blob_id string | repo_name string | path string | length_bytes int64 | score float64 | int_score int64 | text string | is_english bool |
|---|---|---|---|---|---|---|---|
7e031eb448258d9eaa83f2f7526e6a8c82d66451 | Xiankai-Wan/Python-Quiz-Summarize | /Python-语法/Part2/2-17_创建程序段4.py | 2,026 | 4.375 | 4 | # 练习：组合 scores_to_rating 函数
# 先前所有帮助函数都构建得不错！现在可以将所有这些函数组合到一起，使其成为一个整体进行运行。
#
# 在该编程练习中，将编写的每个帮助函数放入编码窗口，然后在编写函数 scores_to_rating 时使用。请注意，此函数以五个分数（可能不是数字）开始，取中间三个分数的平均值，并将该平均值转换为返回的书面评分。
#
# 需要确保将一个函数的输出正确传递到 scores_to_rating 函数体的下一个输入。一次添加一个，同时还需添加有用的注释，而且调用 print 进进行测试，将有助于控制错误！
def convert_to_numeric(score):
"""Covert the score to a float
"""
return float(score)
def sum_of_middle_three(score1,score2,score3,score4,score5):
"""
Find the sum of the middle three numbers out of the five given.
"""
max_score = max(score1,score2,score3,score4,score5)
min_score = min(score1,score2,score3,score4,score5)
sum = score1 + score2 + score3 + score4 + score5 - max_score - min_score
return sum
def score_to_rating_string(score):
rating = "Excellent"
if 0 <= score < 1:
rating = "Terrible"
elif 1 <= score < 2:
rating = "Bad"
elif 2 <= score < 3:
rating = "OK"
elif 3 <= score < 4:
rating = "Good"
else:
rating = rating
return rating
def scores_to_rating(score1,score2,score3,score4,score5):
"""
Turns five scores into a rating by averaging the
middle three of the five scores and assigning this average
to a written rating.
"""
score11 = convert_to_numeric(score1)
score22 = convert_to_numeric(score2)
score33 = convert_to_numeric(score3)
score44 = convert_to_numeric(score4)
score55 = convert_to_numeric(score5)
average_score = sum_of_middle_three(score11,score22,score33,score44,score55)/3
rating = score_to_rating_string(average_score)
return rating
| false |
6a7baeecd50abc0b4f7df9229f1cf6b4c0924ede | Xiankai-Wan/Python-Quiz-Summarize | /访问文件与网络/Part1/1-2_元组.py | 760 | 4.125 | 4 | # 练习：天和小时
# 尝试编写一个使用元组返回多个值的函数。编写 hours2days 函数，传入一个整数类型的参数，该参数表示一个以小时为单位的时间段。该函数应该返回一个元组，用天和小时为单位表示传入的时间段，不足一天的时间用小时表示。例如，39 个小时表示 1 天 15 个小时，所以函数返回的应该是 (1,15)。
#
# 这些例子演示了该函数的使用：
# >>> hours2days(24) # 24 hours is one day and zero hours
# (1, 0)
# >>> hours2days(25) # 25 hours is one day and one hour
# (1, 1)
# >>> hours2days(10000)
# (416, 16)
def hours2days(input_hours):
day = int(input_hours/24)
hour = input_hours%24
return day,hour
print(hours2days(24))
| false |
7d898892b8d486b5639235ee8e5d5a6f8f186b44 | dkbozkurt/Global_AI_Hub_Python_Programming | /Global AI Hub #1.py | 1,209 | 4.15625 | 4 | #!/usr/bin/env python
# coding: utf-8
# In[2]:
#Python as a calculator
# In[3]:
3+5
# # Functions
# In[4]:
print(4+6)
# In[6]:
print(7)
# In[5]:
print("Hello World")
# In[7]:
#commenting
# In[8]:
"""
It is just a note.
3 2 1 1 2 3
"""
# # Variable Assignment
# In[9]:
n=300
# In[10]:
n
# In[11]:
print(n)
# In[12]:
type(n)
# In[13]:
m=n
# In[14]:
print(m)
# In[16]:
m=400
# In[17]:
print(type(m))
# In[20]:
hi= "World"
# In[21]:
print(hi)
# In[22]:
print(type(hi))
# In[24]:
t=3.5
print(type(t))
# In[26]:
t= True
print(type(t))
# In[25]:
t,f=True, False
print(type(t))
# In[29]:
print(len(hi)) #veri uzunlugunu yazdırma
# In[37]:
name="Dogukan"
surname = "Bozkurt"
print(f"Hi {name} {surname}") # F-String
# # Computations
# In[38]:
5**2
# In[41]:
"35"+"67"
# In[42]:
"35"* 3
# In[43]:
"hello"*4
# In[44]:
x=10
# In[45]:
print(x+2)
# In[46]:
print(x**2) #x degisiklige ugramaz sadece etkiler.
# In[50]:
print(x%3)
# In[47]:
print(x-2)
# In[48]:
y=13
# In[49]:
print(y/2)
# In[51]:
z=5
# In[52]:
z=z+1 # = z++ , z+=1
# In[53]:
z
# In[55]:
z*=2
print(z)
| false |
f4e93b4983e001ef62240ee2f29b0eceebaf8384 | AngelBlack888/public | /raznoe/4_set_frozenset.py | 1,757 | 4.1875 | 4 |
# =================================
# Set (множество)
# =================================
print()
print('-= set =-')
# Множество — «контейнер», содержащий не повторяющиеся элементы в случайном порядке.
a = set()
print(a) # set()
b = set(['a', 'b', 'c', 'c', 'a', 'e'])
print(b)
c = set('hello')
print(c)
d = {'a', 'b', 'c', 'd'}
print(d)
e = {i ** 2 for i in range(10)} # генератор множеств
print(e)
f = {} # А так получится словарь
print(type(f)) # <class 'dict'>
# Операции с множествами
print (len(e))
print ('b' in b)
# s == t
c1 = {'e', 'l', 'o', 'h'}
print (c==c1)
# s.issubset(t) s <= t
print(c<=c1)
# s.issuperset(t) s >= t
print(c>={'h'})
# s.union(t, …) s | t
print(b | d)
# s.intersection(t, …) s & t
print(b & d)
# s.difference(t, …) s - t
print(d - b)
# s.symmetric_difference(t) s ^ t
print(d ^ b)
# s.copy()
f = e
g = e.copy()
print('e:', id(e))
print('f:', id(f))
print('g:', id(g))
# Методы, изменяющие множества
# s.update(other, …) s |= t
b |= d
print(b)
# s.intersection_update(t) s &= t
b &= d
print(b)
# s.difference_update(t) s -= t
b -= {'a', 'b'}
print(b)
# s.symmetric_difference_update(t) s ^= t
b ^= c
print(b)
# s.add(elem)
b.add(1)
print(b)
# s.remove(elem)
b.remove('h')
print(b)
# b.remove('z')
# s.discard(elem)
b.discard(1)
print(b)
b.discard('z') # ошибки не возникает
# s.pop()
print(b.pop())
print(b)
# s.clear()
b.clear()
print(b)
# frozenset
a = frozenset('hellow')
b = set('hellow')
print(a==b)
print(type(a-b))
print(type(b|a))
b.add('q')
# a.add('q') # вызовет ошибку
| false |
db152888584f85403424f21c0bb6ef79efb75caf | mihau1987/Python_basics | /Trello_Zadania/03_Instrukcje_Warunkowe/02_Zadania_dodatkowe/Zadanie_slodycze.py | 1,215 | 4.15625 | 4 | slodycze = ['paczek', 'drozdzowka', 'gniazdko', 'jagodzianka', 'szarlotka', 'muffin']
'''question = input("Czego sobie życzysz? ")
if question in slodycze:
print("Podany produkt jest dostepny")
else:
print("Niestety nie posiadamy artykulu na stanie")
q1 = input("Podaj produkt no 1: ")
q2 = input("Podaj produkt no 2: ")
q3 = input("Podaj produkt no 3: ")
if q1 in slodycze:
print("Podane produkty sa dostepne")
else:
print("Niestety nie posiadamy artykulu na stanie")
if q2 in slodycze:
print("Podane produkty sa dostepne")
else:
print("Niestety nie posiadamy artykulu na stanie")
if q3 in slodycze:
print("Podane produkty sa dostepne")
else:
print("Niestety nie posiadamy artykulu na stanie")
q1 = input("Podaj produkt no 1: ")
q2 = input("Podaj produkt no 2: ")
q3 = input("Podaj produkt no 3: ")
if q1 and q2 and q3 in slodycze:
print("Podane produkty sa dostepne")
else:
print("Niestety nie posiadamy produktow na stanie")'''
q = input("Podaj nazwę produktu a dowiesz sie ile sztuk posiadamy: ")
if q in slodycze:
print("Na stanie mamy 12 sztuk podanego produktu")
else:
print("Niestety w tym momencie nie posiadamy produktu na stanie")
| false |
04615dd74772aafcd67fe47862bb41368f428b8c | mihau1987/Python_basics | /Trello_Zadania/03_Instrukcje_Warunkowe/02_Zadania_dodatkowe/Exercise_01.py | 1,189 | 4.125 | 4 | '''
Napisz program, który zawiera listę zwierząt.
Poproś użytkownika o 3 zwierzęta, a następnie sprawdz czy:
- wprowadzone napisy są zwierzętami z listy
- zawierają taką samą liczbę liter
- czy zawierają taka samą liczbę liter A
- czy kończą się na tę samą literę
'''
#1. Tworzę listę zwierząt
#2. Tworzę input z zapytaniem
#3. Tworzę warunek który sprawdza czy napisy są zwierzętami z listy
animals = ["cat", "dog", "aligator", "anaconda", "albatros", "lemur", "amur"]
ani_1 = input("Enter animal no 1 name: ")
ani_2 = input("Enter animal no 2 name: ")
ani_3 = input("Enter animal no 3 name: ")
if ani_1 and ani_2 and ani_3 in animals:
print("Animals are in the list")
if len(ani_1) == len(ani_2) == len(ani_3):
print("Lenght is the same")
if ani_1.count("a") and ani_2.count("a") and ani_3.count("a"):
print("They all have letter A")
if ani_1[-1] == ani_2[-1] == ani_3[-1]:
print("They all end's with the same letter")
else:
print("These animals are not on the list")
#Coś mi się wydaje że skopałem temat inputów, może powinny być oddzielone przecinkiem, czyli metoda split się kłania?
| false |
65b15a041ebd8a1cfc5ff1ca227bf741a9ee45c7 | mihau1987/Python_basics | /Udemy/Kurs_Rafal_Modulo/Ify/Calculator_exercise.py | 437 | 4.15625 | 4 | choice = input("* - multiplication, / - division, + - adding, - - subtrackt, ** - exponentiation: ")
a = int(input("First: "))
b = int(input("Second: "))
if (choice == "*"):
print(a*b)
elif (choice == "/"):
if (b == 0):
print("Dont you dare!")
else:
print(a/b)
elif (choice == "+"):
print(a+b)
elif (choice == "-"):
print(a-b)
elif (choice == "**"):
print(a**b)
else:
print("Wrong choice!")
| false |
8dab454b9b2d3b9fa8e1c7979b921cce2ac269c2 | Tcrumy/Startin-Out-with-Python-2nd-Edition-by-Tony-Gaddis | /Chapter 6/Programming Challenges/1. Feet to Inches.py | 648 | 4.4375 | 4 | # One foot equals 12 inches. Write a function named feet_to_inches that accepts
# a number of feet as an argument, and returns the number of inches in that many
# feet. Use the function in a program that prompts the user to enter a number of
# feet and then displays the number of inches in that many feet.
def main():
feet = float(input("Enter an amount in feet: "))
inches = feet_to_inches(feet)
print(feet, "Feet is", inches, "inches")
# This function will accept a float argument for feet
# It will return the amount of inches in that many feet
def feet_to_inches(feet):
return feet * 12
main()
| true |
22d557acd05c452e8efa0c2a8e5ff15671079373 | Tcrumy/Startin-Out-with-Python-2nd-Edition-by-Tony-Gaddis | /Chapter 7/Programming Challenges/6. Average Numbers.py | 889 | 4.53125 | 5 | # Assume that a file containing a series of integers is named numbers.txt and
# exists on the computer’s disk. Write a program that calculates the average of
# all the numbers stored in the file.
def main():
# Open the file
num_file = open("numbers.txt")
# Read the first line of the file
line = num_file.readline()
# Initialize an accumulator for the sum of the integers in the file
# and an accumulator to the total quantity of numbers stored in the file
sum = 0
num_of_nums = 0
while line != "":
# Convert the number string into an int
num = int(line)
# Add the number to sum and add 1.0 to num_of_nums
sum += num
num_of_nums += 1.0
# Read the next line
line = num_file.readline()
# Display the sum
print("Average:", sum/num_of_nums)
main()
| true |
aba55620a2f212e5e180726919191a05cd7e97f4 | Tcrumy/Startin-Out-with-Python-2nd-Edition-by-Tony-Gaddis | /Chapter 5/Programming Challenges/5. Average Rainfall.py | 1,292 | 4.59375 | 5 | # Write a program that uses nested loops to collect data and calculate the
# average rainfall over a period of years. The program should first ask for the
# number of years. The outer loop will iterate once for each year. The inner
# loop will iterate twelve times, once for each month. Each iteration of the
# inner loop will ask the user for the inches of rainfall for that month. After
# all iterations, the program should display the number of months, the total
# inches of rainfall, and the average rainfall per month for the entire period.
# Prompt user for number of years
years = int(input("Enter how many years of data was collected: "))
# Define accumulator variable for rainfall total
total_rainfall = 0
# Calculate the number of months of data collected
months_of_data = 12 * years
# Write for loop to perform ranfall calculations
for year in range(years):
for month in range(1,13):
print("Enter rainfall data for year", year + 1, "month", month, ": ", end = "")
rainfall_this_month = float(input())
total_rainfall += rainfall_this_month
print(months_of_data, "Months of data have been collected.")
print("Total rainfall: ", total_rainfall)
print("Month Average: ", total_rainfall / months_of_data)
| true |
5540af3451ef8967f49af825150fc02b574ef60f | Tcrumy/Startin-Out-with-Python-2nd-Edition-by-Tony-Gaddis | /Chapter2/Programming Challenges/6. Sales Tax.py | 1,410 | 4.21875 | 4 | # Write a program that will ask the user to enter the amount of a purchase.
