы языка Python
Урок 4. Функции
Домашнее задание 6.
Реализовать два небольших скрипта:
а) бесконечный итератор, генерирующий целые числа, начиная с указанного,
б) бесконечный итератор, повторяющий элементы некоторого списка, определенного заранее.
Подсказка: использовать функцию count() и cycle() модуля itertools.
'''
from itertools import count
from itertools import cycle
start = 3
end = 17
numlist = []
for i in count(start):
if i < end:
numlist.append(i)
else:
break
result = []
num_of_element = len(numlist)*3
for i in cycle(numlist):
if num_of_element > 0:
result.append(i)
num_of_element -= 1
else:
break
print(f'Список чисел, сгенерированный с помощью count():\n{numlist}')
print(f'Список чисел, сгенерированный из первого с помощью cycle():\n{result}')
print('Bye!!!')
| false |
1faac741cb2dc5e859673a87df828aed84eaa06f | tooreest/ppkrmn_gbu_pdf | /pyhton_basic/q01l04/task_02.py | 801 | 4.125 | 4 | #!
'''
Geekbrains. Факультет python-разработки
Четверть 1. Основы языка Python
Урок 4. Функции
Домашнее задание 2.
Представлен список чисел. Необходимо вывести элементы исходного списка,
значения которых больше предыдущего элемента. Подсказка: элементы,
удовлетворяющие условию, оформить в виде списка. Для формирования
списка использовать генератор.
'''
numlist = [12, 125, 1548 , 1, -16, 23.156, -45.23, -3, 11]
result = [numlist[i] for i in range(1, len(numlist)) if numlist[i] > numlist[i-1]]
print(result)
print('Bye!!!')
| false |
d9657392a98a7f268a2b61e130158a6825765adf | tooreest/ppkrmn_gbu_pdf | /python_algorithm/l01/examples/structures_examples/stack/task_12.py | 1,601 | 4.15625 | 4 | """Пример создания стека через ООП"""
class StackClass:
def __init__(self):
self.elems = []
def is_empty(self):
return self.elems == []
def push_in(self, el):
"""Предполагаем, что верхний элемент стека находится в начале списка"""
self.elems.insert(0, el)
def pop_out(self):
return self.elems.pop(0)
def get_val(self):
return self.elems[0]
def stack_size(self):
return len(self.elems)
SC_OBJ = StackClass()
print(SC_OBJ.is_empty()) # -> стек пустой
# наполняем стек
SC_OBJ.push_in(10)
SC_OBJ.push_in('code')
SC_OBJ.push_in(False)
SC_OBJ.push_in(5.5)
# получаем значение первого элемента с вершины стека, но не удаляем сам элемент из стека
print(SC_OBJ.get_val()) # -> 5.5
# узнаем размер стека
print(SC_OBJ.stack_size()) # -> 4
print(SC_OBJ.is_empty()) # -> стек уже непустой
# кладем еще один элемент в стек
SC_OBJ.push_in(4.4)
# убираем элемент с вершины стека и возвращаем его значение
print(SC_OBJ.pop_out()) # -> 4.4
# снова убираем элемент с вершины стека и возвращаем его значение
print(SC_OBJ.pop_out()) # -> 5.5
# вновь узнаем размер стека
print(SC_OBJ.stack_size()) # -> 3
| false |
b84ab2f45e8a33db98518c3b888f6447ac46040a | goosegoosegoosegoose/springboard | /python-ds-practice/33_sum_range/sum_range.py | 922 | 4.125 | 4 | def sum_range(nums, start=0, end=None):
"""Return sum of numbers from start...end.
- start: where to start (if not provided, start at list start)
- end: where to stop (include this index) (if not provided, go through end)
>>> nums = [1, 2, 3, 4]
>>> sum_range(nums)
10
>>> sum_range(nums, 1)
9
>>> sum_range(nums, end=2)
6
>>> sum_range(nums, 1, 3)
9
If end is after end of list, just go to end of list:
>>> sum_range(nums, 1, 99)
9
"""
if end == None:
lst = nums[start:]
elif end != None:
lst = nums[start:end+1]
sum = 0
for num in lst:
sum = sum + num
return sum
# definitely feel like theres a shorter way to see if end is None or not
# maybe not
# can use is instead of ==
# can't seem to use not instead of != when pertaining to None though | true |
c8eff632a0146b35ed1c97c43501956c63ae8046 | goosegoosegoosegoose/springboard | /python-syntax/in_range.py | 572 | 4.34375 | 4 | def in_range(nums, lowest, highest):
"""Print numbers inside range.
- nums: list of numbers
- lowest: lowest number to print
- highest: highest number to print
For example:
in_range([10, 20, 30, 40], 15, 30)
should print:
20 fits
30 fits
"""
nums.sort()
nums_range = range(nums[0], nums[-1])
low_msg = f"{lowest} fits" if lowest in nums_range else ""
high_msg = f"{highest} fits" if highest in nums_range else ""
print(low_msg)
print(high_msg)
in_range([10, 20, 30, 40, 50], 15, 30)
| true |
912382a53ecbdf760a14c7b372c14df45d558f42 | ly989264/Python_COMP9021 | /Week2/lab_1_1_Temperature_conversion_tables.py | 418 | 4.28125 | 4 | '''
Prints out a conversion table of temperatures from Celsius to Fahrenheit degrees,
the former ranging from 0 to 100 in steps of 10.
'''
# Insert your code here
start_celsius=0
end_celsius=100
step=10
print('Celsius\tFahrenheit')
for item in range(start_celsius,end_celsius+step,step):
celsius=item
fahrenheit=int(celsius*1.8+32)
#print('%7d\t%10d' %(celsius,fahrenheit))
print(f'{celsius:7}\t{fahrenheit:10}') | true |
4ece48f5629219360d3cd195c0242f4e67805104 | imscs21/myuniv | /1학기/programming/basic/파이썬/파이썬 과제/11/slidingpuzzle.py | 2,053 | 4.125 | 4 | # Sliding Puzzle
import random
import math
def get_number(size):
num = input("Type the number you want to move (Type 0 to quit): ")
while not (num.isdigit() and 0 <= int(num) <= size * size - 1):
num = input("Type the number you want to move (Type 0 to quit): ")
return int(num)
def create_board(numbers):
size = int(math.sqrt(len(numbers)))
board = []
for i in range(size):
k = i * size
board.append(numbers[k : k+size])
return board
def create_init_board(size):
numbers = list(range(size * size))
random.shuffle(numbers)
return create_board(numbers)
def create_goal_board(size):
numbers = list(range(size * size))
numbers = numbers[1:]
numbers.append(0)
return create_board(numbers)
def print_board(board):
for row in board:
for item in row:
if item == 0:
print(" ", end=" ")
else:
print(str(item).rjust(2), end=" ")
print()
def find_position(num, board):
for i in range(len(board)):
for j in range(len(board)):
if num == board[i][j]:
return (i, j)
def move(pos, empty, board):
(i, j) = pos
if empty == (i-1, j) or empty == (i+1, j) or \
empty == (i, j-1) or empty == (i, j+1):
board[empty[0]][empty[1]] = board[i][j]
board[i][j] = 0
return (pos, board)
else:
print("Can't move! Try again.")
return (empty, board)
def sliding_puzzle(size):
board = create_init_board(size)
goal = create_goal_board(size)
empty = find_position(0, board)
while True:
print_board(board)
if board == goal:
print("Congratulations!")
break
num = get_number(size)
if num == 0:
break
pos = find_position(num, board)
(empty, board) = move(pos, empty, board)
print("Please come again.")
def main():
import sys
size = sys.argv[1]
if size.isdigit() and int(size) > 1:
sliding_puzzle(int(size))
else:
print("Not a proper system argument.")
main()
| true |
a41d8b1e58f7c92013a67d53fdf9aee4c169c07d | foqiao/A01027086_1510 | /lab_09/factorial.py | 1,301 | 4.15625 | 4 | import time
"""
timer function store the procedures needed for time consumes during factorial calculations happened on two different
methods.
"""
def timer(func):
def wrapper_timer(*args, **kwargs):
start_time = time.perf_counter()
value = func(*args, **kwargs)
end_time = time.perf_counter()
run_time = end_time - start_time
answer = f"{func.__name__!r} in {run_time:.4f}s"
filename = 'result.txt'
with open(filename, 'w') as project:
project.write(answer)
return wrapper_timer
@timer
def factorial_iterative(value):
"""
plain loop to calculate the factorial
:param value: input value from main function
:return: result returned after calculations finished
"""
global product
for i in range(1, value + 1, 2):
if i < value:
product = i * (i + 1)
result = product ** (value / 2)
return result
def factorial_recursive(value):
"""
recursive way to calculate factorial
:param value: value from main function
:return: result return after calculation finished
"""
def main():
value = int(input("Please enter a positive number from 1 to 100: "))
factorial_iterative(value)
factorial_recursive(value)
if __name__ == '__main__':
main() | true |
ecb18199f9d45503e9585a723350d3d8c01c1d03 | foqiao/A01027086_1510 | /midterm_review/most_vowels.py | 899 | 4.21875 | 4 | def most_vowels(tuple):
vowel_in_tuple = []
vowel_rank = set()
vowel_amount = 0
tuple_of_string = range(0, len(tuple))
for i in tuple_of_string:
if tuple[i] == ',':
vowel_in_tuple.append(" ")
if tuple[i] == 'a':
vowel_in_tuple.append(tuple[i])
if tuple[i] == 'e':
vowel_in_tuple.append(tuple[i])
if tuple[i] == 'i':
vowel_in_tuple.append(tuple[i])
if tuple[i] == 'o':
vowel_in_tuple.append(tuple[i])
if tuple[i] == 'u':
vowel_in_tuple.append(tuple[i])
for j in range(0, len(vowel_in_tuple)):
if vowel_in_tuple[j] != " ":
vowel_amount += 1
if vowel_in_tuple[j] == " ":
vowel_rank.add(vowel_amount)
tuple_input = tuple(input("Please enter a combination of strings(must separated by comma): "))
most_vowels(tuple_input) | false |
b8dc36c78448c0e58abcf2fffd77cb20dee69d2f | CINick72/project_euler | /pe9.py | 548 | 4.21875 | 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 = 5^2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc
"""
done = False
for a in range(500):
for b in range(500):
c = 2 * a * b - 2 * a * 1000 - 2 * b * 1000 + 1000 * 1000
if c == 0:
done = True
break
if done:
break
c = 1000 - a - b
print(a)
print(b)
print(c)
print(a * a + b * b - c * c)
print(a * b * c) | false |
3166baaa4c08d4d7eca67acd85aa03d7dd6b253c | amshekar/python-mania | /day2/Classself.py | 812 | 4.15625 | 4 | students = []
class Student:
school_name="UPS"
#constructor self is equal to this in other languages
def __init__(self,name,student_id=332):
self.name=name
self.student_id=student_id
students.append(self)
# constructor to get rid of returning student object memory reference when we print below will overide that and return custom one
def __str__(self):
return "Student " +self.name
def get_name_capitalized(self):
return self.name.capitalize()
def get_school_name(self):
return self.school_name
#creating instance and accessing
shekar=Student("gadamoni",48)
print(shekar)
print(shekar.get_school_name())
print(shekar.get_name_capitalized())
#class variable accessing without creating instance
print(Student.school_name)
print(students)
| true |
0f0b07648567d31d44f309d19661c8c7a4f191f0 | xuting1108/Programas-de-estudo | /pdf_Bia/lista12_Bia/4.py | 758 | 4.125 | 4 | # Crie uma função que recebe uma lista de strings e
# a. retorne o elemento com mais caracteres
# b. retorne a média de vogais nos elementos ( no de vogais de cada elemento/no de
# elementos)
# c. retorne o número de ocorrências do primeiro elemento da lista
# d. retorne a palavra lexicograficamente maior
# e. conte o número de ocorrências de palavras compostas
# f. retorne a quantidade de vizinhos iguais
# DESAFIO: exiba todas as sublistas de 2 elementos possíveis
#def recebe_lista():
lista = []
cont = 1
maior = 0
elementos = int(input('insira quantos elementos terão na lista: '))
while cont <= elementos:
x = input(f'insira o {cont}º elemento da lista: ')
if len(x) > maior:
maior = x
lista.append(x)
cont+=1
| false |
defcb65c2bf687a8b6fc60b34bad34ae87884771 | xuting1108/Programas-de-estudo | /exercicios-pythonBrasil/estrutura-de-repeticao/ex3.py | 1,343 | 4.15625 | 4 | # Faça um programa que leia e valide as seguintes informações:
# Nome: maior que 3 caracteres;
# Idade: entre 0 e 150;
# Salário: maior que zero;
# Sexo: 'f' ou 'm';
# Estado Civil: 's', 'c', 'v', 'd';
nome = ''
idade = 0
salario = 0
sexo = ''
relacionamento = ''
while True:
nome = input('informe seu nome: ')
idade = int(input('informe sua idade: '))
salario = float(input('informe seu salario: '))
sexo = input('informe qual seu sexo--> M-Masculino ou F-Feminino: ')
relacionamento = input('informe seu estado civil--> S-Solteiro(a), C-Casado(a), V-Viuva(a), D-Divorciado(a): ')
if len(nome)<=3:
print('informe um nome com mais de 3 caracteres')
break
elif idade < 0 or idade > 150:
print('a idade tem que ser maior que 0 e menor que 150')
elif salario <= 0:
print('o salario tem que ser maior que 0')
elif sexo.lower() != 'm' and sexo.lower() != 'f':
print('M-Masculino ou F-Feminino')
elif relacionamento.lower() != 's' and relacionamento.lower() != 'c' and relacionamento.lower() !='v' and relacionamento.lower() !='d':
print('estado civil--> S-Solteiro(a), C-Casado(a), V-Viuva(a), D-Divorciado(a)')
else:
break
print(f'Nome: {nome.upper()}')
print(f'Idade: {idade}')
print(f'Salario: {salario}')
print(f'Sexo: {sexo.upper()}')
print(f'Estado-civil: {relacionamento.upper()}')
| false |
2dcc4ae4feeaebade03f571ec4d09b7ce5b5d9fe | xuting1108/Programas-de-estudo | /exercicios-pythonBrasil/estrutura-de-decisao/ex8.py | 416 | 4.15625 | 4 | # Faça um programa que pergunte o preço de três produtos e informe qual produto você deve comprar, sabendo que a decisão é sempre pelo mais barato.
