blob_id string | repo_name string | path string | length_bytes int64 | score float64 | int_score int64 | text string | is_english bool |
|---|---|---|---|---|---|---|---|
56e63c8573d5763914cc034fa746d8f591e804d0 | hannavw/dsf_exercises | /snack_names_2.py | 1,402 | 4.25 | 4 | # NEW
# Make a dictionary:
friends = [
{
"friend": "Marleen",
},
{
"friend": "Femke",
},
{
"friend": "Nienke",
},
]
# Ask user for input
index = 0
while index < 3:
name_length = len(friends[index]["friend"])
friends[index]["length"] = name_length
snack = input(f"{friends[index]['friend']}, what's your favourite snack? ")
friends[index]["snack"] = snack
index = index + 1
index = 0
for item in friends:
print(f"{friends[index]['friend']}, you're favourite snack is {friends[index]['snack']}.")
print(f"You're name has {friends[index]['length']} characters. ")
index = index + 1
# OLD
## Put my friends in a list
#friends = ["Marleen", "Femke", "Nienke"]
#snacks = []
#
## Ask each friend for snack and add snack to a list
#index = 0
#for friend in friends:
# snack = input(friend + ", What's your favourite snack? ")
# snacks.append(snack)
# index = index + 1
#
## Loop through friends and place a snack from the snack-list after each friend.
#index = 0
#for friend in friends:
# length = len(friends[index])
# snack = snacks[index]
# print(friend + ", you're favourite snack is: " + snack + ", and you're name has " + str(length) + " characters.")
# index = index + 1 | false |
e19960770dbbfd23b640cf6812d02e8d87666faf | Kajal2000/Python_Udemy_Course | /ex_7.py | 1,628 | 4.5 | 4 | from collections import OrderedDict
# Question 1: Create an empty dictionary called life_book of type OrderedDict.
# From Python 3.6 onwards dictionaries maintain the order in which they were
# created, however, this coding environment uses an earlier version of Python 3.
# Therefore, use OrderedDict in this case, example my_dict = OrderedDict().
# Once created, use the print function to print it.
## Write your code below, 2 lines
life_book = OrderedDict()
print(life_book)
# End question 1
# Question 2: Add the following key, value pair to the life_book dictionary
# 'pet' -> 'dog', so the key is pet and the value is dog and then print it.
## Write your code below, 2 lines
life_book["pet"]="dog"
print(life_book)
# End question 2
# Question 3: Add the following key, value pairs to the dictionary
# 'second_pet' -> 'cat'
# 'first_child' -> 'boy'
# 'second_child' -> 'girl'
# Once added, print the dictionary.
## Write your code below, variable lines
life_book["second_pet"]="cat"
life_book["first_child"]="boy"
life_book["second_child"]="girl"
print(life_book)
# End question 3
# Question 4: Update the value associated with key 'pet' to be 'hamster', then
# print the dictionary.
## Write your code below, 2 lines
life_book["pet"]='hamster'
print(life_book)
# End question 4
# Question 5: Given the dictionary below, use the items() method and save
# the value of all the key value pairs to the variable courses_iterable.
my_courses = {'a':'python', 'b':'javascript', 'c':'ruby on rails', 'd':'machine learning', 'e':'ai'}
courses_iterable = my_courses.items()
## Write your code below, 1 line
# End question 5
| true |
b824f56dc6fbc4daa9238f9570b69f4690eaf5eb | yiyisf/python_tools | /simple/random_chars.py | 573 | 4.125 | 4 | import random
"""
生成随机字符串，至少包含一个大写/小写/数字
"""
char_set = "abcdefghijklmnopqrstuvwxyz"
upper_char_set = char_set.upper()
pw_len = 8
pwlist = []
for i in range(pw_len//3):
pwlist.append(char_set[random.randrange(len(char_set))])
pwlist.append(upper_char_set[random.randrange(len(upper_char_set))])
pwlist.append(str(random.randrange(10)))
for i in range(pw_len- len(pwlist)):
pwlist.append(char_set[random.randrange(len(char_set))])
print("".join(pwlist))
random.shuffle(pwlist)
pwstr = "".join(pwlist)
print(pwstr) | false |
6f0a8f2072e8dec5407499f04777e0d887bbb9ae | duttashi/wrangler | /python-3/experiments/expr_spin_certain_words_in_string.py | 2,493 | 4.34375 | 4 | """
Question: Write a function that takes in a string of one or
more words, and returns the same string, but with all five or
more letter words reversed (Just like the name of this Kata).