# The program should then compute the state and county sales tax. Assume the
# state sales tax is 4 percent and the county sales tax is 2 percent. The
# program should display the amount of the purchase, the state sales tax, the
# county sales tax, the total sales tax, and the total of the sale(which is the
# sum of the amount of purchase plus the total sales tax).
# Hint: use the value 0.02 to represent 2 percent, and 0.04 to represent 4 percent.
# Create two variables to hold the county and state sales tax rate respectivly
county_tax_rate = .02
state_tax_rate = .04
# Prompt user to enter amount of purchase. This will be the subtotal since
# only one item/purchase amount is being asked for
subtotal = float(input("Enter Purchase Amount: "))
# Calculate county and state tax
county_tax = subtotal * county_tax_rate
state_tax = subtotal * state_tax_rate
# Calculate total sales tax
total_sales_tax = county_tax + state_tax
# Calculate total
total = subtotal + county_tax + state_tax
# Display information to the user formated to 2 decimal places
print("Subtotal: ", format(subtotal, ",.2f"))
print("State tax: ", format(state_tax, ",.2f"))
print("County tax:", format(county_tax, ",.2f"))
print("Total tax: ", format(total_sales_tax, ",.2f"))
print("Total: ", format(total, ",.2f"))
| true |
4c3bb32c9e3fd296b507f7f6b173c6e9f36dd0dd | Tcrumy/Startin-Out-with-Python-2nd-Edition-by-Tony-Gaddis | /Chapter 7/Programming Challenges/7. Random Number File Writer.py | 626 | 4.28125 | 4 | # Write a program that writes a series of random numbers to a file. Each random
# number should be in the range of 1 through 100. The application should let the
# user specify how many random numbers the file will hold.
import random
def main():
rand_num_file = open("randomNumbers.txt","w")
# Ask user how many random numbers to generate
nums_to_generate = int(input("Enter the quantity of numebrs to generate: "))
# Write user requested quantity of random numbers to file
for num in range(nums_to_generate):
rand_num_file.write(str(random.randint(0,100)) + "\n")
main()
| true |
34fca6b8c74ac727ee54a98947ec783ffb0090b9 | Tcrumy/Startin-Out-with-Python-2nd-Edition-by-Tony-Gaddis | /Chapter 4/Programming Challenges/4. Magic Dates.py | 1,030 | 4.53125 | 5 | # The date June 10, 1960, is special because when it is written in the
# following format, the month times the day equals the year:
# 6/10/60
# Design a program that asks the user to enter a month (in numeric form),
# a day, and a two digit year. The program should then determine whether the
# month times the day equals the year. If so, it should display a message saying
# the date is magic. Otherwise, it should display a message saying the date is
# not magic.
# Prompt user for month as a digit 1-12
month = int(input("Enter the month as a digit 1-12: "))
# Prompt user for day as a digit 1-31
day = int(input("Enter the day as a digit 1-31: "))
# Prompt the user for the year as a two digit number
year = int(input("Enter the year as a two digit number ex.(1960 = 60): "))
# Use an if-else structure to determine if
# the date is a magic date and display this
# to the user
if month * day == year:
print("The date is magic.")
else:
print("The date is not magic.")
| true |
0af906a373f8a62f8bbd15215ca64ae3a4e2396f | Tcrumy/Startin-Out-with-Python-2nd-Edition-by-Tony-Gaddis | /Chapter 4/Programming Challenges/2. Areas of Rectangles.py | 1,075 | 4.4375 | 4 | # The area of a rectangle is the rectangle’s length times its width. Write a
# program that asks for the length and width of two rectangles. The program
# should tell the user which rectangle has the greater area, or if the areas
# are the same.
# Prompt user for length and width of the first rectangle
length1 = int(input("Enter the length of rectangle 1: "))
width1 = int(input("Enter the width of rectangle 1: "))
# Prompt user for length and width of the second rectangle
length2 = int(input("Enter the length of rectangle 2: "))
width2 = int(input("Enter the width of rectangle 2: "))
# Calculate the areas of each triangle
area1 = length1 * width1
area2 = length2 * width2
# Use an if else structure to determing which
# rectangle has the largest area, and display
# that information to the user
if area1 > area2:
print("Rectangle 1 is larger.")
elif area2 > area1:
print("Rectangle 2 is larger")
# If this else is executed, the rectangles are the same size
else:
print("Rectangle 1 and rectangle 2 are the same size")
| true |
4dd50af0c14fa9da080ea1cd951f1d4960c224e9 | Tcrumy/Startin-Out-with-Python-2nd-Edition-by-Tony-Gaddis | /Chapter 5/Algorithm Workbench/8.py | 324 | 4.46875 | 4 | # Write code that prompts the user to enter a positive nonzero number and
# validates the input.
number = int(input("Enter a positive nonzero number: "))
# Validate the input
while number <= 0:
print("Number was not positive/nonzero, try again")
number = int(input("Enter a positive nonzero number: "))
| true |
dce0e4734a850a9ed0ee2b0fe18171c7fee1a7ca | Tcrumy/Startin-Out-with-Python-2nd-Edition-by-Tony-Gaddis | /Chapter 5/Algorithm Workbench/9.py | 330 | 4.25 | 4 | # Write code that prompts the user to enter a number in the range of 1 through
# 100 and validates the input.
number = int(input("Enter a number 1-100: "))
# Validate the input
while number < 1 or number > 100:
print("Number was not in range 1-100, try again")
number = int(input("Enter a number 1-100: "))
| true |
24c43eda8a7fb649df6af835e378411bd86ae44b | 99003760/coding_writing | /Hackerrank/Python/Math.py | 1,356 | 4.21875 | 4 | #! /usr/bin/env python
# -*- coding: utf-8 -*-
# @Author: xuezaigds@gmail.com
# @Last Modified time: 2016-05-04 20:42:07
'''
You are given a complex z. Your task is to convert it to polar coordinates.
'''
import cmath
com = raw_input()
pha = cmath.phase(complex(com))
r = abs(complex(com))
print "{0}\n{1}".format(r, pha)
'''
ABC is a right triangle, 90° at B. Therefore, ∡ABC=90°.
Point M is the midpoint of hypotenuse AC.
You are given the lengths AB and BC.
Your task is to find ∡MBC in degrees.
'''
import math
AB = int(raw_input())
BC = int(raw_input())
print str(int(round(math.degrees(math.atan2(AB, BC))))) + '°'
'''
You are given a positive integer N.
Your task is to print a palindromic triangle of size N.
For example, a palindromic triangle of size 5 is:
1
121
12321
1234321
123454321
You can't take more than two lines.
The first line (a for-statement) is already written for you.
You have to complete the code using exactly one print statement.
'''
for i in range(1, int(raw_input()) + 1):
print ((10 ** i - 1) // 9)**2
'''
You are given a positive integer N.
Print a numerical triangle of height N−1 like the one below:
1
22
333
4444
55555
......
Can you do it using only arithmetic operations,
a single for loop and print statement?
Use no more than two lines.
'''
for i in range(1, input()):
print ((10 ** i - 1) // 9) * i
| true |
d1fbcca104542dd0bba86c0de5028783220d88c0 | salvadorraymund/Python | /Algorithms/Sort/merge_sort.py | 1,797 | 4.125 | 4 | from random import randint
from timeit import repeat
def run_sorting_algo(algorithm, array):
setup_code = f"from __main__ import {algorithm}" \
if algorithm != "sorted" else ""
stmt = f"{algorithm}({array})"
times = repeat(setup=setup_code, stmt=stmt, repeat=3, number=10)
print(f"Algorithm: {algorithm}. Minimum execution time:{min(times)}")
def merge(left, right):
if len(left) == 0:
return right
if len(right) == 0:
return left
result = []
index_left = index_right = 0
# instead of using while len(result) < len(left) + len(right)
while True:
if left[index_left] <= right[index_right]:
result.append(left[index_left])
index_left += 1
else:
result.append(right[index_right])
index_right += 1
if index_right == len(right):
result += (left[index_left:])
break
if index_left == len(left):
# instead of using append, the remaining items in left/right were added to the list result
result += (right[index_right:])
# if break will be removed, index_left and index_right will continuously increase, thereby causing list index
# to be out of range
break
return result
def merge_sort(array):
if len(array) < 2:
return array
midpoint = len(array) // 2
left = array[:midpoint]
right = array[midpoint:]
# print(left)
# print(right)
return merge(left=merge_sort(array[:midpoint]), right=merge_sort(array[midpoint:]))
array = [8, 2, 4, 6, 5]
print(merge_sort(array))
array_length = 10000
if __name__ == "__main__":
array = [randint(0, 1000) for i in range(array_length)]
run_sorting_algo(algorithm="merge_sort", array=array) | true |
762c57193b924ce19bf45889c826cbdee91489af | salvadorraymund/Python | /Real Python/Web Scraping/reg_ex_primer.py | 1,391 | 4.375 | 4 | import re
# search for the pattern ab*c in the expression ac
# Regular expression ab*c matches any part of the string that begins with an "a"
# and ends with a "c", and has zero instances of "b" between them
# print(re.findall("ab*c", "ac"))
# print(re.findall("ab*c", "abcd"))
# print(re.findall("ab*c", "acc"))
# print(re.findall("ab*c", "abac"))
# print(re.findall("ab*c", "abdc"))
pattern = "ab*c"
# Pattern matching is case sensitive
ex1 = re.findall(pattern, "ABC")
print(ex1)
# pass in re.IGNORECASE as third argument to ignore case sensitivity
ex1 = re.findall(pattern, "ABC", re.IGNORECASE)
print(ex1)
# using period(.) to stand for any single character
# in the example below, finding all the strings that contains the letter "a" and "c" separated by a single character
ex2 = re.findall("a.c", "abc")
print(ex2)
# period(.) can only stand for 1 character
ex3 = re.findall('a.c', 'abbc')
print(ex3)
# The pattern .* inside a regular expression stands for any character repeated any number of times.
pattern2 = "a.*c"
ex4 = re.findall(pattern, "abc")
print(ex4)
ex5 = re.findall(pattern, "abbc")
print(ex5)
match_results = re.search(pattern, 'ABC', re.IGNORECASE)
print(match_results.group())
string = "Everything is <replaced> if it's in <tags>"
my_string = re.sub("<.*>", "ELEPHANTS", string)
print(my_string)
string2 = re.sub("<.*?>", "ELEPHANTS", string)
print(string2)
| true |
b55342707183858f4babd1aa9a54d888c6ef24d8 | salvadorraymund/Python | /Dates and Timels/formatting.py | 553 | 4.1875 | 4 | from datetime import datetime
# %Y/%y - Year, %A/%a - weekday, %B/%b - month, %d - day of month
now = datetime.now()
print(now.strftime("The current year is: %Y"))
print(now.strftime("%d, %a %B, %y"))
# %c - local date and time, %x - local date, %X - local time
print(now.strftime("Local date and time: %c"))
print(now.strftime("Local date: %x"))
print(now.strftime("Local time: %X"))
# %I/%H - 12/24 hour, %M - minute, %S - second
# %p - locale's AM/PM
print(now.strftime("Current time: %I:%M:%S %p"))
print(now.strftime("Current time: %H:%M:%S"))
| false |
1e3375fc53ea8c23a7433c75d87f4c4a929ee28d | tushar-rishav/Algorithms | /Archive/Contests/Codechef/Contest/Others/CodeStorm/grade.py | 490 | 4.125 | 4 | #! /usr/bin/env python
def power(base, exponent):
remaining_multiplicand = 1
result = base
while exponent > 1:
remainder=exponent % 2
if remainder > 0:
remaining_multiplicand = remaining_multiplicand * result
exponent = (exponent - remainder) / 2
result = result * result
return result * remaining_multiplicand
def main():
t=input()
while t:
n=input()
res=power(7,n)%(10**9+7)
print res
t-=1
if __name__=="__main__":
main()
| true |
70f2c7824b157cfe8ca27f0610e5bba00feb488c | jgnotts23/CMEECourseWork | /Week2/Code/lc1.py | 1,459 | 4.59375 | 5 | #!/usr/bin/env python3
""" A demonstration of the use of loops and list comprehensions
to output data from a tuple of tuples """
__appname__ = 'lc1.py'
__author__ = 'Jacob Griffiths (jacob.griffiths18@imperial.ac.uk)'
__version__ = '0.0.1'
### Constants ###
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),
)
### Functions ###
# List comprehensions
# Latin names
print("Bird data imported:")
print(birds)
print("\nFinding latin names with list comprehension...")
ln_lc = set([item[0] for item in birds])
print(ln_lc)
# Common names
print("\nFinding common names with list comprehension...")
cn_lc = set([item[1] for item in birds])
print(cn_lc)
# Mean body masses
print("\nFinding mean body masses with list comprehension...")
mbm_lc = set([item[2] for item in birds])
print(mbm_lc)
# Conventional loops
# Latin names
print("\nFinding latin names with a loop...")
ln_loop = set()
for values in birds:
ln_loop.add(values[0])
print(ln_loop)
# Common names
print("\nFinding common names with a loop...")
cn_loop = set()
for values in birds:
cn_loop.add(values[1])
print(cn_loop)
# Mean body masses
print("\nFinding mean body masses with a loop...")
