blob_id
string | repo_name
string | path
string | length_bytes
int64 | score
float64 | int_score
int64 | text
string | is_english
bool |
|---|---|---|---|---|---|---|---|
3a138f569de9b1e4d5e2227e1ee290489eb5883c
|
emir-naiz/first_git_lesson
|
/Courses/1 month/4 week/day 1/Задача №5 setify number.py
| 353
| 4.125
| 4
|
def setify(number_list):
number_list.sort()
empty_list = []
for number in empty_list:
if number not in empty_list:
empty_list.append(number)
return empty_list
print(setify(1,2,3,3,4,4,6,6,5,5))
print(set(([1,2,3,3,4,4,6,6,5,5]))) # при помощи set можно удалить дублирующие цифры
| false
|
2355ee5a02dcbc66d33649b618d1c57e356e9ada
|
jamj2000/Programming-Classes
|
/Math You Will Actually Use/Week6_Sympy.py
| 725
| 4.3125
| 4
|
' "pip install sympy" before you do any of this '
import sympy # import the calculator with variable buttons :)
from sympy import Symbol
# -------------------------- first, solve the simple equation 2*x=2
x = Symbol('x') # create a "symoblic variable" x
equation = 2*x - 2 # define the equation 2*x = 2
solution = sympy.solve(equation, x) # solve for x
print(solution) # print the list of all solutions
# -------------------------- next, lets do the equation of a circle
y = Symbol('y')
equation = x**2 + y**2 - 1**2 # x**2 + y**2 = r**2 where r =1
solution = sympy.solve(equation, y) # solve for y
print(solution)
# notice how one is the TOP half of the circle
# and the other in the list is the BOTTOM half
| true
|
0b6dae16fec055ad0eccb3d651b02f5643540e81
|
jamj2000/Programming-Classes
|
/Intro To Python/semester1/Week10_Challenge0.py
| 1,798
| 4.34375
| 4
|
'''
This is a deep dive into FUNCTIONS
What is a function?
A function is a bunch of commands bundled into a group
How do you write a function?
Look below!
CHALLENGES: (in order of HEAT = difficulty)
Siberian Winter: Change the link of the Philly function
Canada in Spring: Make the link an input to Philly, not a constant
Spring in California: Convert Antonia's inputs to named inputs
Summer in California: Write a new function which adds two numbers
Rocket Engine: Convert these functions into a CLASS
'''
import webbrowser
def Franko():
print('Hello from Franko, the simple function')
def Philly():
print('Hello from Philly, the funky phunction')
print('You see, you can do a lot of stuff in one function')
print('Like taking the square root of 1,245 to the power of pi!')
import math
sqrt1245_pi = math.sqrt(1245)**math.pi
print('Which is exactly... ', sqrt1245_pi)
print('You can even do fun stuff in here... Check this out...')
webbrowser.open('http://www.jonnyhyman.com')
def Antonia(argument1, argument2, TheyDontAllNeedToSayArgumentThough):
print('Hello from Antonia, the fun function with fun inputs')
print('These are my arguments:')
print(argument1)
print(argument2)
print(TheyDontAllNeedToSayArgumentThough)
webbrowser.open('https://media.giphy.com/media/R8n7YlPHe34dy/giphy.gif')
def Kevin(named_argument = 'Wut', other_one = 'Hah', herp = 'derp' ):
print('Hello from Kevin the function with NAMED inputs...')
print('My inputs were...', named_argument, other_one, herp)
#input(" ----> Notice how none of the functions run until you call them"
# " //// [ENTER] to continue")
print('_______________') # some space to be able to read easier
Antonia()
| true
|
d3f6517aeea44395a9796ee175fc8ef0dcaacaa1
|
csmdoficial/aulas-python
|
/curso/venv/bin/exercicios/desafio 033.py
| 537
| 4.15625
| 4
|
"""faça um programa qu leia três números e mostre qual é o menor e o maior """
a = float(input("Digite um numero:"))
b = float(input("Digite outro numero:"))
c = float(input('Digite o ultimo numero:'))
#verificando quem é o menor
menor = a
if b < a and b < c:
menor = b
if c < a and c < b:
menor = c
#verificando quem é o maior
maior = a
if b > a and b > c:
maior = b
if c > a and c > b:
maior = c
print('O menor valor digitado foi {:.0f}'.format(menor))
print('O maior valor digitado foi {:.0f}'.format(maior))
| false
|
e543c991e7ad17b2a657274bb1a5483636fc0d4f
|
csmdoficial/aulas-python
|
/curso/venv/bin/exercicios/desafio 025.py
| 265
| 4.15625
| 4
|
"""crie um programa que leia o nome de uma pessoa e diga se ela tem 'silva' no nome"""
nome = str(input('Digite um nome:')).upper().strip()
print('SILVA'in nome)
n = str(input('Digite um nume:')).upper().strip()
print('Seu nome tem Silva? {}'.format('SILVA' in n))
| false
|
6fb1b7fff352717f4654ee2f38548c0fff13cfb7
|
iulyaav/coursework
|
/stanford-algo-courses/matrix_multiplication.py
| 1,177
| 4.25
| 4
|
def matrix_add(X, Y):
pass
def matrix_multiply(X, Y):
"""
Let X = | A B | and Y = | E F |
| C D | | G H |
We have the products:
P1 = A(F-H)
P2 = (A+B)H
P3 = (C+D)E
P4 = D(G-E)
P5 = (A+D)(E+H)
P6 = (B-D)(G+H)
P7 = (A-C)(E+F)
Claim: X * Y = | (P5 + P4) - P2 + P6 P1 + P2 |
| P3 + P4 (P1 + P5) - (P3 + P7) |
Nb - This only supports matrice of nxn where n is even
"""
if len(X) == len(Y) == 2:
return [
[X[0][0]*Y[0][0] + X[0][1]*Y[1][0], X[0][0]*Y[0][1] + X[0][1]*Y[1][1]],
[X[1][0]*Y[0][0] + X[1][1]*Y[1][0], X[1][0]*Y[0][1] + X[1][1]*Y[1][1]]
]
def print_matrix(n):
tmp = []
for i in range(n):
# print("{} row:".format(i+1))
tmp.append([int(x) for x in input().split(' ')])
return tmp
if __name__ == "__main__":
print("Let's multiply matrices of n=")
n = int(input())
print("Let's write the first matrix")
first = print_matrix(n)
print("Let's write the second matrix")
second = print_matrix(n)
print(matrix_multiply(first, second))
| false
|
ad291e7fa25471945efc95465a4589cd89a44e7a
|
mihir-liverpool/RockPaperScissor
|
/rockpaperscissor.py
| 841
| 4.125
| 4
|
import random
while True:
moves = ['rock', 'paper', 'scissor']
player_wins = ['paperrock', 'scissorpaper', 'rockscissor']
player_move = input('please choose between rock paper or scissor: ')
computer_move = random.choice(moves)
if player_move not in moves:
print('please choose an option between rock paper and scissor')
elif computer_move == player_move:
print ('your move was ', player_move)
print ('computers move was', computer_move)
print('That is a draw, please try again')
elif player_move+computer_move in player_wins:
print('your move was ', player_move)
print('computers move was', computer_move)
print('you win')
else:
print('your move was ', player_move)
print('computers move was', computer_move)
print('you lose')
| true
|
1dbd6a073c1fbdae4bdc435d7a678e49e8709ab3
|
sombra721/Python-Exercises
|
/Regular Expression_ex.py
| 1,503
| 4.125
| 4
|
'''
The script will detect if the regular expression pattern is contained in the files in the directory and it's sub-directory.
Suppose the directory structure is shown as below:
test
|---sub1
| |---1.txt (contains)
| |---2.txt (does not contain)
| |---3.txt (contains)
|---sub2
| |---1.txt (contains)
| |---2.txt (contains)
| |---3.txt (contains)
|---sub3
|---|---sub3-1
The result will be: {"sub1": 2, "sub2": 3,"sub3": 0,"sub3\sub3-1": 0}
Running the script with the two arguments:
1. Tha path to be traversal.
2. Ragular Expression pattern.
e.g.:
python re_traversal.py C:/test ("^[a-zA-Z]+_TESTResult.*")
'''
import sys
import os
import re
import pylab
def draw_graph(result):
names = result.keys()
counts = result.values()
pylab.xticks(range(len(names)), names)
pylab.plot(range(len(names)), counts, "b")
pylab.show()
def main(argv):
result = {}
for path, subdirs, files in os.walk(argv[1]):
for subdir in subdirs:
result[os.path.join(path, subdir)[len(argv[1])+1:]] = 0
for name in files:
if re.search(argv[2], open(os.path.join(path, name), 'r').read()):
result[os.path.join(path, name)[len(argv[1])+1:-len(name)]] = result.get(os.path.join(path, name)[len(argv[1])+1:-len(name)], 0) + 1
if result:
print result
else:
print "No match detected."
draw_graph(result)
if __name__ == '__main__':
main(sys.argv)
| true
|
8ae7109db20182e8e0367a39c03fa9e4dd04dba5
|
ryanthemanr0x/Python
|
/Giraffe/Lists.py
| 561
| 4.25
| 4
|
# Working with lists
# Lists can include integers and boolean
# friends = ["Kevin", 2, False]
friends = ["Kevin", "Karen", "Jim"]
print(friends)
# Using index
friends = ["Kevin", "Karen", "Jim"]
# 0 1/-2 2/-1
print(friends[0])
print(friends[2])
print(friends[-1])
print(friends[-2])
print(friends[1:])
friends1 = ["Kevin", "Karen", "Jim", "Oscar", "Toby"]
# 0 1 2 3 4
# Will grab all the elements upto, but not including last element
print(friends1[1:4])
friends[1] = "Mike"
print(friends[1])
| true
|
6b82f94e0a3149bcb55199f4d3d3d860732aa475
|
ryanthemanr0x/Python
|
/Giraffe/if_statements.py
| 631
| 4.40625
| 4
|
# If Statements
is_male = False
is_tall = False
if is_male or is_tall:
print("You are a male or tall or both")
else:
print("You are neither male nor tall")
is_male1 = True
is_tall1 = True
if is_male1 and is_tall1:
print("You are a tall male")
else:
print("You are either not male or tall or both")
is_male2 = True
is_tall2 = False
if is_male2 and is_tall2:
print("You are a tall male")
elif is_male2 and not(is_tall2):
print("You are a short male")
elif not(is_male2) and is_tall2:
print("You are not a male and are tall")
else:
print("You are either not male or tall or both")
| true
|
a97dadaf6560e38dafc3078ce3bbc4a849c66431
|
kiyotd/study-py
|
/src/6_asterisk/ex1.py
| 606
| 4.1875
| 4
|
list_ = [0, 1, 2, 3, 4]
tuple_ = (0, 1, 2, 3, 4)
print(list_) # [0, 1, 2, 3, 4]
print(tuple_) # (0, 1, 2, 3, 4)
# *
# unpack 展開する
print(*list_) # 0 1 2 3 4
print(*tuple_) # 0 1 2 3 4
# **
# 辞書などをキーワード引数として渡すことができる
my_dict = {"sep": " / "}
print(1, 2, 3, **my_dict) # 1 / 2 / 3
print(1, 2, 3, sep=" / ") # と同じ
# 関数を定義するとき
# 与えられた実引数はタプルにまとめられる
def sum_func(*nums):
total_ = 0
for num in nums:
total_ += num
return total_
print(sum_func(1, 2, 3, 4, 5)) # 15
| false
|
2d259320a70264f55f837a84741a5a1b2abe6fd5
|
xhweek/learnpy
|
/.idea/ly-01.py
| 525
| 4.21875
| 4
|
#循环打印
#print('* ' * 5)
"""
for i in range(0,4):
for j in range(0,5):
print("* ",end=" ")
print()
"""
for i in range(4):
for j in range(5):
#print(i,"-===========",j)
if i == 0 or i == 3 or j == 0 or j == 4:
print("* ",end = " ")
else:
print(" ",end = " ")
print("")
for i in range(5):
for j in range(i+1):
if i == 4 or j == 0 or j == i:
print("* ",end=" ")
else:
print(" ",end = " ")
print()
| false
|
9e0a944f2100cab01f4e4cad3921c8786998681b
|
ikramsalim/DataCamp
|
/02-intermediate-python/4-loops/loop-over-lists-of-lists.py
| 446
| 4.25
| 4
|
"""Write a for loop that goes through each sublist of house and prints out the x is y sqm, where x is the name of the room
and y is the area of the room."""
# house list of lists
house = [["hallway", 11.25],
["kitchen", 18.0],
["living room", 20.0],
["bedroom", 10.75],
["bathroom", 9.50]]
# Build a for loop from scratch
for room, area in (house):
print("the " + str(room) + " is " + str(area) + " sqm")
| true
|
dcf1b439aa3db2947653f9197e961d4969f92095
|
ycayir/python-for-everybody
|
/course2/week5/ex_09_05.py
| 782
| 4.25
| 4
|
# Exercise 5: This program records the domain name (instead of the
# address) where the message was sent from instead of who the mail came
# from (i.e., the whole email address). At the end of the program, print
# out the contents of your dictionary.
# python schoolcount.py
#
# Enter a file name: mbox-short.txt
# {'media.berkeley.edu': 4, 'uct.ac.za': 6, 'umich.edu': 7,
# 'gmail.com': 1, 'caret.cam.ac.uk': 1, 'iupui.edu': 8}
handle = open('../../files/mbox-short.txt')
counts = dict()
for line in handle:
line = line.strip()
if not line.startswith('From'): continue
words = line.split()
if len(words) < 2: continue
email = words[1]
emailParts = email.split('@')
domain = emailParts[1]
counts[domain] = counts.get(domain, 0) + 1
print(counts)
| true
|
16d0cc5afda917ebc335553cc6d338014b19f87b
|
aflyk/hello
|
/pystart/lesson1/l1t2.py
| 467
| 4.4375
| 4
|
"""
Пользователь вводит время в секундах. Переведите время в часы,
минуты и секунды и выведите в формате чч:мм:сс. Используйте форматирование строк.
"""
user_time = int(input("Введите кол-во секунд: "))
user_time = user_time % (60*60*24)
print(f"{user_time // 3600}:{(user_time % 3600) // 60}:{(user_time % 3600) % 60 } ")
| false
|
254210227b8c1c1ce4ddca88fb6a889a1c0e7c47
|
aflyk/hello
|
/pystart/lesson6/l6t2.py
| 1,224
| 4.3125
| 4
|
"""
Реализовать класс Road (дорога), в котором определить атрибуты: length (длина),
width (ширина). Значения данных атрибутов должны передаваться при создании
экземпляра класса. Атрибуты сделать защищенными. Определить метод расчета массы
асфальта, необходимого для покрытия всего дорожного полотна. Использовать формулу:
длина*ширина*масса асфальта для покрытия одного кв метра дороги асфальтом,
толщиной в 1 см*число см толщины полотна. Проверить работу метода.