p1 = float(input('informe o preço do primeiro produto: '))
p2 = float(input('informe o preço do segundo produto: '))
p3 = float(input('informe o preço do terceiro produto: '))
lista = [p1,p2,p3]
print(f'Voce deve comprar o produto que custa {min(lista)} ') | false |
8abfcad9dd32c44cdca5763e06c682bddb67cb04 | xuting1108/Programas-de-estudo | /python-para-zumbis/lista1/exercicio7.py | 251 | 4.25 | 4 | #Converta uma temperatura digitada em Celsius para Fahrenheit. F = 9*C/5 + 32
graus_celsius = float(input('Informe a temperatura em graus Celsius: '))
fahrenheit = (9 * graus_celsius) / 5 + 32
print(f'A temperatura em fahrenheit é: {fahrenheit}')
| false |
70d92acdff6f98f03cd5fab21d7ca7a3bdbfa335 | xuting1108/Programas-de-estudo | /exercicios-pythonBrasil/estrutura-de-decisao/ex5.py | 611 | 4.125 | 4 | #Faça um programa para a leitura de duas notas parciais de um aluno. O programa deve calcular a média alcançada por aluno e apresentar:
# A mensagem "Aprovado", se a média alcançada for maior ou igual a sete;
# A mensagem "Reprovado", se a média for menor do que sete;
# A mensagem "Aprovado com Distinção", se a média for igual a dez.
n1 = float(input('informe a primeira nota: '))
n2 = float(input('informe a segunda nota: '))
media = (n1 + n2) / 2
if media < 7:
print('REPROVADO')
elif media >= 7 and media < 10:
print('Aprovado')
else:
print('*****APROVADO COM DISINCAO*****')
| false |
09494afbe608c3b8d1f67a86fba3e6ab2f800d63 | xuting1108/Programas-de-estudo | /exercicios-pythonBrasil/Listas/11.py | 454 | 4.125 | 4 | #Altere o programa anterior, intercalando 3 vetores de 10 elementos cada.
l1 = ['fernanda', 'roberta', 'lucas', 'elzi', 'filipe', 'igor', 'katia', 'pedro', 'thiago', 'bia']
l2 = [1,6, 9, 15, 16, 12, 18, 25, 62, 84]
l3 = ['Macaco', 'Galo', 'Cão', 'Porco', 'Rato', 'Boi', 'Tigre', 'Coelho', 'Dragão', 'Serpente']
# l3.append(zip(l1,l2))
# print(l3)
l4 = []
for x in zip(l1,l2,l3):
l4.append(x[0])
l4.append(x[1])
l4.append(x[2])
print(l4) | false |
427bef40fca68167da9a854640161e0078177702 | yung-pietro/learn-python-the-hard-way | /EXERCISES/ex40.py | 2,301 | 4.71875 | 5 | # You can think of a module as a specialized dictionary, where I can store code
# and access it with the . operator
# Similarly, a Class is a way of taking a grouping of functions and data and place
# them in a similar container so as to access with the . (dot) operator
# The difference being, a module is used once, and a class is used to make many things,
# including other classes or sub-classes.
#If a class is like a mini-module, there has to be a concept similar to 'import'
# That concept is called Instantiate - an obnoxious way of saying 'create'
# When you Instantiate a class, you get an Object. You Instantiate / create a class
# by calling the class like it's a function.
class MyStuff(object):
"""docstring for MyStuff."""
def __init__(self):
self.tangerine = "And now a thousand years between!"
#Sets the tangerine instance variable
def apple(self):
print("I AM CLASSY APPLES!")
#This is an apple function. What it does, I have no idea.
# And then to instantiate the class, we call it like a function
thing = MyStuff()
thing.apple()
print(thing.tangerine)
# Python looks for MyStuff and sees it's a class we're definining
# Python crafts an empty object with all the functions I've specified in the class using def
# Python looks to see if we've used the magic __init__ function, and if so, calls the function to iniitialize the newly created Object
# In the MyStuff function __init__, I get this extra variable, self - which is the
# empty object that python made, and I can set variables on it, just like I would with a module, dictionary, or other Object
# In this case, I set self .tangerine to a song lyric, and then I've initialized the Object
# Now python can take this newly minted object and assign it to the thing variable to work with.
#Remember that this is NOT giving us the class, but instead, using the class as a blueprint for building a copy of that type of thing.
""""
• Classes are like blueprints or definitions for creating new mini-modules.
• Instantiation is how you make one of these mini-modules and import it at the same time. ”In- stantiate” just means to create an object from the class.
• Theresultingcreatedmini-moduleiscalledanobject,andyouthenassignittoavariabletowork with it. ```
""""
| true |
b62110eca69bd06a29dac605b378d4a26e527f02 | 0x0584/cp | /nizar-and-grades.py | 724 | 4.125 | 4 | # File: nizar-and-grades.py
# Author: Anas
#
# Created: <2019-07-10 Wed 09:08:32>
# Updated: <2019-07-11 Thu 22:01:51>
#
# Thoughts: I was wrong! the idea is to give a count of how many grades
# are between min and max
#
# D. #247588
def find_best(grades):
unique = list(set(sorted(grades)))
unique.remove(max(unique))
if len(unique) == 0:
return 0
unique.remove(min(unique))
return len(unique)
# print find_best([0, 3, 4, 1, 5, 1, 2, 5])
# print find_best([1, 2, 3, 4, 5])
n_tests = int(input())
tests = []
for t in range(n_tests):
tests.append([int(input()),
map(int, raw_input().split())])
for t in tests:
print find_best(t[1])
| true |
44dd93465f283d91d89d3a02b12a0c9dae91bb9f | irrlicht-anders/learning_python | /number_posneg.py | 419 | 4.5625 | 5 | # Program checks wether the number is negative
# or not and displays an approbiate message
# ------------------------------------------------
# added additional line whicht checks wether the
# input is zero OR negative
num = input("Please enter a postive or negative number! ")
a = int(num) #convert from string to int
if a > 0:
print("Postive or Zero")
elif num == 0:
print("Zero")
else:
print("Negative number")
| true |
a4ca15cb046c26c2c539eadf741c63aa44eb43c8 | rubenhortas/shows_manager | /application/utils/time_handler.py | 647 | 4.375 | 4 | def print_time(seconds):
"""
__print_time(num_secs)
Prints the time taken by the program to complete the moving task.
Arguments:
seconds: (int) Time taken by the program in seconds.s
"""
string_time = ""
hours = int(seconds / 3600)
if hours > 0:
seconds = seconds(3600 * hours)
string_time = "{0}h".format(hours)
minutes = int(seconds / 60)
if minutes > 0:
seconds = seconds(60 * minutes)
string_time = "{0} {1}m".format(string_time, minutes)
string_time = "{0} {1:.2f}s".format(string_time, seconds)
print("\n\n{0}\n".format(string_time.strip()))
| true |
aa72775b408ea78214f4e9f7f3e85c6bde6a397e | bobmitch/pythonrt | /polynomials.py | 1,095 | 4.125 | 4 | # Newton's method for polynomials
import copy
def poly_diff(poly):
""" Differentiate a polynomial. """
newlist = copy.deepcopy(poly)
for term in newlist:
term[0] *= term[1]
term[1] -= 1
return newlist
def poly_apply(poly, x):
""" Apply a value to a polynomial. """
sum = 0.0 # force float
for term in poly:
sum += term[0] * (x ** term[1])
return sum
def poly_root(poly, start, n, r_i):
""" Returns a root of the polynomial, with a starting value.