mbm_loop = set()
for values in birds:
mbm_loop.add(values[2])
print(mbm_loop) | true |
79c943c6d2aacb349e7440bb35e8a8ee32c4bfb9 | the-hobbes/misc | /smallest_snippet.py | 2,571 | 4.125 | 4 | #!/usr/bin/python
'''
Smallest snippet problem:
Find the smallest snippet that contains all the search key words. For example:
input_list = ['a', 'car', 'has', 'a', 'dog']
search_words = ['a', 'dog', 'has']
Possible output:
[0, 4]
[2, 4]
Desired output (the shortest snippet) is [2, 4]
'''
input_list = ['a', 'car', 'has', 'a', 'dog']
search_words = ['a', 'dog', 'has']
def exponential_solution():
# this is the naive solution, that has O(n^2) running time
solution = []
shortest_result = [-1, -1]
shortest_length = -1
for i in range(len(input_list)):
copy = search_words[:]
if input_list[i] in copy:
shortest_result[0] = i # save away the first hit we've got
copy.remove(input_list[i]) # and remove it from the list words to look for, since we've already found it
for j in range(i, len(input_list)): # now, look through the rest of the list,
if input_list[j] in copy: # we've located another keyword
copy.remove(input_list[j]) # remove the keyword from the copied search terms
shortest_result[1] = j # update the ending bound of the snippet
if len(copy) < 1:
# we've found all the values we are searching for, now we see if this snippet is the shortest we've found so far
length_of_snippet = shortest_result[1] - shortest_result[0]
if shortest_length == -1 or length_of_snippet <= shortest_length:
shortest_length = length_of_snippet
solution.append(shortest_length)
print 'Length of smallest snippet: ' + str(shortest_length)
print 'The start and end of smallest snippet: ' + str(sorted(solution))
def linear_solution():
# this is the clever solution, with O(n) running time
indecies = {}
for item in search_words: # make a dictionary of the search terms
indecies[item] = -1
for i in range(len(input_list)): # each time we find a search term, update its position in the dictionary
if input_list[i] in indecies:
indecies[input_list[i]] = i
# now we have the positions of the shortest snippet, we need to pick out the max and min
min_index = min(indecies.values())
max_index = max(indecies.values())
print 'Length of smallest snippet: ' + str(max_index - min_index)
print 'The start and end of smallest snippet: [%d, %d]' % (min_index, max_index)
def main():
print '------------Exponential Solution------------'
exponential_solution()
print '\n------------Linear Solution------------'
linear_solution()
if __name__ == '__main__':
main() | true |
f530c79e55ad41914a21202de56aaf1a2f8ad028 | Sourabhsk870/5TH-SEM-NOTES | /18CS55/Sample Programs/Code/prog2.py | 225 | 4.3125 | 4 | # WAP TO READ TWO NUMBERS AND FIND THE LARGEST ONE
print("Enter two numbers:")
a = int (input() )
b = int (input() )
if a == b:
print("Both are equal")
else:
if a<b:
print(b," is greater")
else:
print(a," is greater")
| false |
3d8001d043f311922c572254e4c52884969b5e59 | vivekkhimani/DataStructuresAndAlgorithmsPractice | /StacksAndQueues/Applications_And_Interview/ThreeInOne.py | 2,404 | 4.28125 | 4 | #Implementing three stacks using a single array (similar implementation can be used for k stacks in one array. The number of elements n in an array would have to be divided with k, which will give size of individual stack)
#We will use in-built Python list objects as an array
#Initializing an array of fixed size 20 just for the sake of convenience
#We will assume it just to be an integer stack
stackArray = [0]*18
#Next we define sizes for each of the stack
#Let's assume the size of each is 6, therefore, the first stack is allocated indices 0-5, the second stack is allocated indices 6-11, third one is 12-17!
class Stack1:
def __init__(self,pointer=0):
self.__pointer = pointer
def getPointer(self):
return self.__pointer
def incPointer(self):
self.__pointer += 1
def decPointer(self):
self.__pointer -= 1
def isEmpty(self):
return (self.__pointer == 0)
def isFull(self):
return (self.__pointer == (0 + len(stackArray)/3))
def push(self,newData):
if self.isFull():
print("Overflow")
return
stackArray[self.__pointer] = int(newData)
self.incPointer()
def pop(self):
if self.isEmpty():
print("Underflow")
return
retVal = stackArray[self.__pointer-1]
stackArray[self.__pointer] = 0
self.decPointer()
return retVal
def returnTop(self):
return stackArray[self.__pointer-1]
def __str__(self):
i = 0
retString = ""
while i < self.__pointer:
retString+=str(stackArray[i])+"->"
i+=1
return retString+"NULL"
#The remaining two Stacks can be implemented as different classes with same methods in same manner. The only difference will be the starting point of the pointer and the ending point. Remember, the ending point of the previous stack will be the starting point of the next one, i.e.
#Stack1 => Start = 0, End = len(arrayStack)/3 - 1
#Stack2 => Start = len(arrayStack)/3, End = (len(arrayStack)/3 * 2) - 1
#Stack3 => Start = len(arrayStack)/3 * 2, End = len(arrayStack) - 1
#The same logic can be applied for n stacks in an array!
if __name__ == "__main__":
firstStack = Stack1()
firstStack.push(1)
firstStack.push(2)
print(firstStack)
firstStack.pop()
print(firstStack)
firstStack.push(3)
firstStack.push(10)
firstStack.push(12)
firstStack.push(13)
firstStack.push(15)
print(firstStack.isEmpty())
print(firstStack.isFull())
firstStack.push(14)
print(firstStack)
print(firstStack.returnTop())
| true |
b3d8558939540fed47f84811116d3cb43b71871f | vivekkhimani/DataStructuresAndAlgorithmsPractice | /LinkedLists/Implementation_And_Operations/CircularLinkedList.py | 2,263 | 4.15625 | 4 | #This document demonstrates how to construct a Circular List using Python and perform various operations on the Linked List.
class Node:
def __init__(self,data=None,nextNode=None):
self.__data = data
self.__nextNode = nextNode
def getNext(self):
return self.__nextNode
def getData(self):
return self.__data
def setNext(self,newNext):
self.__nextNode = newNext
def setData(self,newData):
self.__data = newData
def __str__(self):
return str(self.getData())
class CircularLinkedList:
def __init__(self,head=None):
self.__head = head
def insertHead(self,Data):
self.__head = Node(Data)
self.__head.setNext(self.__head)
def insertBeforeHead(self,Data):
current = self.__head
while current.getNext() != self.__head:
current = current.getNext()
current.setNext(Node(Data))
current = current.getNext()
current.setNext(self.__head)
self.__head = current
def insertEnd(self,Data):
current = self.__head
while current.getNext() != self.__head:
current = current.getNext()
current.setNext(Node(Data))
current = current.getNext()
current.setNext(self.__head)
def deleteEnd(self):
current = self.__head
while current.getNext().getNext() != self.__head:
current = current.getNext()
current.setNext(self.__head)
def deleteHead(self):
current = self.__head
while current.getNext() != self.__head:
current = current.getNext()
current.setNext(self.__head.getNext())
self.__head = self.__head.getNext()
def countElements(self):
if self.__head == None:
return str(0)
current = self.__head
counter = 1
while current.getNext() != self.__head:
current = current.getNext()
counter+=1
return str(counter)
def printList(self):
current = self.__head
printString = ""
while current.getNext() != self.__head:
printString+=str(current) + " -> "
current = current.getNext()
printString+=str(current) + " ->" + str(current.getNext())+"(HEAD)"
return printString
#Testing
if __name__ == "__main__":
myList = CircularLinkedList()
myList.insertHead(5)
myList.insertEnd(10)
myList.insertEnd(20)
myList.insertEnd(50)
myList.insertBeforeHead(39)
myList.deleteEnd()
myList.deleteHead()
print(myList.countElements())
print(myList.printList()) | true |
0732767b7f22c3554fd22da9738a6bbfd9ddc3a2 | Awesome94/Algo_Python | /stringRotation.py | 884 | 4.21875 | 4 | def stringRotation(string1, string2):
n = len(string1)
indx = 0
if n != len(string2): return False
for i in range(n):
if string2.count(string1) == 1:
break
indx+=1
return string1[:indx]+string2[:-indx] == string1
# print(stringRotation("waterbottle", "erbottlewat"))
def areRotations(string1, string2):
size1 = len(string1)
size2 = len(string2)
temp = ''
if size1 != size2:
return 0
temp = string1 + string1
print(temp, string2, temp.count(string2))
if (temp.count(string2)> 0):
return 1
else:
return 0
# Driver program to test the above function
string1 = "ACCD"
string2 = "ACDA"
if areRotations(string1, string2):
print("Strings are rotations of each other")
else:
print("Strings are not rotations of each other")
# This code is contributed by Bhavya Jain
| true |
6ee56447f2e591c445356f8dacac6d00092f3e38 | Awesome94/Algo_Python | /selectionSort.py | 399 | 4.15625 | 4 | def selection(array):
"""
selection sort finds min index and swaps the numbers in the array accordingly
"""
for x in range(len(array)):
min_index = x
if array[min_index] > array[x]:
min_index = x
array[min_index], array[x] = array[x], array[min_index]
return array
array = [1, 12, 4, 21, 8, 42, 121]
print(selection(array)) | true |
543a6851d238ea4f34581d20bc0e9f91b4cf936c | RViveka/simple_prog | /max_num.py | 343 | 4.25 | 4 | def max_num(num1,num2,num3):
if num1 >= num2 and num1>=num3:
print(num1,"is the max numnber")
elif num2>=num3:
print(num2,"is the max number")
else:
print(num3,"is the max number")
a=int(input("Enter the number1"))
b=int(input("Enter the number2"))
c=int(input("Enter the number3"))
max_num(a,b,c) | true |
2c8e30c5efea42cef2df3eeb0098de905236fcae | bhaskarmamillapalli/Python_Examples | /Python_Practice/Strings/String_Functions.py | 407 | 4.21875 | 4 | # this program is to concatenate the strings
s="Hello World"
s1="Guido Van Rossum"
print(s)
print(s1)
print("Concatenated string::",s+" "+s1)
# Repetition -- multiplication of string.
"letter z is printed 10 times."
letter ="z"
print(letter*10)
'String built in methods.'
print(s.upper())
print(s.lower())
print(s.split())
print(s1.split())
print("Split the string with letter o",s.split("o"))
print(s1.split("o"))
| true |
8f4ffbb8ef3547047ab10fed8da1e1303fa0ee62 | meifu2027/PythonLearn | /20180815-03/conditional.py | 1,245 | 4.4375 | 4 | # 条件判断
age = 20
if age >= 18:
print("your age is", age)
print("adult")
age = 3
if age >= 18:
print("your age is", age)
print("adult")
else:
print("teenage")
age = 10
if age >= 18:
print("your age is", age)
print("adult")
elif age >=6:
print("child")
else:
print("teenage")
# 只要x是非零数值、非空字符串、非空list等，就判断为True，否则为False。
x = 10
if x:
print("true")
# input()返回的数据类型是str，str不能直接和整数比较，
# 必须先把str转换成整数。
# Python提供了int()函数来完成这件事情
s = input("birth:")
birth = int(s)
if birth < 2000:
print("00前")
else:
print("00后")
# practice
# 小明身高1.75，体重80.5kg。请根据BMI公式（体重除以身高的平方）帮小明计算他的BMI指数，并根据BMI指数：
# 低于18.5：过轻
# 18.5-25：正常
# 25-28：过重
# 28-32：肥胖
# 高于32：严重肥胖
height = 1.75
weight = 80.5
bmi = weight / (1.75 * 1.75)
print(bmi)
if bmi < 18.5:
print("过轻")
elif bmi >= 18.5 and bmi < 25:
print("正常")
elif bmi >= 25 and bmi < 28:
print("过重")
elif bmi >= 28 and bmi < 32:
print("肥胖")
else:
print("严重肥胖") | false |
acfb8546dc2e6b4c62f2ef34b00df24417fbc3c4 | AungKyawZaww9/python | /chapter 7 python/Defination_of_class_data2.py | 778 | 4.125 | 4 | from Definition import Date
class Employee:
def __init__(self, firstName, lastName, birthMonth, birthDay,
birthYear, hireMonth, hireDay, hireYear):
self.birthDate = Date(birthMonth, birthDay, birthYear)
self.hireDate = Date(hireMonth, hireDay, hireYear)
self.lastName = lastName
self.firstName = firstName
print("Employee constructor: %s, %s"\
% (self.lastName, self.firstName))
def __del__(self):
print("Employee object about to be destroyed: %s, %s"\
% (self.lastName, self.firstName))
def display(self):
print("%s, %s" % (self.lastName, self.firstName))
print("Hired:", self.hireDate.display())
print("Birthdate: ", self.birthDate.display()) | false |
7b2081e92fe73ced6787897bf25ef99e60384370 | annettemathew/count_strings | /count_strings.py | 740 | 4.125 | 4 | #3. Count the number of strings where the string length is 2 or more and the first and
#last character are the same from a given list of strings.
#Sample List : ['abc', 'xyz', 'aba', '1221']
#Expected Result : 2
#Accept a list of comma-separated strings
#Iterate through list, and check length >= 2
#Compare first and last character & increment counter if equal
input_str = input("Please enter a string: ")
str_list = [x for x in input_str.split(',') if x.strip()]
counter = 0
for element in range(0, len(str_list)):
if(len(str_list[element]) >= 2):
str = str_list[element] #initializing a new string equal to the current list item
if(str[0] == str[len(str) - 1]):
counter += 1
print(counter) | true |
716f5e1096a400c7e4ea186bb8c2e7bb9f0dd63b | YuriPellini/Athon.Py | /Lista 2/Exe_8.py | 728 | 4.78125 | 5 | '''******************************************
ATHON
Programa de Introdução a Linguagem Python
Disiplina: Lógica de Programação
Professor: Francisco Tesifom Munhoz
Data: Primeiro Semestre 2021
*********************************************
Atividade: Lista 2 (Ex 3)
Autor: Yuri Pellini
Data: 19 de Maio de 2021
Comentários:
******************************************'''
#Entrada
X=float(input("Coloque o tamanho do lado do triângulo em cm:"))
Y=float(input("Coloque o segundo valor:"))
Z=float(input("Coloque o último valor:"))
# Saida
if(X==Y and Y==Z):
print("Isso é um triângulo equilátero")
else:
if(X==Y or Y==Z or X==Z):
print("Isso é um triângulo isósceles")
else:
print("Isso é um triângulo escaleno") | false |
1f8b0b8f424ff5c82ae3e9b2dd3e407c109ad0dc | KF10/ichw | /pyassign2/currency.py | 1,856 | 4.15625 | 4 | #!/usr/bin/env python3
"""currency.py: provide a 'exchange' function which return the
amount of another currency when you want to convert a certain
amount of currency to another.
__author__ = "Kuangwenyu"
__pkuid__ = "1800013245"
__email__ = "w.y.kuang@pku.edu.cn"
"""
from urllib.request import urlopen
import json
def exchange(currency_from, currency_to, amount_from):
"""Returns: amount of currency received in the given exchange.
In this exchange, the user is changing amount_from money in
currency currency_from to the currency currency_to. The value
returned represents the amount in currency currency_to.
The value returned has type float.
Parameter currency_from: the currency on hand
Precondition: currency_from is a string for a valid currency code
Parameter currency_to: the currency to convert to
Precondition: currency_to is a string for a valid currency code
Parameter amount_from: amount of currency to convert
Precondition: amount_from is a float
"""
doc = urlopen('http://cs1110.cs.cornell.edu/2016fa/a1server.php?from={0}&to={1}&amt={2}'.
format(currency_from, currency_to, amount_from))
docstr = doc.read()
doc.close()
jstr = docstr.decode('ascii')
ret_dict = json.loads(jstr) # 将jstr转为字典
convert = ret_dict['to']
mylist = convert.split()
amount_to = float(mylist[0])
return amount_to
def test_exchange():
"""test the 'exchange' function.
"""
assert(17.13025 == exchange('USD', 'CNY', 2.5))
assert(2.1589225 == exchange('USD', 'EUR', 2.5))
assert(0.018484513053739 == exchange('AOA', 'AUD', 3.7))
def test_all():
"""test all cases.