Например: 20м*5000м*25кг*5см = 12500 т
"""
class Road:
def __init__(self, lenght, width):
self._lenght = lenght
self._width = width
def asf(self, weight=25, depth=5):
return f"{(self._lenght * self._width * weight * depth)/1000} т"
if __name__ == "__main__":
small_road = Road(20, 5000)
print(small_road.asf())
| false
|
c3c1bf04376f728f6470a055e56c878ceec2d3b3
|
mjdall/leet
|
/spiral_matrix.py
| 2,106
| 4.1875
| 4
|
def spiral_order(self, matrix):
"""
Returns the spiral order of the input matrix.
I.e. [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]
becomes [1, 2, 3, 6, 9, 8, 7, 4, 5].
:type matrix: List[List[int]]
:rtype: List[int]
"""
if not matrix or matrix is None:
return []
output = []
# holds the limits for when we change axis direction
curr_x_lim = len(matrix[0])
curr_y_lim = len(matrix)
# holds the total number of entries in the array
# as well as the number of total hops we need to do
total_hops = curr_x_lim * curr_y_lim
# the current directions we are travelling in the array
# if x_direction is not 0 then y_direction will be 0, vice versa
x_direction = 1
y_direction = 0
# the total number of steps we need to take in the current direction
steps_in_direction = curr_x_lim
# keep going until we've traversed each node
x, y = 0, 0
while total_hops != 0:
# append the current spot in the matrix
output.append(matrix[y][x])
# increment variables keeping track of position
steps_in_direction -= 1
total_hops -= 1
# we're done travelling in this direction
if steps_in_direction == 0:
# if we were travelling in x direction
if x_direction:
# y direction follows what x direction just was
y_direction = x_direction
# set x dir to zero
x_direction = 0
# the next time we traverse in x direction, we dont go as far
curr_x_lim -= 1
# set the number of steps we're expecting to take next
steps_in_direction = curr_y_lim - 1
else:
# x direction changes direction to y
x_direction = -y_direction
y_direction = 0
curr_y_lim -= 1
steps_in_direction = curr_x_lim
# get the next spot in the array
x += x_direction
y += y_direction
return output
| true
|
3dd6609f9805f7c8bbf9949b56584a2c044475a3
|
mjdall/leet
|
/count_elements.py
| 668
| 4.28125
| 4
|
def count_elements(arr):
"""
Given an integer array arr, count element x such that x + 1 is also in arr.
If there're duplicates in arr, count them seperately.
:type arr: List[int]
:rtype: int
"""
if arr is None or len(arr) == 0:
return 0
found_numbers = {}
for num in arr:
found_numbers.setdefault(num, 0)
found_numbers[num] += 1
nums_in_arr = found_numbers.keys()
total_count = 0
for num in nums_in_arr:
if num + 1 in found_numbers:
total_count += found_numbers[num]
return total_count
| true
|
334da6ffae18a403cc33ae03c1600238a6a45ddd
|
denemorhun/Python-Problems
|
/Hackerrank/Strings/MatchParanthesis-easy.py
| 1,891
| 4.40625
| 4
|
# Python3 code to Check for balanced parentheses in an expression
# EASY
'''
Given a string containing just the characters '(', ')', '{', '}', '[' and ']',
determine if the input string is valid.
An input string is valid if:
Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.
Example 1:
Input: "()"
Output: true
Example 2:
Input: "()[]{}"
Output: true
Example 3:
Input: "(]"
Output: false
Example 4:
Input: "([)]"
Output: false
Example 5:
Input: "{[]}"
Output: true
Solution:
# position 0 -> []
# position 1 -> []
# position 2 -> ()
Use a stack to push all left paranthesis to open list.
If the item equals the value in both open and closed lists, pop it.
If we finish with an empty stack, paranthesis are balanced.
'''
def isValid(input) -> bool:
# convert string into a list
input = list(input)
# code: 0, 1, 2
open_list = ["[","{","("]
close_list = ["]","}",")"]
# No paranthesis means matching case
if list is None:
return True
stack = []
# scroll through each char and add open paranthesis to the stack.
for i in input:
if i in open_list:
stack.append(i)
elif i in close_list:
# get the paranthesis code: 0, 1, 2
code = close_list.index(i)
if len(stack) > 0 and open_list[code] == stack[len(stack)-1]:
stack.pop()
# If all left paranthesis have been removed from stack we've found all the matches
if len(stack) == 0:
return True
else:
return False
if __name__ == '__main__':
input = ["khkjhkjhj", "#@#$@#$@", "()", "((([[[{}]]])))", ")(", "((]]", ""]
for i in input:
result = isValid(i)
print("All paranthesis are matching.", i if result else "Mismatch", i)
| true
|
bb708c316a2f7041f23c61036ff825153ad424a2
|
denemorhun/Python-Problems
|
/Grokking the Python Interview/Data Structures/Stacks and Queues/q_using_stack.py
| 1,149
| 4.21875
| 4
|
from typing import Deque
from stack import MyStack
# Push Function => stack.push(int) //Inserts the element at top
# Pop Function => stack.pop() //Removes and returns the element at top
# Top/Peek Function => stack.get_top() //Returns top element
# Helper Functions => stack.is_empty() & stack.isFull() //returns boolean
class NewQueue:
def __init__(self):
self.main_stack = MyStack()
# Write your code here
# Inserts Element in the Queue
def enqueue(self, value):
self.main_stack.push(value)
return True
# Removes Element From Queue
def dequeue(self):
# Write your code here
temp_stack = MyStack()
while not self.main_stack.size() == 1:
temp_stack.push(self.main_stack.pop())
popped_item = self.main_stack.pop()
while not temp_stack.is_empty():
self.main_stack.push(temp_stack.pop())
return popped_item
def __str__(self):
return str(self.main_stack)
nq = NewQueue()
nq.enqueue(1)
nq.enqueue(2)
nq.enqueue(3)
nq.enqueue(4)
nq.enqueue(5)
val = nq.dequeue()
print(nq)
print(val)
| true
|
a34383ee463d3c9749caf0e52c10e3e114fe02a7
|
denemorhun/Python-Problems
|
/AlgoExperts/LinkedList/remove_duplicates_from_linked_list.py
| 1,693
| 4.15625
| 4
|
'''
A Node object has an integer data field, , and a Node instance pointer, , pointing to
another node (i.e.: the next node in a list).
A removeDuplicates function is declared in your editor, which takes a pointer to the
node of a linked list as a parameter. Complete removeDuplicates so that it deletes
any duplicate nodes from the list and returns the head of the updated list.
Note: The pointer may be null, indicating that the list is empty. Be sure to
reset your pointer when performing deletions to avoid breaking the list.
Input Format
Constraints
The data elements of the linked list argument will always be in non-decreasing order.
Your removeDuplicates function should return the head of the updated linked list.
Sample Input
6
1
2
2
3
3
4
Sample Output
1 2 3 4
Explanation
, and our non-decreasing list is . The values and both occur twice in the list
, so we remove the two duplicate nodes. We then return our updated (ascending) list
'''
# This is an input class. Do not edit.
class LinkedList:
def __init__(self, value):
self.value = value
self.next = None
# Even though this class says linkedlist, it's actually Node
def removeDuplicatesFromLinkedList(linkedList):
# actually this is the head node
currentNode = linkedList
print("Starting with head node")
while currentNode is not None:
nextNode = currentNode.next
print(currentNode.value)
# if next node is equal to current node's value repeat loop
while nextNode is not None and nextNode.value == currentNode.value:
print('duplicate')
nextNode = nextNode.next
currentNode.next = nextNode
currentNode = currentNode.next
return linkedList
| true
|
07bab891b92ca248e39056da104d2a9afdff953c
|
denemorhun/Python-Problems
|
/Hackerrank/Arrays/validate_pattern.py
| 1,708
| 4.34375
| 4
|
'''Validate Pattern from array 1 to array b
Given a string sequence of words and a string sequence pattern, return true if the sequence of words matches the pattern otherwise false.
Definition of match: A word that is substituted for a variable must always follow that substitution. For example, if "f" is substituted as "monkey" then any time we see another "f" then it must match "monkey" and any time we see "monkey" again it must match "f".
Examples
input: "ant dog cat dog", "a d c d"
output: true
This is true because every variable maps to exactly one word and vice verse.
a -> ant
d -> dog
c -> cat
d -> dog
input: "ant dog cat dog", "a d c e"
output: false
This is false because if we substitute "d" as "dog" then you can not also have "e" be substituted as "dog".
a -> ant
d, e -> dog (Both d and e can't both map to dog so false)
c -> cat
input: "monkey dog eel eel", "e f c c"
output: true
This is true because every variable maps to exactly one word and vice verse.
e -> monkey
f -> dog
c -> eel
'''
def is_it_a_pattern(arr, patt):
# matching_values dictionary
mv = {}
#isPattern = True
if len(arr) != len(patt):
return False
for i in range(len(arr)):
if arr[i] not in mv.keys():
mv[arr[i]] = patt[i]
else:
if mv[arr[i]] != patt[i]:
return False
return True
if __name__ == '__main__':
color1 = ["red", "green", "green"]
patterns1 = ["a", "b", "b", "d"]
print( "True" if is_it_a_pattern(color1, patterns1) else "False")
color2 = ["blue", "green", "blue", "blue"]
patterns2 = ["blue", "b", "c", "c"]
print( "True" if is_it_a_pattern(color2, patterns2) else "False")
| true
|
7ff5903c490ef8c34099db9252ec42918f2e850b
|
tcano2003/ucsc-python-for-programmers
|
/code/lab_03_Functions/lab03_5(3)_TC.py
| 1,028
| 4.46875
| 4
|
#!/usr/bin/env python3
"""This is a function that returns heads or tails like the flip of a coin"""
import random
def FlipCoin():
if random.randrange(0, 2) == 0:
return ("tails")
return ("heads")
def GetHeads(target):
"""Flips coins until it gets target heads in a row."""
heads = count = 0
while heads < target:
count += 1
if FlipCoin() == 'heads':
heads += 1
# print("heads") test code to show when heads occurs
else: # 'tails'
heads = 0
# print("tails") test code to show when tails occurs
return count
def GetAverage(number_of_experiments, target):
"""Calls GetHeads(target) that number_of_experiments
times and reports the average."""
total = 0
for n in range(number_of_experiments):
total += GetHeads(target)
print ("Averaging %d experiments, it took %.1f coin flips to get %d in a row." % (number_of_experiments, total/number_of_experiments, target))
GetAverage(100, 3)
| true
|
3b1315a4ffacc91f29a75daa3993a1cb2aab4133
|
tcano2003/ucsc-python-for-programmers
|
/code/lab_03_Functions/lab03_6(3).py
| 2,442
| 4.46875
| 4
|
#!/usr/bin/env python3
"""Introspect the random module, and particularly the randrange()
function it provides. Use this module to write a "Flashcard"
function. Your function should ask the user:
What is 9 times 3
where 9 and 3 are any numbers from 0 to 12, randomly chosen.
If the user gets the answer right, your function should print
"Right!" and then return a 1. If the answer is wrong, print
"Almost, the right answer is 27" and return a 0.
Write a function called TestAndScore(n) that calls your
Flashcard function n times and prints the percentage of right
answers like this, "Score is 90". It also returns this
percentage.
Make another function called GiveFeedback(p) that receives a
percentage, 0 - 100. If p is 100 it prints "Perfect!", if it's
90-99 it prints "Excellent", 80-89 prints "Very good", 70-79
prints "Good enough", and <= 69, "You need more practice".
Test all that in your program, calling TestAndScore(10) and
then pass the returned value into GiveFeedback().
Make a new function called "Praise" that takes no arguments.
It prints one of (at least) 5 phrases of Praise, chosen
randomly. It might print, "Right On!", or "Good work!",
for example. Call this Praise() function from your
Flashcard() function whenever your user gets the answer right.
"""
import random
def FlashCard():
n1 = random.randrange(13)
n2 = random.randrange(13)
answer = n1 * n2
print ("What is %d times %d? " % (n1, n2))
while True:
try:
response = int(input())
except ValueError:
print ("%d times %d is a number. What number? " % (
n1, n2))
continue
if response == answer:
print ("Right!",
Praise())
return 1
print ("Almost, the right answer is %d." % answer)
return 0
def GiveFeedback(p):
if p == 100:
print ('Perfect!')
elif p >= 90:
print ('Excellent')
elif p >= 80:
print ('Very good')
elif p >= 70:
print ('Good enough')
else:
print ('You need more practice')
def Praise():
pats = ("That's great!", "You're right!",
"Good work!", "Right On!", "Superb!")
print (random.choice(pats))
def TestAndScore(n):
score = 0
for time in range(n):
score += FlashCard()
score = int(100.0 * score/n)
print ("Score is %d." % score)
return score
GiveFeedback(TestAndScore(10))
| true
|
008fa26d54dc77f783b3b60f5e2f85c83535a3fd
|
tcano2003/ucsc-python-for-programmers
|
/code/lab_14_Magic/circle_defpy3.py
| 1,823
| 4.1875
| 4
|
#!/usr/bin/env python3
"""A Circle class, acheived by overriding __getitem__
which provides the behavior for indexing, i.e., [].
This also provides the correct cyclical behavior whenever
an iterator is used, i.e., for, enumerate() and sorted().
reversed() needs __reversed__ defined.
"""
#need to have def __getitem__(<<obj>> , <<3>>): in order to call obj[3], otherwise will fail
class Circle:
def __init__(self, data, times): #want to prevent infinite results with a circle class
"""Put the 'data' in a circle that goes around
'times' times."""
self.data = data
self.times = times
def __getitem__(self, i): #i is index given by the caller
"""circle[i] --> Circle.__getitem__(circle, i)."""
l_self = len(self)
if i >= self.times * l_self:
raise IndexError ("Circle object goes around %d times" % (self.times)) #raise IndexError
return self.data[i % l_self] # return answer
def __len__(self):
return len(self.data) #given the len() of the data
def main():
circle = Circle("around", 3)
print ("Works with circle[i], for i > len(circle) too:")
for i in range(3 * len(circle) + 1):
try:
print ("circle[%2d] = %s" % (i, circle[i]))
except IndexError as e:
print(e)
break
print ("Works with sorted:")
print (sorted(circle))
print ("Works for loops:")
small_circle = Circle("XO", 2)
for i, elementi in enumerate(small_circle):
print ("small_circle[%d] = %s" % (i, elementi))
print ("Works for nested loops:")
for i, elementi in enumerate(small_circle):
for j, elementj in enumerate(small_circle):
print ("%3d:%3d -> %s%s" % (i, j, elementi, elementj))
if __name__ == '__main__':
main()
| true
|
32ecb9b4cfe42fc63007e72ac3c8778fa8121ad5
|
huang-zp/pyoffer
|
/my_queue.py
| 646
| 4.125
| 4
|
class MyQueue:
def __init__(self):
"""
Initialize your data structure here.