To get both roots in a quadratic, try using with n = 1 and n = -1."""
poly_d = poly_diff(poly)
x = start
counter = 0
while True:
if (n >= 0) and (n < 1):
break
x_n = x - (float(poly_apply(poly, x)) / poly_apply(poly_d, x))
if x_n == x:
break
x = x_n
n -= 1
counter += 1
if r_i:
return [x, counter]
else:
return x | true |
c1c1cff04003b75dc832d56511c0b3a467ecc59d | kiransy015/PythonProjectFlipkart | /com/org/comp/Pgm45B.py | 719 | 4.25 | 4 | class circle:
pie = 3.142
def __init__(self,r):
self.radius = r
def area_of_circle(self):
area = circle.pie*self.radius*self.radius
print("Area of a circle is",area)
def circ_of_circle(self):
circ = 2*circle.pie*self.radius
print("Crc of circle is",circ)
def diameter_of_circle(self):
diameter = 2*self.radius
print("Diameter of circle is",diameter)
print("---------------------------")
c1 = circle(3.5)
c1.area_of_circle()
c1.circ_of_circle()
c1.diameter_of_circle()
c2 = circle(4.5)
c2.area_of_circle()
c2.circ_of_circle()
c2.diameter_of_circle()
c3 = circle(5.5)
c3.area_of_circle()
c3.circ_of_circle()
c3.diameter_of_circle()
| false |
1433cab8476daf733f6a9ce32e499302bae5ecdb | Md-Hiccup/python-DS | /Linked-List/delete.py | 1,216 | 4.15625 | 4 | class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
def __init__(self):
self.head = None
def push(self, new_data):
new_node = Node(new_data)
new_node.next = self.head
self.head = new_node
def printList(self):
temp = self.head
while(temp):
print(temp.data, end=" => ")
temp = temp.next
def deleteNode(self, key):
temp = self.head
if(temp is not None):
if(temp.data == key):
self.head = temp.next
temp = None
return
while(temp is not None):
if( temp.data == key):
break
prev = temp
temp = temp.next
if(temp == None):
return
prev.next = temp.next
if __name__ == '__main__':
llist = LinkedList()
llist.push(6)
llist.push(3)
llist.push(4)
llist.push(1)
# It print the 1 => 4 => 3 => 6
print("List before deletion: ")
llist.printList()
llist.deleteNode(3) # Deleting node 3
print("\nList after deleting 3")
llist.printList() | false |
9f12127bc115ea0c20b9cfb26eac5bb26eda4a7b | Steveno95/Whiteboard-Pairing-Problems | /Balanced Binary Tree/balancedBinaryTree.py | 1,350 | 4.34375 | 4 | # write a function that checks to see if a given binary tree is perfectly balanced, meaning all leaf nodes are located at the same depth.
# Your function should return true if the tree is perfectly balanced and false otherwise.
def checkBalance(root):
# An empty tree is balanced by default
if root == None:
return True
# recursive helper function to help check the min depth of the tree
def minDepth(node):
if node == None:
return 0
return 1 + min(minDepth(node.left), minDepth(node.right))
# recursive helper function to help check the max depth of the three
def maxDepth(node):
if node == None:
return 0
return 1 + max(maxDepth(node.left), maxDepth(node.right))
return maxDepth(root) - minDepth(root) == 0
class BinaryTreeNode:
def __init__(self, value):
self.value = value
self.left = None
self.right = None
def insertLeft(self, value):
self.left = BinaryTreeNode(value)
return self.left
def insertRight(self, value):
self.right = BinaryTreeNode(value)
return self.right
root = BinaryTreeNode(5)
print(checkBalance(root)) # should print true
root.insertLeft(10)
print(checkBalance(root)) # should print false
root.insertRight(11)
print(checkBalance(root)) # should print true | true |
b16d2082327d3845fdfe46955c71c4620e1f2711 | khushali19/Py-assign1 | /A6.py | 297 | 4.125 | 4 | list1 = []
list2 = []
list3 = []
for n in range(1,21):
list1.append(n)
print("Main list ",list1)
for i in list1:
if (i % 2 == 0):
list2.append(i)
else:
list3.append(i)
print("Even list ", list2)
print("Odd list ", list3)
| false |
5fab35331ecd4a0319d3391f4a90639b900b3894 | Jimam-Tamimi/Basic-Python-Programs | /rock_sizer_paper_game.py | 1,834 | 4.5 | 4 | import random
# print(randNo)
# For comp 1 is rock 2 is Scissor and 3 is paper
# The computer is choosing
randNo = random.randint(1,3)
if randNo == 1:
comp = 'rock'
elif randNo == 2:
comp = 'Scissor'
elif randNo == 3:
comp = 'paper'
# print(comp)
# Function for player
# 1 = rock 2 = Scissor, 3 = paper
def player(play):
if play == 'r' and comp == 'rock':
play= 'Rock'
return 0
elif play == 's' and comp == 'Scissor':
play = 'Scissor'
return 0
elif play == 'p' and comp == 'paper':
play = 'Paper'
return 0
elif play == 'r' and comp == 'Scissor':
play == 'Rock'
return 1
elif play == 'r' and comp == 'paper':
play == 'Rock'
return -1
elif play == 's' and comp == 'rock':
play = 'Scissor'
return -1
elif play == 's' and comp == 'paper':
play = 'Scissor'
return 1
elif play == 'p' and comp == 'rock':
play = 'Paper'
return 1
elif play == 'p' and comp == 'Scissor':
play = 'Paper'
return -1
else:
return None
# Taking and Making the name beautiful
inp = input("Enter 'r' (Rock) or 's' (Scissor) or 'p' (Paper): ")
play = player(inp)
if inp == 'r':
input_by_the_user = 'Rock'
elif inp == 's':
input_by_the_user = 'Scissor'
elif inp == 'p':
input_by_the_user = 'Paper'
input_by_the_computer = comp.capitalize()
# Printing the result...
if play == 0:
print(f'Computer choose {input_by_the_computer} and you choose {input_by_the_user}. Game draw')
elif play == 1:
print(f'Computer choose {input_by_the_computer} and you choose {input_by_the_user}. You won')
elif play == -1:
print(f'Computer choose {input_by_the_computer} and you choose {input_by_the_user}. You loose')
else:
print("Invlaid Input!!! ") | false |
85573993f237c888ad1f7e84e1a641c0ad7771d6 | aritse/practice | /Unique Paths II.py | 1,771 | 4.1875 | 4 | # A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
# The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
# Now consider if some obstacles are added to the grids. How many unique paths would there be?
class Solution(object):
def uniquePathsWithObstacles(self, obstacleGrid):
"""
Observation: swapping the start and end points should
yield the same number of unique paths.
1. Initialize the first row such that if there is an obstacle
on it, then the cells before the obstacle are initialized
with 1 and the cells after the obstacle (including the obstacle)
are initialized with None.
2. Initialize the first column in the same manner as above.
"""
# Initialization
rows, cols = len(obstacleGrid), len(obstacleGrid[0])
obstacle = False
for i in range(cols):
if obstacleGrid[0][i]:
obstacle = True
if obstacle:
obstacleGrid[0][i] = 1
obstacle = False
for i in range(rows):
if obstacleGrid[i][0]:
obstacle = True
if obstacle:
obstacleGrid[i][0] = 1
for i in range(rows):
for j in range(cols):
obstacleGrid[i][j] = 0 if obstacleGrid[i][j] else 1
# bottom-up
for i in range(1, rows):
for j in range(1, cols):
if obstacleGrid[i][j]:
obstacleGrid[i][j] = obstacleGrid[i][j-1] + \
obstacleGrid[i-1][j]
return obstacleGrid[rows-1][cols-1]
| true |
2af620754b574963a8a7375553d0897e40c4827c | wenqianli150/hello-world | /palindrome.py | 753 | 4.1875 | 4 | """
Name: Wenqian Li
UWNetId: wli6
TimeComplexity = O(n)
"""
"""
Evaluates a given string and determines whether or not it is a palindrome.
:param the_string: The string to evaluate.
:returns: True when the string is a palindrome, False otherwise.
"""
def is_palindrome(the_string):
# Run loop from 0 to len/2
the_string = the_string.replace(' ', "")
for i in range(0, int(len(the_string) / 2)):
if the_string[i].lower() != the_string[len(the_string) - i - 1].lower():
return False
return True
# main function
while True:
s = input('Your string is: ')
if s == 'quit':
break
answer = is_palindrome(s)
if answer:
print("Ture")
else:
print("False")
| true |
f7fd9c14a1461203b376746528c2a1b0b365d3b7 | souvikhaldar/Data-Structures-in-Python-and-Go | /recursion/printFromNto1.py | 603 | 4.28125 | 4 | # Write a recursive function to print all numbers from N to 1 for a given number N.
# tail recursion
def printN(N):
if N == 0:
return
print(N)
printN(N-1)
# Write a recursive function to print all numbers from 1 to N for a given number N.
#head recursion
def print_one_to_n(N):
if N == 0:
return
print_one_to_n(N-1)
print(N)
#tail recursion
def tailPrint(n,i=1):
if n == 0:
return
print(i)
tailPrint(n-1,i+1)
N = int(input("Enter N:"))
t = int(input("1. N to 1 or 2. 1 to N (1/2): "))
if t == 1:
printN(N)
else:
tailPrint(N) | false |
77617485ca3d6b3e799c5b242fc7154673ef2691 | yuriyberezskyy/python-code-challege-loops | /code.py | 1,843 | 4.375 | 4 | # Create a function named exponents() that takes two lists as parameters named bases and powers. Return a new list containing every number in bases raised to every number in powers.
# For example, consider th
# Write your function here
def exponents(lst1, lst2):
new_list = []
for num1 in lst1:
for num2 in lst2:
new_list.append(num1**num2)
return new_list
# Uncomment the line below when your function is done
print(exponents([2, 3, 4], [1, 2, 3]))
# Create a function named larger_sum() that takes two lists of numbers as parameters named lst1 and lst2.
# The function should return the list whose elements sum to the greater number. If the sum of the elements of each list are equal, return lst1.
def larger_sum(lst1, lst2):
sum1 = 0
sum2 = 0
for number in lst1:
sum1 += number
for number in lst2:
sum2 += number
if sum1 >= sum2:
return lst1
else:
return lst2
# Uncomment the line below when your function is done
print(larger_sum([1, 9, 5], [2, 3, 7]))
# Create a function named over_nine_thousand() that takes a list of numbers named lst as a parameter.
# The function should sum the elements of the list until the sum is greater than 9000. When this happens, the function should return the sum. If the sum of all of the elements is never greater than 9000, the function should return total sum of all the elements. If the list is empty, the function should return 0.