"""
test_exchange()
print('All tests passed')
def main():
"""main module
"""
test_all()
if __name__ == '__main__':
main()
| true |
6b981828e3b06a05a99f11a543ff178202e1768b | mcelaric/PythonDataRepresentations | /List Mutations/quiz.py | 2,775 | 4.1875 | 4 | # Question 1
# What slicing returns [5,7,9]
my_list = [1,3,5,7,9]
print(my_list[2:5]) # YES
print(my_list[2:]) # YES
# Question 2
# What returns a tuple of length 1?
print((1,)) # YES
print(tuple([1])) # YES
# Question 3
# Why does following code snippet raise an error in Python?
# instructors = ("Scott", "Joe", "John", "Stephen")
# instructors[2 : 4] = []
# print(instructors)
# Answer - tuples are immutable
# TypeError: 'tuple' object does not support item assignment
# Question 4
# Given a non-empty list my_list,
# which item in the list does the operation my_list.pop() remove?
print("pop operation", my_list.pop())
my_list = [1,3,5,7,9]
# Answer
print("index -1", my_list[-1])
# Question 5
# What output does the following code snippet print to the console?
my_list = [1, 3, 5, 7, 9]
my_list.reverse()
print(my_list.reverse())
#None
"""
Given a list fib = [0, 1], write a loop that appends the sum of the last two items in
fib to the end of fib.
What is the value of the last item in fib after twenty iterations of this loop?
Enter the answer below as an integer.
As a check, the value of the last item in fib after ten iterations is 89.
"""
def append_sum(lst,n):
for i in range(0,n):
last_item = lst[-1]
second_last_item = lst[-2]
combined = last_item + second_last_item
lst.append(combined)
#return(lst)
return lst.pop()
fib = [0,1]
print(append_sum(fib,20))
# Question 7
"""
One of the first examples of an algorithm was the Sieve of Eratosthenes.
This algorithm computes all prime numbers up to a specified bound.
The provided code below implements all but the innermost loop for this algorithm
in Python.
Review the linked Wikipedia page and complete this code.
Implement the Sieve of Eratosthenes
https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
Running your completed code should print two numbers in the console.
The first number should be 46
"""
import math
primes = []
number = 200
for i in range(2,number+1):
primes.append(i)
i = 200
#from 2 to sqrt(number)
while(i <= int(math.sqrt(number))):
#if i is in list
#then we gotta delete its multiples
if i in primes:
#j will give multiples of i,
#starting from 2*i
for j in range(i*2, number+1, i):
if j in primes:
#deleting the multiple if found in list
primes.remove(j)
i = i+1
print (primes)
def compute_primes(bound):
"""
Return a list of the prime numbers in range(2, bound)
"""
answer = list(range(2, bound))
for divisor in range(2, bound):
# Remove appropriate multiples of divisor from answer
pass
return answer
print(len(compute_primes(200)))
print(len(compute_primes(2000))) | true |
6963e37edd9fc6001544ea8f8174beded61567df | mcelaric/PythonDataRepresentations | /Lists/define_access_lists.py | 975 | 4.3125 | 4 | """
List Literals
"""
lst1 = [1, 5, 9, -32]
print(lst1)
lst2 = ["dog", "cow", "horse"]
print(lst2)
emptylst = []
print(emptylst)
"""
Range & List Indexing
"""
lst = list(range(10))
# First element
print(lst[0])
# Third element
print(lst[2])
# Length
print(len(lst))
# Last element
print(lst[9])
print(lst[len(lst) - 1])
print(lst[-1])
"""
List Slicing
"""
lst = list(range(10))
# Slice with 3 elements
print(lst[4:7])
"""
Negative indices may be used in slicing, and they have exactly the same meaning that they did when
used as indices. You can even mix positive and negative indices in a slice.
"""
# Slice with the last 2 elements of the list
print(lst[8:10])
print(lst[8:])
print(lst[-2:])
# Slice with the first 4 elements of the list
print(lst[0:4])
print(lst[:4])
"""
Rather than causing errors, if there are no elements in the list between the indices in the slice,
then an empty list is produced.
"""
# Empty slices
print(lst[20:25])
print(lst[7:3]) | true |
3bba72b9df6fbdcc20528dc7ec4b98d9360d1d46 | erdituna/python | /basic examples/format_namefunc.py | 552 | 4.15625 | 4 | def format_name(first_name , last_name):
if first_name == "":
format_name = str(last_name)+", "+str(first_name)
elif last_name =="":
format_name = str(last_name)+", "+str(first_name)
elif first_name != "" and last_name != "":
format_name = str(last_name)+", "+str(first_name)
else:
format_name = str(last_name)+", "+str(first_name)
return format_name
print(format_name("Ernst","Hemingway"))
print(format_name("","Madonna"))
print(format_name("Voltaire",""))
print(format_name("","")) | false |
b0dd03379603ac1a85949b499a3286e85ef2d24e | AmeyMhaskar/new_repository | /amey.python.py | 289 | 4.21875 | 4 | #Program to iterate through a list using indexing
genre=['pop','rock','jazz']
#iterate over the list using index for in range(len(genre)):
print("I LIKE",genre[1])
digits=[0,1,5]
for i in digits:
print(i)
else:
print("no item left")
range(0,10)
print(range(10))
| true |
93cd105809b704154df50b830c2463fb0a2c6e53 | pnguyen44/python_code_challenges | /summation.py | 442 | 4.28125 | 4 | # 8 kyu -Grasshopper - Summation
# Summation
# Write a program that finds the summation of every number between 1 and num. The number will always be a positive integer greater than 0.
#
# For example:
#
# summation(2) -> 3
# 1 + 2
#
# summation(8) -> 36
# 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8
def summation(num):
return reduce(lambda x,y: x + y, range(num + 1))
# Alternative Solution:
# def summation(num):
# return sum(range(1,num+1))
| true |
68eed5ec07790385fee83fc4fb14aac197a69d73 | vinnav/Automate-with-Python | /1-CrashCourse/5-commaCode.py | 571 | 4.25 | 4 | spam = ["apples", "bananas", "tofu", "cats"]
# Write a function that takes a list value as an argument and returns a string
# with all the items separated by a comma and a space, with and inserted before
# the last item. For example, passing the previous spam list to the function would
# return 'apples, bananas, tofu, and cats'. But your function should be able to work
# with any list value passed to it.
def commaCode(list):
for i in range(len(list)):
print(list[i], end="")
if i < len(list)-1:
print(", ", end="")
commaCode(spam)
| true |
c463a1fbcaa6aeb1d7b4ad42ea85c70a6afe3ec6 | Sankalp679/Competitive-Programming | /Leet Code/30 Days of Code/week_03/construct_binary_search_tree_from_preorder.py | 759 | 4.125 | 4 | """
Return the root node of a binary search tree that matches the given preorder traversal.
"""
# Definition for a binary tree node.
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
def bstFromPreorder(self, preorder: List[int]) -> TreeNode:
def insert_node(root,val):
if root is None:
return TreeNode(val)
elif root.val > val:
root.left = insert_node(root.left,val)
elif root.val < val:
root.right = insert_node(root.right,val)
return root
root = None
for i in preorder:
root = insert_node(root,i)
return root | true |
2488827fbb2e856791fe7eace892ed527136c69c | Grosbin/python-2021 | /positiveOrNegative.py | 210 | 4.21875 | 4 | value = int(input("Enter the value: "))
if value < 0:
print("The value is negative: ", value)
elif value > 0:
print("The value is positive: ", value)
else:
print("The value is cero: " , value)
| true |
b8440e2fd7e0d2e34f9620cb9e84fea717692834 | Grosbin/python-2021 | /numeroPalindromo.py | 245 | 4.1875 | 4 | n = int(input("Enter the number palindrieme"))
tmp = n
reverse = 0
while n >0:
dig = n % 10
reverse = reverse*10+dig
n = n//10
if tmp == reverse:
print("Values are palindrieme")
else:
print("Values are not palindrieme")
| false |
d79fe6d261b588cb3883f873fbc033e14728891c | hubbm-bbm101/lab5-exercise-solution-b2210356166 | /b2210356166/ex2.py | 226 | 4.1875 | 4 | address=input("write e-mail")
if "@" in address:
if '.' in address:
print("it is a valid e-mail")
else:
print("it is not a valid e-mail")
else:
print("it is not a valid e-mail")
| true |
1a90b10303d76cd69d5807a123543f6dd5dcecbf | bhelga/university | /python/Lab8/Lab8_1.py | 895 | 4.3125 | 4 | # 8.1.1
name = input("Введiть назву роману та натиснiть Enter:\t")
author = input("Введiть iм'я автора роману та натиснiть Enter:\t")
print(f"Письменник – автор роману:\t{author} – автор роману {name}")
# 8.1.2
first = input("Введіть перший рядок:\t")
second = input("Введіть другий рядок:\t")
if len(first) < 3 or len(second) < 2:
print("Error!")
else:
third = first[1] + first[2] + second[len(second) - 2]
print("Third is: " + third)
# 8.1.3
word = "голова"
print("Our word:\t" + word)
word += word[4]
word += word[1:4]
word += word[0]
word += word[5]
word = word[6:12]
print("Our final word:\t" + word)
# 8.1.4
user_str = input("Введіть ваш рядок:\t")
user_str = user_str.replace("ах", "ух")
print("Finished:\t" + user_str) | false |
cd815d6653188c2ec4ca1e1594225a039e794763 | bhelga/university | /python/Lab14/try7_1.py | 722 | 4.21875 | 4 | def prime_gen(n):
primes = [2]
# починаємо з трійки, бо двійка вже в масиві
nextPrime = 3
while nextPrime < n:
isPrime = True
i = 0
# the optimization here is that you're checking from
# the number in the prime list to the square root of
# the number you're testing for primality
squareRoot = int(nextPrime ** .5)
while primes[i] <= squareRoot:
if nextPrime % primes[i] == 0:
isPrime = False
i += 1
if isPrime:
primes.append(nextPrime)
# only checking for odd numbers so add 2
nextPrime += 2
print(primes)
prime_gen(1000) | false |
2d98befaa1604fa38c07b3874a813b9f3dfee5b0 | jameswmccarty/AdventOfCode2019 | /day04.py | 2,495 | 4.21875 | 4 | #!/usr/bin/python
"""
--- Day 4: Secure Container ---
You arrive at the Venus fuel depot only to discover it's protected by a password. The Elves had written the password on a sticky note, but someone threw it out.
However, they do remember a few key facts about the password:
It is a six-digit number.
The value is within the range given in your puzzle input.
Two adjacent digits are the same (like 22 in 122345).
Going from left to right, the digits never decrease; they only ever increase or stay the same (like 111123 or 135679).
Other than the range rule, the following are true:
111111 meets these criteria (double 11, never decreases).
223450 does not meet these criteria (decreasing pair of digits 50).
123789 does not meet these criteria (no double).
How many different passwords within the range given in your puzzle input meet these criteria?
Your puzzle input is 136760-595730.
-- Part Two ---
An Elf just remembered one more important detail: the two adjacent matching digits are not part of a larger group of matching digits.
Given this additional criterion, but still ignoring the range rule, the following are now true:
112233 meets these criteria because the digits never decrease and all repeated digits are exactly two digits long.
123444 no longer meets the criteria (the repeated 44 is part of a larger group of 444).
111122 meets the criteria (even though 1 is repeated more than twice, it still contains a double 22).
How many different passwords within the range given in your puzzle input meet all of the criteria?
Your puzzle input is still 136760-595730.
"""
def valid1(password):
paired = False
for pair in ['00','11','22','33','44','55','66','77','88','99']:
if pair in str(password):
paired = True
if paired:
for i in range(5):
if int(str(password)[i]) > int(str(password)[i+1]):
return False
return True
return False
def valid2(password):
paired = False
for pair in ['00','11','22','33','44','55','66','77','88','99']:
if pair in str(password) and pair+pair[0] not in str(password):
paired = True
if paired:
for i in range(5):
if int(str(password)[i]) > int(str(password)[i+1]):
return False
return True
return False
if __name__ == "__main__":
# Part 1 Solution
total = 0
for i in range(136760,595730+1):
if valid1(i):
total += 1
print(total)
# Part 2 Solution
total = 0
for i in range(136760,595730+1):
if valid2(i):
total += 1
print(total)
| true |
ca6ad1cb025429610a8cc95d7c4db1cf4230e5ec | junipernineaj/python-reeducation | /Chapter2/summary2.py | 663 | 4.28125 | 4 | # 2-8. Number Eight:
# Write addition, subtraction, multiplication, and division operations that each result in the number 8.
# Be sure to enclose your operations in print() calls to see the results.
# You should create four lines that look like this:
print(5+3)
print(10-2)
print((2+2)*2)
print((1+3)*2)
print((10+5+1)/2)
# Your output should simply be four lines with the number 8 appearing once on each line.
# 2-9. Favorite Number: Use a variable to represent your favorite number.
# Then, using that variable, create a message that reveals your favorite number. Print that message.
favourite_number=26
print (f"My favourite number is {favourite_number}") | true |
b158cfca77c9366b956c99d28bdbc210453f1357 | joseangel-sc/CodeFights | /Challenges/iqAddress.py | 901 | 4.125 | 4 | def iqAddress(n):
r = ""
n=float(n)
while n!=1:
r = str(n%10.5) + r
n = math.ceil(n/2)
return "1"+r
'''Have you ever heard of an IQ-address? For the given integer n, it is calculated as follows:
Let result = "".
If n = 1, prepend "1" to the beginning of result and return it as an answer.
Prepend n % 10.5 to the beginning of result.
Divide n by 2 with rounding up to the nearest integer.
Go to step 2.
Given an integer n, your task is to return IQ-address generated from it.
Example
For n = 21, the output should be
iqAddress(n) = "12.03.06.00.50.0".
Here's why:
21% 10.5 = 0.0
11% 10.5 = 0.5
6 % 10.5 = 6.0
3 % 10.5 = 3.0
2 % 10.5 = 2.0
Thus, the answer is "1"+"2.0"+"3.0"+"6.0"+"0.5"+"0.0" = "12.03.06.00.50.0".
Input/Output
[time limit] 4000ms (py)
[input] integer n
Constraints:
0 ≤ n ≤ 105.
[output] string
The IQ-address generated from n.'''
| true |
2948a131295b9fecf0bcb71841e0d91cf3feb844 | joseangel-sc/CodeFights | /Arcade/BookMarket/ProperNounCorrection.py | 621 | 4.1875 | 4 | #Proper nouns always begin with a capital letter, followed by small letters.
#Correct a given proper noun so that it fits this statement.