Strings passed in will consist of only letters and spaces.
Spaces will be included only when more than one word is present.
Examples: spinWords( "Hey fellow warriors" ) => returns "Hey wollef sroirraw"
spinWords( "This is a test") => returns "This is a test"
spinWords( "This is another test" )=> returns "This is rehtona test"
This question was asked on Codewars website.
PROGRAM LOGIC
--------------------
Declare two empty lists called list_str_to_words and list_spin_words.
split the string into words on basis of spaces and save to a list_str_to_words.
Iterate over the list and count the character length of each word.
If the word length is greater than 5 then reverse the word and
write to list_spin_words
if the word length is not greater tha 5, add the word to
list_spin_words
---------------------
Created on Wed Oct 7 15:35:34 2020
@author: Ashish
"""
# declare global string variable
some_string = "Hey fellow warriors"
# declare global lists
list_str_to_words, list_spin_words, rev_word_list = [], [], []
print(list_spin_words)
# split string into a list of words
list_str_to_words = some_string.split()
# print(list_str_to_words)
# find word length
for word in list_str_to_words:
# print(str)
if(len(word) > 5):
# print(word)
# add word to list
list_spin_words.append(word)
else:
rev_word = " ".join(reversed(word))
list_spin_words.append(rev_word)
str_to_return = " ".join(list_spin_words)
# print(list_spin_words)
print("Original string is: ", some_string)
print("Conditional reversed string is: ", str_to_return)
# create a custom function using the above code logic
# function will accept a strig as input and return the reverse string
def conditional_reverse_string(some_string):
list_str_to_words = some_string.split()
for word in list_str_to_words:
# print(str)
if(len(word) > 5):
# print(word)
# add word to list
list_spin_words.append(word)
else:
rev_word = " ".join(reversed(word))
list_spin_words.append(rev_word)
str_to_return = " ".join(list_spin_words)
return(str_to_return)
# Invoke the function
print(conditional_reverse_string(some_string))
| true |
9123eb5c6bb10745e8d294379fcf2e5f568d2906 | cesareferrari/Exercises | /recursion/reverse_string_solution.py | 365 | 4.46875 | 4 | """
Write a function that takes a string and outputs the string reversed.
Use recursion (call the function inside the function).
"""
def reverse_string(string):
if len(string) == 0:
return string
else:
return reverse_string(string[1:]) + string[0]
string = "Hello, World!"
result = reverse_string(string)
print(result == "!dlroW ,olleH")
| true |
6012d6a46724303b50ec846471f5f3c0ffb75bfa | kevfallon97/algorithms_and_data_structures | /array_sequences/anagram_check.py | 770 | 4.15625 | 4 | '''
ANAGRAM CHECK
PROBLEM STATEMENT
Given two strings, check to see if they are anagrams.
An anagram is when the two strings can be written using the exact same letters
(so you can just rearrange the letters to get a different phrase or word).
For example:
"public relations" is an anagram of "crap built on lies."
"clint eastwood" is an anagram of "old west action"
Note: Ignore spaces and capitalization.
'''
# SOLUTION 1
def anagram_1(s1, s2):
s1 = s1.replace(' ', '').lower()
s2 = s2.replace(' ', '').lower()
return sorted(s1) == sorted(s2)
# SOLUTION 2
def anagram_2(s1, s2):
s1 = list(s1.replace(' ', '').lower())
s2 = list(s2.replace(' ', '').lower())
for letter in s1:
if letter in s2:
s2.remove(letter)
else:
return False
return True
| true |
b371703aaa423d32be96d8a99047f214619d3276 | kevfallon97/algorithms_and_data_structures | /array_sequences/array_pair_sum.py | 788 | 4.25 | 4 | '''
ARRAY PAIR SUM
PROBLEM STATEMENT
Given an integer array, output all the unique pairs that sum to a specific value k.
So the input:
pair_sum([1,3,2,2], 4)
would return 2 pairs:
(1,3)
(2,2)
Note: For testing purposes, return the number of pairs identified
'''
# SOLUTION
def pair_sum(nums, k):
pairs = set() # hold unique pairs in set
for i, numA in enumerate(nums[:-1]): # for every elem in list, check for pair with remaining elements
for numB in nums[i+1:]:
if numA + numB == k: # check if pair sum to k
pairs.add((numA, numB)) # add valid pair provided they are unique
return pairs # return set of unique pairs
nums = [1,3,2,2,2]
pairs = pair_sum(nums, 4)
print(f"Number of pairs: {len(pairs)}")
print(f"Unique pairs: {pairs}")
| true |
7f15ec5260ba537d9de81a7bb3450feae56eb083 | ParaBeIlum/Python_Homeworks | /GeekBrains/Урок 1. Введение в Алгоритмизацию и простые алгоритмы на Python/lesson_1_task_1.py | 975 | 4.25 | 4 | # 1. Выполнить логические побитовые операции «И», «ИЛИ» и др. над числами 5 и 6.