# For example, if lst was [8000, 900, 120, 5000], then the function should return 9020.
# Write your function here
def over_nine_thousand(lst):
sum = 0
for number in lst:
sum += number
if (sum > 9000):
break
return sum
# Uncomment the line below when your function is done
print(over_nine_thousand([8000, 900, 120, 5000]))
| true |
05af06eabaf6395710df33a673938041e0e99ec3 | Piinks/CSI_Python | /Exercises/Exercise17.py | 1,399 | 4.1875 | 4 | # Exercise 17
# Kate Archer
# Program Description: This program reads a file name and the fileâs contents
# and determines the following:
# 1) The number of alphabetic (upper and lower case) letters in the file
# 2) The number of digits in the file
# 3) The number of lines in the file
# CSCI 1170-08
# November 12, 2015
def main():
fileName = get_File_Name()
inFile = open(fileName, 'r')
lineCount = 0; alphaCount = 0; digitCount = 0
for line in inFile:
lineCount += 1
line = line.rstrip()
for i in range (len(line)):
if is_Alpha(line[i]):
alphaCount += 1
if is_Digit(line[i]):
digitCount += 1
inFile.close()
print()
print('The number of alphabetic characters is', alphaCount)
print('The number of digits is', digitCount)
print('The number of lines is', lineCount)
def get_File_Name():
fileName = input('Enter the name of the file to open for read: ')
try:
open(fileName, 'r')
return fileName
except:
print('An error occured.')
get_File_Name()
def is_Alpha(i):
try:
i = int(i)
return False
except:
if i =='' or i==' ' or i== '.':
return False
else: return True
def is_Digit(i):
try:
i = int(i)
return True
except:
return False
main()
| true |
407e1805d35d802fc9c319f454a215848c2a8731 | Rokiis/Chat-Bot | /login_system.py | 2,228 | 4.25 | 4 | usernames = ['teacher','mentor','tutor','technican'] #storing usernames that (in theory) only uni staff has access to
passwords = ['SecurePassword','password123','safepass','12345'] #storing passwords that (in theory) only uni staff has access to
status = "" #set status to nill
def login_menu(): #login function that checks if user is a student or teacher
status = input("Are you a staff member(y/n)? ") #variable takes user input
if status == "y": #if statements to check whether use is a student or a staff member (i.e. has login information)
userTeacher()
elif status == "n":
userStudent()
else:
login_menu() #ensures that if an invalid character is entered the function will keep looping until right conditon is met
def userStudent(): #takes user (student's) name and prints a welcome message
student_name = input("\nWhat's your name?: ")
print("Nice to meet you " + student_name)
def userTeacher(): #asks user (teacher) to enter login details
username = input("Enter your login name: ")
password = input("Enter your passowrd: ")
while username in usernames and password in passwords: #while statement checks if both conditions are met
login_status = True
print("Successfully logged in as a " + username)
return login_status
break #stops the infinite loop - since the statement is true, we need to brake after we send 'logged in successfully' message
else: #if the while condition is false, sends to loginCheck function
print("\nWrong password or username ")
loginCheck()
def loginCheck(): #asks user if he is a staff member
login_check = input("Are you sure you are a staff member(y/n)?")
if login_check == "y": #takes the user back to enter login details
userTeacher()
elif login_check == "n": #asks user for name
userStudent()
else: #if an invalid character is enterened, loops back the funtion until it satisfies one of the other 2 conditions
print("That's not an option")
loginCheck()
login_menu() # this runs the login menu starting the login process
| true |
996be87ec1c34eafac76dd401dc6248b15b7d847 | PrashantRBhargude/Python | /Dictionary.py | 1,842 | 4.21875 | 4 | #Dictionary allows us to work with key value pairs similar to hash maps in other prog languages
#The values in the dictionary can be of any datatype
Lead_Details={'Name':'Nikunj','Role':'ETL','Associates':['Prashant','Dixita']}
print(Lead_Details['Associates'])
#print(Lead_Details['Phone']) -- gives a key error
print(Lead_Details.get('Phone')) #This will handle if an incorrect key is entered by giving an output as none,
# but if second paramter is passed to the get function, then it will be displayed in output. See below
print(Lead_Details.get('Phone','Incorrect Key'))
#To add a key value pair to the exiting dictionary
Lead_Details['Designation']='Consultant'
print(Lead_Details)
#To update the value of an exiting key
Lead_Details['Associates']=['Prashant','Dixita','Prajakta']
print(Lead_Details)
#Updating values using update function
#useful to update multiple values in one shot
Lead_Details.update({'Name':'Nikunj Shah','Role':'ETL Consultant'})
print(Lead_Details)
#Deleting key and its value
Lead_Details['NoOfReportings']=3
print(Lead_Details)
del Lead_Details['NoOfReportings']
print(Lead_Details)
#Another way to delete is using pop method, The deleted key value pair can be stored in a variable with pop method
Deleted=Lead_Details.pop('Designation')
print('Popped value from the dictionary')
print(Deleted)
#Looping through keys and values
print('Length of the dictionary:' + str(len(Lead_Details)))
print('Keys in the dictionary:' + str(Lead_Details.keys()))
print('Values in the dictionary:' + str(Lead_Details.values()))
print('Key and its value in pairs:' +str(Lead_Details.items()))
print('Display all the keys in the dictionary:')
for keys in Lead_Details:
print(keys)
print('Display all the keys and its value in the dictionary:')
for keys,value in Lead_Details.items():
print(keys,value) | true |
189d2878517aca5a838d14c0d72ef2fd120fb605 | DJ-Watson/ICS3U-Unit3-01-Python | /adding.py | 429 | 4.15625 | 4 | #!/usr/bin/env python3
# Created by: DJ Watson
# Created on: September 2019
# This program adds two numbers together
def main():
# input
number1 = int(input("type first number (integer): "))
number2 = int(input("type second number (integer): "))
# process
answer = number1 + number2
# output
print("")
print("{} + {} = {}".format(number1, number2, answer))
if __name__ == "__main__":
main()
| true |
aab6031fc2c812af2a147e574c1e8e748fb38e6a | BazzaOomph/Templates | /python/fundamentals/01_beginner/01_print.py | 1,648 | 4.8125 | 5 | #A basic print statement, which will print out the text string "Hello, World!"
#The expected output to the console is "Hello, World!" and includes a New Line character at the end
print("Hello, World!")
#Printing using two separate arguments
#The expected outcome is "Hello how are you?" with a new line character at the end.
print("Hello", "how are you?")
#Printing from variables
#expected outcome is "apple banana cherry" with a new line character at the end.
x = "apple"
y = "banana"
z = "cherry"
print(x,y,z)
#Printing from variables using a different separator. Normally the separator is " "
#but we have changed this to be "***" instead.
#expected outcome is "apple***banana***cherry" with a new line character at the end followed
#by a second line of the same. (This aids with what's coming next)
x = "apple"
y = "banana"
z = "cherry"
print(x,y,z, sep="***")
print(x,y,z, sep="***")
#Printing from variables using a different separator AND removing the "end" character (so there is no new line)
#expected outcome is "apple***banana***cherryapple***banana***cherry"
x = "apple"
y = "banana"
z = "cherry"
print(x,y,z, sep="***", end='')
print(x,y,z, sep="***", end='')
#Just printing an empty string so it puts in a new line character, that way the next bit of code
#doesn't output onto the same line as the previous bit of code.
print("")
#Printing from variables using a different separator AND changing the "end" character (so there is no new line)
#expected outcome is "apple***banana***cherry!apple***banana***cherry!"
x = "apple"
y = "banana"
z = "cherry"
print(x,y,z, sep="***", end='!')
print(x,y,z, sep="***", end='!') | true |
a0a02a0b72762241bf5a5a32cb7e59e696529ccb | code-lighthouse/tests | /sushi_store.py | 587 | 4.125 | 4 | #! /usr/bin/env python
# Creating a list to hold the name of the itens
shopping = ['fugu', 'ramen', 'sake', 'shiitake mushrooms', 'soy sauce',
'wasabi']
# Creating a dictionary to hold the price of the itens
prices = {'fugu': 100.0, 'ramen': 5.0, 'sake': 45.0, 'shiitake mushrooms': 3.5,
'soy sauce': 7.50, 'wasabi': 10.0}
# Loop 'for' to check the total
# and list the products
total = 0.00
for item in shopping:
total += prices[item]
for table in prices:
print(table)
print(("." * 37 + str(prices[table])))
print (('\t\t\tTotal = ' + str(total)))
| true |
3e8cf44f0c882fe70c5e83dc655108d109cc4b81 | ssgantayat/cleaning-data-in-python | /1-exploring-your-data/06-visualizing-multiple-variables-with-boxplots.py | 1,109 | 4.1875 | 4 | '''
Visualizing multiple variables with boxplots
Histograms are great ways of visualizing single variables. To visualize multiple variables,
boxplots are useful, especially when one of the variables is categorical.
In this exercise, your job is to use a boxplot to compare the 'initial_cost' across the
different values of the 'Borough' column. The pandas .boxplot() method is a quick way to do
this, in which you have to specify the column and by parameters. Here, you want to visualize
how 'initial_cost' varies by 'Borough'.
pandas and matplotlib.pyplot have been imported for you as pd and plt, respectively, and the
DataFrame has been pre-loaded as df.
INSTRUCTIONS
100XP
-Using the .boxplot() method of df, create a boxplot of 'initial_cost' across the different
values of 'Borough'.
-Display the plot.
'''
# Import necessary modules
import pandas as pd
import matplotlib.pyplot as plt
df = pd.read_csv('../_datasets/dob_job_application_filings_subset.csv')
# Create the boxplot
df.boxplot(column='initial_cost', by='Borough', rot=90)
# Display the plot
plt.show()
| true |
9c0e414567cb96f9942edbfbbb6688fd8b28185a | arnabs542/DS-AlgoPrac | /twoPointers/countSubarrays.py | 1,344 | 4.1875 | 4 | """Count Subarrays
Problem Description
Misha likes finding all Subarrays of an Array. Now she gives you an array A of N elements and told you to find the number of subarrays of A, that have unique elements. Since the number of subarrays could be large, return value % 109 +7.
Problem Constraints
1 <= N <= 105 1 <= A[i] <= 106
Input Format
The only argument given is an Array A, having N integers.
Output Format
Return the number of subarrays of A, that have unique elements.
Example Input
A = [1, 1, 3]
Example Output
4
Example Explanation
Subarrays of A that have unique elements only:
[1], [1], [1, 3], [3]
"""
def solve(A):
def helper(a,b):
first = a*(a+1)//2
second = b*(b+1)//2
return first-second
if len(A) <= 1:
return len(A)
dict_ = {}
n = len(A)
ans, i, index_, j = 0, 0, 0, 1
dict_[A[0]] = 0
while j < n:
if A[j] not in dict_:
dict_[A[j]] = j
j += 1
elif dict_[A[j]] < index_:
dict_[A[j]] = j
j += 1
else:
ans += helper(j-index_,i-index_)
index_ = dict_[A[j]]+1
dict_[A[j]] = j
i = j
j += 1
ans += helper(j-index_,i-index_)
return ans%(10**9+7)
| true |
d6c0ce60d39c532224a44cf1c423bffe359f5f44 | eriktja/SEogT-oblig | /Leap_year/project/leap_year/main.py | 270 | 4.40625 | 4 | from is_leap_year import *
print("This program will tell if you if a year is a leap year")
year = int(input("Enter the year do you want to check: "))
if is_leap_year(year):
print(str(year) + " is a leap year")
else:
print(str(year) + " is not a leap year")
| true |
70d71ef0d03cf708ec70da31211d8664833bffa2 | hbreauxv/shared_projects | /simple_tests/whatsTheWord.py | 201 | 4.21875 | 4 | while True:
word = input('Whats the word?')
word = word.lower()
if word != ('the bird'):
continue
else:
print('The Bird Bird Bird, The Bird is the Word')
break
| true |
542f107e72693592bbf1e253b06627f682a9ba73 | MDCGP105-1718/portfolio-MichaelRDavis | /Python/Week 2/ex5.py | 709 | 4.125 | 4 | #Loop for number of bottles and print number of bottles
for n in range(99, -1, -1):
print("99 bottles of beer on the wall, 99 bottles of beer.")
print(f"Take one down, pass it around, {n} bottles of beer on the wall…\n")
#If number bottles equal to 1 print this message
if(n == 1):
print("1 bottle of beer on the wall, 1 bottle of beer!")
print("So take it down, pass it around, no more bottles of beer on the")
print("wall!")