"""
self.stack_in = []
self.stack_out = []
def push(self, x: int) -> None:
"""
Push element x to the back of queue.
"""
self.stack_in.append(x)
def pop(self) -> int:
"""
Removes the element from in front of queue and returns that element.
"""
if not self.stack_out:
while self.stack_in:
self.stack_out.append(self.stack_in.pop())
return self.stack_out.pop()
else:
return self.stack_out.pop()
| true
|
aee36a7523ad8fc3de06ef7b047aff60fa3ab0d6
|
ADmcbryde/Collatz
|
/Python/recursive/coll.py
| 2,842
| 4.25
| 4
|
# CSC330
# Assignment 3 - Collatz Conjecture
#
# Author: Devin McBryde
#
#
#
def collatzStep(a):
counter = 0
if (a == 1):
return counter
elif ( (a%2) == 1) :
counter = collatzStep(a*3+1)
else:
counter = collatzStep(a/2)
counter = counter + 1
return counter
#The program is programmed as a function since local references are faster
# than global variable references and speeds up the inner while loop
def collatz():
#stores the top ten highest values and the steps to 1
# and initializes the array to zeroes
maxValues = [[0 for x in range(2)]for y in range(10)]
#stores the top ten highest values and the steps to 1
minVal = 0
col = long(0)
#initialize x for the loop
x = 2
#Main loop that goes through all values between 2 and 5000000000
# Top value has the L suffix since literals are interpreted as integers
# This is a while loop instead since range(x) would produce an
# array that would take over 20-40 gigabytes to store, depending
# on if the array would store 4 or 8 byte integers. This can
# also be solved by using xrange() instead, which uses iterators
# instead, but its behavior is less well understood
while(x < 1000000):
alreadyexists = False
#reset the next two values for the new number
#col holds the value of the iterated number
col = x
#count tracks the number of iterations total
count = 0
count = collatzStep(col)
#Here we avoid having a value with a duplicate number of steps using the boolean flag
#Need to add loop
#Here we check if the count is larger than the smallest count recorded and add if it is
for z in range(10):
if (count == maxValues[z][0]):
alreadyexists = True
if (count > maxValues[minVal][0] and not(alreadyexists)):
#here we replace the value of the smallest count
for y in range(10):
if(y == minVal):
maxValues[y][0] = count
maxValues[y][1] = x
#we now reset the minVal to look for the new lowest count value
minVal = 0
#search for the smallest count size in maxValues
for y in range(10):
if(maxValues[y][0] < maxValues[minVal][0]):
minVal = y
#increment x for the while loop to go to the next
x += 1
#Now we perform a basic selection sort on the step count before printing
for i in range(0,9):
minValue = maxValues[i][0];
minColNum = maxValues[i][1];
minLocale = i;
for j in range(i+1,10):
if(minValue < maxValues[j][0]):
minValue = maxValues[j][0];
minColNum = maxValues[j][1];
minLocale = j;
tempVal = maxValues[i][0];
tempNum = maxValues[i][1];
maxValues[i][0] = minValue;
maxValues[minLocale][0] = tempVal;
maxValues[i][1] = minColNum;
maxValues[minLocale][1] = tempNum;
#print the maxValues array
for i in range(10):
print "Value: " , maxValues[i][1] , " Steps Taken: " , maxValues[i][0]
collatz()
| true
|
0d075a750c9745eb97a2577db4b7c7cf2407d903
|
ergarrity/coding-challenges
|
/missing-number/missing.py
| 1,009
| 4.21875
| 4
|
"""Given a list of numbers 1...max_num, find which one is missing in a list."""
def missing_number(nums, max_num):
"""Given a list of numbers 1...max_num, find which one is missing.
*nums*: list of numbers 1..[max_num]; exactly one digit will be missing.
*max_num*: Largest potential number in list
>>> missing_number([7, 3, 2, 4, 5, 6, 1, 9, 10], 10)
8
"""
nums_dict = {}
# Creates keys for all possible numbers in list and sets all values to False
for i in range(max_num):
nums_dict[i+1] = False
# Iterates over nums and changes value to True for each num key
for num in nums:
nums_dict[num] = True
# Iterates over nums_dict and returns key with false value (this will be
# missing number)
for key in nums_dict:
if nums_dict[key] == False:
return key
if __name__ == '__main__':
import doctest
if doctest.testmod().failed == 0:
print("\n*** ALL TESTS PASS. NICELY DONE!\n")
| true
|
8717208247c2d4f9eb24522c1b54ec33ce41789c
|
kwozz48/Test_Projects
|
/is_num_prime.py
| 613
| 4.1875
| 4
|
#Asks the user for a number and determines if the number is prime or not
number_list = []
def get_integer(number = 'Please enter a number: '):
return int(input(number))
def is_number_prime():
user_num = get_integer()
number_list = list(range(2, user_num))
print (number_list)
i = 0
a = 0
while i < user_num - 2:
if user_num % number_list[i] == 0:
a += 1
i += 1
else:
i += 1
if a > 0:
print ('The number is not prime')
else:
print ('The number is prime')
return ()
answer = is_number_prime()
| true
|
7b84faec3e2daad6d731610e6d4967f1e3859537
|
yucheno8/newPython6.0
|
/Day01/p_07输出和格式字符串一起使用.py
| 479
| 4.125
| 4
|
'''
格式 化字符串并输出
'''
# 占位符形式的字符串
# 1
a = 10
print('现在要打印一个数字，值是 %d' % a)
# 2
print('name: %s age: %d' % ('Tom', 12))
# 占位符的常用 格式
print('%d' % 1)
print('%5d' % 1)
print('%05d' % 1)
print('%-5d' % 1)
print('%3d' % 12345)
print('%.2f' % 3.1415926)
print('%.3f' % 1)
# f-string
name = 'Tom'
age = 11
print(f'name: {name} age：{age} score: {99}')
s = f'name: {name} age：{age} score: {99}'
print(s)
| false
|
3c47b56deda38352aa8cd571d65a29c76d40827e
|
BRTytan/Unipexercicios
|
/IPE/Listas/ex04-04.py
| 330
| 4.15625
| 4
|
"""
Exercicio 04-04:Escreva um programa que mostre os números de 1 até um numero digitado pelo usuário,
mas, apenas os números impares.
"""
###################################################################################################
n1 = int(input("Digite um número: "))
x = 1
while x <= n1:
print(x)
x += 2
| false
|
f04b4b2d639091ef6d2a21bf0a40604936664cc4
|
BRTytan/Unipexercicios
|
/IPE/inicias/ex01-02.py
| 382
| 4.34375
| 4
|
"""Exercicio 02: Resolva as expressões matematicas manualmente no caderno, apos, converta as seguintes expressões
matematicas para que possam ser calculadas usando o interpretador python(confirmando o resultado encontrado)"""
a = (10+20)*30
b = (4 ** 2) / 30
c = (9**4+2) * (6-1)
print('Resultado de A é: {} \nResultado de B é: {:.2f} \nResultado de C é: {}'.format(a, b, c))
| false
|
0eb55829a0aee6d7136f91550e89b9bb738f4e73
|
Scertskrt/Ch.08_Lists_Strings
|
/8.1_Months.py
| 722
| 4.40625
| 4
|
'''
MONTHS PROGRAM
--------------
Write a user-input statement where a user enters a month number 1-13.
Using the starting string below in your program, print the three month abbreviation
for the month number that the user enters. Keep repeating this until the user enters 13 to quit.
Once the user quits, print "Goodbye!"
months = "JanFebMarAprMayJunJulAugSepOctNovDec"
'''
done = False
month = int(input("Please pick a month(1-12) or quit "))
months = "JanFebMarAprMayJunJulAugSepOctNovDec"
while done == False:
if month > 12 or month < 1:
print("Goodbye!")
done = True
break
end = month*3
print(months[end-3:end])
month = int(input("Please pick a month(1-12) or quit(13) "))
| true
|
539fadb2fc145e48a70950cd77d804d77c9cec07
|
roachaar/Python-Projects
|
/national debt length.py
| 1,937
| 4.5
| 4
|
##############################################################################
# Computer Project #1: National Debt
#
# Algorithm
# prompt for national debt and denomination of currency
# user inputs the above
# program does simple arithmetic to calculate two pieces of information
# 1: national debt as a height in miles of stacked currency
# 2: this height compared to the distance between the earth and moon
# lastly, the program displays these values to the user
##############################################################################
national_debt_str = input('Enter the national debt: ') #user inputs his or her\
#nation's national debt
denomination_str = input('Enter a denomination of currency: ') #user\
#inputs a chosen denomination of the same currency as the national debt is\
#given in
national_debt_float = float(national_debt_str) #we identify these as the values\
#in which they are meant to be
denomination_int = int(denomination_str) #again
BILL_HEIGHT=.0043 #in inches
INCHES_PER_FOOT= 12 #inches per foot
FEET_PER_MILE=5280 #feet per mile
AVG_DISTANCE_BETWEEN_EARTH_AND_MOON=238857 #in miles
height_of_national_debt_float = (national_debt_float/denomination_int)*\
(BILL_HEIGHT)/(INCHES_PER_FOOT)/(FEET_PER_MILE) #we use dimensional analysis to\
#get the height in miles
compared_to_moon_float= (national_debt_float/denomination_int)*(BILL_HEIGHT)/\
(INCHES_PER_FOOT)/(FEET_PER_MILE)/(AVG_DISTANCE_BETWEEN_EARTH_AND_MOON) #same\
#thing here, only we now divide by the distance between the earth and the moon
print('The height of your national debt in miles of your chosen denomination of\
currency is:',height_of_national_debt_float)
print('The ratio of this height and the distance between the earth and the moon\
is:' ,compared_to_moon_float) #these 2 values are printed for the user to see
| true
|
3597d031334cadd8da79740431f5941c2adb38c5
|
BeryJAY/Day4_challenge
|
/power/power.py
| 531
| 4.34375
| 4
|
def power(a,b):
#checking data type
if not((isinstance(a,int) or isinstance(a,float)) and isinstance(b,int)):
return "invalid input"
#The condition is such that if b is equal to 1, b is returned
if(b==1):
return(a)
#If b is not equal to 1, a is multiplied with the power function and called recursively with arguments as the base and power minus 1
if(b!=1):
return(a*power(a,b-1))
#the final result is then output
if __name__ == '__main__':
print("Result:",power(2,5))
| true
|
511d35743234e26a4a93653af24ee132b3a62a7a
|
AntoanStefanov/Code-With-Mosh
|
/Classes/1- Classes.py
| 1,465
| 4.6875
| 5
|
# Defining a list of numbers
numbers = [1, 2]
# we learned that when we use the dot notation, we get access to all methods in list objects.
# Every list object in Python has these methods.
# numbers.
# Wouldn't that be nice if we could create an object like shopping_cart and this object would have methods
# like this:
# shopping_cart.add()
# shopping_cart.remove()
# shopping_cart.get_total()
# Another example:
# Would be nice if we could have a point object with methods.
# point.draw()
# point.move()
# point.get_distance() * between point x and point y
# Here come classes
# A class is a blueprint for creating new objects.
# Throughout the course you have heard the term class.
# For example: let's define a variable and set it to integer and print it's type
x = 1
print(type(x))
# We see a class of int
# So in Python we have a class called int for creating integers. Similarly, we have classes for creating booleans,
# lists, dictionaries and so on.
# EVERY OBJECT WE HAVE IN PYTHON IS CREATED USING A CLASS WHICH IS A BLUEPRINT FOR CREATING OBJECTS OF THAT TYPE.
#
# In this section you're going to learn how to create custom classes like customer, shopping_cart, point and so on.
# Before that let's define a few terms.
# 1. Class: blueprint for creating new objects
# 2. Object: instance of a class
# Example:
# Class: Human (this class will define all the attributes of humans) then we could create objects
# Objects: John, Mary, ...
| true
|
a8f7dceb7daec1a665e215d9d99eeb620e73f398
|
AntoanStefanov/Code-With-Mosh
|
/Exceptions/7- Cost of Raising Exceptions.py
| 1,849
| 4.5
| 4
|
# As I explained in the last lecture, when writing your own functions,
# prefer not to raise exceptions, because those exceptions come with a price.
# That's gonna show you in this lecture.
# From the timeit module import function called timeit
# with this function we can calculate the execution time of some code.
# This is how it works. Imagine you want to calculate the execution time of
# this code.
# We define a variable code1 and set it to string
# This string should include our code so using triple quotes code
# for multiple lines. One piece of code inside string with triple quotes
from timeit import timeit
code1 = '''
def calculate_xfactor(age):
if age <= 0:
raise ValueError('Age cannot be 0 or less.')
return age / 10
try:
calculate_xfactor(-1)
except ValueError as error:
pass
'''
# Now we call timeit
# - first arg = python code , second arg(keyword arg) = number of exectutions
print('first code=', timeit(code1, number=10000))
# This function returns the exectution time of the code after 10 000 repetitions.
# DIfferent approach ### 4 times faster
code2 = '''
def calculate_xfactor(age):
if age <= 0:
None
return age / 10
xfactor = calculate_xfactor(-1)
if xfactor == None:
pass
'''
print('second code=', timeit(code2, number=10000))
# that's so because we execute the code 10 000 times
# 1 time execution no difference
# If you are building a simple application for a few users raising exceptions
# in your functions is not going to have a bad impact on the performance of your app.
# But if you are building application where performance and scaleability is important,
# then it's better to raise exceptions when you really have to .
# if you can handle the situation with a simple if statement, think twice for exceptions.
# Raise exceptions if YOU REALLY HAVE TO !
| true
|
1f7f8382c688f22fe10e8057185ce55307c90b1d
|
AntoanStefanov/Code-With-Mosh
|
/Exceptions/4- Cleaning Up.py
| 706
| 4.375
| 4
|
# There are times that we need to work with external resources like files,
# network connections, databases and so on. Whenever we use these resources,
# after, after we're done we need to release them.
# For example: when you open a file, we should always close it after we're done,
# otherwise another process or another program may not be able to open that file.
try:
file = open('name')
age = int(input('Age: '))
xfactor = 10 / age
except (ValueError, ZeroDivisionError):
print('You didnt enter a valid age.')
else:
print("No exceptions were thorwn.")
finally: # always executed clause with or without exception
# and we use it to release external resources
file.close()
| true
|
91dc1d2f54c0cf7ad75ff9f9449b5c46051c4b4c
|
pedrolisboaa/pythonpro-pythonbirds
|
/exercicios/busca_linear.py
| 616
| 4.125
| 4
|
"""
Crie um programa que recebe uma lista de inteiros e um valor que deve ser buscado.
O programa deve retornar o índice onde o valor foi encontrado, ou -1, caso não encontre o valor.