#Example
#For noun = "pARiS", the output should be
#properNounCorrection(noun) = "Paris";
#For noun = "John", the output should be
#properNounCorrection(noun) = "John".
#Input/Output
#[time limit] 4000ms (py)
#[input] string noun
#A string representing a proper noun with a mix of capital and small Latin letters.
#Constraints:
#1 ≤ noun.length ≤ 10.
#[output] string
#Corrected (if needed) noun.
def properNounCorrection(noun):
return noun[0].upper()+noun[1:].lower()
| true |
27cec83b2b8d6f35a19b67a8fcaccbe2b866d7ca | joseangel-sc/CodeFights | /Arcade/BookMarket/IsMAC48Address.py | 1,406 | 4.28125 | 4 | #A media access control address (MAC address) is a unique identifier assigned to network interfaces for communications on the physical network segment.
#The standard (IEEE 802) format for printing MAC-48 addresses in human-friendly form is six groups of two hexadecimal digits (0 to 9 or A to F), separated by hyphens (e.g. 01-23-45-67-89-AB).
#Example
#For inputString = "00-1B-63-84-45-E6", the output should be
#isMAC48Address(inputString) = true;
#For inputString = "Z1-1B-63-84-45-E6", the output should be
#isMAC48Address(inputString) = false;
#For inputString = "not a MAC-48 address", the output should be
#isMAC48Address(inputString) = false.
#Input/Output
#[time limit] 4000ms (py)
#[input] string inputString
#Constraints:
#15 ≤ inputString.length ≤ 20.
#[output] boolean
#true if inputString corresponds to MAC-48 address naming rules, false otherwise.
def isMAC48Address(inputString):
if len(inputString)!=17:
return False
inputStringSplit = inputString.split('-')
if len(inputStringSplit)!=6:
return False
for i in range(0,len(inputStringSplit)):
if len(inputStringSplit[i])!=2:
return False
if (("0"<=inputStringSplit[i][0]<="9" or "A"<=inputStringSplit[i][0]<="F") and ("0"<=inputStringSplit[i][1]<="9" or "A"<=inputStringSplit[i][1]<="F")):
return True
else:
return False
| true |
76ad0916806228157457a646b305b335bcb2ae87 | bawilc01/py_loop_sample_code | /loop_example.py | 2,133 | 4.15625 | 4 | # Looping through numbers in python
# will print numbers 1 - 10 using for loop
# i = 0
#
# for i in range(10):
# # same as i = i + 1
# # i = 0, i = 0 + 1, which = 1.
# # first number to print will be 1
#
# i+=1
#
# # print i until reaching the last number in the range
# print(i)
#
#
# # will print numbers 0 - 9 using for loop
# i = 0
#
# for i in range(10):
# # print i until reaching the last number in the range
# # nothing has been done to i yet, so it will print first value of i,
# # which is 0
# print(i)
#
#
# # same as i = i + 1
# # i = 0, i = 0 + 1, which = 1.
# # first number to print will be 1
#
# i += 1
# THIS WAS A REAL INTERVIEW QUESTION ON A TECHNICAL INTERVIEW
# prints the sum of the last sum + sum for 10 numbers using for loop
# i interates by 1, all the way up to 10
# example, first answer is 0 + 1 = 1; sum starts at 0, sum increments by 1, then sum = 1
# next answer is 1 + 1
# sum = 0
# # sum will change
#
# i = 0
# # i is numbers 1 - 10
#
# for i in range(10):
# sum = sum + i
# # or sum += 1
# # first answer 0 + 1 = 1
#
# print(sum)
# sum is now = 1
# starting over at the loop, we are now going to move to the second number (i) in the range, which is 2.
# sum = 1, i = 2
# the next number printed is new sum = 1 (current sum) + 2 (value of i)
# sum is now = 3 and will be printed
# next number is new sum = 3 (current sum) + 3 ( value of i)
# sum is now = 6
# the next 10 sums to be printed will appear in the console until all i's up to 10 are interated through
# print numbers i * 1 and interates i by 1 before returning to loop to screen using while loop until i equals 10
# will print numbers 0 - 10
# if you remove the = and make it while i < 10, then 0 - 9 will print to the screen because once i hits 10, the loop will stop because 10 is not less than 10
i = 0
while i<=10:
print(i*1)
i=i+1
# create an array or list of things and print all to the screen with a loop
# data = {"Brittney Coble", "some string", "another string"}
#
# for data in list:
# print(data)
| true |
ea5d8df4e30e111f2511f5db7ef4ace67d0a5615 | hikyru/Python | /FindingOddNumbers_2_22.py | 840 | 4.1875 | 4 | # -*- coding: utf-8 -*-
"""
Created on Sat Feb 22 7:28 2020
@author: KatherineYu
"""
# Input a Number
x = int(input("Enter a Number: "))
# Find all odd numbers in the number
print("Finding all odd numbers in", x, "!")
# m is a variable used for counting the number of digits printed
m = 0
for i in range(1, x + 1, 1):
# if there are remainers after i is divided by 2, that means i is an odd number
# because even numbers are perfectly divisible
if i % 2 == 1:
print(i, end='\t')
# for every time an odd number is printed, m a number is stored into m
m += 1
if m % 10 == 0:
# if m is perfectly divisible by 10, that means there are at least ten odd numbers in x
# a new line will be added
print() # will only run every every time there are ten more digits added to m
| true |
c2d6d7c9d1dd7633f2c10492fad2e0a176361ccb | ikshwak/udacity-DS-assign3 | /problem_4.py | 2,052 | 4.15625 | 4 | def sort_012(input_list):
"""
Given an input array consisting on only 0, 1, and 2, sort the array in a single traversal.
Args:
input_list(list): List to be sorted
"""
if input_list == None:
return []
start = 0
end = len(input_list)-1
mid = 0
while mid <= end:
if input_list[mid] == 0:
input_list[start],input_list[mid] = input_list[mid],input_list[start]
start += 1
mid += 1
elif input_list[mid] == 1:
mid += 1
elif input_list[mid] == 2:
input_list[end],input_list[mid] = input_list[mid],input_list[end]
end -= 1
else:
print("invalid 012 list")
return []
return input_list
def test_function(test_case):
sorted_array = sort_012(test_case)
print("Solution: " + str(sorted_array))
print("Library sort: " + str(sorted(test_case)))
"""
TEST CASE 1
"""
test_function([0, 0, 2, 2, 2, 1, 1, 1, 2, 0, 2])
"""
Result:
Solution: [0, 0, 0, 1, 1, 1, 2, 2, 2, 2, 2]
Library sort: [0, 0, 0, 1, 1, 1, 2, 2, 2, 2, 2]
"""
"""
TEST CASE 2
"""
test_function([2, 1, 2, 0, 0, 2, 1, 0, 1, 0, 0, 2, 2, 2, 1, 2, 0, 0, 0, 2, 1, 0, 2, 0, 0, 1])
"""
Result:
Solution: [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2]
Library sort: [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2]
"""
"""
TEST CASE 3
"""
test_function([0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2])
"""
Result:
Solution: [0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2]
Library sort: [0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2]
"""
"""
TEST CASE 4
"""
test_function([0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 4, 2, 2, 2, 2, 2, 3])
"""
Result:
invalid 012 list
Solution: []
Library sort: [0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 4]
"""
"""
TEST CASE 5
"""
test_function([])
"""
Result:
Solution: []
Library sort: []
"""
| true |
eca5ae226d964703e3a0c91ee43581d601edd3c2 | clair3st/python-prework | /lpthw-exercises/ex11.py | 695 | 4.125 | 4 | # ex11: Asking questions
# comma at the end of the print statements so the print doesn't
# end the line with the new line character and go to the next line.
# raw_input is an inbuilt python function, it reads a line from
# input and converst to a string and retunrs that
print "How old are you?",
age = raw_input()
print "How tall are you?",
height = raw_input()
print "How much do you weigh?",
weight = raw_input()
print "So, you're %r old, %r tall and %r heavy." % (age, height, weight)
print "What is your favourite color?",
age = raw_input()
print "What is your favourite food?",
food = raw_input()
print "What is your 2 favourite numbers?",
number1 = raw_input()
number2 = raw_input()
| true |
664221f1a7fea12548cc719acc0df9ed218324ce | pengliangs/python-learn | /基础语法/day2/2.序列.py | 1,032 | 4.25 | 4 | # 序列是 python 中的基本数据结构的一种
# 序列分类：
# 可变序列：
# > 列表（list）
# 不可变序列：
# > 字符串（str）
# > 元组 （tuple）
username = "张师傅"
print(username[:2])
# list修改
usernames = ["张三","李四","王五","赵六"]
print("修改前",username)
usernames[0]="zhangs"
print("修改后",usernames)
# 切片赋值
# usernames[:2] = 123
# 切片修改必须使用对应序列进行赋值,如果赋值元素个数大于匹配个数则插入元素
usernames[:2] = ["zs","lis"]
print("切片修改",usernames)
# 设置切片后修改赋值，值元素个数必须跟切片数匹配
usernames[::2] = ["a","b"]
print("切片修改步长",usernames)
# 删除列表元素
del usernames[0]
print("删除第一个元素",usernames)
# 通过切片删除
del usernames[:1]
print("切片删除",usernames)
# 修改操作只能是可变序列
s = "hello"
# s[0] = "H"
# 如果要改变则需要将其转为可变序列
ss = list(s)
print(ss)
| false |
08012e5618ba7a9157ea532c6eff58b431155b9b | pengliangs/python-learn | /基础语法/day4/3.对象的初始化.py | 841 | 4.1875 | 4 | class Person:
# 在类中可以定义一些特殊方法（魔术方法）
# 特殊方法都是__开头，__结尾的方法
# 特殊方法不需要我们自己调用，不要自己去尝试调用外部是可以调用到的
# __init__ 有点像java中的构造函数，在对象创建的时候被调用
# 创建对象流程
# p1 = Person()
# 1.创建一个变量
# 2.在内存中创建一个新对象
# 3.__init__(self) 方法执行
# 4.将对象的id赋值给变量
def __init__(self,name):
# 初始化一个属性
#self.name = ""
self.name = name
print("Person.__init__执行")
def hello(self):
print("你好~ 我是:%s"%self.name)
print("Person中代码块，执行")
# p1 = Person()
# p1.name = "p1"
# p1.hello()
p2 = Person("p2")
p2.hello() | false |
5e3e5af970159cc30d5bd5600f4c146a475e9f90 | TadayoshiCarvajal/Intro-to-Programming-Tutorial | /episode09.py | 1,291 | 4.375 | 4 | '''
Here is the Python script that corresponds to the 9th episode in
my YouTube tutorial series, Intro to Programming in Python.
Follow me at KaizenMachina on YouTube for daily content.
'''
'''
From the last video:
'''
firstName = 'Jon'
lastName = 'Snow'
fullName = firstName + ' ' + lastName
'''
Sequences are a positional ordering of elements.
'Jon' is NOT the same as 'Noj' or 'ojN'!
How do we access the various elements (characters) of a string?
Answer: Indexing
'''
print(fullName[0])
print(fullName[1])
print(fullName[2])
print(fullName[3])
print(fullName[4])
print(fullName[5])
print(fullName[6])
print(fullName[7])
'''
We use the len functio to get the length of a sequence (# of elements)
'''
print(len(firstName))
print(len(lastName))
print(len(fullName))
long_word = 'supercalifragilisticexpialidocious'
print(len(long_word))
'''
Use length of the string to access the last index:
'''
print(long_word[len(long_word) - 1])
#fullName[len(fullName)] #this will give us an error since 8 is out of bounds.
'''
long_word[len(long_word) - 1] this is pretty tedious
a more concise way is to use negative indexes.
long_word[len(long_word) - 1] is the same as long_word[-1]
'''
print(long_word[-1])
#fullName[-len(fullName)-1] #this will give us an error since -1 is out of bounds. | true |
2364c53ba0411024f81e21041d02eae9d9e613a5 | daniela-mejia/Python-Net-idf19- | /PhythonAssig/5-06-19 Assigment7/loan.py | 682 | 4.21875 | 4 | #Determine the monthly payment for a car loan.Example output
#Daniela Mejia
def main ():
loan_Amount = float(input('Enter the amount of your loan: '))
loan_interest = float(input('Enter the interest rate (%): '))
loan_years = int(input('Enter the numbers of years for this loan: '))
repayment_years = loan_years * 12
repayment_interest = loan_interest/100
#Formula
#Month_payment = L[i(1+i)n] / [(1+i)n-1]
monthlyRepay =(loan_Amount * repayment_interest * (1+repayment_interest) * repayment_years / ((1+repayment_interest) * repayment_years - 1))
print (' You have to pay monthly : ', (monthlyRepay))
#print (monthlyRepay)
main () | true |
176de592151a412ca29ede7f749c6d11d92c5959 | daniela-mejia/Python-Net-idf19- | /PhythonAssig/4-25-19 Assigment5/4-25 Assigment1(Lottery).py | 655 | 4.21875 | 4 | # Design a program that generates 7 random digit lottery number.
import random
#Initialise an empty list that will be used to store the 6 lucky numbers!
lotNumbers = []
for i in range (0,7):
number = random.randint(0,9)
#Check if this number has already been picked and ...
while number in lotNumbers:
# ... if it has, pick a new number instead
number = random.randint(0,9)
#Now that we have a unique number, let's append it to our list.
lotNumbers.append(number)
#Sort the list in ascending order
lotNumbers.sort()
#Display the lsit on screen:
print("LUCKY! Lottery numbers are: ")
print(lotNumbers)
| true |
a62686e9025fec93281ca7ba22e8ea7cb77dbbdd | daniela-mejia/Python-Net-idf19- | /PhythonAssig/5-15-19 Assigment 12/evendigits.py | 480 | 4.1875 | 4 | # This Program finds how many even numbers are in the input number by the user
# Start defining a function
def Even(a):
return a % 2 == 0
def countEven():
count = 0
string_of_Num = int(input("Please enter a numbers separated by spaces: "))
list_of_Num = string_of_Num.split(' ')
for number in list_of_Num:
num = int(number)
if Even(num) == True:
count += 1
print ("There are" + str (count) + "even numbers.")