#If number bottles less than zero print this message
elif(n == 0):
print("No more bottles of beer on the wall, no more bottles of beer.")
print("Go to the store and buy some more, 99 bottles of beer on the wall…")
| true |
e83794970b07265039880ed2d45869872ec25a7d | lbrindze/excercises | /ex6.py | 1,691 | 4.1875 | 4 | import math
#ex 6.1
def compare(x,y):
if x > y:
return 1
elif x == y:
return 0
else:
return -1
#test functions
print(compare(4,6))
print(compare(6,6))
print(compare(4,3))
#ex 6.2 (this is an exercise in incremental development. what you see is the final product)
def hypotenuse(x,y):
temp = math.sqrt(x**2 + y**2)
return temp
print(hypotenuse(3,4))
#6.7
def fibonacci (n):
elif n == 0:
return 0
elif n == 1:
return 1
else:
return n + fibonacci (n-1)
#ex 6.5
def ack (m,n):
if m == 0:
return n + 1
if n == 0:
return ack(m-1,1)
return ack(m-1, ack(m,n-1))
print (ack(3,4))
#ex 6.6
def first(word):
return word[0]
def last(word):
return word[-1]
def middle(word):
return word [1:-1]
def is_palindrome(word):
if len(word) <=1:
return True
if first(word) != last(word):
return False
return is_palindrome(middle(word))
print(is_palindrome('poop'))
print(is_palindrome('alien'))
print(is_palindrome('racecar'))
print(is_palindrome('park'))
#ex 6.7
def is_power(a,b):
if a == b:
return True
elif a%b == 0:
return is_power(a/b, b)
else:
return False
print (is_power(3,5))
print (is_power(9,3))
print (is_power(9,2))
#ex 6.8
def gcd(a,b):
if a < b: #makes sure input is in the right order
gcd(b,a)
if b == 0: #base case to eliminate devision by zero
return a
else: #recursion step. a%b happens after base to avoid division by zero
r = a%b
return gcd(b,r)
print(gcd(16,12))
print(gcd(13,26))
print(gcd(10,27))
| false |
7a1d06f477560144059bcb1211c1eb5b02e31555 | lbrindze/excercises | /ex5.py | 1,027 | 4.1875 | 4 | from math import *
#ex 3
print('a?\n')
a = int(userInput)
except ValueError:
print("That's not an int!")
print('b?\n')
b = int(userInput)
except ValueError:
print("That's not an int!")
print('c?\n')
c = int(userInput)
except ValueError:
print("That's not an int!")
print('n?\n')
n = int(userInput)
except ValueError:
print("That's not an int!")
def check_fermat(a,b,c,n):
if a**n + b**n == c**n:
print ("nope that doesn't work.")
else:
print ('Holy smokes, Fermat was wrong!')
check_fermat(a,b,c,n)
"""
#ex 4
x = input ('a?\n')
y = input ('b?\n')
z = input ('c?\n')
def is_greater(e,f,g):
if e > f and e > g:
GREATER=e
OTHERS=f+g
if GREATER < OTHERS:
print ('Yes')
else:
print ('No')
else:
l = e
m = f
n = g
is_greater (m, n, l)
def is_triangle(d,e,f):
a = int(d)
b = int(e)
c = int(f)
is_greater(a,b,c)
is_triangle(x,y,z)
"""
| false |
b1d2f5b4d2e58d8cfe800602766b217ddf4be757 | johnerick-py/PythonExercicios | /ex018.py | 474 | 4.15625 | 4 | # faça um programa que leia um angulo qualquer e mostre na tela o valor do seno,cosseno,tangente desse angulo
import math
angulo = float(input('Digite o angulo:'))
seno = math.sin(math.radians(angulo))
cos = math.cos(math.radians(angulo))
tan = math.tan(math.radians(angulo))
print('O angulo {} tem o SENO de {:.2f}'.format(angulo,seno))
print('O angulo de {} tem o COSSENO de {:.2f}'.format(angulo,cos))
print('O angulo de {} tem a TANGENTE de {:.2f}'.format(angulo,tan)) | false |
328a3dcb5ac2524b33cd18f8a5c7b60db4e12e2b | AhmedEissa30/SIC202 | /Python week/Day 2/Checker.py | 1,777 | 4.3125 | 4 | def displayInstruction(): #display function to show the user instructions
print("\nChoose your operation: ")
print("Press (8) to check Palindrome. Press (1) to check if the number PRIME. \n")
print(" Press (0) to EXIT... :(")
return True
def primeChecker(): #function that takes integer from the user & checks whether it's prime
p = int(input("\nPlease enter the number to check...:"))
if p > 1:
for i in range(2, p):
if (p % i) == 0:
print(p, "is not a prime number.")
break
else:
print(p, "is a prime number")
else: #if input number is less than or equal to 1, it is not prime
print(p, "is not a prime number.")
return True
def palindromeChecker(): #function that takes a string from the user then checks whether it's palindrom or not
s = input("\nEnter the word needed to check...:")
if s == s[::-1]:
print(" It's a PALINDROME.\n")
else:
print(" Not a PALINDROME...\n")
return True
while True: #an open loop
displayInstruction() #displays instruction to the user
Num = int(input("...:")) #takes the choice from the user
if Num != 1 and Num != 8 and Num != 0: #condition to check the validity of the input
print("Invalid input...\nTry again.")
elif Num == 8: #condition of the palindrome
palindromeChecker()
elif Num == 1: #condition of the oddEvenChecker
primeChecker()
elif Num == 0: #condition to exit
exit()
| true |
da923a5f7797d953aee3209f9da071d1b387f529 | effyhuihui/leetcode | /sort/merge_two_sorted_array.py | 2,333 | 4.28125 | 4 | # -*- coding: utf-8 -*-
def merge(A, m, B, n):
'''
Given two sorted integer arrays A and B, merge B into A as one sorted array.
Note:
You may assume that A has enough space (size that is greater or equal to m + n)
to hold additional elements from B. The number of elements initialized in A and
B are m and n respectively.
Catch:
The catch here is to manage the operation in O(M+N) time, since "insert" will
take O(M*N) time, and to make the merge IN PLACE.
Thoughts:
其实很简单，要避免insert的话，一般就是从后往前进行插入。 这里既然A已经是一个很大的array了，
而且B和A的occupied的长度都知道（分别是m，n），那么最后merge完以后的array长度也就是已知的m+n。
所以只需要把AB的current last element来进行比较，把其中较大的放到A列的current末尾即可。 要是两个数列
不是同时sort完，再把最后剩下的数组的的0 到 current last 的所有元素依次放入A数组的0：current last即可。
注意：
1. 这里所有带有“current”的变量都是需要maintain的。
2. 其实就是不停的在replace A数组的第N个元素，直到其中一个数组终结。
'''
# compare from the end element of the two arraies
pos_b, pos_a = n, m
## cur starts at the end of the future sorted array
cur = m+n- 1
while pos_b and pos_a:
if A[pos_a-1] >= B[pos_b-1]:
A[cur] = A[pos_a-1]
pos_a -= 1
else:
A[cur] = B[pos_b-1]
pos_b -= 1
cur -= 1
if pos_b:
A[:cur+1] = B[:cur+1]
return A[:m+n]
class Solution_secondround:
# @param {integer[]} nums1
# @param {integer} m
# @param {integer[]} nums2
# @param {integer} n
# @return {void} Do not return anything, modify nums1 in-place instead.
def merge(self, nums1, m, nums2, n):
nums1 += [0 for i in range(n)]
cur1, cur2 = m-1, n-1
curnew = m+n-1
while cur1 >=0 and cur2 >=0:
if nums1[cur1] >= nums2[cur2]:
nums1[curnew] = nums1[cur1]
cur1 -= 1
else:
nums1[curnew] = nums2[cur2]
cur2 -= 1
curnew -= 1
while cur2>=0:
nums1[:curnew+1] = nums2[:cur2+1]
return nums1
A = [1,5,7,9,11, 0,0,0,0,0,0,0,0,0]
B = [2,4,7,9,12,14]
x = merge(A, 5, B, 6)
print x
| false |
58b9f2c8704396d6302bfcb8223e034781a104fb | effyhuihui/leetcode | /uncategoried/rotateArray.py | 2,407 | 4.25 | 4 | __author__ = 'effy'
#-*- coding: utf-8 -*-
'''
Rotate an array of n elements to the right by k steps.
For example, with n = 7 and k = 3,
the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].
Note:
Try to come up as many solutions as you can,
there are at least 3 different ways to solve this problem.
hint： 看起来好多rotate的问题都可以转化成局部reverse的问题。
'''
'''
with extra space
'''
class Solution:
# @param {integer[]} nums
# @param {integer} k
# @return {void} Do not return anything, modify nums in-place instead.
def rotate(self, nums, k):
l = len(nums)
copy = [i for i in nums]
new_index= l-k
old_index = 0
while old_index<l:
if new_index>=l:
new_index = 0
nums[old_index] = copy[new_index]
old_index += 1
new_index += 1
x = Solution()
a = [1,2,3,4,5,6,7]
x.rotate(a,3)
'''
no extra space
以n - k为界，分别对数组的左右两边执行一次逆置；然后对整个数组执行逆置。
例如：n=7,k=3
[1,2,3,4,5,6,7]
先对[1,2,3,4] [5,6,7]分别reverse
[4,3,2,1,7,6,5]
再reverse整个list
[5,6,7,1,2,3,4]
'''
class Solution_opt:
# @param {integer[]} nums
# @param {integer} k
# @return {void} Do not return anything, modify nums in-place instead.
def rotate(self, nums, k):
def reverse(start,end):
mid = (start+end)//2
i,j = start,end
while i<=mid:
tmp = nums[i]
nums[i] = nums[j]
nums[j] = tmp
i+=1
j-=1
n = len(nums)
k = k%n
reverse(n-k,n-1)
reverse(0,n-k-1)
reverse(0,n-1)
x = Solution_opt()
a = [1,2,3,4,5,6]
x.rotate(a,11)
'''
将数组元素依次循环向右平移k个单位
'''
class Solution:
# @param nums, a list of integer
# @param k, num of steps
# @return nothing, please modify the nums list in-place.
def rotate(self, nums, k):
n = len(nums)
idx = 0
distance = 0
cur = nums[0]
for x in range(n):
next = (idx + k) % n
temp = nums[next]
nums[next] = cur
idx = next
cur = temp
distance = (distance + k) % n
if distance == 0:
idx = (idx + 1) % n
cur = nums[idx]
| true |
6e754d747744f574a26e56f861e828e2f49602e6 | GoYMS/localhost_python | /爬虫/案例/v20.py | 969 | 4.1875 | 4 | """
python中正则模块是re
使用大致步骤 ；
1.compile函数将正则表达式的字符串变为一个Pattern对象
2.通过Pattern对象的一些列方法对文本进行匹配，匹配结果是一个Match对象
3.用Match对象的方法，对结果进行模拟
"""
import re
#\d表示数字，后边+表示这个数字可以出现一次或者多次
#r表示后边的是原生字符串，后边不需要转义
s=r"\d+"
#返回Pattern对象
pattern = re.compile(s)
m = pattern.match("one123two13213")
#默认匹配从头部开始，返回一个match的值
print(m)
#后边两个数字表示查找的范围
m = pattern.match("one123two13213",3,8)
print(m)
#group表示直接将match打印出来
print(m.group())
#表示从那个位置开始
print(m.start(0))
#表示从那个位置结束
print(m.end(0))
#表示所取出来的结果在字符串中的范围，(自我理解，包左不包右)
print(m.span(0))
| false |
162f59fb97253adce1ff39b40fa641ddd1cfcf24 | jcbrockschmidt/project_euler | /p010/solution.py | 697 | 4.125 | 4 | #!/usr/bin/env python3
from time import time
def sum_primes_below(n):
""" Calculates the sum of all primes below `n`. """
primes = [2, 3]
num = primes[-1] + 2
while num < n:
is_prime = True
limit = int(num**0.5)
for p in primes:
if p > limit:
break
if num % p == 0:
is_prime = False
break
if is_prime:
primes.append(num)
num += 2
return sum(primes)
if __name__ == '__main__':
start = time()
solu = sum_primes_below(2e6)
elapse = time() - start
print('Solution: {}'.format(solu))
print('Solution found in {:.8f}s'.format(elapse))
| true |
132ed1d71d5f8406474b25de8de3f55dad2def45 | tebannz/pruebadedesarrollopy | /Prueba.py | 2,834 | 4.15625 | 4 | # Deberán crear un programa, el cual deberá recibir un parámetro n ingresado por el usuario, y mostrar los primeros n pares.