"""
tamanho_lista = int(input('Digite o tamanho da sua lista:'))
lista = []
for i in range(tamanho_lista):
inserir_lista = int(input('Digite o número que deseja inserir na lista: '))
lista.append(inserir_lista)
buscar_na_lista = int(input('Digite número que deseja buscar na lista: '))
if buscar_na_lista in lista:
print(f'O {buscar_na_lista} está no range {lista.index(buscar_na_lista)}')
else:
print(-1)
| false
|
a12230a5edc331e0989940c8ecd1243b9415aba9
|
daria-andrioaie/Fundamentals-Of-Programming
|
/a12-911-Andrioaie-Daria/main.py
| 2,983
| 4.46875
| 4
|
from random import randint
from recursive import recursive_backtracking
from iterative import iterative_backtracking
def print_iterative_solutions(list_of_numbers):
"""
The function calls the function that solves the problem iteratively and then prints all the found solutions.
:param list_of_numbers: the list of numbers for which we want to determine all the possibilities to insert between
them the operators + and – such that by evaluating the expression the result is positive.
"""
list_of_solutions = iterative_backtracking(list_of_numbers)
number_of_solutions = len(list_of_solutions)
if number_of_solutions == 0:
print('There are no solutions for the given set of numbers')
else:
for solution in list_of_solutions:
print(solution)
print(str(number_of_solutions) + ' SOLUTIONS: ')
print('\n')
print('\n')
def print_recursive_solutions(list_of_numbers):
"""
The function calls the function that solves the problem recursively and then prints all the found solutions.
:param list_of_numbers: the list of numbers for which we want to determine all the possibilities to insert between
them the operators + and – such that by evaluating the expression the result is positive.
"""
list_of_solutions = []
recursive_backtracking(list_of_solutions, [], list_of_numbers)
number_of_solutions = len(list_of_solutions)
if number_of_solutions == 0:
print('There are no solutions for the given set of numbers')
else:
for solution in list_of_solutions:
print(solution)
print(str(number_of_solutions) + ' SOLUTIONS: ')
print('\n')
print('\n')
def initialise_numbers(number_of_elements):
"""
The function computes a random sequence of natural number.
:param number_of_elements: the total number of elements in the sequence
:return: the sequence of numbers, represented as a list
"""
list_of_numbers = []
for iteration in range(number_of_elements):
natural_number = randint(1, 100)
list_of_numbers.append(natural_number)
return list_of_numbers
def print_menu():
print('1. Solve recursively.')
print('2. Solve iteratively')
print('3. Exit.')
print('\n')
def start():
available_options = {'1': print_recursive_solutions, '2': print_iterative_solutions}
number_of_elements = int(input('Enter a natural number greater than 9: '))
list_of_numbers = initialise_numbers(number_of_elements)
not_finished = True
while not_finished:
print_menu()
user_option = input('Enter option: ')
if user_option in available_options:
available_options[user_option](list_of_numbers)
elif user_option == '3':
not_finished = False
else:
print('Bad command')
start()
| true
|
41891f9a090ead7873bf5c006f423238a48f05db
|
daria-andrioaie/Fundamentals-Of-Programming
|
/a12-911-Andrioaie-Daria/iterative.py
| 2,158
| 4.5
| 4
|
from solution import is_solution, to_string
def find_successor(partial_solution):
"""
The function finds the successor of the last element in the list.
By "successor" of an element we mean the next element in the list [0, +, -].
:param partial_solution: array containing the current partial solution
:return: True, if the last element has a successor, False, otherwise
"""
if partial_solution[-1] == 0:
partial_solution[-1] = '+'
return True
elif partial_solution[-1] == '+':
partial_solution[-1] = '-'
return True
return False
def iterative_backtracking(list_of_numbers):
"""
The function uses the concept of backtracking to search the whole space of solutions, and keeps the found solutions
in a list that will be returned.
:param list_of_numbers: the list of numbers for which we want to determine all the possibilities to insert between
them the operators + and – such that by evaluating the expression the result is positive.
:return: the resulted solutions
"""
list_of_solutions = []
partial_solution = [0]
while len(partial_solution):
current_element_has_successor = find_successor(partial_solution)
if current_element_has_successor:
if is_solution(partial_solution, list_of_numbers):
printable_solution = to_string(partial_solution, list_of_numbers)
list_of_solutions.append(printable_solution)
elif len(partial_solution) == len(list_of_numbers) - 1: # the solution array is full, but does not
# represent a solution
partial_solution.pop()
else:
partial_solution.append(0) # go to the next element in order to complete
# the solution
else:
# go back with one step in the partial solution
partial_solution.pop()
return list_of_solutions
| true
|
a43a94fda2b1a6ff937b0ab5df42f12fff90bde1
|
sal-7/OMIS485
|
/Python_chapter 4/A_temperature.py
| 1,206
| 4.125
| 4
|
""" /***************************************************************
OMIS 485 P120 - Ch4 Spring 2019
Programmer: Faisal Alharbi, Z-ID 1748509
Date Due:04/07/2018
Tutorials from Chapter 4 - Temperature Exercise.
***************************************************************/"""
#!/usr/bin/env python3
import temperature as temp
def display_menu ():
print ("MENU")
print ("1. Fah. to Cel.")
print ("2. Cel. to Fah.")
print ()
def convert_temp():
option = int (input ("Enter a menu option: "))
if option == 1:
f = int (input ("Enter degrees Fahrenheit: "))
c = temp.to_celsius (f)
c = round (c,2)
print ("Degrees Celsius:", c)
elif option == 2:
c = int (input ("Enter degrees Celsius: "))
f = temp.to_fahrenheit (c)
f = round (f, 2)
print ("Degrees Fahrenheit:", f)
else:
print ("Enter a valid menu number")
def main ():
display_menu()
again = "y"
while again.lower() == "y":
convert_temp()
print ()
again = input ("Convert another temperature? (Y/N:) ")
print ("See you!")
if __name__ == "_main_":
main()
main()
| false
|
48f7e3ee1d356eda254447e928321e6d7a8402c8
|
samsolariusleo/cpy5python
|
/practical02/q07_miles_to_kilometres.py
| 661
| 4.3125
| 4
|
# Filename: q07_miles_to_kilometres.py
# Author: Gan Jing Ying
# Created: 20130207
# Modified: 20130207
# Description: Program that converts miles to kilometres and kilometres to miles
# before printing the results
# main
# print headers
print("{0:6s}".format("Miles") + "{0:11s}".format("Kilometres") +
"{0:11s}".format("Kilometres") + "{0:6s}".format("Miles"))
# define for loop
for miles in range(1,11):
convert_to_kilo = miles * 1.609
convert_to_miles = (miles + 3) * 5 / 1.609
print("{0:<6}".format(miles) + "{0:<11.3f}".format(convert_to_kilo) +
"{0:<11}".format((miles + 3) * 5) +
"{0:<6.3f}".format(convert_to_miles))
| true
|
2e1e1897d6e0c4f92ee6815c9e1bb607dee10024
|
samsolariusleo/cpy5python
|
/practical02/q12_find_factors.py
| 562
| 4.4375
| 4
|
# Filename: q12_find_factors.py
# Author: Gan Jing Ying
# Created: 20130207
# Modified: 20130207
# Description: Program that displays the smallest factors of an integer.
# main
# prompt for integer
integer = int(input("Enter integer: "))
# define smallest factor
factor = 2
# define a list
list_of_factors = []
# find factors
while factor < integer:
while integer % factor != 0:
factor = factor + 1
list_of_factors.append(factor)
integer = integer//factor
# return results
print("The smallest factors are " + str(list_of_factors)[1:-1] + ".")
| true
|
750d0ff689da9aa8aea56e2e9b248df47ed51057
|
samsolariusleo/cpy5python
|
/practical02/q05_find_month_days.py
| 956
| 4.625
| 5
|
# Filename: q05_find_month_days.py
# Author: Gan Jing Ying
# Created: 20130207
# Modified: 20130207
# Description: Program that displays number of days in the month of a particular
# year.
# main
# prompt for month
month = int(input("Enter month: "))
# prompt for year
year = int(input("Enter year: "))
# define list of months
allmonths = ["January", "February", "March", "April", "May", "June",
"July", "August", "September", "October", "November", "December"]
# define list of days
alldays = ["31", "28", "31", "30", "31", "30",
"31", "31", "30", "31", "30", "31", "29"]
# check if month is feb
# if month is feb, then check if year is leap year
# if both are true, days is 29
# else, days are normal
if month == 2 and year % 4 == 0:
print(allmonths[month-1] + " " + str(year) + " has " + alldays[-1] +
" days")
else:
print(allmonths[month-1] + " " + str(year) + " has " + alldays[month-1] +
" days")
| true
|
cce715b6938e2f5c16d6ee158ebfa7e15b4e0645
|
samsolariusleo/cpy5python
|
/practical02/q11_find_gcd.py
| 725
| 4.28125
| 4
|
# Filename: q11_find_gcd.py
# Author: Gan Jing Ying
# Created: 20130207
# Modified: 20130207
# Description: Program to find the greatest common divisor of two integers.
# main
# prompt user for the two integers
integer_one = int(input("Enter first integer: "))
integer_two = int(input("Enter second integer: "))
# find out which is the smaller of integer_one and integer_two
if integer_one > integer_two:
d = integer_two
elif integer_one < integer_two:
d = integer_one
else:
print("The greatest common divisor is " + str(integer_one) + ".")
exit()
# find greatest divisor
while integer_one % d != 0 or integer_two % d != 0:
d = d - 1
# return results
print("The greatest common divisor is " + str(d) + ".")
| true
|
8254638df080d0a3b76a0dddd42cf41e393dfed7
|
samsolariusleo/cpy5python
|
/practical01/q1_fahrenheit_to_celsius.py
| 453
| 4.28125
| 4
|
# Filename: q1_fahrenheit_to_celsius.py
# Author: Gan Jing Ying
# Created: 20130122
# Modified: 20130122
# Description: Program to convert a temperature reading from Fahrenheit to Celsius.
#main
# prompt to get temperature
temperature = float(input("Enter temperature (Fahrenheit): "))
# calculate temperature in Celsius
temperature = (5/9) * (temperature - 32)
# display result
print("Temperature in Celsius: {0:.2f}".format(temperature))
| true
|
8e1570c6e03ec4817ab65fdba077df7bad1e97da
|
justintrudell/hootbot
|
/hootbot/helpers/request_helpers.py
| 490
| 4.46875
| 4
|
import itertools
def grouper(iterable, n):
"""
Splits an iterable into groups of 'n'.
:param iterable: The iterable to be split.
:param n: The amount of items desired in each group.
:return: Yields the input list as a new list, itself containing lists of 'n' items.
"""
"""Splits a list into groups of three."""
it = iter(iterable)
while True:
chunk = tuple(itertools.islice(it, n))
if not chunk:
return
yield chunk
| true
|
dfb6d90f75f7c2fbe370185856549fdf6c089927
|
gbaweja/consultadd_assignments
|
/assignment1/A1-6.py
| 654
| 4.1875
| 4
|
# Govinda Baweja
# Assignment 1
# Date: 04-26-2019
# Nested if else statement
i = 10
if (i == 10):
# First if statement
print ("\ni is 10")
if (i < 15):
print ("\ni is smaller than 15")
# Nested - if statement
# Will only be executed if statement above
# it is true
if (i < 12):
print ("\ni is smaller than 12 too")
else:
print ("\ni is greater than 15")
# Ladder if else Statement
i = 20
if (i == 10):
print ("\ni is 10")
elif (i == 15):
print ("\ni is 15")
elif (i == 20):
print ("\ni is 20")
else:
print ("\ni is not of my choice")
| false
|
4182316191f1b325cb7640dd77c1acc023045c82
|
ZhbitEric/PythonLearning
|
/Python_work/_20ds_reference.py
| 310
| 4.1875
| 4
|
# 引用
shopList = ['apple', 'mango', 'carrot', 'banana']
myList = shopList
del shopList[0]
print(shopList)
print(myList)
# 这样的引用就没有指向同一个对象了
myList = shopList[:] # 切片操作，相当于循环复制拿元素出来
print(myList)
del myList[0]
print(shopList)
print(myList)
| false
|
73d37bf82e89c293e0e0fd86e99d74ec79b3b275
|
SRAH95/Rodrigo
|
/input_statement.py
| 1,748
| 4.15625
| 4
|
"""message = input("Tell me something, and I will repeat it back to you: ")
print(message)"""
########################################################################################
'''name = input("Please enter your name: ")
print("Hi, " + name.title() + "!")'''
########################################################################################
prompt = "If you tell us who you are, we can personalize the message you see."
prompt += "\nWhat is your first name? "
name = input(prompt)
print("\nHello, " + name.title() + "!")
########################################################################################
'''age = input("How old are you? ")
age = int(age)
print(age >= 18) '''
########################################################################################
'''height = input("How tall are you, in inches? ")
height = int(height)
if height >= 36:
print("\nYou're tall enough to ride!")
else:
print("\nYou'll be able to ride when you're a little older.")'''
########################################################################################
"""prompt = "\nEven or Odd number."
prompt += "\nPlease, enter a number: "
odd = input(prompt)
odd = int(odd)
if odd % 2 == 0:
print("---> " + str(odd) + " is an even number.")
else:
print("---> " + str(odd) + " is an odd number.")"""