| true |
31fb656137a5cc30fe2958fb5dcd5087474b7071 | daniela-mejia/Python-Net-idf19- | /PhythonAssig/4-25-19 Assigment6/Employeeclass.py | 657 | 4.15625 | 4 | #Daniela Mejia 04/25/2019
#python object oriented programming Employee Class
class Employee:
def __init__(self, name, Id, dep, job):
self.name = name
self.Id = Id
self.dep = dep
self.job = job
#method
def fulldata(self):
return 'Name:{},Id: {}, Department{},Job: {}'.format(self.name,self.Id,self.dep,self.job)
#data instance variables
emp1 = Employee('Susan Meyers','45899','Accounting','Vice President')
emp2 = Employee('Mark Jones','39119','IT','Programmer')
emp3 = Employee('Joy Rogers','81774','Manufacturing','Engineer')
print(emp1.fulldata())
print(emp2.fulldata())
print(emp3.fulldata())
| false |
e07a29aa104446c327b18fdaa0dceaf756937a40 | MarinaGoto/python-basic | /Turtle_Graphics/bigX.py | 1,057 | 4.5625 | 5 | # P 1. p. 244
#Give a set of instructions for controlling the turtle to draw a line from the top-left corner of the screen
#to the bottom-right corner, and from the top-right corner to the bottom-left corner,
#thereby making a big X on the screen. There should be no other lines drawn on the screen.
import turtle
# set window size
turtle.setup(800, 600)
# get reference to turtle window
window = turtle.Screen()
window.title('Drawing an X')
#get default turtle and hide
the_turtle = turtle.getturtle()
the_turtle.hideturtle()
#shape
the_turtle.shape('turtle')
the_turtle.fillcolor('black')
#color
turtle.colormode(255)
the_turtle.pencolor(169, 71, 169)
#size
the_turtle.pensize(41)
#resize
the_turtle.resizemode('user')
the_turtle.turtlesize(3, 3)
#create square (absolute postioning)
the_turtle.penup()
the_turtle.setposition(-400, 300)
the_turtle.pendown()
the_turtle.setposition(400, -300)
the_turtle.penup()
the_turtle.setposition(400,300)
the_turtle.pendown()
the_turtle.setposition(-400,-300)
the_turtle.speed(1)
turtle.exitonclick() | true |
76ba3708402ebbb24a5875b25caa19664357c9fa | MarinaGoto/python-basic | /Simple programs/T2.py | 2,336 | 4.125 | 4 | # get month and year
month = int(input('Please write a month (1-12): '))
year = int(input('Please write a year (1800-2099): '))
# determine if leap year
if (year % 4 == 0) and (not (year % 100 == 0) or (year % 400 == 0)):
leap_year = True
else:
leap_year = False
##################################################################################################3
# Stage 2
# Determining the Day of the Week (using the day of the week algorithm)
century_digits = year // 100
year_digits = year % 100
# century_digits = int(str(year)[:2])
# year_digits = int(str(year)[-2:])
value = year_digits + (year_digits // 4)
if century_digits == 18:
value = value + 2
if century_digits == 20:
value = value + 6
if (month == 1) and not leap_year:
value = value + 1
elif month == 2:
if leap_year:
value = value + 3
else:
value = value + 4
elif (month == 3) or (month == 11):
value = value + 4
elif month == 5:
value = value + 2
elif month == 6:
value = value + 5
elif month == 8:
value = value + 3
elif month == 10:
value = value + 1
elif (month == 9) or (month == 12):
value = value + 6
# for this program, we will need to determine only the day of the week
# for the first day of any given month
# the day value in the day of the week algorithm is hard-coded to 1
day_of_the_week = (value + 1) % 7
# determine week day name
if day_of_the_week == 1:
dayName = 'Sunday'
elif day_of_the_week == 2:
dayName = 'Monday'
elif day_of_the_week == 3:
dayName = 'Tuesday'
elif day_of_the_week == 4:
dayName = 'Wednesday'
elif day_of_the_week == 5:
dayName = 'Thursday'
elif day_of_the_week == 6:
dayName = 'Friday'
elif day_of_the_week == 0:
dayName = 'Saturday'
# determine month name
if month == 1:
monthName = 'January'
if month == 2:
monthName = 'February '
if month == 3:
monthName = 'March'
if month == 4:
monthName = 'April'
if month == 5:
monthName = 'May'
if month == 6:
monthName = 'June'
if month == 7:
monthName = 'July'
if month == 8:
monthName = 'August '
if month == 9:
monthName = 'September'
if month == 10:
monthName = 'October '
if month == 11:
monthName = 'November'
if month == 12:
monthName = 'December'
# display results
print('The first day of ', monthName, 'is', dayName) | false |
b0dccffa020a21a621ac176da8b6d415d5b27c27 | MarinaGoto/python-basic | /Simple programs/ordered3Function.py | 368 | 4.40625 | 4 | #Write a Python function named ordered3 that is passed three integers,
#and returns true if the three integers are in order from smallest to largest,
#otherwise it returns false.
def ordered3(integer):
if integer[0] < integer[2]:
print('True')
else:
print('False')
###############################
integ = [8,0,11]
ordered3(integ) | true |
ac0680af71765fdda7b96d0bf4390983c281305f | MarinaGoto/python-basic | /Simple programs/Temperature Conversion Program.py | 1,049 | 4.40625 | 4 |
# coding: utf-8
# In[1]:
#Temperature Conversion Program (Fahrenheit to Celsius)
#This program will convert a temperature entered in Fahrenheit
#to the equivalent degrees in Celsius
#program greeting
print('This program will convert degrees Fahrenheit to degrees Celsius')
#get temperature in Fahrenheit
fahren = float(input('Enter degrees Fahrenheit: '))
#calc degrees Celsius
celsius = (fahren - 32) * 5/9
#output degrees Celsius
print(fahren, 'degrees Fahrenheit equals',
format(celsius, '.1f'), 'degrees Celsius')
# In[ ]:
#Temperature Conversion Program (Celsius to Fahrenheit)
#This program will convert a temperature entered in Celsius
#to the equivalent degrees in Fahrenheit
#program greeting
print('This program will convert degrees Celsius to degrees Fahrenheit ')
#get temperature in Fahrenheit
celsius = float(input('Enter degrees Celsius: '))
#calc degrees Celsius
fahren = (celsius * 9/5) +32
#output degrees Celsius
print(celsius, 'degrees Celsius equals',
format(fahren, '.1f'), 'degrees Fahrenheit')
| true |
a4feacf6c4a2f22d166803f1fbbca953a33319d3 | edarakchiev/Functions | /Exercise/04.negative_vs_positive.py | 775 | 4.125 | 4 | def negative_numbers(numbers):
return [n for n in numbers if n < 0]
def positive_numbers(numbers):
return [num for num in numbers if num >= 0]
def print_negative_bigger():
print("The negatives are stronger than the positives")
def print_positive_bigger():
print("The positives are stronger than the negatives")
def positive_vs_negative(pos_num, neg_num):
if pos_num >= abs(neg_num):
return True
return False
nums = [int(el) for el in input().split()]
negative_num = sum(negative_numbers(nums))
positive_num = sum(positive_numbers(nums))
print(f"{negative_num}")
print(f"{positive_num}")
if positive_vs_negative(positive_num, negative_num):
print_positive_bigger()
else:
print_negative_bigger() | false |
a9c779035d7b9c5c9cf8aa45dda84db478bf0ccf | gaohank/python-coding-proj | /02_algorithm/02_leetcode/sum_2_values.py | 1,825 | 4.375 | 4 | """
给出两个非空的链表用来表示两个非负的整数。
其中，它们各自的位数是按照 逆序 的方式存储的，并且它们的每个节点只能存储 一位 数字。
如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。
您可以假设除了数字 0 之外，这两个数都不会以 0 开头。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/add-two-numbers
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
"""
# Definition for singly-linked list.
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
n = l1
i = 1
num_l1 = 0
# get num of l1
while n:
num_l1 = num_l1 + n.val * i
i = i * 10
n = n.next
m = l2
j = 1
num_l2 = 0
# get num of l2
while m:
num_l2 = num_l2 + m.val * j
j = j * 10
m = m.next
str_num = str(num_l1 + num_l2)
str_num = str_num[::-1]
res = list_result = ListNode(0)
for s in str_num:
list_result.next = ListNode(int(s))
list_result = list_result.next
return res.next
if __name__ == '__main__':
s = Solution()
l1 = ListNode(3)
l1_2 = ListNode(4)
l1_3 = ListNode(2)
l1_2.next = l1_3
l1.next = l1_2
l2 = ListNode(4)
l2_1 = ListNode(6)
l2_2 = ListNode(4)
l2_1.next = l2_2
l2.next = l2_1
l3 = s.addTwoNumbers(l1, l2)
print(l3.val)
print(l3.next.val)
print(l3.next.next.val)
| false |
df7028948b7d894b0a0bef88ece106788a555d26 | wyh2023/Software-Eng-Computing-I | /PythonCode/verification.py | 402 | 4.15625 | 4 |
def password_verification(user, pwd):
"""
:param user: The username stored in the computer.
:param pwd: The password for the user stored in the computer.
"""
input_user = input('What is the user name?')
input_pwd = input('What is the password?')
if input_pwd == pwd and input_user == user:
print('Welcome!')
else:
print("I don't know you.")
| true |
483f0059ed617267e0ca7b28aa15814e07fda842 | vatsalgamit/CodingBat_Python_Problems | /Practice_Examples_Python.py | 918 | 4.40625 | 4 | """
The parameter weekday is True if it is a weekday,
and the parameter vacation is True if we are on vacation.
We sleep in if it is not a weekday or we're on vacation.
Return True if we sleep in.
sleep_in(False, False) → True
sleep_in(True, False) → False
sleep_in(False, True) → True
"""
def sleep_in(weekday,vacation):
if not weekday or vacation:
return True
return False
print(sleep_in(False,True))
"""
We have two monkeys, a and b, and the parameters a_smile and b_smile indicate if each is smiling.
We are in trouble if they are both smiling or if neither of them is smiling.
Return True if we are in trouble.
monkey_trouble(True, True) → True
monkey_trouble(False, False) → True
monkey_trouble(True, False) → False
"""
def monkey_trouble(a_smile,b_smile):
if a_smile == b_smile:
return True
return False
print(monkey_trouble(True,True))
| true |
45984492066285c083760ac97cc51fca64600164 | unitrium/02180-Introduction-to-AI-SP21-BOARD-GAME-ASSIGNMENT | /code/game/player.py | 2,781 | 4.3125 | 4 | """Class for defining a player."""
from typing import List, Tuple
from abc import ABC, abstractmethod
class Action:
x: int
y: int
direction: int
def __init__(self, x: int, y: int, direction: int) -> None:
self.x = x
self.y = y
self.direction = direction
def direction_position(self) -> Tuple[int, int]:
"""Determines the position of the black tile."""
if self.direction == 0:
return (self.x, self.y-1)
elif self.direction == 1:
return (self.x+1, self.y)
elif self.direction == 2:
return (self.x, self.y+1)
else:
return (self.x-1, self.y)
class Player(ABC):
"""A player that can interact with a board."""
white: bool
def __init__(self, white: bool) -> None:
self.white = white
@abstractmethod
def receive(self, board: "Board") -> None:
"""Receive the state of the board."""
pass
class Human(Player):
"""A player who is also human."""
def display(self, state: List[List[int]]) -> None:
"""Displays a state to the human when it's their turn.
White tiles are represented by X, black tiles are
represented by O."""
print("Your turn, the board looks like this:")
print()
print(" 0 1 2 3 4 5 6 7 8 9 10")
number = 0
for line in state:
printed_line = str(number)
if number < 10:
printed_line += " "
for col in line:
if col is None:
printed_line += "_ "
elif col == 0:
printed_line += "X "
else:
printed_line += "O "
print(printed_line[0:-1])
number += 1
def ask_action(self) -> Action:
"""Asks the human for a move."""
correct_move = False
while not correct_move:
try:
print("What is your move ?")
x = int(input("Start by giving the x coordinate of the white part."))
y = int(input("Now give the y coordinate of the white part."))
direction = int(input(
"The direction of the black part? 0 is up, 1 is right, 2 is down, 3 is left"))
correct_move = True
except ValueError:
print("Error please enter an integer")
return Action(x, y, direction)
def receive(self, board: "Board") -> None:
move_accepted = False
while not move_accepted:
self.display(board.state)
if board.receive(self.ask_action()):
move_accepted = True
else:
print("Your move isn't possible. Please choose again.")
| true |
463cd52817d0e8bbf3dfb3649eaa205f9f1d7bc3 | rakaar/OOP-Python-Notes | /3-methods.py | 1,795 | 4.25 | 4 | '''
regular methods: methods that take instance as an argument by default
class methods: methods that take class as an argument by default
static methods: just ordinary functions
decorators: things needed to define method type unless its a regular method
alternative constructors: class methods that are capable of creating instances other than __init__
'''
class Employee:
# class variable
bonus = 100
def __init__(self, firstName, lastName, salary):
self.firstName = firstName
self.lastName = lastName
self.salary = salary
def fullName(self):
return self.firstName + ' ' + self.lastName
def increase_salary(self): # a regular method
self.salary = self.salary + self.bonus
# a function that takes class as a argument: class method
@classmethod # this is a decorator which tells that this is a class method
def change_class_bonus(cls, increment):
cls.bonus += increment
# a ordinary function in a class
@staticmethod # decorator required to tell that this is a regular function
def subtract(x, y):
return x - y
# a alternative constructor: able to create an instance
@classmethod
def from_string(cls, the_string):
first, last, salary = the_string.split('-')
return cls(first, last, salary)
# Suppose I want a method that takes class as an argument an increases the bonus
# one thing that can be simply done is
Employee.bonus = 200
print(Employee.bonus) # 200
Employee.change_class_bonus(800)
print(Employee.bonus) # 200+800 = 1000
print(Employee.subtract(1, 2)) # 1-2 = -1
# using the alternative constructor
alt_emp = Employee.from_string('alt-emp-100')
print(alt_emp.firstName) # alt
print(alt_emp.lastName) # emp
print(alt_emp.salary) # 100
| true |
6f298ee034c8c5042e533f5d2cc15744ff3f3432 | BrandonLim8890/python | /projects/numbers/005next-prime.py | 636 | 4.15625 | 4 | '''
Name: next-prime.py
Purpose: Have the program look for the next prime number until the user stops
Author: Brandon Lim
'''
def isPrime(number):
for i in range(2, int(number ** 0.5 + 1)):
if number % i is 0:
return False
return True
def nextPrime(number=2):
user = ''
print('Hit enter to keep finding prime numbers')
while user is '':
while not isPrime(number):
number += 1
print(str(number))
number += 1
user = str(input())
print('Enter a starting number(Defualt value is 2):')
val = input()
if val is '':
nextPrime()
else:
nextPrime(int(val)) | true |
f39f6201fca53c33b92841c21390c81b00f3ce9f | nss-cohort-36/011420_orientation_args-n-kwargs | /index.py | 1,016 | 4.21875 | 4 | # args and kwargs
# args - takes any number of positional argument values and adds them to a list
def list_fave_colors(name, *args):
for color in args:
print(f"My name is {name}, and one of my fave colors is {color}")
# Pass in any number of arguments after the name ("Fred" or "Tia", and the function works, thanks to *args)
list_fave_colors("Fred", "red", "green", "blue", "orange")
list_fave_colors("Tia", "puce", "goldenrod")
#kwargs -- takes any number of keyword arguments and adds them to a dictinary
def make_family(name, **tacos):
family_stuff = tacos.items()
family_str = f"We are the {name} family. "
for title, person in family_stuff:
family_str += f"The {title} is {person}. "
return family_str
# Now we can make families of any size, and have flexibility to have those family members have whatever role we want
family = make_family("Shepherd", mom="Anne", dad="Joe", dog="Murph")
other_family = make_family("Parker", aunt="May", nephew="Peter")
print(family)
print(other_family)
| true |
c474daf715193c3f1b4b1e0cbd1abdf61e0b6685 | heltonavila/MC102 | /lab04/lab04_jogo_da_velha.py | 1,234 | 4.1875 | 4 | # O programa criado consiste num jogo da velha. Para a realização do mesmo é solicitado a entrada de 9 caracteres, sendo eles escolhidos entre 3 opções: X,O ou -.