# Выполнить над числом 5 побитовый сдвиг вправо и влево на два знака.
operation = input('Введите тип битовой операции (унарные/бинарные): ')
if operation == 'унарные':
print(f'Побитовый сдвиг числа 5 вправо на 2 знака: 5 >> 2 = {5 >> 2}')
print(f'Побитовый сдвиг числа 5 влево на 2 знака: 5 << 2 = {5 << 2}')
elif operation == 'бинарные':
print(f'Побитовая операция "AND" над числами 5 и 6: 5 & 6 = {5 & 6}')
print(f'Побитовая операция "OR" над числами 5 и 6: 5 | 6 = {5 | 6}')
print(f'Побитовая операция "XOR" над числами 5 и 6: 5 ^ 6 = {5 ^ 6}')
| false |
6d1801b115990e2a3ef9dcd8a165121e5b87a5c1 | ParaBeIlum/Python_Homeworks | /GeekBrains/Урок 1. Введение в Алгоритмизацию и простые алгоритмы на Python/lesson_1_task_4.py | 944 | 4.1875 | 4 | # 4. Пользователь вводит две буквы. Определить, на каких местах
# алфавита они стоят, и сколько между ними находится букв.
print("Введите 2 строчных буквы английского алфавита, будет вычислено, какое место "
"в алфавите занимает каждая буква и сколько букв находится между ними")
letter1 = input("Введите первую букву: ").lower()
letter2 = input("Введите вторую букву: ").lower()
letter1pos = ord(letter1) - 96
letter2pos = ord(letter2) - 96
posDiff = abs(letter1pos - letter2pos) - 1
print(f'Буква "{letter1}" занимает место {letter1pos}\n'
f'Буква "{letter2}" занимает место {letter2pos}\n'
f'Между ними букв: {posDiff}')
| false |
04e5801a64567b5278cdb6a31b86f5d9992f74fe | aluisq/Python | /estrutura_repeticao/ex13.py | 379 | 4.15625 | 4 | numero = abs(int(input("Digite um número: ")))
while numero % 2 != 0:
print("""
__________________________
Número inválido
__________________________
""")
numero = abs(int(input("Digite um número: ")))
if numero % 2 == 0:
for i in range(0,numero + 2, 2):
print(i)
else:
print("Algo está errado.")
| false |
ae005f3eee91d52dac933f43a6cd9ed9afdd364e | LibraZYJ/python-learn | /基本语法/Day14/元组.py | 1,014 | 4.21875 | 4 | """
Python中的元祖与列表类型也是一种容器数据类型，可以用一个变量（对象）来存储多个数据，
不同之处在于元祖的元素不能修改，在前面的代码中我们已经不止一次使用过元祖了，顾名思义
我们把多个元素组合到一起就形成了一个元祖，所以它和列表一样可以保存多条数据。
@Date 2020.4.7
"""
# 定义元祖
t = ('Zhao', 20, True, '南京')
print(t)
# 获取元祖中的元素
print(t[0])
print(t[3])
# 遍历元祖中的值
for member in t:
print(member)
# 重新给元祖赋值
# 发生异常: TypeError
# 变量t重新引用了新的元祖原来的元祖被垃圾回收
t = ('QQ', 22, True, '江苏南京')
print(t)
# 将元祖转换成列表
person = list(t)
print(person)
# 列表是可以修改它的元素的
person[0] = 'Li'
person[1] = 25
print(person)
# 将列表转换成元祖
print(tuple(person))
fruits_list = ['apple', 'banana', 'orange']
fruits_tuple = tuple(fruits_list)
print(fruits_tuple)
| false |
2dbc889c1895c3e7f030cce40d0157ca0ece2ed0 | eduardknezovic/chess-blindfold-trainer | /app/main.py | 1,169 | 4.34375 | 4 |
"""
In this we are to create an app that will be used to check our knowledge of square colors of the chessboard.