def obtenerNumero():
return int(input("Ingrese un número: "));
# Ahora deberán crear el programa, donde no se considere el cero. Si , la salida del programa deberá ser:
def mostrarPares(numero):
for i in range(numero*2, ++2): #aqui se recorre la primera función desde 0
print(i); #aqui estará mostrando de 2 en 2
# Crea programa donde se sumen todos los valores impares desde 0 hasta n, donde n es ingresado por el usuario.
def mostrarParesSinCero(numero):
for i in range(2, (numero*2)+2, ++2):
print(i);
# Crear programa, donde el usuario ingresa dos valores, el límite inferior (min) y superior(max) del intervalo para realizar la suma de los impares.
def mostrarSumaImpares(numero):
resultado = 0;
for i in range(numero):
if(i%2 != 0):
resultado += i;
print(resultado);
# Función que suma los valores impares desde 0 hasta n, donde n es el valor ingresado por usuario.
def mostrarSumaImparesLimites(inferior, superior):
resultado = 0;
for i in range(inferior, superior):
if(i%2 != 0):
resultado += i;
print(resultado);
# Función que calcula el Fibonacci de n elementos.
def fibonacci(elementos):
if elementos < 2:
return elementos;
else:
return fibonacci(elementos-1) + fibonacci(elementos-2);
# Función que muestra la sucesión de fibonacci n veces.
def mostrarFibonacci(ocurrencias):
for i in range(ocurrencias+1):
print(fibonacci(i));
# Función que ejecuta los programas. Basicamente lo que hice en ésta prueba de admisión fué
#crear funciones que retornaran en distintas funciones de los programas requeridos , traté de hacerlo lo más "atómico" posible
def ejecutarProgramas():
print("Solo pares, Parte 1:");
n = obtenerNumero(); #aqui se guarda el numero que ingresó el usuario por parámetro para luego ser llamado a la funcion del ejercicio 1
mostrarPares(n); #aqui mostraran todos los numeros
print("Solo pares, Parte 2:");
mostrarParesSinCero(n); #esta vez empezará desde el 2 y no del 0
print("Suma impares, Parte 1:");
n2 = obtenerNumero();
mostrarSumaImpares(n2); #se pasa el n2 por parametro
print("Suma impares, Parte 2:"); #aqui esta la suma de impares parte 2
inferior = obtenerNumero();
superior = obtenerNumero();
mostrarSumaImparesLimites(inferior, superior);
print("Secuencia de Fibonacci");
elementosFibonacci = obtenerNumero(); #elementos de Fibonacci
mostrarFibonacci(elementosFibonacci);
ejecutarProgramas();
| false |
c289fec7d585d06c696bd94d9c0b8590cfa30110 | ibbocus/oop_polymorphism | /polymorphism.py | 839 | 4.65625 | 5 | """
This is the parent/base/super class
"""
class Planet:
def __init__(self, mass, spin, magnetic_field):
self.mass = mass
self.spin = spin
self.magnetic_field = magnetic_field
# this is a subclass of planets, which includes the parent class attributes as well as attributes specific to the child class
class Goldilocks(Planet):
def __init__(self, mass, spin, magnetic_field, life):
# By using super, we don't need to explicitly initialize the parent class attributes
super().__init__(mass, spin, magnetic_field)
self.life = True
# Creation of our objects, first object refers to the parent class and the second object
# refers to the child class
mars = Planet(200, 50000, 15)
earth = Goldilocks(300, 25000, 500, True)
print(mars.mass)
print(earth.mass)
print(earth.life)
| true |
3d658085a747514cdadebcbf57479e6a57901d81 | SperZ/PythonFirstPractice | /array.py | 372 | 4.25 | 4 | arr1 = [1,2,3,4,5]
arr2 = [4,5,6,7,8]
# combines to arrays together to create one array with all elements of both arrays
arr3 = arr1 + arr2;
print(arr3);
#
arr_repeat = [];
arr_repeat.append("eat");
arr_repeat.append("sleep");
arr_repeat.append("repeat");
arr_repeat.pop();# removes the item from the array/list at the give index
print(arr_repeat);
#print(popped);
| true |
d9e075e78cf95649c0a1b2fc7df2febff46b2e2b | grayey/hackerrank-playful-py | /recursion.py | 520 | 4.21875 | 4 | #!/bin/python3
"""
Task
Write a factorial function that takes a positive integer, N as a parameter and prints the result of N!(N factorial).
"""
import math
import os
import random
import re
import sys
# Complete the factorial function below.
def factorial(n):
new_n = n-1;
return 1 if new_n == 0 else n * factorial(new_n)
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')
n = int(input())
result = factorial(n)
fptr.write(str(result) + '\n')
fptr.close()
| true |
5012f094648baf39d78856f1ce9906c1e45d9f8e | wmontanaro/AdventOfCode2017 | /day3.py | 2,436 | 4.34375 | 4 | def is_down(k):
n = 0
while ((2*n + 1)**2 - n) < k:
n += 1
if (2*n + 1)**2 - n == k:
return n
return False
def is_right(k):
n = 0
while ((2*n - 1)**2 + n) < k:
n += 1
if (2*n - 1)**2 + n == k:
return n
return False
def is_left(k):
n = 0
while ((2*n)**2 + (n+1)) < k:
n += 1
if (2*n)**2 + (n+1) == k:
return n
return False
def is_up(k):
n = 0
while (2*n)**2 - (n-1) < k:
n += 1
if (2*n)**2 - (n-1) == k:
return n
return False
def is_plus(k):
if k == 1:
return 0
if is_down(k):
return is_down(k)
if is_right(k):
return is_right(k)
if is_left(k):
return is_left(k)
if is_up(k):
return is_up(k)
return False
'''figure out if up or down, move opposite to plus'''
'''how to figure out if up or down?
can figure out its octant by starting with the floor of its square root
e.g. 18 lies between upper left diag (n=2) and left plus
e.g. 47 -> 6, n=3; so UL is 37, L is 40, LL is 43, D is 46, D+1 is 47
so to summarize up/down:
floor sqrt // 2 -> n
(368078 -> 303)
this is the ring the number is on
if (2n)**2 + 1 <= num < (2n)**2 + 1 + n, then it is up
if (2n)**2 + 1 + n < num < (2n)**2'''
import math
def check_square(k):
v = math.sqrt(k)
if int(v) == v: #v perfect square
return(int(v)-1)
return False
def get_vert(k):
#if check_square(k):
#return "square"
v = math.sqrt(k)
if math.floor(v) % 2 == 1:
lower_square = math.floor(v)
upper_square = math.floor(v)+2
else:
lower_square = math.floor(v)-1
upper_square = math.floor(v)+1
#print("lower_square: " + str(lower_square))
#print("upper_square: " + str(upper_square))
n = (lower_square + 1) // 2
#print("n: " + str(n))
left = 4*(n**2) + 1 + n
#print("left: " + str(left))
right = 4*(n**2) + 1 - (3*n)
#print("right: " + str(right))
if right < k < left:
return "above 1"
return "below 1"
def step_down(k):
pass
def tfunc(k):
if is_plus(k):
return is_plus(k)
vert = get_vert(k)
steps = 0
if vert == "above 1":
while not is_plus(k):
k = step_down(k)
steps += 1
return steps + is_plus(k)
if vert == "below 1":
while not is_plus(k):
k = step_up(k)
steps += 1
| false |
592648f17627d10b2154f94c3dc20cc8136722d9 | Nick2253/project-euler | /Python/euler007/euler007.py | 1,233 | 4.15625 | 4 | import math
def isPrime(num):
'''Determines if num is prime
input:
numList = list of ints
maxNum = int
output:
sumTotal = int sum of all multiples of numbers in numList less than maxNum
'''
isPrime = True
i = 2 #We don't care if num is divisible by 1
if num <= 1:
return False
while (i < (1 + math.sqrt(num)) and i != num): #We add 1 to the sqrt to preclude rounding issues
if num%i == 0:
isPrime = False
break
i += 1
return isPrime
def nPrime(maxIndex):
'''Determines the maxIndexth prime
input:
maxIndex = int
output:
prime at maxIndex
'''
i = 0
num = 0
while i < maxIndex:
num += 1
if isPrime(num):
i += 1
return num
if __name__ == "__main__":
print "isPrime(1) = " + str(isPrime(1))
print "isPrime(2) = " + str(isPrime(2))
print "isPrime(10) = " + str(isPrime(10))
print "isPrime(23) = " + str(isPrime(23))
print "nPrime(6) = " + str(nPrime(6))
print "nPrime(10001) = " + str(nPrime(10001)) | true |
98796dc9365babd222ddd97216558daac34d7c9d | FlyingJ/kaggle-learn-python | /roulette_probabilites.py | 1,901 | 4.1875 | 4 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Mon Jan 7 17:44:46 2019
@author: jason
"""
def conditional_roulette_probs(history):
"""Given a history of roulette spins (as list of integer values)
return a dictionary where the keys are numbers on the roulette wheel, and
the values are dictionaries mapping numbers on the wheel to probabilities,
such that `d[n1][n2]` is an estimate of the probability that the next spin
will land on n2, given that the previous spin landed on n1.
Example:
conditional_roulette_probs([1, 3, 1, 5, 1])
> {1: {3: 0.5, 5: 0.5},
3: {1: 1.0},
5: {1: 1.0}
}
"""
pass
def next_spins(history):
'''Given a history of roulette spins, return a dictionary where the keys
are spins and the values are lists of spins immediately following.