########################################################################################
"""people = input("\nHow many people are going to dinner toningth?\n ")
people = int(people)
if people <= 8:
print("There is an avilable table, wait for the waitress please")
else:
print("You have to wait for an avilable table, sorry")"""
########################################################################################
| true
|
ba982e9794b3cbaaa034584cbb1f2017068f5ce5
|
Pranalihalageri/Python_Eduyear
|
/day5.py
| 516
| 4.125
| 4
|
1.
list1 = [5, 20, 4, 45, 66, 93, 1]
even_count, odd_count = 0, 0
# iterating each number in list
for num in list1:
# checking condition
if num % 2 == 0:
even_count += 1
else:
odd_count += 1
print("Even numbers in the list: ", even_count)
print("Odd numbers in the list: ", odd_count)
2.
list1 = [10, 20, 4, 45, 99]
# sorting the list
list1.sort()
# printing the last element
print("Largest element is:", list1[-1])
print("minimum element is:", *list1[:1])
| true
|
258d4ea28ffe6b9a85b2ecd34a34214ccb3a0682
|
FredericoVieira/listas-python-para-zumbis
|
/Lista 02/lista2ex5.py
| 433
| 4.15625
| 4
|
# -*- coding: utf-8 -*-
num1 = int(raw_input('Digite o valor do primeiro inteiro: '))
num2 = int(raw_input('Digite o valor do segundo inteiro: '))
num3 = int(raw_input('Digite o valor do terceiro inteiro: '))
list = [num1, num2, num3]
num_max = max(list)
num_min = min(list)
print('O maior número entre %d, %d, %d é %d') % (num1, num2, num3, num_max)
print('O menor número entre %d, %d, %d é %d') % (num1, num2, num3, num_min)
| false
|
b0398f9d1751611c505aa530849336dbb7f3ef00
|
Ran05/basic-python-course
|
/ferariza_randolfh_day4_act2.py
| 1,153
| 4.15625
| 4
|
'''
1 Write a word bank program
2 The program will ask to enter a word
3 The program will store the word in a list
4 The program will ask if the user wants to try again. The user will input
Y/y if yes and N/n if no
5 If yes, refer to step 2.
6 If no, Display the total number of words and all the words that user entered.
'''
<<<<<<< HEAD
=======
1 Write a word bank program
2 The program will ask to enter a word
3 The program will store the word in a list
4 The program will ask if the user wants to try again. The user will input
Y/y if yes and N/n if no
5 If yes, refer to step 2.
6 If no, Display the total number of words and all the words that user entered.
'''
>>>>>>> 4f0a6fb087d8f2ddb4b87fea98c0c42a7cbbfbc8
wordList = []
a = (str(input("Please Enter a word: ")))
wordList.append(a)
question = input("Do you want to add more words?: ")
<<<<<<< HEAD
if question == True:
print(str(input("Please Enter a word: ")))
else: wordList.append(a)
=======
while question == "Y" or "Yes":
print(a)
if question == "N" or "no":
break
print(wordList)
>>>>>>> 4f0a6fb087d8f2ddb4b87fea98c0c42a7cbbfbc8
| true
|
92151c4a1ceca1910bd60785b2d5d030559cd241
|
niteshrawat1995/MyCodeBase
|
/python/concepts/classmethods.py
| 1,481
| 4.125
| 4
|
# Class methods are methods wich take class as an argument (by using decorators).
# They can be used as alternate constructors.
class Employee:
num_of_emps = 0
raise_amount = 1.04
def __init__(self, first, last, pay):
self.first = first
self.last = last
self.pay = pay
self.email = first + '.' + last + '@company.com'
Employee.num_of_emps += 1
def full_name(self):
return '{} {}'.format(self.first, self.last)
def apply_raise(self):
self.pay = self.pay * self.raise_amount
@classmethod
def set_raise_amount(cls, amount):
cls.raise_amount = amount
@classmethod
def from_string(cls, emp_str):
first, last, pay = emp_str.split('-')
# need to always return the cls(Employee) object.
return cls(first, last, pay)
@staticmethod
def is_workday(day):
if day.weekday() == 5 or day.weekday() == 6:
return False
else:
return True
emp_1 = Employee('Nitesh', 'Rawat', 20000)
emp_2 = Employee('Vikas', 'Sharma', 30000)
# Employee.set_raise_amount(1.05)
# print(Employee.raise_amount)
# print(emp_1.raise_amount)
# print(emp_2.raise_amount)
#emp_str_1 = 'John-Doe-70000'
#emp_str_2 = 'Amy-Smith-70000'
#emp_str_3 = 'Angier-Santos-000'
emp_str_1 = Employee.from_string('John-Doe-70000')
# print(emp_str_1.full_name())
import datetime
my_date = datetime.date(2018, 5, 27)
print(Employee.is_workday(my_date))
| true
|
5f109dfef4214b33302401e473d9b115a65bffa5
|
niteshrawat1995/MyCodeBase
|
/python/concepts/getters&setters&deleters.py
| 1,284
| 4.15625
| 4
|
# getters,setters and deleters can be implemented in python using property decorators.
# property decorators allows us to define a mehtod which we can access as an attribute.
# @property is the pythonic way of creating getter and setter.
class Employee(object):
def __init__(self, first, last, pay):
self.first = first
self.last = last
self.pay = pay
# getter-like
@property
def email(self):
print('getting email')
return '{}.{}@company.com'.format(self.first, self.last)
# getter-like
@property
def fullname(self):
print('getting fullname')
return '{} {}'.format(self.first, self.last)
# setter-like
@fullname.setter
def fullname(self, name):
print('setting fullname')
first, last = name.split(' ')
self.first = first
self.last = last
# deleter-like
@fullname.deleter
def fullname(self):
prin('deleting fullname')
self.first = None
self.last = None
print('Delete name!')
emp1 = Employee('Nitesh', 'Rawat', '20000')
# we now cannot set the attribute like this:
#emp1.email = 'Nitesh.chopra@gmail.com'
# To do the above task we need a setter
print(emp1.fullname)
# print(emp1.email)
emp1.fullname = 'Time Tom'
print(emp1.fullname)
| true
|
95484e7fe8d71eaa0fbcd723b67eddeafd17f535
|
niteshrawat1995/MyCodeBase
|
/python/practise/DS/Doubly_LinkedList.py
| 937
| 4.34375
| 4
|
# Implementing doubly linked list:
class Node(object):
"""Node will have data ,prev and next"""
def __init__(self, data):
self.data = data
self.prev = None
self.next = None
class LinkedList(object):
"""Linked list will have head attribute."""
def __init__(self):
self.head = None
def append(self, data):
# create new node:
new_node = Node(data)
# check if linked list is empty:
if self.head is None:
self.head = new_node.prev
return
while self.head.next != None:
self.head = self.head.next
self.head.next = new_node.prev
new_node.prev = self.head
def print_list(self):
while self.head != None:
print(str(self.head.data))
self.head = self.head.next
dll = LinkedList()
dll.append('A')
dll.append('B')
dll.append('C')
dll.append('D')
dll.print_list()
| false
|
e78334b2ae75714172fad80bf81e8c76497f7cb5
|
scantea/hash-practice
|
/hash_practice/exercises.py
| 2,554
| 4.3125
| 4
|
def grouped_anagrams(strings):
""" This method will return an array of arrays.
Each subarray will have strings which are anagrams of each other
Time Complexity: O(n)
Space Complexity: O(1)
"""
freq_hash = {}
for word in strings:
key = ''.join(sorted(word))
if key not in freq_hash.keys():
freq_hash[key] = []
freq_hash[key].append(word)
values_list = []
for value in freq_hash.values():
values_list.append(value)
return values_list
def top_k_frequent_elements(nums, k):
""" This method will return the k most common elements
In the case of a tie it will select the first occuring element.
Time Complexity: O(n)
Space Complexity: O(n)
"""
if nums == []:
return nums
freq_table = {}
for num in nums:
if num in freq_table:
freq_table[num] += 1
else:
freq_table[num] = 1
max_freq = 0
common_elemnts = []
for key, value in freq_table.items():
if freq_table[key] > max_freq:
max_freq = freq_table[key]
common_elemnts.append(key)
elif freq_table[key] == max_freq:
common_elemnts.append(key)
return common_elemnts
def valid_sudoku(table):
""" This method will return the true if the table is still
a valid sudoku table.
Each element can either be a ".", or a digit 1-9
The same digit cannot appear twice or more in the same
row, column or 3x3 subgrid
Time Complexity: O(n)3
Space Complexity: O(n)
"""
for col in range(0, len(table)):
column_hash = {}
for square in table[col]:
if square != ".":
if square not in column_hash:
column_hash[square] = 1
else:
return False
for row in range(3):
for col in range(3):
block_dict = {}
block = [
table[3*row][3*col:3*col+3], # finding index in sudoku board
table[3*row+1][3*col:3*col+3],
table[3*row+2][3*col:3*col+3]
]
for segment in block:
for elem in segment:
if elem != ".":
if elem in block_dict:
return False
else:
block_dict[elem] = 1
return True
| true
|
5b119b5d6ee2dfeb455b174a2b2332de7cd7a6a7
|
sivaram143/python_practice
|
/conditions/ex_01.py
| 202
| 4.3125
| 4
|
#!/usr/bin/python
# program to check whether a given no is even or ood
num = input("Enter any number:")
if num % 2 == 0:
print("{0} is even".format(num))
else:
print("{0} is odd".format(num))
| true
|
de43dc0d3c246bfdd7a124e2942be5ebc717cb9d
|
sivaram143/python_practice
|
/operators/arithmetic.py
| 706
| 4.375
| 4
|
#!/usr/bin/python
# Arithmetic Operators: +, -, *, /, %, **(exponent:power)
num1 = input("Enter first number:")
num2 = input("Enter second number:")
print("************* Arithmetic Operators *************")
print("Addition(+):{0}+{1}={2}".format(num1,num2,num1+num2))
if num1 > num2:
res = num1 - num2
else:
res = num2 - num1
print("Subtraction(-):{0}-{1}={2}".format(num1,num2,res))
print("Multiplication(*):{0}*{1}={2}".format(num1,num2,num1*num2))
if num2 == 0:
res = "Infinity"
else:
res = num1 / num2
print("Division(/):{0}/{1}={2}".format(num1,num2,res))
print("Modulus(%):{0}%{1}={2}".format(num1,num2,res))
print("Exponential(**):{0}**{1}={2}".format(num1,num2,num1**num2))
| false
|
fdc9e5dc5a169d1178cf3efa9e1f70a2f42576a1
|
rcoady/Programming-for-Everyone
|
/Class 1 - Getting Started with Python/Assignment4-6.py
| 982
| 4.375
| 4
|
# Assignment 4.6
# Write a program to prompt the user for hours and rate per hour using raw_input to
# compute gross pay. Award time-and-a-half for the hourly rate for all hours worked
# above 40 hours. Put the logic to do the computation of time-and-a-half in a function
# called computepay() and use the function to do the computation. The function should
# return a value. Use 45 hours and a rate of 10.50 per hour to test the program
# (the pay should be 498.75). You should use raw_input to read a string and float()
# to convert the string to a number. Do not worry about error checking the user input
# unless you want to - you can assume the user types numbers properly.
def computepay(h, r):
if h > 40:
overtime = h - 40
total = (40 * r) + ((overtime * r) * 1.5)
else:
total = (h * r)
return total
hrs = raw_input("Enter Hours:")
hrs = float(hrs)
rate = raw_input("Enter Rate:")
rate = float(rate)
p = computepay(hrs, rate)
print p
| true
|
e94e9bba980e60c3c6bbb96ef5c221a3852e9941
|
Shakleen/Problem-Solving-Codes
|
/Hacker Rank/Language Proficiency/Python/3. Strings/13. sWAP cASE.py
| 276
| 4.125
| 4
|
def swap_case(s):
L = list(s)
for pos in range(len(L)):
if L[pos].isalpha:
L[pos] = L[pos].lower() if L[pos].isupper() else L[pos].upper()
return ''.join(L)
if __name__ == '__main__':
s = input()
result = swap_case(s)
print(result)
| false
|
c3be60d54a88d10b85264b02689c43ffc9ec8b9e
|
Hr09/Python_code
|
/cicerciper.py
| 1,240
| 4.21875
| 4
|
message=input("Enter a message:")
key=int(input("How many Charachters should we shift (1-26)"))
secret_message=""
for char in message:
if char.isalpha():
char_code=ord(char)
char_code+=key
if char.isupper():
if char_code > ord('Z'):
char_code-=26
elif char_code<ord('A'):
char_code+=26
else:
if char_code > ord('z'):
char_code -= 26
elif char_code < ord('a'):
char_code += 26
secret_message+=chr(char_code)
else:
secret_message +=char
print("Encrypted:",secret_message)
key=-key
orig_message=" "
for char in secret_message:
if char.isalpha():
char_code=ord(char)
char_code+=key
if char.isupper():
if char_code > ord('Z'):
char_code -= 26
elif char_code < ord('A'):
char_code += 26
else:
if char_code > ord('z'):
char_code -= 26
elif char_code < ord('a'):
char_code += 26
orig_message+=chr(char_code)
else:
orig_message+=char
print("Decrypted :",orig_message)
| false
|
57abcfd1f015392cec8b05af62fdf4a1ff06aa35
|
Luisarg03/Python_desktop
|
/Apuntes_youtube/operador_in.py
| 512
| 4.15625
| 4
|
print("Eleccion de asignaturas")
print("Asignaturas disponibles: Bioquimica - Fisiologia - Salud mental")
asignatura=input("Elige la asignatura: ")
if asignatura in ("Bioquimica", "Fisiologia", "Salud mental"):
print("La asignatura elegida es "+asignatura)
else:
print("La asignatura elegida no esta contemplada")
#Python es case sensitive, para que tome las palabras como mayuscula o minuscula se usa:
#lower() transforma la palabra en minusculas
#upper() trasnforma la palabra en mayuscula
| false
|
381467c9ac0ae671463340dee5d4b34297918d14
|
Luisarg03/Python_desktop
|
/Apuntes_youtube/practica_lista.py
| 1,537
| 4.3125
| 4
|
lista1=["maria", "luis", "natalia", "juan", "marcos", "pepe"]#el indice empieza contar desde el cero, ej "luis" seria posicion 2 indice 1
print([lista1])
print(lista1[2])#indico que solo quiero que aparezca el indice 2
print(lista1[-1])#el menos hace que el indice se cuente de derecha a izquierda, en este caso solo aparece el ultimo
print(lista1[:3])#indico que solo se muestren los tres primeros indices, no hace falta por el "cero"
print(lista1[2:])#empiezo desde el indice 2 en adelante hasta el final
lista1.append("carlos")#agregar elemento a la lista, pero lo agrega al final de esta
print([lista1])
lista1.insert(4,"pedro")#inserto el elemento en el indice que indico
print(lista1[:])
lista1.extend(["fabio","lucas","leandro"])#permite agrerar varios elementos a la vez
print(lista1[:])
print(lista1.index("lucas"))#indica en que indice se encuentra el elemento
print("pepe" in lista1)#true o false, indice si el elemento esta o no dentro de la lista
print("peponi" in lista1)
lista2=["vaso", 456, 3.34, True, False, "camila"]#puedo enlistar distintos tipos de variable
lista2.remove(True)#elimino un elemento de la lista indicando el nombre correspondiente
print(lista2[:])
lista2.pop()#elimina el ultimo elemento de la lista
print(lista2[:])
lista3=["cocina", "pokemon"]
lista4=lista1+lista2+lista3#suma las listas, un concatenador es, pero crea una lista mayor por eso se usa otra variable
print(lista4[:])
print(lista1[:]*2) #el multiplicador sirve de repetidor, repite la lista la veces que se indique
| false
|
73b0b07826794a242dde684f94f946581c180949
|
Dszymczk/Practice_python_exercises
|
/06_string_lists.py
| 803
| 4.40625
| 4
|
# Program that checks whether a word given by user is palindrome
def is_palindrome(word):
return word == word[::-1]
word = "kajak" # input("Give me some word please: ")
reversed_word = []
for index in range(len(word) - 1, -1, -1):
reversed_word.append(word[index])
palindrome = True
for i in range(len(word)):
if word[i] != reversed_word[i]:
palindrome = False
if palindrome:
print("Word is palindrome")
# string based program <- much faster way
print("\n\nstring based program")
reversed_word = word[::-1]
print(reversed_word)
if reversed_word == word:
print("Word is palindrome")
# Shorter code
if word == word[::-1]:
print("\n\nWord is palindrome")
# Using function
if is_palindrome(word):
print("\n\n(3) Word is palindrome")
| true
|
6fcd96b8c44668ccac3b4150d9b6c411b325bcc0
|
mm/adventofcode20
|
/day_1.py
| 2,436
| 4.375
| 4
|
"""AoC Challenge Day 1
Find the two entries, in a list of integers, that sum to 2020
https://adventofcode.com/2020/day/1
"""
def find_entries_and_multiply(in_list, target):
"""Finds two entries in a list of integers that sum to a
given target (also an integer), and then multiply those afterwards.