"""
import random
def get_y_or_n():
value = input("Leave empty if white, enter anything for black:")
if value == "":
return True
else:
return False
def generate_set_of_squares():
result = []
is_white = True
for letter in ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']:
is_white = not is_white
for number in range(1, 9):
square = (letter, number)
item = (square, is_white)
result.append(item)
is_white = not is_white
return result
def main():
all_squares = generate_set_of_squares()
result_to_text = {True: "WHITE", False: "BLACK"}
user_outcome_to_text = {True: "WIN!", False: "FAIL!"}
while True:
square, is_white = random.choice(all_squares)
string_square = square[0] + str(square[1])
print(string_square)
user_is_white = get_y_or_n()
print(f"{user_outcome_to_text[user_is_white==is_white]} The square {string_square} is {result_to_text[is_white]}")
if __name__ == "__main__":
main() | true |
6069ced4148ac08f8c362dc32f7bc6ef5e370eba | Suyen-Shrestha/Python-Assignment-II | /Solution9.py | 728 | 4.125 | 4 |
def binary_search(sequence,item):
lower = 0
upper = len(sequence) - 1
while lower <= upper:
mid_index = (lower+upper) // 2
if sequence[mid_index] == item:
return mid_index
elif sequence[mid_index] < item:
lower = mid_index + 1
else:
upper = mid_index - 1
return -1
sample_li = [5,9,6,23,54,68,17,32]
print(f'The sample list: {sample_li}')
sample_li.sort() # sorting is require for performing binary search.
print(f'The list after sorting: {sample_li}')
print(f'The index of "17" in sorted list using binary search: {binary_search(sample_li,17)}')
print(f'The index of "2" in sorted list using binary search: {binary_search(sample_li,2)}') | true |
779abb883726bc1adfb10aef750f7865c144f14b | James-E-Sullivan/BU-MET-CS521 | /Module-4/HW_10.1_Sullivan_James.py | 2,001 | 4.25 | 4 | """
10.1 (Assign grades)
Write a program that reads a list of scores and then assigns grades based on
the following scheme:
A if score >= best - 10
B if score >= best - 20
C if score >= best - 30
D if score >= best - 40
F otherwise
"""
def grade_score(student_score, max_score):
"""
Assigns a grade based on a student's score and the maximum score
:param student_score: The student score(int or float)
:param max_score: The maximum score (int or float)
:return student_grade:
"""
if student_score >= max_score - 10:
student_grade = 'A'
elif student_score >= max_score - 20:
student_grade = 'B'
elif student_score >= max_score - 30:
student_grade = 'C'
elif student_score >= max_score - 40:
student_grade = 'D'
else:
student_grade = 'F'
return student_grade
def find_best_score(input_scores):
"""
Evaluates user input scores and returns the greatest score
:param input_scores: List of scores (integers)
:return best_score: Returns greatest score in the list
"""
# Starts best score at 0. Assumes no negative scores are possible.
best_score = 0
# Iterates through scores and updates best_score if a better score found
for student_score in input_scores:
new_score = student_score
if new_score > best_score:
best_score = new_score
return best_score
while True:
try:
score_list = [int(input_score) for input_score in input(
'Enter scores (integers separated by spaces): ').split()]
student_number = 0
for score in score_list:
grade = grade_score(score, find_best_score(score_list))
print('Student', student_number, 'score is', score,
'and grade is', grade)
student_number += 1
break
# User prompted again if they enter invalid input
except ValueError:
print("Could not convert input to integer. Please try again.")
| true |
00c33f31786aecdb7653ad1bf73bcc3d59934bec | James-E-Sullivan/BU-MET-CS521 | /Module-3/HW_5.1_Sullivan_James.py | 1,347 | 4.15625 | 4 | """
(Count positive and negative numbers and compute the average of numbers)
Write a program that reads an unspecified number of integers, determines
how many positive and negative values have been read, and computes the
total and average of the input values (not counting zeroes). Your
program ends with the input '0'. Display the average as a floating-point
number.