# Sendo assim, é verificada a igualdade dos termos por linha, coluna e diagonal.
# Caso a igualdade seja verdadeira para os caracteres X ou O, o programa apresenta como saída o termo que torna válida essa igualdade, seguido da palavra "venceu".
# Caso não se confirme a igualdade, a saída do programa é a palavra "empate".
v1 = input()
v2 = input()
v3 = input()
v4 = input()
v5 = input()
v6 = input()
v7 = input()
v8 = input()
v9 = input()
if (v1 == v2 == v3) or (v1 == v5 == v9) or (v1 == v4 == v7): # Os comandos a seguir testam a igualdade dos termos por linha, coluna e diagonal.
if (v1 == "X") or (v1 == "O"):
print(str(v1) + " venceu")
elif (v5 == v2 == v8) or (v5 == v4 == v6):
if (v5 == "X") or (v5 == "O"):
print(str(v5) + " venceu")
elif (v9 == v6 == v3) or (v9 == v8 == v7):
if (v9 == "X") or (v9 == "O"):
print(str(v9) + " venceu")
elif (v7 == v5 == v3):
if (v7 == "X") or (v7 == "O"):
print(str(v7) + " venceu")
else: # Caso a igualdade seja falsa, o programa imprime "empatou".
print("empatou")
| false |
18d996fd62f260065a15039544dd12bd9722805b | davetek/linearAlgebraRefresher | /linalgL2Q8.py | 2,066 | 4.34375 | 4 | #for lesson 2 of Linear Algebra Refresher course, part 1, Quiz 8
# covering vector magnitude and direction
#import support module containing vector operation functions
import vectorOperations
print "Lesson 2: 1.8 Quiz exc 1"
#declare the vectors to perform operations on
inputVectors = [[7.887, 4.138], [-8.802, 6.776]]
print "dot product of " + str(inputVectors) + ":"
#get and print the vector magnitude
print vectorOperations.dotProductOfVectors(inputVectors)
print ""
print "Lesson 2: 1.8 Quiz exc 2"
#declare the vectors to perform operations on
inputVectors = [[-5.955, -4.904, -1.874], [-4.496, -8.755, 7.103]]
print "dot product of " + str(inputVectors) + ":"
#get and print the vector magnitude
print vectorOperations.dotProductOfVectors(inputVectors)
print ""
print "Lesson 2: 1.8 Quiz exc 3"
#declare the vectors to perform operations on
inputVector3a = [3.183, -7.627]
inputVector3b = [-2.668, 5.319]
print "angle between " + str(inputVector3a) + " and " + str(inputVector3b) + ":"
#compute and print the angle between the input vectors
print "radians: " + str(vectorOperations.angleBetweenVectors(inputVector3a, inputVector3b))
print ""
print "Lesson 2: 1.8 Quiz exc 4"
#declare the vectors to perform operations on
inputVector4a = [7.35, 0.221, 5.188]
inputVector4b = [2.751, 8.259, 3.985]
print "angle between " + str(inputVector4a) + " and " + str(inputVector4b) + ":"
#compute and print the angle between the input vectors, passing True for optional third parameter to output degrees
print "degrees: " + str(vectorOperations.angleBetweenVectors(inputVector4a, inputVector4b, True))
print ""
#test zero vector
#declare the vectors to perform operations on
inputVectorZero = [0, 0, 0]
inputVector4b = [2.751, 8.259, 3.985]
print "angle between " + str(inputVectorZero) + " and " + str(inputVector4b) + ":"
#compute and print the angle between the input vectors, passing True for optional third parameter to output degrees
print "degrees: " + str(vectorOperations.angleBetweenVectors(inputVectorZero, inputVector4b, True))
print ""
| true |
9dcbf1ca0b86cde7ad3bd956ae583ee9a06cad2d | mrAvi07/data_structures_algorithms | /2. Sorting Algorithms/selection_sort.py | 645 | 4.15625 | 4 | """
Selection Sort
"""
def find_smallest(arr):
smallest = arr[0]
smallest_index = 0
for i in range(1, len(arr)):
if arr[i] < smallest:
smallest = arr[i]
smallest_index = i
return smallest_index
def selection_sort(arr):
new_arr = []
for i in range(len(arr)):
smallest = find_smallest(arr)
new_arr.append(arr.pop(smallest))
return new_arr
if __name__=="__main__":
print(" Unsorted List ")
number_list = [17, 3, 25, 4, 9, 10, 2]
print(number_list)
print("selection sort ")
new_arr = selection_sort(number_list)
print(new_arr)
| true |
3471f166f59483988681af0c0bd3e721dc3bde46 | Gabospa/python-crud | /sort.py | 1,649 | 4.125 | 4 |
secuencia = [5,2,4,1]
def bubble_sort(array):
n = len(array)
for i in range(n):
for j in range(n-i-1):
if array[j] > array[j+1]:
array[j], array[j+1] = array[j+1], array[j]
return array
def merge_sort(array, low, high):
if low < high:
#divide and conquer
length = len(array)
mid = (low + high)/2
merge_sort(array, low, mid)
merge_sort(array, mid+1,high)
def merge(arr, l, m, h):
#calcula la mitad del sub arreglo
n1 = m-l+1
n2 = h-m
#crea sub arreglo 1
for i in range(n1):
new_arr1[i] = arr[l+1]
#crea sub arreglo 2
for j in range(n2):
new_arr2[j] = arr[m+j+1
for k in range(length):
if
def quick_sort(array):
pass
def selector_ordenamiento(num):
if num == 1:
secuencia_ordenada = bubble_sort(secuencia)
print(secuencia_ordenada)
print('Metodo ordenado por Bubble Sort')
elif num == 2:
secuencia_ordenada = merge_sort(secuencia, low, high)
print(secuencia_ordenada)
print('Metodo ordenado por Merge Sort')
else:
secuencia_ordenada = quick_sort(secuencia)
print(secuencia_ordenada)
print('Metodo ordenado por Quick Sort')
def run():
print('1. Bubble Sort \n2. Merge Sort \n3. Quick Sort')
seleccion = int(input('Seleccione metodo ordenamiento: '))
print(f'La secuencia original es {secuencia}')
selector_ordenamiento(seleccion)
if __name__ == '__main__':
run() | false |
bafb6c09ecd0017428441e109733ebcb189863ad | spkibe/calc | /calc.py | 606 | 4.15625 | 4 | operation = input('operation type: ').lower()
num1 = input("First number: ")
num2 = input("First number: ")
try:
num1, num2 = float(num1), float(num2)
if operation == 'add':
result = num1 + num2
print(result)
elif operation == 'subtract':
result = num1 - num2
print(result)
elif operation == 'multiply':
result = num1 * num2
print(result)
elif operation == 'divide':
result = num1 / num2
print(result)
else:
print('You didi choose the right operation')
except:
#
print("Impoper numbers or Operation") | true |
4483f2cd521246a0f333a06908b66dbead1a865b | kmoreno08/Programming_Projects | /Chapter2_PP#4.py | 2,260 | 4.375 | 4 | #Chapter 2_Programming Project - Weird Multiplication
'''You will be implementing the Russian Peasant method for multiplication.
a) Write a program to find the product of two intergers.
b) Modify your program so that it repeatedly asks whether you want to find
another product.'''
#Entry message
print("Welcome to the russian peasant method for multiplication.")
print("This program will show each step and what process took place.")
print("To exit: type 'exit' at anytime.")
print()
#Loop to keep asking user for input
while True:
innerLoop = True
numberA = input("What is your first integer? : ")
numberA = numberA.lower()
#exit program by user
if str(numberA) == 'exit':
break
numberA = int(numberA)
numberB = input("What is your second integer? : ")
numberB = numberB.lower()
#exit program by user
if str(numberB) == 'exit':
break
numberB = int(numberB)
#Zero sum, empty comment and message for table
sum = 0
comment = ""
message = ""
print("__" * 21)
#Headline
print(f'{"A":>5}|{"B":>5}|{"Comment":<30}|')
while innerLoop:
#If B is odd
if numberB % 2 == 1:
#If B is zero, exit inner loop to print answer
if numberB == 0:
break
comment = "Add A to the product, B is odd"
#Table
message = f'{numberA:>5}|{numberB:>5}|{comment:<20}|'
print(message)
#Calculation if B is odd
numberB = int(numberB/2)
sum = sum + numberA
numberA = numberA * 2
else:
#exit inner loop to print answer
if numberB == 0:
break
comment = "Ignore this A value, B is even"
#Table
message = f'{numberA:>5}|{numberB:>5}|{comment:<20}|'
print(message)
#Calculation if B is even
numberA = numberA * 2
numberB = int(numberB/2)
#Answer
print("__" * 21)
print("Your total is " + str(sum))
print("**" * 21)
#Exit message
print("Thank you for trying the Russian method multiplication calculator.")
print("Have a great day!")
| true |
369cb12749638b0613a81d2d86172dad72c00ba0 | Amidala1/GBPython | /Lesson3/Lesson_3_1.py | 973 | 4.21875 | 4 | """1. Реализовать функцию, принимающую два числа (позиционные аргументы) и выполняющую их деление. Числа запрашивать
у пользователя, предусмотреть обработку ситуации деления на ноль.
"""
def division(num1, num2):
"""
Функция для расчета деления двух чисел
"""
try:
return round(num1/num2, 2)
except ZeroDivisionError:
return "Деление на ноль!"
print("*** Программа деления двух чисел ***")
while True:
try:
number1 = int(input("Введите первое число: "))
number2 = int(input("Введите второе число: "))
except ValueError:
print("Введите числовое значение")
else:
print(division(number1, number2))
break
| false |
0a6a97d35f72d46753285d3b29f9df118009cbee | Amidala1/GBPython | /Lesson4/Lesson_4_2.py | 1,019 | 4.34375 | 4 | """
2. Представлен список чисел. Необходимо вывести элементы исходного списка, значения которых больше предыдущего элемента.
Подсказка: элементы, удовлетворяющие условию, оформить в виде списка. Для формирования списка использовать генератор.
Пример исходного списка: [300, 2, 12, 44, 1, 1, 4, 10, 7, 1, 78, 123, 55].
Результат: [12, 44, 4, 10, 78, 123].
"""
from random import randint
start_number = 1
end_number = 1000
el_count = 9
numbers_list = [randint(start_number, end_number) for x in range(el_count)]
new_numbers_list = [value for index, value in enumerate(numbers_list) if (index > 0) and
(numbers_list[index] > numbers_list[index - 1])]
print(f"Исходный список: {numbers_list}\nОбработанный список: {new_numbers_list}")
| false |
1ea2adeb0c26ef4d01beb83f4393902b79a35678 | krrishsgk/Data-Structures | /LeetCode/04-NumberPalindrome.py | 930 | 4.28125 | 4 | """
Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.
Example 1:
Input: 121
Output: true
Example 2:
Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
Example 3:
Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
Follow up:
Coud you solve it without converting the integer to a string?
"""
class Solution:
def isPalindrome(self, number: int) -> bool:
string = str(number)
palindrome = True
counter = 0
while counter < len(string) and palindrome == True:
if string[counter] == string[-(counter+1)]:
pass
else:
palindrome = False
break
counter = counter + 1
return palindrome
| true |
20186fc7f5e9b93c5bea045b3a094399e3b99957 | krrishsgk/Data-Structures | /LeetCode/2-ReverseInteger.py | 743 | 4.25 | 4 | """
Given a 32-bit signed integer, reverse digits of an integer.
Example 1:
Input: 123
Output: 321
Example 2:
Input: -123
Output: -321
Example 3:
Input: 120
Output: 21
Note:
Assume we are dealing with an environment which could only store integers within
the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this
problem, assume that your function returns 0 when the reversed integer
overflows.
"""
class Solution:
def reverse(self, x: int) -> int:
sign = (x>0)-(x<0)
x = abs(x)
digits = [0]
while(x > 0):
digits.append(x%10)
x = x // 10
reverse = sign*int(''.join(map(str, digits)))
return reverse*(-2**31 < reverse < 2**31)
| true |
ca0b23af364b9ce99ae61898adfa1525c2fca981 | mengyuqianxun/Python-Algorithm-interview | /ch2_stack_queue/3.py | 1,412 | 4.25 | 4 | class Stack:
#模拟栈
def __init__(self):
self.items=[]
#判断栈是否为空
def isEmpty(self):
return len(self.items)==0
#返回栈的大小
def size(self):
return len(self.items)
#返回栈顶元素
def top(self):
if not self.isEmpty():
return self.items[len(self.items)-1]
else:
return None
#弹栈
def pop(self):
if len(self.items)>0:
return self.items.pop()
else:
print(u"栈已经为空")
return None
#压栈
def push(self,item):
self.items.append(item)
#把栈底元素移动到栈顶
def moveBottomToTop(s):
if s.isEmpty():
return
top1 = s.top()
s.pop()
if not s.isEmpty():
#递归处理不包含栈顶元素的子栈
moveBottomToTop(s)
top2 = s.top()
s.pop()
#交换栈顶元素与子栈栈顶元素
s.push(top1)
s.push(top2)
else:
s.push(top1)
def reverse_stack(s):
if s.isEmpty():
return
moveBottomToTop(s)
top1 = s.top()
s.pop()
reverse_stack(s)
s.push(top1)
if __name__ == "__main__":
s = Stack()
s.push(1)
s.push(2)
s.push(3)
s.push(4)
s.push(5)
reverse_stack(s)
print(u"翻转后出栈顺序为")
while not s.isEmpty():
print(s.top())
s.pop() | false |
8ba957ec1958047c49a57a4fcf096ba0a7fe1d9b | arpitrajput/20_Days_of_Code | /Day_12_Compound_Interest.py | 412 | 4.125 | 4 | # Python3 program to find compound interest for given values.
def compound_interest(principle, rate, time):
CI = principle * (pow((1 + rate / 100), time)) - principle
print("Compound interest is", CI)
principal = float(input("Enter the principal amount: \n"))
rate = float(input("Enter the interest rate: \n"))
time = float(input("Enter the time in years: \n"))
compound_interest(principal,rate,time)
| true |
c2b946b650e88d82d8c2f166218e70f9e871e294 | truongnguyenlinh/procedural_python | /A1/random_game.py | 1,126 | 4.1875 | 4 | # Linh Truong
# A01081792
# 01-25-2019
import random
def rock_paper_scissors():
"""Generate a game of rock, paper or scissors with the computer.