Example:
print(next_spins([1, 3, 1, 5, 1]))
> {1: [3, 5], 3: [1], 5: [1]}
'''
next_spins = {}
for index, spin in enumerate(history[:-1]):
# initialize list of next spins if not exist
if spin not in list(next_spins):
next_spins[spin] = []
next_spins[spin].append(history[index + 1])
return next_spins
def next_spin_probabilities(next_spins):
'''Given a dictionary where keys are spins and values are lists of spins
immediately following, return a dictionary where keys are spins and values
are dictionaries of next spins and their probabilities.
Example:
print(next_spin_probabilites({1: [3, 5], 3: [1], 5: [1]}))
> {1: {3: 0.5, 5: 0.5}, 3: {1: 1.0}, 5: {1: 1.0}}
'''
next_spin_probabilities = {}
for spin in list(next_spins):
total_events = len(next_spins[spin])
spin_probabilites = {}
for next_spin in next_spins[spin]:
if next_spin not in list(spin_probabilities):
spin_probabilities | true |
c353caa38ab6021b987657320054050a65a7e365 | Nobodylesszb/python_module | /Algorithms/itertools/itertools_accumulate.py | 427 | 4.34375 | 4 | #该accumulate()函数处理输入iterable，
# 将第n和第n + 1项传递给函数并生成返回值而不是任何一个输入。
# 用于组合这两个值的默认函数会将它们相加，
# 因此accumulate()可用于生成一系列数字输入的累积和
from itertools import *
print(list(accumulate(range(5))))
print(list(accumulate('abcde')))
"""
output:
[0, 1, 3, 6, 10]
['a', 'ab', 'abc', 'abcd', 'abcde']
""" | false |
a88ee14815298cea657eb0de403a77b6a3d2569a | Nobodylesszb/python_module | /Algorithms/itertools/itertools_product_repeat.py | 860 | 4.21875 | 4 | from itertools import *
#要计算序列与自身的乘积，请指定输入应重复的次数
def show(iterable):
for i, item in enumerate(iterable, 1):
print(item, end=' ')
if (i % 3) == 0:
print()
print()
print('Repeat 2:\n')
print(list(product(range(3), repeat=2)))
show(list(product(range(3), repeat=2)))
print('Repeat 3:\n')
show(list(product(range(3), repeat=3)))
"""
Repeat 2:
[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]
(0, 0) (0, 1) (0, 2)
(1, 0) (1, 1) (1, 2)
(2, 0) (2, 1) (2, 2)
Repeat 3:
(0, 0, 0) (0, 0, 1) (0, 0, 2)
(0, 1, 0) (0, 1, 1) (0, 1, 2)
(0, 2, 0) (0, 2, 1) (0, 2, 2)
(1, 0, 0) (1, 0, 1) (1, 0, 2)
(1, 1, 0) (1, 1, 1) (1, 1, 2)
(1, 2, 0) (1, 2, 1) (1, 2, 2)
(2, 0, 0) (2, 0, 1) (2, 0, 2)
(2, 1, 0) (2, 1, 1) (2, 1, 2)
(2, 2, 0) (2, 2, 1) (2, 2, 2)
""" | false |
b934a8fa0d994cc3ae00ab2fe7a8e7d867db7dd8 | zayzayzayzay/Introduction_python | /Comrehension_part2.py | 757 | 4.28125 | 4 | #Dictionary comprehension
word = "letters"
letter_count = {letter: word.count(letter) for letter in word}
print(letter_count)
letter_counts = {letter: word.count(letter) for letter in set(word)}
print(letter_counts)
#Set Comprehensions
a_set = {number for number in range(1,6) if number % 3 == 1}
print(a_set)
#generator comprehension (there is no tuple comprehension)
number_thing = (number for number in range(1,7))
print(type(number_thing))
#way of iterate over generator
for number in number_thing:
print(number)
#list way
number_things = (number for number in range(1,7))
number_list= list(number_things)
print(number_list)
#a generator doesn't keep its value in memory
#it can be run only once
number_list= list(number_things)
print(number_list)
| true |
b6d148a52d674bad4b4561b26d26deaebd5d9e9b | zayzayzayzay/Introduction_python | /class2.py | 1,829 | 4.125 | 4 | #name mangling for privacy
class Duck(): #class definition
def __init__(self,input_name): #constructor
self.__name = input_name
@property #getter
def name(self):
print('inside the getter')
return self.__name
@name.setter #setter
def name(self,input_name):
print('inside the setter')
self.__name = input_name
fowl = Duck('howard')
print(fowl.name)
fowl.name = 'donald'
print(fowl.name)
# you can’t access the __name attribute
#print(fowl.__name)
# you can access the __name attribute
print(fowl._Duck__name)
#method types
class A():
count = 0
def __init__(self):
A.count += 1
def exclaim(self):
print("I'am an A!")
@classmethod #class method
def kids(cls):
print("A has ",cls.count, " little objects.")
easy_a = A()
breezy_a = A()
wheezy_a = A()
A.kids()
#staticmethod
class CoyoteWeapon():
@staticmethod
def commercial():
print('This CoyoteWeapon has been brought to you by Acme')
CoyoteWeapon.commercial()
#Duck Typing
class Quote():
def __init__(self, person, words):
self.person = person
self.words = words
def who(self):
return self.person
def says(self):
return self.words + '.'
class QuestionQuote(Quote):
def says(self):
return self.words+ '?'
class ExclamationQuote(Quote):
def says(self):
return self.words + '!'
hunter = Quote('Elmer Fudd', "I'm hunting wabbits")
print(hunter.who(), 'says:', hunter.says())
hunted1 = QuestionQuote('Bugs Bunny', "What's up, doc")
print(hunted1.who(), 'says:', hunted1.says())
hunted2 = ExclamationQuote('Daffy Duck', "It's rabbit season")
print(hunted2.who(), 'says:', hunted2.says())
#a class having no relation with the previous one
class BabblingBrook():
def who(self):
return 'Brook'
def says(self):
return 'Babble'
brook = BabblingBrook()
def who_says(obj):
print(obj.who(), 'says', obj.says())
who_says(hunter)
who_says(brook)
| false |
c6e9e1f1724312e6ec7213807ae5533a78a3da8a | cs-fullstack-2019-fall/python-classobject-b-cw-marcus110379 | /cw.py | 2,534 | 4.125 | 4 | def main():
problem1()
problem2()
problem3()
# Create a class Dog. Make sure it has the attributes name, breed, color, gender. Create a function that will print all attributes of the class. Create an object of Dog in your problem1 function and print all of it's attributes.
class Dog:
def __init__(self, name, breed, color, gender):
self.name = name
self.breed = breed
self.color = color
self.gender = gender
def printAll(self):
print(f"{self.name}, {self.breed},{self.color},{self.gender}") # !! : use string formatting to take advantage of f strings
def problem1():
myDog = Dog("Rocky", "poodle", "black","male")
myDog.printAll()
# Problem 2:
# We will keep having this problem until EVERYONE gets it right without help
# Create a function that has a loop that quits with the equal sign. If the user doesn't enter the equal sign, ask them to input another string.
def user():
userInput = input("enter a string")
while userInput != "=":
userInput = input("enter another string")
def problem2():
user()
# Problem 3:
# Create a class Product that represents a product sold online. A product has a name, price, and quantity.
# The class should have several functions:
# a) One function that can change the name of the product.
# b) Another function that can change the name and price of the product.
# c) A last function that can change the name, price, and quantity of the product.
# Create an object of Product and print all of it's attributes.
class Product:
def __init__(self, name, price, quantity):
self.name = name
self.price = price
self.quantity = quantity
def funcA(self, name): # !! : this is a bad function name
self.name = name
def funcB(self, name, quantity): # !! : this is a bad function name AND change the name and price of the product
self.name = name
self.quantity= quantity
def funcC(self, name, price, quantity): # !! : this is a bad function name
self.name = name
self.price = price
self.quantity = quantity
def problem3():
product1 = Product("genie", "$25", "1")
print(product1.name, product1.price, product1.quantity)
product1.funcA("test1a")
print(product1.name, product1.price, product1.quantity)
product1.funcB("test2a", "test2b")
print(product1.name, product1.price, product1.quantity)
product1.funcC("test3a", "test3b", "test3c")
print(product1.name, product1.price, product1.quantity)
main() | true |
9a60a94ac5ddd89bf5aab0c03d5a5bada2d0dc8f | Rastwoz/python101 | /3. Data Structures/Challenges and Solutions/Dictionaries/homework2.py | 921 | 4.53125 | 5 |
#create dictionary with Hello in each language
translator = {"French":"Bonjour",
"Spanish":"Hola",
"Italian":"Ciao",
"German": "Guten Tag",
"Indian": "Namaste"
}
#if user enters a language in dictionary, translate the word. Otherwise, let them add it
#if user enters quit, exit program
while True:
selection = input("Select your language? ").title()
if selection == "Quit":
break
elif selection in translator:
print(translator.get(selection))
elif selection not in translator:
print("Thats a new language. Lets add it")
new_language = input(f"What does 'Hello' mean in {selection}: ")
translator[selection] = new_language
#alternative way of adding a language
#translator.update({selection:new_language})
| true |
5bb5f7ed6d3aa6f4cbf0d311543506363dc6f589 | yoontrue/Python_work | /python_day03/day03ex102_class.py | 784 | 4.125 | 4 | # 앞에서 사용한 딕셔너리 구조를 class로 변경
# 클래스 선언은 class 키워드를 이용한다.
'''
class 클래스명(상속) :
생성자메소드
멤버메소드
멤버필드
'''
class People:
# 생성자 메소드 선언, 생성자와 멤버메소드는 self 매개변수가 선언 되어야한다.
def __init__(self, name):
self.name = name
def setName(self, name):
self.name = name
def getName(self):
return self.name
def greeting(self):
return self.name + '님 안녕!'
# 객체 생성 시 자바처럼 new 사용하지 않는다
kim = People('kim')
pList = [kim, People('lee'), People('park')]
pList[0].setName('hong')
for person in pList:
print('인사 >>> ', person.greeting())
| false |
0df5dc3aeeb90af857f68060156591461bbdb2eb | hugoreyes83/scripts | /fibonacci.py | 268 | 4.28125 | 4 | def fibonacci(n):
'''function returns fibonacci sequence up to provided number'''
fibonacci_list = [0,1,1]
for i in range(3,n+1):
fibonacci_list.insert(i,fibonacci_list[i-1]+fibonacci_list[i-2])
return fibonacci_list[n-1]
print(fibonacci(10))
| false |
8905adda00560a3cf143b8faba96d4f17b4429af | priestd09/project_euler | /e_16.py | 497 | 4.125 | 4 | #!/usr/bin/env python
"""
##---
# jlengrand
#Created on : Fri Jan 13 15:24:59 CET 2012
#
# DESCRIPTION : Solves problem 16 of Project Euler
2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 2^1000?
##---
"""
def sum_power_2(value):
"""
Returns the sum of the digits of 2^value
"""
return sum([int(el) for el in list(str(pow(2, value)))])
if __name__ == '__main__':
print "Answer : %d" % (sum_power_2(1000))
| true |
d0a181c79d71fa82a8fdc193768ebc54ace30937 | priestd09/project_euler | /e_4.py | 979 | 4.25 | 4 | #!/usr/bin/env python
"""
#---
Julien Lengrand-Lambert
Created on : Wed Jan 11 14:42:54 CET 2012
DESCRIPTION : Solves problem 4 of Project Euler
A palindromic number reads the same both ways. The largest palindrome made
from the product of two 2-digit numbers is 9009 = 91 99.