"""
positives = 0
negatives = 0
count = 0
total = 0
while True:
# User prompted for integer
user_int = eval(input("Enter an integer, the input ends if it is 0: "))
# If user enters 0 before entering an integer, user prompted again
if user_int == 0 and count == 0:
print("You didn't enter any number")
continue
# If user enters 0 after entering at least 1 int, loop exits
elif user_int == 0 and count > 0:
break
# If user enters int, count incremented, int added to total, and
# positives or negatives incremented, depending on value.
else:
count += 1
total += user_int
if user_int > 0:
positives += 1
elif user_int < 0:
negatives += 1
# Prints positives, negatives, total, and average values
print("The number of positives is", positives)
print("The number of negatives is", negatives)
print("The total is", total)
print("The average is", total / count)
| true |
8e3cd86bbb8bd9b80c248992fc9a30c2f13a16f3 | chriscent27/calculator | /calculator.py | 740 | 4.1875 | 4 | def addition(num1, num2):
return num1 + num2
def subtraction(num1, num2):
return num1 - num2
def multiplication(num1, num2):
return num1 * num2
def division(num1, num2):
return num1 / num2
def Calculate():
num1, operation, num2 = (input("Enter operation eg(2 + 5): ").split())
num1 = int(num1)
num2 = int(num2)
if(operation == '+'):
result = addition(num1, num2)
elif(operation == '-'):
result = subtraction(num1, num2)
elif(operation == '*'):
result = multiplication(num1, num2)
elif(operation == '/'):
result = division(num1, num2)
else:
result = 'INVALID CHOICE'
print("Result is: ", result)
#tes
if __name__ == '__main__':
Calculate()
| false |
0a081246bfd34447d50703260786251be84cd860 | kash991064/pythontest1 | /akash8.py | 499 | 4.125 | 4 | #set---is a collection of data
#s1 = set()
#print(type(s1))
#s_from_list =set([1,2,3,4])
#print(s_from_list)
#print(type(s_from_list))
# or
#list1 = (110,232,363,414,525)
#list_set1 = set(list1)
#print(list_set1)
# how to add elements in set--(set add only unique value)
s1 = set()
s1.add(1)
s1.add(23)
print(s1)
# another example
s1 = set()
s1.add(1)
s1.add(23)
s2 = s1.union({23,45,66,77})
print(s1,s2)
# intersction
s1 = set()
s1.add(1)
s2 = s1.intersection({1,45,66,77})
print(s1,s2)
| true |
00a077ed9b2ec9c41824738cb088f0d3455715b8 | ibnahmadCoded/how_to_think_like_a_computer_scientist_Chapter_11 | /collission_checker.py | 2,282 | 4.1875 | 4 | from study import *
class Rectangle:
""" A class to manufacture rectangle objects """
def __init__(self, posn, w, h):
""" Initialize rectangle at posn, with width w, height h """
self.corner = posn
self.width = w
self.height = h
def __str__(self):
return "({0}, {1}, {2})".format(self.corner, self.width,
self.height)
def grow(self, delta_width, delta_height):
""" Grow (or shrink) this object by the deltas """
self.width += delta_width
self.height += delta_height
def move(self, dx, dy):
""" Move this object by the deltas """
self.corner.x += dx
self.corner.y += dy
def area(self):
"""Returns area of rectangle object"""
return self.width * self.height
def perimeter(self):
"""Returns perimeter of rectangle object"""
return 2 * (self.width + self.height)
def flip(self):
"""swaps the height and the width of the object"""
w = self.width
self.width = self.height
self.height = w
def contains(self, point):
a = point.x >= self.corner.x and point.x < self.width
b = point.y >= self.corner.y and point.y < self.height
return a and b
def same_coordinates(self, rectangle):
return (self.corner.x == rectangle.corner.x) and (self.corner.y == rectangle.corner.y)
def get_all_points(self):
"""returns all points in rectangle object"""
points = []
a = list(range(self.corner.x, self.width + 1))
b = list(range(self.corner.y, self.height + 1))
for i in a:
for j in b:
p = (i, j)
points.append(p)
return points
def collision(self, rectangle):
"""checks if the object collides with another rectangle"""
if self.same_coordinates(rectangle):
return True
else:
points = rectangle.get_all_points() #gets all points in d rectangle
for point in points:
if self.contains(point): #if point falls in d object, there's collision
return True
return False
| false |
51655d62767619c3c32bd5015e9eeb655481088e | mh-yesilyurt/globalaihub_ | /last_homework.py | 1,463 | 4.15625 | 4 |
for i in range (0,3):
name=input("Enter your Name and Surname: ")
if name=="Hakan Yesilyurt":
print("Welcome")
break
elif i==3 and name!="Hakan Yesilyurt":
print("Please try again later.")
exit()
courses=[]
for k in range(0,5):
lesson=input("Enter the lesson you want to take (if you don't want to take more lessons, press Q): ")
if lesson.upper()=="Q":
break
else:
courses.append(lesson)
if len(courses)<3:
print("You failed in a class.")