"""
two_numbers = None
# We can use the idea of "complements" here. First, we'll start
# an indices dict, which will have key = the number itself, and the
# value equal to its index.
indices = {}
for i in range(0, len(in_list)):
# Get the "complement" -- for this current value, what is the
# extra value I need to add up to my target?
complement = target - in_list[i] # e.g if in_list[i] = 1995, target = 2020, => complement = 25
# Does this value exist in my list? This is where the indices dict
# comes in handy!
if complement in indices:
# Return the two numbers!
two_numbers = (in_list[i], in_list[indices[complement]])
else:
# Otherwise, add it to our indices list. It might match
# up well with another number!
indices[in_list[i]] = i
if two_numbers:
print(f"Two numbers found which add up to {target}: {two_numbers}")
return two_numbers[0]*two_numbers[1]
else:
return None
def find_three_entries_for_target(in_list, target):
"""Finds three integers which add up to a given target (also an integer),
and multiply them afterwards.
"""
three_numbers = None
# What we can do (sort of brute-force-y) is fix a number as we go through the
# list, set a new target and then run the *two-integer* version of this problem
# on the rest of the list.
for i in range(0, len(in_list)):
new_target = target - in_list[i]
# Now, perform the two-integer solution on the *rest* of the list
# (pretend this number isn't even there)
two_number_product = find_entries_and_multiply(in_list[i+1:], new_target)
if two_number_product:
return in_list[i] * two_number_product
def input_file_to_list(filepath):
num_list = []
with open(filepath, 'r') as in_file:
for line in in_file:
num_list.append(int(line.strip()))
return num_list
in_numbers = input_file_to_list('inputs/day_1.txt')
print(find_three_entries_for_target(in_numbers, 2020))
| true
|
7252e716f0bb533d1a612728caf027276883e0ef
|
git4rajesh/python-learnings
|
/String_format/dict_format.py
| 800
| 4.5625
| 5
|
### String Substitution with a Dictionary using Format ###
dict1 = {
'no_hats': 122,
'no_mats': 42
}
print('Sam had {no_hats} hats and {no_mats} mats'.format(**dict1))
### String Substitution with a List using Format ###
list1 = ['a', 'b', 'c']
my_str = 'The first element is {}'.format(list1)
print(my_str)
### List extraction
my_str = 'The first element is {0}, the second element is {1} and third element is {2}'.format(*list1)
print(my_str)
### String Substitution with a Tuple using Format ###
tuple1 = ('one', 'second', 'third')
my_str = 'The first element is {0}, the second element is {1} and third element is {2}'.format(*tuple1)
print(my_str)
### String Substitution with a String variable using Format ###
my_name = 'Rajesh'
my_str = 'Hi {0}'.format(my_name)
print(my_str)
| true
|
79618644a020eaa269bc7995d650afed3043b411
|
ravitej5226/Algorithms
|
/backspace-string-compare.py
| 1,312
| 4.15625
| 4
|
# Given two strings S and T, return if they are equal when both are typed into empty text editors. # means a backspace character.
# Example 1:
# Input: S = "ab#c", T = "ad#c"
# Output: true
# Explanation: Both S and T become "ac".
# Example 2:
# Input: S = "ab##", T = "c#d#"
# Output: true
# Explanation: Both S and T become "".
# Example 3:
# Input: S = "a##c", T = "#a#c"
# Output: true
# Explanation: Both S and T become "c".
# Example 4:
# Input: S = "a#c", T = "b"
# Output: false
# Explanation: S becomes "c" while T becomes "b".
# Note:
# 1 <= S.length <= 200
# 1 <= T.length <= 200
# S and T only contain lowercase letters and '#' characters.
class Solution(object):
def backspaceCompare(self, S, T):
"""
:type S: str
:type T: str
:rtype: bool
"""
clean_S=[]
clean_T=[]
for i in range(len(S)):
if(S[i]=='#'):
clean_S.pop() if len(clean_S)>0 else ''
else:
clean_S.append(S[i])
for i in range(len(T)):
if(T[i]=='#'):
clean_T.pop() if len(clean_T)>0 else ''
else:
clean_T.append(T[i])
return "".join(clean_S)=="".join(clean_T)
s=Solution()
print(s.backspaceCompare("ab##","c#d#"))
| true
|
08815d5371e53a75f10ec4d2b0b9bba1747a6fa6
|
ravitej5226/Algorithms
|
/zigzag-conversion.py
| 1,458
| 4.15625
| 4
|
# The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
# P A H N
# A P L S I I G
# Y I R
# And then read line by line: "PAHNAPLSIIGYIR"
# Write the code that will take a string and make this conversion given a number of rows:
# string convert(string text, int nRows);
# convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".
class Solution(object):
def convert(self, s, numRows):
"""
:type s: str
:type numRows: int
:rtype: str
"""
output=[];
if s=='':
output=''
return output
if numRows==1:
return s
for i in range(0,numRows):
start_index=i;
alt=True;
temp=0
if start_index<len(s):
output.append(s[start_index]);
else:
break;
while start_index<len(s):
if temp>0:
output.append(s[start_index]);
if(alt):
temp=2*(numRows-1-i);
else:
temp=2*(i);
start_index=start_index+temp;
alt=not alt;
return ''.join(output)
s=Solution();
print(s.convert('PAYPALISHIRING',3))
| true
|
5fa0a0ab95042208eb2bef0dc47498c34056dda6
|
arthuroe/codewars
|
/6kyu/sort_the_odd.py
| 2,963
| 4.25
| 4
|
'''
You have an array of numbers.
Your task is to sort ascending odd numbers but even numbers must be on their places.
Zero isn't an odd number and you don't need to move it. If you have an empty array, you need to return it.
Example
sortArray([5, 3, 2, 8, 1, 4]) == [1, 3, 2, 8, 5, 4]
'''
import sys
def sort_array(source_array):
print(source_array)
# Traverse through all array elements
for i in range(len(source_array)):
if source_array[i] % 2 != 0:
min_idx = i
for j in range(i, len(source_array)):
if source_array[j] % 2 != 0:
if source_array[min_idx] > source_array[j]:
min_idx = j
else:
continue
# Swap the found minimum element with the first element
source_array[i], source_array[min_idx] = source_array[min_idx], source_array[i]
print(source_array)
return source_array
sort_array([64, 25, 12, 22, 11, 5, 8, 3])
print()
sort_array([4, 25, 1, 2, 10, 7, 8, 3])
print()
sort_array([5, 3, 1, 8, 0])
print()
sort_array([5, 3, 2, 8, 1, 4])
print()
sort_array([])
'''----other solutions----'''
def sort_array(arr):
odds = sorted((x for x in arr if x % 2 != 0), reverse=True)
return [x if x % 2 == 0 else odds.pop() for x in arr]
def sort_array(source_array):
odds = iter(sorted(v for v in source_array if v % 2))
return [next(odds) if i % 2 else i for i in source_array]
def sort_array(numbers):
evens = []
odds = []
for a in numbers:
if a % 2:
odds.append(a)
evens.append(None)
else:
evens.append(a)
odds = iter(sorted(odds))
return [next(odds) if b is None else b for b in evens]
from collections import deque
def sort_array(array):
odd = deque(sorted(x for x in array if x % 2))
return [odd.popleft() if x % 2 else x for x in array]
def sort_array(source_array):
return [] if len(source_array) == 0 else list(map(int, (','.join(['{}' if a % 2 else str(a) for a in source_array])).format(*list(sorted(a for a in source_array if a % 2 == 1))).split(',')))
def sort_array(source_array):
odd = sorted(list(filter(lambda x: x % 2, source_array)))
l, c = [], 0
for i in source_array:
if i in odd:
l.append(odd[c])
c += 1
else:
l.append(i)
return l
def sort_array(source_array):
# retrieve the odd values
sorted_array = sorted([value for value in source_array if value % 2 != 0])
# insert the even numbers in the original place
for index, value in list(enumerate(source_array)):
if value % 2 == 0:
sorted_array.insert(index, value)
return sorted_array
'''
----tests-----
Test.assert_equals(sort_array([5, 3, 2, 8, 1, 4]), [1, 3, 2, 8, 5, 4])
Test.assert_equals(sort_array([5, 3, 1, 8, 0]), [1, 3, 5, 8, 0])
Test.assert_equals(sort_array([]),[])
'''
| true
|
78576dc109ab336280899a740fbc2f3797563e52
|
bryansilva10/CSE310-Portfolio
|
/Language_Module-Python/Shopping-Cart_Dictionary/cart.py
| 1,423
| 4.21875
| 4
|
#var to hold dictionary
shoppingCart = {}
#print interface
print("""
Shopping Options
----------------
1: Add Item
2: Remove Item
3: View Cart
0: EXIT
""")
#prompt user and turn into integer
option = int(input("Select an option: "))
#while user doesn't exit program
while option != 0:
if option == 1:
#add item
#prompt user for item and qty
item = input("Enter an item: ")
#if item already exists in cart
if item in shoppingCart:
print("Item already in cart")
qty = int(input("Enter quantity: "))
#update qty
shoppingCart[item] += qty
#if it does not
else:
qty = int(input("Enter quantity: "))
#add qty to the item key
shoppingCart[item] = qty
elif option == 2:
#remove item
# prompt user for itme to be removed
item = input("Enter an item: ")
#remove it
del (shoppingCart[item])
elif option == 3:
#loop through items in cart
for item in shoppingCart:
#print each item
print(item, ":", shoppingCart[item])
elif option != 0:
#in case user enters invalid option
print("Please enter a valid option")
option = int(input("\n\nSelect an option: "))
#when loop breaks and exits
else:
print("You closed the program...")
| true
|
9b72b3dd6d3aba0798f18f06ab37bdf0c39a339a
|
xu2243051/learngit
|
/ex30.py
| 858
| 4.125
| 4
|
#!/usr/bin/python
#coding:utf-8
#================================================================
# Copyright (C) 2014 All rights reserved.
#
# 文件名称：ex30.py
# 创 建 者：许培源
# 创建日期：2014年12月09日
# 描 述：
#
# 更新日志：
#
#================================================================
import sys
reload(sys)
sys.setdefaultencoding('utf8')
people = 130
cars = 40
buses = 15
if cars > people:
print "We should take the cars."
elif cars < people:
print "We should not take the cars."
else:
print "We can't decide."
if buses > cars:
print "That's too many buses."
elif buses < cars:
print "Maybe we could take the buses."
else:
print "We still can't decide."
if people > buses:
print "Alright, let's just take the buses."
else:
print "Fine, let's stay home then."
| true
|
3c4cb233be63715f662d6f81f76feae91e51ac85
|
cupofteaandcake/CMEECourseWork
|
/Week2/Code/lc1.py
| 1,787
| 4.375
| 4
|
#!/usr/bin/env python3
"""A series of list comprehensions and loops for creating sets based on the bird data provided"""
__appname__ = 'lc1.py'
__author__ = 'Talia Al-Mushadani (ta1915@ic.ac.uk)'
__version__ = '0.0.1'
__license__ = "License for this code"
birds = ( ('Passerculus sandwichensis','Savannah sparrow',18.7),
('Delichon urbica','House martin',19),
('Junco phaeonotus','Yellow-eyed junco',19.5),
('Junco hyemalis','Dark-eyed junco',19.6),
('Tachycineata bicolor','Tree swallow',20.2),
)
## Conventional loop for creating list containing birds' latin names
birds_latin = set()
for species in birds:
birds_latin.add(species[0]) #searches for and adds all of the first variables for each species, which is the latin name
print(birds_latin)
## Conventional loop for creating list containing birds' common names
birds_common = set()
for species in birds:
birds_common.add(species[1]) #searches for and adds all of the second variables for each species, which is the common name
print(birds_common)
## Conventional loop for creating list containing birds' mean body masses
birds_mean_body_mass = set()
for species in birds:
birds_mean_body_mass.add(species[2]) #searches for and adds all of the third variables for each species, which is the mean body mass
print(birds_mean_body_mass)
## List comprehension for creating list containing birds' latin names
birds_latin_lc = set([species[0] for species in birds])
print(birds_latin_lc)
## List comprehension for creating list containing birds' common names
birds_common_lc = set([species[1] for species in birds])
print(birds_common_lc)
## List comprehension for creating list containing birds' mean body masses
birds_mbm_lc = set([species[2] for species in birds])
print(birds_mbm_lc)
| true
|
0b4d452a26c2b44684c2eae50ba622c56dd97f4f
|
kriti-ixix/python2batch
|
/python/Functions.py
| 614
| 4.125
| 4
|
'''
Functions are of two types based on input:
- Default
- Parameterised
Based on return type:
- No return
- Some value is returned
'''
'''
#Function definition
def addTwo(first, second):
#first = int(input("Enter first number: "))
#second = int(input("Enter second number: "))
third = first + second
print("The sum is: ", third)
print("Addition!")
#Function calling
addTwo(5, 10)
addTwo(30, 20)
x = 40
y = 50
addTwo(x, y)
'''
def addTwo(first, second):
third = first + second
return third
x = addTwo(5, 10)
y = addTwo(20, 30)
print("The sums are ", x, " ", y)
| true
|
2bbe8852b53fe5d099afbbd8a310704c04c0423b
|
mecosteas/Coding-Challenges
|
/count_words.py
| 1,260
| 4.375
| 4
|
"""
Given a long text string, count the number of occurrences of each word. Ignore case. Assume the boundary of a word is whitespace - a " ", or a line break denoted by "\n". Ignore all punctuation, such as . , ~ ? !. Assume hyphens are part of a word - "two-year-old" and "two year old" are one word, and three different words, respectively.
Return the word counts as a string formatted with line breaks, in alphanumeric order.