"""
random_num = random.randint(0, 2)
options = ['rock', 'paper', 'scissors']
random_choice = options[random_num]
guess = input("Rock, paper or scissors?").strip().lower()
if (guess != "rock") and (guess != "paper") and (guess != "scissors"):
print("Incorrect value entered.")
elif (guess == "paper") and (random_choice == "rock"):
print("The computer entered rock so you've won!")
elif (guess == "scissors") and (random_choice == "paper"):
print("The computer entered paper so you've won!")
elif (guess == "rock") and (random_num == "scissors"):
print("The computer entered scissors so you've won!")
elif guess == random_choice:
print("The computer also entered", random_choice, "so it was a tie.")
else:
print("The computer entered", random_choice, "so you've lost.")
def main():
"""Execute the program."""
rock_paper_scissors()
if __name__ == '__main__':
main()
| true |
6f2af4445305821d674ca2888ac325eead368c38 | candytale55/list-comprehension-challenges-Python-3 | /greetings.py | 275 | 4.375 | 4 | # Create a new list named greetings that adds "Hello, " in front of each name in the list names.
names = ["Elaine", "George", "Jerry", "Cosmo"]
greetings = ["Hello, " + x for x in names]
print(greetings)
# ['Hello, Elaine', 'Hello, George', 'Hello, Jerry', 'Hello, Cosmo']
| false |
67ff4b0237870ed57bfa8c65698e8d9d4eb28c7f | Environmental-Informatics/python-learning-the-basics-RedpandaTheNinja | /Exercise_3.3.py | 1,236 | 4.375 | 4 | #Kevin Lee
#Assin 1
#examples from text book
# print('+','-','-','-','-','+','-','-','-','-','+')
# print('+', end=' ')
# print('-')
# print('|', ' ', ' ', ' ',' ', '|', ' ', ' ', ' ',' ' ,'|')
#creating 2x2
#first create each side by string of charc
string1 = "+----+----+"
string2 ="| | |"
# print(string1)
#create a function 2 different aruments as inputs
#cprint as much as problem asks to
def easy(arg1, arg2):
print(arg1)
print(arg2)
print( arg2 )
print( arg2 )
print( arg2 )
print( arg1 )
print(arg2)
print( arg2 )
print( arg2 )
print( arg2 )
print( arg1 )
easy(string1,string2)
#double the length from previous stringg
#since it is being called later, overwirte can be performed
string1 = "+----+----+----+----+"
string2 ="| | | | |"
def easy2(arg1, arg2):
print( arg1 )
print( arg2 )
print( arg2 )
print( arg2 )
print( arg2 )
print( arg1 )
print( arg2 )
print( arg2 )
print( arg2 )
print( arg2 )
print( arg1 )
print( arg2 )
print( arg2 )
print( arg2 )
print( arg2 )
print( arg1 )
print( arg2 )
print( arg2 )
print( arg2 )
print( arg2 )
print( arg1 )
easy2( string1 , string2 )
| true |
55d50be16cfd0fa6606a4fab7974eec5eb33c32b | liliarc96/Aula-Python | /Exercicios_2/positivo_negativo.py | 670 | 4.1875 | 4 | '''
• Faça um programa que lê 10 números, calcula e
exibe:
• O triplo de cada número;
• A mensagem “É positivo”, caso o número digitado seja
positivo, ou “É negativo”, caso seja negativo;
'''
cont = 1
#Em while
while (cont <= 10):
numero = int(input("Digite um número:"))
numero = numero*3
print("O triplo desse número é",numero)
if (numero >= 0):
print("É positivo!")
else:
print("É negativo!")
cont += 1
cont_2 = 0
#Em for
for cont_2 in range (0,10):
numero = int(input("Digite um número:"))
numero = numero*3
print("O triplo desse número é",numero)
if (numero >= 0):
print("É positivo!")
else:
print("É negativo!")
| false |
fdcc61259401bb0fe0f23c59e36a95fa4d334eaa | liliarc96/Aula-Python | /Exercicios_2/mercadinho.py | 1,075 | 4.15625 | 4 | '''
• Faça um programa que repita as seguintes tarefas,
até que o código 0 seja digitado:
•Leia o código do produto;
•Leia a quantidade adquirida;
•Se o código for 1, escreva 'Caderno - R$ 12.00';
•Se for 2, escreva 'Régua - R$ 2.50';
•Se for 3, escreva 'Borracha - R$ 0.25';
•Se for 4, escreva 'Mochila - R$ 50.00';
•Calcule e exiba o total a ser pago (valor * quantidade);
'''
codigo = 5
valor_total = 0.00
quantidade = 0
print(''' CÓDIGO 1 = Caderno - R$ 12.00;
CÓDIGO 2 = Régua - R$ 2.50;
CÓDIGO 3 = Borracha - R$ 0.25;
CÓDIGO 4 = Mochila - R$ 50.00;
SAIR - 0
''')
while codigo != 0:
codigo = int(input("\nDigite o código do produto: "))
if codigo == 1:
valor = 12.00
elif codigo == 2:
valor = 2.50
elif codigo == 3:
valor = 0.25
elif codigo == 4:
valor = 50.00
else:
valor = 0.00
if codigo > 0 and codigo < 5:
quantidade = int(input("\nDigite a quantidade adquirida: "))
else:
quantidade = 0
valor_total = (valor_total + (valor * quantidade))
print("\nO valor total da compra foi de R$",valor_total)
| false |
798484ec330a30ff093df934a8f70878cc38801c | AriHiggs89/INFOTC-2040-CHALLENGES | /Python Challenges/animalclass.py | 1,053 | 4.28125 | 4 | # This is the program that defines an animal's class
# By: Ariana Higgins
import random
class Animal:
# function that defines what happens when a new animal object is created
def __init__(self, animal_type, animal_name):
self.__animal_type = animal_type
self.__name = animal_name
# generates a random number between 1 and 3
random_number = random.randint(1, 3)
# this gives the animal it's mood based on the random number generated
if random_number == 1:
self.__animalmood = "happy"
elif random_number == 2:
self.__animalmood = "hungry"
elif random_number == 3:
self.__animalmood = "sleepy"
# function that returns the animal type
def get_animal_type(self):
return self.__animal_type
# function that returns the animal name
def get_name(self):
return self.__name
# function that returns animal mood
def check_mood(self):
return self.__animalmood
| true |
c47ad8c8f14f73ba726a9ed664f72cba10622e6d | Nandita-20/pythoncode | /miscellanous_finctions.py | 1,650 | 5.0625 | 5 | #Range() function provides us list of number to iterate andis used with loop
'''print("print negative values on reverse order\n")
for i in range(-7,-3):
print(i)
print("print negative to positive values on reverse order\n")
for i in range(-7,3):
print(i)
print("print values in reverse order by jumping r values\n")
for i in range(7,1,-2):
print(i)
print("print values upto the given limit\n")
for i in range(8):
print(i)
print("print values from lower to upper limit m-n\n")
for i in range(1,8):
print(i)'''
#######################################################################################
#zip() : is a built-in Python function that gives us an iterator of tuples
'''Like a .zip file holds real files within itself, a zip holds real data within.
It takes iterable elements as input and returns an iterator on them (an iterator of tuples).
It evaluates the iterables left to right.'''
#zip function using multiple argument of same length
'''a=[1,2,3]
b=['nandita','sangita','ankita']
print(dict(zip(a,b)))'''
'''for i in zip(['red','green'],['rose','leaves'],['%','#','$']):
print(i)'''
#zip function using multiple argument of same length
#Note: When we provide multiple lists of different lengths, it stops at the shortest one.
'''for i in zip(['red','green'],['rose','leaves'],['%']):
print(i)'''
#op: ('red', 'rose', '%')
#zip() function using single argument
'''for i in zip([1,2,3]):
print(i)'''
#unzipping() function in python
'''z=zip(['red','green'],['rose','leaves'],['%','#'])
a,b,c=zip(*z)
print(a,b,c)'''
#O/P: Now it is unpacked: ('red', 'green') ('rose', 'leaves') ('%', '#')
| true |
363d5bdc3d0430f439f2b7865f17f1a612d875c2 | farjadfazli/code-challenges | /binarysearch_rotate-by-90-degrees-counter-clockwise.py | 531 | 4.25 | 4 | """
Rotate by 90 Degrees Counter-Clockwise
Given a two-dimensional square matrix, rotate it 90 degrees counter-clockwise.
Example 1
Input
matrix = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]
Output
[
[3, 6, 9],
[2, 5, 8],
[1, 4, 7]
]
"""
class Solution:
def solve(self, matrix):
# first we zip the matrix in order to combine every ith element of each list
zipped = zip(*matrix)
# then we can reverse this to obtain the desired rotation
return (list((reversed(list(zipped))))) | true |
58b41bfbc73e9ff14f1d5a2b74a25e7901abc1e5 | farjadfazli/code-challenges | /reverse-integer.py | 1,078 | 4.125 | 4 | """
Given a 32-bit signed integer, reverse digits of an integer.
Example 1:
Input: 123
Output: 321
Example 2:
Input: -123
Output: -321
Example 3:
Input: 120
Output: 21
Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
Source: https://leetcode.com/problems/reverse-integer/
"""
class Solution:
def reverse(self, x: int) -> int:
negative = False
if (x < 0):
negative = True
reversed = 0
current = x
if (negative):
current *= -1
while current > 0:
reversed *= 10
calculation = current % 10
reversed += calculation
import math
current = current // 10
print(reversed)
if (negative):
reversed *= -1
if (reversed > 2147483647 or reversed < -2147483648):
return 0
return reversed | true |
f35b661e6f549c3b00e5a7d865b43082fb0b5a0a | farjadfazli/code-challenges | /palindrome-number.py | 1,316 | 4.28125 | 4 | """
Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.
Follow up: Could you solve it without converting the integer to a string?
Example 1:
Input: x = 121
Output: true
Example 2:
Input: x = -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
Example 3:
Input: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
Example 4:
Input: x = -101
Output: false
Constraints:
-231 <= x <= 231 - 1
Source: https://leetcode.com/problems/palindrome-number/
"""
class Solution:
def isPalindrome(self, x: int) -> bool:
# we will determine if the integer is a palindrome by reversing it
# and seeing if reversed integer == integer
if x < 0:
return False
reversed = 0
current = x
while current > 0:
reversed *= 10
calculation = current % 10
reversed += calculation
current = current // 10
if reversed == x:
return True
return False
# alternate solution by converting integer to string:
# if str(x) == "".join(reversed(str(x))):
# return True
| true |
d84a0fb6c8a41613165765593791dd1a86080b6e | farjadfazli/code-challenges | /binarysearch_transpose-of-a-matrix.py | 975 | 4.1875 | 4 | """
Transpose of a Matrix
Given an integer square (n by n) matrix, return its transpose. A transpose of a matrix switches the row and column indices. That is, for every r and c, matrix[r][c] = matrix[c][r].
For example, given matrix
[1, 2, 3]
[4, 5, 6]
[7, 8, 9]
it becomes
[1, 4, 7]
[2, 5, 8]
[3, 6, 9]
Constraints
n ≤ 100
Example 1
Input
matrix = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]
Output
[
[1, 4, 7],
[2, 5, 8],
[3, 6, 9]
]
"""
class Solution:
def solve(self, matrix):
# we will take the nth element of each sublist and put it in the nth sublist of the new matrix
result = []
length = len(matrix)
counter = 0
while counter < length:
for row in matrix:
result.append(row[counter])
counter += 1
i = 0
final = []
while i < len(result):
final.append(result[i:i+length])
i+=length
return final | true |
f331ba6ce83fe06967d7ed9948d1b0f50eb9a602 | farjadfazli/code-challenges | /binarysearch_counting-dinosaurs.py | 809 | 4.125 | 4 | """
Counting Dinosaurs
You are given a string animals and another string dinosaurs. Every letter in animals represents a different type of animal and every unique character in dinosaurs represents a different dinosaur.
Return the total number of dinosaurs in animals.
Example 1
Input
animals = "abacabC"
dinosaurs = "bC"
Output
3
Explanation
There's two types of dinosaurs "b" and "C". There's 2 "b" dinosaurs and 1 "C" dinosaur. Note we didn't count the lowercase "c" animal as a dinosaur.
"""
class Solution:
def solve(self, animals, dinosaurs):
from collections import Counter
dinosaurs = set(dinosaurs)
counter = 0
animalsDict = Counter(animals)
print(animalsDict)
for i in dinosaurs:
counter += animalsDict[i]
return counter | true |
f4e1b664751849880849d57b522313e4ce1a92b5 | macraiu/software_training | /leetcode/py/414_Third_Maximum_Number.py | 1,103 | 4.125 | 4 | """ Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n). """
def thirdMax(nums):
a = sorted(set(nums))
if len(a) < 3:
return max(a)
else:
return a[-3]
import time
import random
A = []
for i in range(0,10000):
n = random.randint(-10000,10000)
A.append(n)
t0 = time.clock()
print(thirdMax(A))
t1 = time.clock() - t0
print("first ", t1)
def thirdMax2(nums):
# ONE SUBMISSIONS WITH GOOD TIME
# Make a Set with the input.
nums = set(nums)
# Find the maximum.
maximum = max(nums)
# Check whether or not this is a case where
# we need to return the *maximum*.
if len(nums) < 3:
return maximum
# Otherwise, continue on to finding the third maximum.
nums.remove(maximum)
second_maximum = max(nums)
nums.remove(second_maximum)
return max(nums)
t0 = time.clock()
print(thirdMax2(A))
t1 = time.clock() - t0
print("second ", t1)
# FOR HIGH VARIANCE IN THE DISTRIBUTION MINE PERFORMS BETTER | true |
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