chosen_course=input("Enter the course you want to take exams: ")
midterm_grade=input("Enter your Midterm Grade: ")
final_grade=input("Enter your Final Grade: ")
project_grade=input("Enter your Project Grade: ")
course={"Course Name":chosen_course,"Midterm":midterm_grade,"Final":final_grade,"Project":project_grade}
total_grade=int(course["Midterm"])*0.3+int(course["Final"])*0.5+int(course["Project"])*0.2
if total_grade>= 90 :
print(f"You got AA from \"{chosen_course}\" course")
elif total_grade>=70 and total_grade<90:
print(f"You got BB from \"{chosen_course}\" course")
elif total_grade>=50 and total_grade<70:
print(f"You got CC from \"{chosen_course}\" course")
elif total_grade>=30 and total_grade<50:
print(f"You got CC from \"{chosen_course}\" course")
elif total_grade<30:
print(f"You got FF from \"{chosen_course}\" course. You have failed, please work harder next time.")
| true |
847730cc2367f1f94934d4ed241f3a7deea00c0d | JinbeiZame/Study-python-language | /dictionary.py | 2,265 | 4.4375 | 4 | """Dictionary เก็บข้อมูลได้หลาย ๆ ค่าในตัวแปรเดียวกัน แต่คุณสมบัติพิเศษของดิกชันนารีคือมีข้อมูลตั้งแต่ 2 ค่าขึ้นไปและมีความสัมพันธ์กัน
การประกาศตัวแปร การเพิ่มข้อมูล การแสดงผลข้อมูล การกำหนดตำแหน่งที่ต้องการ และการสลับตำแหน่งคีย์ของข้อมูล
"""
#INSERT
people = {'boy':'gonge','girl':'minnie','animal':'dog','natural':'rain'}
people['socialnetwork'] = 'facebook' #added
#print(people)
#result run after : {'boy': 'gonge', 'girl': 'minnie', 'animal': 'dog', 'natural': 'rain', 'socialnetwork': 'facebook'}
#UPDATE
people['animal'] = 'rabbit' #updated
#print(people)
#result run after : {'boy': 'gonge', 'girl': 'minnie', 'animal': 'rabbit', 'natural': 'rain', 'socialnetwork': 'facebook'}
#DISPLAY DATA IN DICTIONARY
company = {'silicon valley':['facebook','yahoo','microsoft','google','apple','instagram','happy'],'facebook':'mark suckerberg','apple':'stave job'}
#print(company.keys())
#result run after : dict_keys(['silicon valley', 'facebook', 'apple'])
#print(company.values())
#result run after : dict_values([['facebook', 'yahoo', 'microsoft', 'google', 'apple', 'instagram', 'happy'], 'mark suckerberg', 'stave job'])
#print(company.items())
#result run after : dict_items([('silicon valley', ['facebook', 'yahoo', 'microsoft', 'google', 'apple', 'instagram', 'happy']), ('facebook', 'mark suckerberg'), ('apple', 'stave job')])
# print(company)
#result run after : {'silicon valley': ['facebook', 'yahoo', 'microsoft', 'google', 'apple', 'instagram', 'happy'], 'facebook': 'mark suckerberg', 'apple': 'stave job'}
day = {'1':'monday','2':'tuesday','3':'wednesday','4':'thursday','5':'friday','6':'saturday','7':'sunday'}
number = day.keys()
day_pop = day.pop('2')
print(day_pop) #delete '2':'tuesday'
print(day)
#result run after : {'1': 'monday', '3': 'wednesday', '4': 'thursday', '5': 'friday', '6': 'saturday', '7': 'sunday'}
| false |
b876d58642577425d6049f1cb02dd95ff8decd99 | djtiedemann/riddles | /spheres-weight/spheresWeight.py | 2,640 | 4.15625 | 4 | # https://fivethirtyeight.com/features/can-you-flip-the-magic-coin/