Example:
"I do not like green eggs and ham,
I do not like them, Sam-I-Am"
Output:
i 2
do 2
not 2
like 2
green 1
eggs 1
and 1
ham 1
them 1
sam-i-am 1
Also Valid:
and 1
do 2
eggs 1
green 1
ham 1
i 2
like 2
not 2
sam-i-am 1
them 1
"""
def count_words(text):
word_table = {}
for word in text.split():
word = word.strip(".,~?!").lower()
if word_table.get(word) == None:
word_table[word] = 1
else:
word_table[word] += 1
return '\n'.join(f'{word} {freq}' for word, freq in word_table.items())
text = "I do not like green eggs and ham, \nI do not like them, Sam-I-Am"
# print(text)
# text_arr = text.split()
# print(text_arr)
# text_arr = [word.strip(".,~?!").lower() for word in text_arr]
# print(text_arr)
# text = ''.join(text_arr)
# print(text)
print(count_words(text))
| true
|
507e4a5ea9054c11509e8d6f678d74aa6c3f545e
|
sleepingsaint/DS-ALG
|
/DS/linkedList.py
| 2,724
| 4.21875
| 4
|
# defining stack element object
class Element(object):
def __init__(self, value):
self.value = value
self.next = None
# defining stack object
class Stack(object):
def __init__(self, head=None):
self.head = head
# helper class functions
# function to add elements
def append(self, new_element):
if self.head:
current = self.head
while current.next:
current = current.next
current.next = Element(new_element)
else:
self.head = Element(new_element)
# function to see the position of the element
def get_position_value(self, position):
current = self.head
counter = 1
if position < 1:
return None
else:
while current and counter <= position:
if counter == position:
return current.value
current = current.next
counter += 1
return None
# function to insert an element at an certain position
# Inserting at position 3 means between the 2nd and 3rd elements
def insert_element(self, index, new_element):
if self.head:
current = self.head
counter = 1
if index == 1:
e = Element(new_element)
e.next, self.head = self.head, e
elif index > 1:
while current and counter < index - 1:
current = current.next
counter += 1
if not current:
print('cannot insert the element')
else:
e = Element(new_element)
e.next, current.next = current.next, e
# function to delete an first node with given value
def delete(self, value):
if self.head:
current = self.head
previous = None
while current.value != value and current.next:
previous = current
current = current.next
if current.value == value:
if previous:
previous.next = current.next
else:
self.head = current.next
else:
return None
# initializing the stack
stck = Stack()
# adding elements
stck.append(1)
stck.append(2)
stck.append(3)
# getting position
# inserting element
stck.insert_element(2, 4)
stck.delete(3)
print(stck.get_position_value(1))
print(stck.get_position_value(2))
print(stck.get_position_value(3))
print(stck.get_position_value(4))
stck.insert_element(500, 6)
print(stck.get_position_value(1))
print(stck.get_position_value(2))
print(stck.get_position_value(3))
| true
|
d5891584688ff83c60bf74fc7611c625f57b14db
|
Sukanyacse/Assignment_2
|
/assignment 2.py
| 254
| 4.1875
| 4
|
numbers=(1,2,3,4,5,6,7,8,9)
count_even=1
count_odd=-1
for value in range(1,10):
if(value%2==0):
count_even=count_even+1
else:
count_odd=count_odd+1
print("Number of even numbers:",count_even)
print("Number of odd numbers:",count_odd)
| true
|
92ff8e67eb22909831b10bfcdd1faedef658825e
|
Lauraparedesc/Algoritmos
|
/Actividades/ejercicios/exlambda.py
| 1,291
| 4.25
| 4
|
#Exponente n de un # dado
exponente = lambda base = 0, exponente = 0 : base**exponente
resultado = exponente (5,2)
print (resultado)
#string
string = lambda cantidad = 0 : print ('♡'*cantidad)
string (60)
#maximo #
listaEdades1 = [18,12,14,13,12,20]
listaEdades2 = [19,47,75,14,12,22]
lambdamaximos = lambda x = [], y = []: print (max(x),max(y))
lambdamaximos (listaEdades1, listaEdades2)
#V o F par
numeroPar = lambda par = 0 : par % 2 == 0
print(numeroPar(23))
print(numeroPar(12))
#V o F impar
numeroImpar = lambda impar = 0 : impar % 2 != 0
print(numeroImpar(23))
print(numeroImpar(12))
#union 2 palabras
unirP = lambda word1 , word2 : word1 + ' ' + word2
print (unirP('Soy','programador'))
#saludo
preguntaNombre = 'Ingrese su nombre por favor : '
nombre = input(preguntaNombre)
saludar = lambda name = '' : print (f'Bienvenid@ {name}, disfruta el programa')
saludar(nombre)
#Letras
palabraX = 'Blessed'
lenPalabra = lambda palabra : len (palabra)
print (lenPalabra(palabraX))
#¿?
showLen = lambda funcion, palabra : print (funcion(palabra))
showLen(lenPalabra, palabraX)
#Triangulo
calcularAreaTriangulo = lambda base = 0, altura = 0: base*altura/2
print(calcularAreaTriangulo(5,8))
#IMC
imc = lambda peso = 0, altura = 1: round(peso/altura**2,3)
print (imc(48,1.58))
| false
|
e644585b9b3a99f72e3ed4fc948ecb26ca0465f0
|
moheed/python
|
/languageFeatures/python_using_list_as_2d_array.py
| 1,976
| 4.5625
| 5
|
#NOTE: creating list with comprehension creates many
#pecularities.. as python treats list with shallow copy...
#for example
arr=[0]*5 #=== with this method, python only creates one integer object with value 5 and
#all indices point to same object. since all are zero initially it doesn't matter.
arr[0]=5 #when we modify arr[0], it creates a new int object with val=5 and makes the zero-index point
#new object.
print(arr) #this works expected.
#however,
arr2=[[0]*5]*5 #expecting 5x5 matrix filled with zero
print(arr2)
arr2[0][0]=1 #expecting only 0,0 to be 1.. but o/p?
print(arr2) #[[1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0]]
#HOW all elements of first column are changed... Weired???
#This is because list works in shallow way... as shown for arr(1-d array)
#however, when we declare a 2-d arr, that means all indices in a given inner [](ie col) point to single integer object 0
#when we update by arr2[0][0]=x, what we are doing is, updating the first element of inner[](ie col)
#since all row-indices for that col point to same object, it updates all first values of given col.
"""
Similarly, when we create a 2d array as “arr = [[0]*cols]*rows” we are essentially the extending the
above analogy.
1. Only one integer object is created.
2. A single 1d list is created and all its indices point to the same int object in point 1.
3. Now, arr[0], arr[1], arr[2] …. arr[n-1] all point to the same list object above in point 2.
The above setup can be visualized in the image below.
One way to check this is using the ‘is’ operator which checks if the two operands refer to the same object.
Best Practice:
If you want to play with matrix run the LOOP and ensure all mxn objects are allocated.!
# Using above second method to create a
# 2D array
rows, cols = (5, 5)
arr=[]
for i in range(cols):
col = []
for j in range(rows):
col.append(0)
arr.append(col)
print(arr)
"""
| true
|
7cc37dc2b2888106efb3c2ee41d61bb86b511f5b
|
moluszysdominika/PSInt
|
/lab02/zad3.py
| 644
| 4.15625
| 4
|
# Zadanie 3
# Basic formatting
text1 = "Dominika"
text2 = "Moluszys"
print("{1} {0}" .format(text1, text2))
# Value conversion
class Data(object):
def __str__(self):
return "Dominika"
def __repr__(self):
return "Moluszys"
print("{0!s} {0!r}" .format(Data()))
# Padding and aligning strings
print("{:>12}" .format("Dominika"))
print("{:12}" .format("Moluszys"))
print("{:_<10}" .format("Domix"))
print("{:^11}" .format("Domix"))
# Truncating long strings
print("{:.15}" .format("konstantynopolitanczykowianeczka"))
# Combining truncating and padding
print("{:15.20}" .format("konstantynopolitanczykiewiczowna"))
| false
|
92a753e7e633025170d55b3ebdb9f2487b3c4fa0
|
HayleyMills/Automate-the-Boring-Stuff-with-Python
|
/Ch6P1_TablePrinter.py
| 1,352
| 4.4375
| 4
|
##Write a function named printTable() that takes a list of lists of strings
##and displays it in a well-organized table with each column right-justified.
##Assume that all the inner lists will contain the same number of strings.
##For example, the value could look like this:
##tableData = [['apples', 'oranges', 'cherries', 'banana'],
## ['Alice', 'Bob', 'Carol', 'David'],
## ['dogs', 'cats', 'moose', 'goose']]
##Your printTable() function would print the following:
##
## apples Alice dogs
## oranges Bob cats
##cherries Carol moose
## banana David goose
#def printTable():
#input is a list of strings
#output is an orgniased table with each column rjust
tableData = [['apples', 'oranges', 'cherries', 'banana'],
['Alice', 'Bob', 'Carol', 'David'],
['dogs', 'cats', 'moose', 'goose']]
rows = len(tableData) #number of lists
cols = len(tableData[0]) #number of cols
maxlength = 0
for i in range(0,(rows)): #loop for number of rows
for x in tableData[i]: #loop for the each item in the row
if maxlength < len(x): #find the max length of the words
maxlength = len(x)
for k in range(0, (cols)):
for v in range(0, (rows)):
print((tableData[v][k]).rjust(maxlength), end = ' ')
print()
| true
|
e317905dca19712d90a62f463a4f782bd22668e5
|
Ahmad-Magdy-Osman/IntroComputerScience
|
/Classes/bankaccount.py
| 1,195
| 4.15625
| 4
|
########################################################################
#
# CS 150 - Worksheet #11 --- Problem #1
# Purpose: Practicing User-Defined Classes.
#
# Author: Ahmad M. Osman
# Date: December 9, 2016
#
# Filename: bankaccount.py
#
########################################################################
class BankAccount:
#Bank Account Class
#Creating account and initiating balance value
def __init__(self, balance=0):
self.balance = balance
#Returning the account's current balance value
def getBalance(self):
return self.balance
#Setting the account's balance to the value of amount
def setBalance(self, amount):
self.balance = amount
#Depositing the value of amount into the account's balance
def deposit(self, amount):
print("$%.2f will be deposited." %(amount))
self.balance += amount
#Withdrawing the value of amount from the account if the balance is sufficient
def withdraw(self, amount):
if self.balance >= amount:
print("$%.2f will be withdrawn." %(amount))
self.balance -= amount
else:
print("Insufficient funds")
#Printing account balance
def __str__(self):
display = "Account Balance: " + ("$%.2f" %(self.balance))
return display
| true
|
b19db0ac2fef12e825b63552fbd0b298fcb632ec
|
Ahmad-Magdy-Osman/IntroComputerScience
|
/Turtle/ex10.py
| 859
| 4.65625
| 5
|
######################################
#
# CS150 - Interactive Python; Python Turtle Graphics Section, Exercise Chapter - Exercise 10
# Purpose: Drawing a clock with turtles
#
# Author: Ahmad M. Osman
# Date: September 22, 2016
#
# Filename: ex10.py
#
#####################################
#Importing turtle module
import turtle
#Initiating window canvas
window = turtle.Screen()
window.bgcolor("lightgreen")
#Initiating clock turtle
clock = turtle.Turtle()
clock.shape("turtle")
clock.color("blue")
clock.hideturtle()
clock.speed(0)
clock.pensize(5)
clock.pu()
clock.setposition(0, 0)
clock.stamp()
#Drawing clock
for i in range(12):
clock.left(30)
clock.forward(125)
clock.pd()
clock.forward(10)
clock.pu()
clock.forward(25)
clock.stamp()
clock.forward(-160)
#Waiting for user to exit through clocking on the turtle program screen
window.exitonclick()
| true
|
bbcefac3f0243ed8df81ed8a4875626b78ab3ca4
|
iangraham20/cs108
|
/labs/10/driver.py
| 976
| 4.5
| 4
|
''' A driver program that creates a solar system turtle graphic.
Created on Nov 10, 2016
Lab 10 Exercise 5
@author: Ian Christensen (igc2) '''
import turtle
from solar_system import *
window = turtle.Screen()
window.setworldcoordinates(-1, -1, 1, 1)
ian = turtle.Turtle()
ss = Solar_System()
ss.add_sun(Sun("SUN", 8.5, 1000, 5800))
ss.add_planet(Planet("EARTH", .475, 5000, 0.6, 'blue'))
try:
ss.add_planet(
Planet(input('Please enter the name of the planet: '),
float(input('Please enter the radius of the planet: ')),
float(input('Please enter the mass of the planet: ')),
float(input('Please enter the distance of the planet from the origin: ')),
input('Please enter the color of the planet: ')))
except ValueError as ve:
print('ValueError occurred:', ve)
except TypeError as te:
print('TypeError occurred:', te)
except:
print('Unknown Error')
#Keep the window open until it is clicked
window.exitonclick()
| true
|
0e75d78ea6d8540a5417a8014db80c0b84d32cc9
|
iangraham20/cs108
|
/projects/07/find_prefix.py
| 1,677
| 4.125
| 4
|
''' A program that finds the longest common prefix of two strings.
October 25, 2016
Homework 7 Exercise 7.3
@author Ian Christensen (igc2) '''
# Create a function that receives two strings and returns the common prefix.
def common_prefix(string_one, string_two):
''' A function that compares two strings, determines the common prefix and returns the result. '''
# Create necessary variables.
prefix_results = ''
character = 0
# Begin while loop to compare strings.
while character <= (len(string_one) - 1) and character <= (len(string_two) - 1):
# Begin if statement to compare the characters of the two strings.
if string_one[character] == string_two[character]:
# Add appropriate values to the variables.
prefix_results += string_one[character]
character += 1
# Once the prefix has been found exit the if statement.
else:
break
# Return the results and exit the function.
return prefix_results
# Create a function that runs tests on the function named common_prefix.
def test_common_prefix():
''' A function that calls common_prefix, tests the results, and prints a boolean expressions. '''
assert(common_prefix('', '') == '')
assert(common_prefix('abc123', '') == '')
assert(common_prefix('abcDefg', 'abcdefg') == 'abc')
assert(common_prefix('abcde5g', 'abcdefg') == 'abcde')
assert(common_prefix('12345f7', '1234567') == '12345')
test_common_prefix()
# Create a user input for common_prefix.
print(common_prefix(input('Please enter first string: '), input('Please enter second string: ')))
| true
|
b3e1ca7c075544d30a3a7cbd65193871f62f7a64
|
iangraham20/cs108
|
/projects/12/polygon.py
| 881
| 4.25
| 4
|
'''
Model a single polygon
Created Fall 2016
homework12
@author Ian Christensen (igc2)
'''
from help import *
class Polygon:
''' This class represents a polygon object. '''
def __init__(self, x1 = 0, y1 = 0, x2 = 0, y2 = 50, x3 = 40, y3 = 30, x4 = 10, y4 = 30, color = '#0000FF'):
''' This is the Constructor method for the Polygon class. '''
self._x1 = x1
self._y1 = y1
self._x2 = x2
self._y2 = y2
self._x3 = x3
self._y3 = y3
self._x4 = x4
self._y4 = y4
self._color = color
def render(self, canvas):
''' This render method allows the polygon object to be displayed on a screen. '''
canvas.create_polygon(self._x1, self._y1, self._x2, self._y2, self._x3, self._y3, self._x4, self._y4, fill = self._color, outline='black')
| false
|
50b03235f4a71169e37f3ae57654b7a37bcb9d10
|
adamsjoe/keelePython
|
/Week 3/11_1.py
| 246
| 4.125
| 4
|
# Write a Python script to create and print a dictionary
# where the keys are numbers between 1 and 15 (both included) and the values are cube of keys.
# create dictonary
theDict = {}
for x in range(1, 16):
theDict[x] = x**3
print(theDict)
| true
|
86e3aa25fd4e4878ac12d7d839669eda99a6ea1c
|
adamsjoe/keelePython
|
/Week 1/ex3_4-scratchpad.py
| 964
| 4.125
| 4
|
def calc_wind_chill(temp, windSpeed):
# check error conditions first
# calc is only valid if temperature is less than 10 degrees
if (temp > 10):
print("ERROR: Ensure that temperature is less than or equal to 10 Celsius")
exit()
# cal is only valid if wind speed is above 4.8
if (windSpeed < 4.8):
print("ERROR: Ensure that wind speed greater than 4.8 km/h")
exit()
# formula for windchill
v = windSpeed**0.16
chillFactor = 13.12 + 0.6215 * temp - 11.37 * v + 0.3965 * temp * v
return chillFactor
def convertToKMH(mphSpeed):
return mphSpeed * 1.609
# test values
temperature = 5
windSpeed = convertToKMH(20)
# call the fuction to calculate wind chill
wind_chill = calc_wind_chill(temperature, windSpeed)
# print a string out
print("The wind chill factor is ", wind_chill, " Celsius")
# print a pretty string out
print("The wind chill factor is ", round(wind_chill, 2), " Celsius")
| true
|
d3ab5cfe7bb1ff17169b2b600d21ac2d7fabbf70
|
adamsjoe/keelePython
|
/Week 8 Assignment/scratchpad.py
| 1,193
| 4.125
| 4
|
while not menu_option:
menu_option = input("You must enter an option")
def inputType():
global menu_option
def typeCheck():
global menu_option
try:
float(menu_option) #First check for numeric. If this trips, program will move to except.