Find the largest palindrome made from the product of two 3-digit numbers.
#---
"""
def largest_palindrom():
"""
Returns the largest palindrom made from the product of two 3-digits number
"""
three_digits = range(100, 1000)
largest = 0
for digit_1 in three_digits:
for digit_2 in three_digits:
val = digit_1 * digit_2
if is_palindrom(val):
if largest < val:
largest = val
return largest
def is_palindrom(number):
"""
returns True if a number is a palindrom, False otherwise
"""
return str(number)[::-1] == str(number)
if __name__ == '__main__':
print "Answer : %d " % (largest_palindrom())
| true |
38f49214e119612b127236bd230f2b11bba088ee | priestd09/project_euler | /e_6.py | 924 | 4.125 | 4 | #!/usr/bin/env python
"""
#---
Julien Lengrand-Lambert
Created on : Wed Jan 11 14:42:54 CET 2012
DESCRIPTION : Solves problem 6 of Project Euler
Find the difference between the sum of the squares of the first one hundred
natural numbers and the square of the sum.
#---
"""
def diff_sum_squares(value):
"""
Returns the difference between the sum of the squares of the first one
hundred natural numbers and the square of the sum.
"""
return squares_sum(value) - sum_squares(value)
def sum_squares(value):
"""
Returns the sum of the square of elements from 1 to value
"""
vals = range(1, value + 1)
return sum([pow(val, 2) for val in vals])
def squares_sum(value):
"""
Returns the square of the sum of elements from 1 to value
"""
return pow(sum(range(1, value + 1)), 2)
if __name__ == '__main__':
val = 100
print "Answer : %d " % (diff_sum_squares(val))
| true |
75bbb63e6278cca0465f5ff4b1d7aebc10ce77ee | ChenlingJ/cj | /hackerrank/lists.py | 1,663 | 4.125 | 4 | # Consider a list (list = []). You can perform the following commands:
#
# insert i e: Insert integer e at position i.
# print: Print the list.
# remove e: Delete the first occurrence of integer e.
# append e: Insert integer e at the end of the list.
# sort: Sort the list.
# pop: Pop the last element from the list.
# reverse: Reverse the list.
#
# Initialize your list and read in the value of n followed by n lines of commands where each command will be of the 7
# types listed above. Iterate through each command in order and perform the corresponding operation on your list.
def read_lines(n: int) -> list[str]:
"""
Read and return `n` lines of input.
"""
return [input() for _ in range(n)]
def parse_commands(lines: list[str]) -> list[tuple[str, list[int]]]:
"""
Process a list of `lines` and turn into command tuples `(command, args)`.
"""
result = []
for line in lines:
command, *args = line.split()
args = list(map(int, args))
result.append((command, args))
return result
def run_commands(commands: list[tuple[str, list[int]]]) -> list[str]:
arr = []
output = []
methods = {
"insert": arr.insert,
"remove": arr.remove,
"append": arr.append,
"sort": arr.sort,
"pop": arr.pop,
"reverse": arr.reverse,
}
for command, args in commands:
if command == "print": # print is not a method of arr.
output.append(str(arr))
else:
methods[command](*args)
return output
if __name__ == "__main__":
N = int(input())
print("\n".join(run_commands(parse_commands(read_lines(N)))))
| true |
6c82dd1a17cb90a870d009e5ea8ac40772dafcf0 | GeekGoutham/Python-Interactive-Challenges | /Unique_String.py | 1,122 | 4.15625 | 4 | #Author = Vignesh Goutham
def check_unique_dict(string_input): #Implmented using dict -- Poor performance as you need to check if there is more than one entry
char_dict = {} #Still O(n) but dict takes more space then list
for character in string_input:
if character in char_dict.keys():
flag = False
return False
else:
char_dict[character] = 1
# print(char_dict)
return True
def check_unique_list(string_input): #Better than dict implementation : Still O(n) complexity
string_check_list=[]
for character in string_input:
if character in string_check_list:
return False
else:
string_check_list.append(character)
return True
def check_unique_set(string_input): #O(n) as well -- better algorithm using set() -- set does not allow duplicates.
return len(set(string_input)) == len(string_input)
if __name__ == '__main__':
string_input = input("Enter the string needed to be checked : ")
result = check_unique_list(string_input)
print(str(result)) | true |
25bf8924786bb17333d9425a722c74755e8c5768 | GeekGoutham/Python-Interactive-Challenges | /String_reverse_inplace.py | 1,306 | 4.28125 | 4 | #Author = Vignesh Goutham
def string_reverse(charlist):
if charlist == None: #strings are immutable in python, we cant do a strict in-place string reverse
print("No character array sent") #In-place algo = where the input is changed to output and a new memory/variable for o/p is not created
return False #in-place is distructive as it destroys the input
current = 0
far_current = len(charlist)-1
while(current != far_current):
charlist[current],charlist[far_current] = charlist[far_current], charlist[current]
current += 1
far_current -= 1
print(charlist)
def string_reverse_v2(string1): #Pythonic way to do with strings (Not in-place)
if string1 == None:
print("No string to reverse")
return False
else:
print(string1[::-1])
def string_reverse_v3(string1): #Using reversed method -- (Not in-place)
if string1 == None:
print("No string to return")
else:
rev = reversed(string1)
print("".join(rev))
if __name__ == "__main__":
charlist = ['v','i','g','n','e','s','h']
print(charlist)
string_reverse(charlist)
print("=====================")
string_reverse_v2("Vignesh")
string_reverse_v3("vignesh") | true |
943e612b6887ec0ad85c23fd2a2c7f4383b8c29a | napte/python-learn | /08_list_for_each.py | 578 | 4.1875 | 4 | def main():
a = [1, 2, 3]
print 'List a = ' + str(a)
print 'Doing for-each loop and printing numbers and their squares\n'
for num in a:
print '\n------\n'
print 'num = ' + str(num)
print str(num) + '^2 = ' + str(num**2)
print '\n------\n'
a = [1, 2, 3, 4, 5, 6, 7, 8]
even = []
print 'List a => ' + str(a)
print 'List even (empty right now) => ' + str(even)
for num in a:
if num%2 == 0:
even.append(num)
print 'even => ' + str(even)
# Standard boilerplate code that calls the main function
if __name__ == '__main__':
main()
| true |
030fee26786b2675ffa42a974a308af70cacc2c8 | minji-o-j/Python | /대학원 파이썬 수업/수업/1027/city.py | 272 | 4.15625 | 4 | city = ['DC','NY','Seoul', 'Tokyo','Paris']
print("what is the capital of S.Korea?")
for i in range(len(city)):
print('{}. {}'.format(i+1,city[i]))
answer = int(input('choose an answer: '))
if answer == 3:
print('correct')
else:
print('incorrect')
| false |
ec6cd149c7835a225125b541c44d6e33d1e69765 | richardcinca/CursuriPY | /ex4.py | 573 | 4.28125 | 4 | iterator=0
reversed_word=""
word = input("Insert word: ")
print("Your word is '{}'".format(word))
length=len(word)
print("The length of the word is {} letters".format(length))
for var in word:
#print("Var is {}".format(var))
#print(word[iterator])
#iterator+=1
#reversed_word=var+reversed_word
#print(reversed_word)
#print("......")
print("{} must be equal to {}".format(word[iterator],var))
iterator = iterator+1
reversed_word=var+reversed_word
print("Reversed word is:{}".format(reversed_word))
input("\n\n P <enter> to exit.") | true |
63f722f1667e843eb0a0d42e2030b215fe65913f | jayadams011/KalPython | /bmi claculator.py | 464 | 4.21875 | 4 | weight = eval(input("Enter weight in pounds: "))
height = eval(input("enter height in inches: "))
KILOGRAMS_PER_POUND = 0.45359237
METERS_PER_INCH = 0.0254
weightInKg = weight * KILOGRAMS_PER_POUND
heightInMt = height * METERS_PER_INCH
bmi = weightInKg / (heightInMt ** 2)
print ("BMI is" , format(bmi, ".2f"))
if bmi < 18.5:
print ("Underweight")
elif bmi < 25:
print ("Normal")
elif bmi < 30:
print ("Overweight")
else:
print ("obese")
| false |
72d3ee8cb18777b39a4d19f8c9ff1338b121681f | jayadams011/KalPython | /2_4_poundsToKilograms.py | 628 | 4.1875 | 4 | """
(Convert pounds into kilograms) Write a program that converts pounds into kilograms. The program
prompts the user to enter a value in pounds, converts it to kilograms, and displays the result. One
pound is 0.454 kilograms. Here is a sample run:
Enter a value in pounds: 55.5
55.5 pounds is 25.197 kilograms
"""
#user inputs number of pounds to convert to Kgs
#create a var that requires input from the user
pounds = eval(input( "Enter the number of pounds you would like to convert to kilograms. -->"))
# convert pounds to kilograms
kilograms = pounds * 0.454
#print output
print( pounds, " is ",kilograms, " kilograms.")
| true |
e4bf2decd4a1b5c8c246d5a50908abbd9e7b9f89 | stevenb92/PythonShizzle | /numchar.py | 310 | 4.34375 | 4 | #Prompts for an input string and then returns the input string
#along with the number of charcters in the string
inputString = ""
length = 0
while length == 0:
print ("Enter an input string:")
inputString = str(input())
length = len(inputString)
print ("{} has {} characters".format(inputString,length)) | true |
3864a6a4ed3038f46d45dd19ab2b41a930b2a8fd | JoshOrndorff/LearnPythonByExample | /Unit6-2DLists/lesson2-WheresWaldo.py | 1,737 | 4.21875 | 4 | # Let's make a 2D list that contains a bunch of people. Since all these strings
# are different lengths it would be easy for this table to look messy. Using
# white space effectively can help the data look more organized.
people = [["Jack", "Abagail", "Waleed" ],
["Rebeca", "Obi", "Orndorff"],
["Simon", "Waldo", "Jamie" ]]
# Our goal is to figure out where in the grid the name "Waldo" appears, and
# print its coordinates. Algorithm-wise that means looking through each row,
# and within each row looking through each name until we see Waldo.
rowCounter = 0
for row in people:
colCounter = 0
for item in row:
if item == "Waldo":
print("Found him at location {}, {}".format(rowCounter, colCounter))
colCounter += 1
rowCounter += 1
# ---------------- Exercises -----------------
# 1. Change the search so that it will find "waldo", "WALDO", and "walDO" as
# Well as just "waldo".
# 2. Change the for loops to while loops. What looping condition should we use
# to make sure we don't miss any rows or items in those rows?
# 3. Modify lines 6 and 7 to add a fourth row of people to the list. Does your
# code still find waldo correctly? What if Waldo is in the bottom row?
# 5. What happens if waldo appears more tha once in this grid?
# 4. ADVANCED: Convert this program to use a function called wheres_waldo that
# takes a grid as an argument, and returns the coordinates. What happens
# if Waldo appears more than once in this grid? What happens if Waldo isn't
# in the grid at all?
#
# Hint: your return line line should look something like:
#
# return rowCounter, colCounter
#
# That line returns a tuple which we haven't formally seen... yet.
| true |
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