# the general idea is to break the spheres up into 3 groups and make sure their weights are the same.
# say we have n spheres. we'll create an array of size n that's made up of 1, 2, 3. each index represents the nth sphere and the value represents the grouping:
# so say we have the following
# sphere number: 123456
# 122322
#
# this represents the following division:
# 1: 1
# 2: 2, 3, 5, 6
# 3: 4
#
# this is a valid solution if 1^3 = 2^3 + 3^3 + 5^3 + 6^3 = 4^3
#
#
# so for each number of spheres, we generate all valid arrays of size n made up of 1,2,3. we then test each of those arrays to see if the combination creates a split where the sum
# is equal. if so, that's the right number
class SphereSolver:
def divideSpheres(self, numSpheres):
print('checking: ' + str(numSpheres))
grouping = []
for i in range(0, numSpheres):
grouping.append(1)
isLastGrouping = False
while(not isLastGrouping):
isValid = self.validateGrouping(grouping)
if(isValid):
print('winner')
(grouping, isLastGrouping) = self.generateNextGrouping(grouping)
# generate the next valid grouping. basically we search from the right to find the first value that isn't a 3, increment that value by 1 and set all values to the right equal to 1
# returns false if it isn't the last grouping and true if it is
def generateNextGrouping(self, grouping):
#print(grouping)
# we can always assign the first sphere to group 1 without changing the problem, so we can go to 0 instead of -1
for index in range(len(grouping) - 1, 0, -1):
if(grouping[index] == 3):
continue
if(grouping[index] == 2):
grouping[index] = 3
if(grouping[index] == 1):
grouping[index] = 2
for i in range(index+1, len(grouping)):
grouping[i] = 1
return (grouping, False)
return (grouping, True)
# a grouping is valid if all of the spheres in each grouping have the same weight
def validateGrouping(self, grouping):
sum1 = 0
sum2 = 0
sum3 = 0
for i in range(0, len(grouping)):
if(grouping[i] == 1):
sum1 = sum1 + (i+1)**3
if(grouping[i] == 2):
sum2 = sum2 + (i+1)**3
if(grouping[i] == 3):
sum3 = sum3 + (i+1)**3
#print('checking ' + str(grouping))
#print('sum 1: ' + str(sum1))
#print('sum 2: ' + str(sum2))
#print('sum 3: ' + str(sum3))
return sum1 == sum2 and sum2 == sum3
sum = 0
solver = SphereSolver()
for i in range (18, 19):
# the weight of the spheres is proportional to the volume which is proportional to the radius cubed
sum = sum + i**3
if sum % 3 == 0:
solver.divideSpheres(i) | true |
1c5dbc3157d316e4c59177ed1004b4d71e1ed12d | partho-maple/coding-interview-gym | /leetcode.com/python/148_Sort_List.py | 1,616 | 4.25 | 4 | # Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
# Recursive approach using merge sort. Time O(nlogn). Space O(n) since it's recursive approach. To make it iterative, implement the merge sort iteratively
class Solution(object):
def sortList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if not head or not head.next:
return head
fastNode, slowNode = head.next, head # divide the list into two parts by fast-slow pointer technique
while fastNode and fastNode.next:
fastNode = fastNode.next.next
slowNode = slowNode.next
secondHalf = slowNode.next
slowNode.next = None # cut down the first part
leftHalf = self.sortList(head)
rightHalf = self.sortList(secondHalf)
return self.mergeTwoLists(leftHalf, rightHalf)
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if not l1 or not l2:
return l1 if l1 else l2
p1, p2, p1Prev = l1, l2, None
while p1 and p2:
if p1.val < p2.val:
p1Prev = p1
p1 = p1.next
else:
if p1Prev:
p1Prev.next = p2
p1Prev = p2
p2 = p2.next
p1Prev.next = p1
if not p1:
p1Prev.next = p2
return l1 if l1.val < l2.val else l2
| true |
6f0762bdd72bf42621795deb59ae70fa4902914b | andre-Hazim/Unit_3-03 | /rock_paper_scissors.py | 1,991 | 4.125 | 4 | # created by Andre
# created on october 3, 2017
# created for isc3u
# created for unit 3-03 daily assignment
# a rock paper scisoors program
import ui
from numpy import random
def rock_touch_up_inside(sender):
#This checks the number of students entered versus the constant (25 stuents)
computer_choice = random.randint(1,3)
#process
if computer_choice == 1:
#output
view['computers_choice_label'].text = "Rock"
view['who_won_label'].text = "It's a tie"
elif computer_choice == 2:
view['computers_choice_label'].text = "Paper"
view['who_won_label'].text = "You lost!"
elif computer_choice == 3:
view['computers_choice_label'].text = "Scissor"
view['who_won_label'].text = "You won!"
def paper_touch_up_inside(sender):
#This checks the number of students entered versus the constant (25 stuents)
computer_choice = random.randint(1,3)
#process
if computer_choice == 1:
#output
view['computers_choice_label'].text = "Rock"
view['who_won_label'].text = "You won!"