if float(menu_option).is_integer() == True: #Checking if integer
menu_option = 'an integer'
else:
menu_option = 'a float' #Note: n.0 is considered an integer, not float
except:
if len(menu_option) == 1: #Strictly speaking, this is not really required.
if menu_option.isalpha() == True:
menu_option = 'a letter'
else:
menu_option = 'a special character'
else:
inputLength = len(menu_option)
if menu_option.isalpha() == True:
menu_option = 'a character string of length ' + str(inputLength)
elif menu_option.isalnum() == True:
menu_option = 'an alphanumeric string of length ' + str(inputLength)
else:
menu_option = 'a string of length ' + str(inputLength) + ' with at least one special character'
| true
|
4720925e6b6d133cdfaf1fd6cf5358813bbec7e3
|
george5015/procesos-agiles
|
/Tarea.py
| 460
| 4.15625
| 4
|
#!/usr/bin/python3
def fibonacci(limite) :
numeroA, numeroB = 0,1
if (limite <= 0) : print(numeroA)
elif (limite == 1) : print(numeroB)
else :
print(numeroA)
while numeroB < limite:
numeroA, numeroB = numeroB, numeroA + numeroB
print(numeroA)
return
numeroLimite = int(input("Ingrese el numero para calcular la serie de Fibonacci :"))
print ("Los numeros de fibonacci para el numero " + str(numeroLimite) + " son: ")
fibonacci(numeroLimite)
| false
|
35612faf608a4cb399dacffce24ab01767475b35
|
imsure/tech-interview-prep
|
/py/fundamentals/sorting/insertion_sort.py
| 775
| 4.15625
| 4
|
def insertion_sort(array):
"""
In place insertion sort.
:param array:
:return:
"""
n = len(array)
for i in range(1, n):
v = array[i]
k = i
j = i - 1
while j >= 0 and array[j] > v:
array[j], array[k] = v, array[j]
k = j
j -= 1
def insertion_sort_2(array):
"""
In place insertion sort. Eliminate unnecessary variable v and k.
:param array:
:return:
"""
n = len(array)
for i in range(1, n):
j = i
while j > 0 and array[j] < array[j-1]:
array[j], array[j-1] = array[j-1], array[j]
j -= 1
array = [5, 4, 3, 2, 1]
insertion_sort(array)
print(array)
array = [5, 4, 6, 7, 1]
insertion_sort(array)
print(array)
| false
|
db160f1b741178589e3b097dfdb4d46f39168350
|
Bngzifei/PythonNotes
|
/学习路线/1.python基础/day09/06-str方法.py
| 608
| 4.21875
| 4
|
"""str()就是可以自定义输出返回值,必须是str字符串"""
class Dog:
def __init__(self, name):
self.name = name
def __str__(self): # 把对象放在print()方法中输出时,就会自动调用str()方法
return '呵呵呵%s' % self.name # 只能返回字符串
# overrides method :覆盖方法 重写了
dog1 = Dog('来福')
print(dog1) # 如果将str()方法注释掉,把对象放在print中输出时,默认输出的是对象的内存地址 <__main__.Dog object at 0x0000015BF223B320>
# dog1.__str__() 这一步是自动调用了,自定义返回值
# ss = type(dog1)
# print(ss)
| false
|
e3d36397e4ba0307b0574963d8bfddbed8d12ef0
|
Bngzifei/PythonNotes
|
/学习路线/1.python基础/day10/09-多继承.py
| 1,608
| 4.34375
| 4
|
"""
C++ 和Python 支持多继承,其他语言不支持
发现规律找规律记,记住区别.初始化只是一次.再次调用init()方法就是和普通的方法调用没区别了.
多态依赖继承.
鸭子类型不依赖继承.
Python中没有重载一说,即使同名函数,方法,出现2次后,第2次的慧覆盖掉第1次的.在一个类里面只能有一个,通过修改参数来进行需要的变化.
同名属性和同名方法只会继承某个父类的一个.继承链中挨的最近的那个.
"""
class Dog:
def eat(self):
print('肉')
def drink(self):
print('水')
class God:
def fly(self):
print('lll')
def eat(self):
print('丹')
# 格式:class 类名(父类1,父类2...) 派生类:延伸的类,就是一个新的类而已 也有的叫法是 class 派生类(父类1,父类2...)
class XTQ(Dog, God):
# 如果想要控制,指定的就要重新写了.
def eat(self):
# super(Dog, self).eat() # 调的是Dog后面那个,也就是God的属性了.多继承时,看着比较混乱,所以不要这么写
God.eat(self) # 多继承时,如果调用指定父类被重写的方法,尽量直接使用指定的父类方法名去调用
xtq = XTQ()
# xtq.fly()
# xtq.drink()
xtq.eat() # 继承链:调用第一个的,有顺序的.
# print(XTQ.__mro__) # 查看指定类的继承链 (<class '__main__.XTQ'>, <class '__main__.Dog'>, <class '__main__.God'>, <class 'object'>) ,实际上注意:Dog和God是没有关系的是并列的
# print(Dog.__mro__) # (<class '__main__.Dog'>, <class 'object'>)
# print(God.__mro__) # (<class '__main__.God'>, <class 'object'>)
| false
|
d8c4501a2f890f7e63f7e08444204882e45cef5c
|
Bngzifei/PythonNotes
|
/学习路线/1.python基础/day03/01-猜拳游戏.py
| 970
| 4.1875
| 4
|
"""
石头1剪刀2布3 猜拳游戏
import random # 导入生成随机数模块
同行的注释时: #号和代码之间是2个空格,#号和注释内容之间是1个空格 这是PEP8的编码格式规范
"""
import random # 导入生成随机数模块
# print(random.randint(1,3)) # 生成1,2,3其中的某一个 以后的才是左闭右开,前小后大,只能生成整数 区间: 数学意义上的()是开区间 ,[]是闭区间.取值规则是:取闭不取开.这里的random.randint(n,m)实际上是n<= 所取得值 <= m.原因是内置函数写死了.
player_num = int(input('请出拳 石头(1)/剪刀(2)/布(3):'))
computer_num = random.randint(1, 3) # 可以在这里直接对一个变量进行随机数值的赋值运算.
if ((player_num == 1 and computer_num == 2) or
(player_num == 2 and computer_num == 3) or
(player_num == 3 and computer_num == 1)):
print('胜利!')
elif player_num == computer_num:
print('平局.')
else:
print('输了...')
| false
|
279e7cd4a60f94b9e7ff20e2d16660850f6a76cb
|
Bngzifei/PythonNotes
|
/学习路线/1.python基础/day06/01-参数混合使用.py
| 1,040
| 4.5
| 4
|
print('参数补充:')
"""
形参:位置参数(就是def func(a,b):),默认参数(就是def func(a = 5,b = 9):),可变参数(就是def func(*args):),字典类型的可变参数(就是def func(**kwargs):)
实参:普通实参,关键字参数
形参使用顺序:位置参数 -->可变参数 -->默认参数 -->字典类型可变参数
实参使用顺序:普通实参 ---> 关键字参数
"""
"""位置参数和默认参数混合使用时,位置参数应该放在默认参数的前面"""
"""位置参数和可变参数混合使用时位置参数在可变参数的前面"""
#
#
# def func1(a, *args):
# print(a)
# print(args)
#
#
# func1(1, 2, 3, 4)
"""默认参数和可变参数混合使用:默认参数应该放在可变参数的后面"""
# 就是来收没人要的实参,可变参数不能用关键字参数的方式指定
def func2(*args, a=10):
print(a)
print(args)
func2(1, 2, 3) # 10
# (1, 2, 3)
func2(1, 2, 3, a=20) # 20,如果需要重新给a赋值,需要重新给a赋值(调用的时候叫关键字参数)
# (1, 2, 3)
| false
|
5149018a67ac854d93a1c553f315b0b44858848a
|
Bngzifei/PythonNotes
|
/学习路线/1.python基础/day05/11-函数的参数补充.py
| 2,020
| 4.15625
| 4
|
"""
形参:有4种类型,位置参数,默认参数(缺省参数),可变参数(不定长 参数).字典类型可变参数
实参:2种类型 普通实参,关键字参数
"""
"""位置参数:普通参数 实参和形参位置数量必须一一对应"""
# def func1(num1, num2):
# print(num1)
#
#
# func1(10, 20)
"""
默认参数(缺省参数): 如果某个形参的值在多次使用时,都是传一样的值,我们可以把此参数改成默认参数
默认参数如果没有传递实参就用默认值,如果传递了实参就使用传递来的实参
默认参数必须放在非默认参数的后面.除非均是默认参数
"""
# def func2(a, b=2):
# print(a)
# print(b)
#
#
# func2(1) # 1,2 实参1 ,形参2(默认参数)
"""关键字参数(实参的时候):
关键字参数是实参
在实参赋值时候,给指定的形参设置指定的真实数据
如果函数中有多个默认参数时,
想给后面的那个参数设置实参,就必须以关键字参数方式进行指定
"""
def func3(a=20, b=30):
print(a)
print(b)
# func3(b=50) # 一定是这种格式
# b = 50就是给指定的形参b赋值50,函数调用的时候这里就不再是b = 30,直接都是b = 50
"""可变参数(不定长参数):可以接收任意数量的实参,并且自动组包成元组
用的不多,这个参数类型
可变参数的本质是将传递的参数包装成了元组
*args:参数名官方指定的,减少沟通成本
args:英文参数的缩写
写一点,运行一点
函数内部使用的时候不需要加*
在形参名args前面加 * 是为了告诉python解释器此参数是一个可变参数
"""
# def func4(*args,a,b,): # 这样是不对的,args会把所有的实参都接收掉,后面的a,b就不会接收到实参了
# def func4(a,b,*args):
# print(args)
#
#
# func4(1,2,3,4) # (1, 2, 3, 4)
# 后天:组包,解包补充说明
# 引用的问题
# 可变类型 - 不可变类型
# 模块的使用(.py文件的使用)
# 学生名片主逻辑
#
#
# 学生名片管理系统(150行左右)
| false
|
b36169cf1712dca276d8bc33bca89412500342d1
|
Bngzifei/PythonNotes
|
/学习路线/1.python基础/day04/09-切片.py
| 834
| 4.1875
| 4
|
"""
切片:取出字符串的一部分字符
字符串[开始索引:结束索引:步长] 步长不写,默认为1
下一个取得索引的字符 = 当前正在取得字符索引 + 步长
其他语言叫截取
步长:1.步长的正负控制字符串截取的方向 2.截取的跨度
"""
str1 = "hello python"
# print(str1[0:5])
# print(str1[:5]) # 如果开始索引为0,可以省略
# str2 = str1[6:12]
# str2 = str1[6:] # 如果结束索引最后,可以省略
# print(str2)
# list1 = [11, 12, 14]
# print(list1[-1]) # 负索引,倒着数,比较方便
#
# str2 = str1[-1:]
# print(str2)
# str2 = str1[::-1] # 倒置,翻转
# str2 = str1[4::-1] # 步长是-1,方向:右->左,跨度1
str2 = str1[4::-2] # 步长为-2,隔一个取一个
print(str2)
# 在列表,元组,字符串中都可以使用
# 可变参数.预习到这里
| false
|
258f80cccb19730b8ad4b8f525b9fcac8ab170f2
|
Bngzifei/PythonNotes
|
/学习路线/2.python进阶/day11/02-闭包.py
| 1,079
| 4.34375
| 4
|
"""
closure:闭包的意思
闭包特点:
1.>函数的嵌套定义,就是函数定义里面有另外一个函数的定义
2.>外部函数返回内部函数的引用<引用地址>
3.>内部函数可以使用外部函数提供的自由变量/环境变量 <顺序是先去找自己的位置参数,看看是否有同名,如果没有就向外扩展一层,继续这个过程.直到找到>
这就是闭包的三个特点
概念:内部函数 + 自由变量 构成的 整体 这是IBM 开发网站的一个说法
理解:内部函数 + 外部函数提供给内部函数调用的参数.
"""
def func(num1):
# 外部函数 的变量称之为自由变量/环境变量 内部函数也可以使用
print('in func', num1)
def func_in(num2):
print('in func_in', (num1 + num2))
# return func_in()# 这样就会调用func_in函数了
# 返回func_in的引用地址 外部就可以进行调用func_in函数了
return func_in
# 因为func_in函数并没有被调用
# func() # in func
# 接收外部函数func 返回值return是 内部函数func_in的引用
f = func(99)
f(100)
| false
|
cdab2900e3495440613dfeea792f09953078fa1a
|
Bngzifei/PythonNotes
|
/学习路线/1.python基础/day03/07-定义列表.py
| 530
| 4.1875
| 4
|
"""
存储多个数据,每个数据称之为元素
格式:[元素1,元素2...]
列表中尽可能存储同类型数据,且代表的含义要一致.实际上可以存储不同类型的数据
获取元素:列表[索引]
常用的标红:整理一下好复习
增.删.改.查
"""
list1 = ['11', '22', 18, 1.75, True]
print(type(list1))
print(list1[4])
l1 = list()
print(l1)
# IndexError: list index out of range
# print(list1[8]) 索引超出list范围,报错
list1[0] = '你大爷' # 修改元素的值
print(list1)
| false
|
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