elif computer_choice == 2:
view['computers_choice_label'].text = "Paper"
view['who_won_label'].text = "It's a tie"
elif computer_choice == 3:
view['computers_choice_label'].text = "Scissor"
view['who_won_label'].text = "You lost!"
def scissor_touch_up_inside(sender):
#This checks the number of students entered versus the constant (25 stuents)
computer_choice = random.randint(1,3)
#process
if computer_choice == 1:
#output
view['computers_choice_label'].text = "Rock"
view['who_won_label'].text = "You lost!"
elif computer_choice == 2:
view['computers_choice_label'].text = "Paper"
view['who_won_label'].text = "You won!"
elif computer_choice == 3:
view['computers_choice_label'].text = "Scissor"
view['who_won_label'].text = "It's a tie"
view = ui.load_view()
view.present('sheet')
| true |
552c18b57ebb23e689c9307c986ea1b2bad01fdc | ylana-mogylova/Python | /Examples/palidrome.py | 480 | 4.34375 | 4 | # if word is palindrome
input_word = "waabbcbbaaw"
def palindrome(input_string):
revert_string = ""
index = len(input_string) - 1
while index >= 0:
revert_string += input_string[index]
index -= 1
if input_string == revert_string:
return True
else:
return False
if __name__ == "__main__":
if palindrome(input_word):
print("word is palindrome")
else:
print("word is not palindrome") | false |
57f37f19ad326a3af0fe1151ccb68e4d5d2a0f1f | ylana-mogylova/Python | /Examples/number_occur_substr_in_str.py | 710 | 4.21875 | 4 | """
Write a function to count how many times the substring appears in the larger String.
"""
large_string = "abcpotqwrpot"
substring = "pot"
def count_substr_occur(input_string, input_substring):
counter = 0
start_index = 0
flag = True
while flag:
find_index = input_string.find(input_substring, start_index)
if find_index == -1:
flag = False
else:
counter += 1
start_index = find_index + len(input_substring)
return counter
if __name__ == "__main__":
print("Substring %s appears %s number(s) in the string %s" % (
substring, count_substr_occur(large_string, substring), large_string)) | true |
5e0acd1f29a4f27e94bc17199f9746c294832d17 | delisco/algorithm | /sorts/bubble_sort.py | 1,177 | 4.15625 | 4 | '''
@__date__ = 2020.02.29
@author = DeLi
'''
import time
def bubble_sort(a_list):
"""
Pure implementation of bubble sort algorithm in Python
:param collection: some mutable ordered collection with heterogeneous
comparable items inside
:return: the same collection ordered by ascending
Examples:
>>> bubble_sort([0, 5, 2, 3, 2])
[0, 2, 2, 3, 5]
>>> bubble_sort([])
[]
>>> bubble_sort([-23, 0, 6, -4, 34])
[-23, -4, 0, 6, 34]
>>> bubble_sort([-23, 0, 6, -4, 34]) == sorted([-23, 0, 6, -4, 34])
True
"""
list_length = len(a_list)
for i in range(list_length-1):
swapped = False
for j in range(list_length-1-i):
if a_list[j] > a_list[j+1]:
swapped = True
a_list[j], a_list[j+1] = a_list[j+1], a_list[j]
# Show the sorting process
# print(a_list)
if not swapped:
break # Stop iteration if the collection is sorted.
return a_list
if __name__ == "__main__":
start = time.process_time()
print(bubble_sort([3, 6, 7, 1]))
print(f"Processing time: {time.process_time() - start}")
| true |
8686d027c1526889ee5d3bc739cc4a7f4ad44e28 | eniskurban98/Python-Projects | /fibonaccinumbers.py | 774 | 4.1875 | 4 | # Fibonacci numbers
while True:
user_input = input('How many steps do you want for create fibonacci series? Type exit to stop:> ')
if user_input == 'exit':
print('You are quitting from the fibonacci numbers program...')
break
else:
try:
a = 0
b = 1
user_input = int(user_input)
fibonaccinumlist = []
if user_input == 0 or user_input < 0:
print('Please enter a positive number')
else:
for i in range(user_input):
fibonaccinumlist.append(a)
a, b = b, a + b
print(f'For first {user_input} step, fibonacci numbers are: {fibonaccinumlist}')
except ValueError:
print('Please provide a valid number.')
| true |
bce23272f3089164b2737adc3ae19e711c681242 | debbycosoi/Activities_Intro_to_Python | /Activity8-MarielCosoi.py | 596 | 4.375 | 4 | # In this file you are to:
# Write a Program to check the greatest among 3 numbers.
# Write a Program to imitate a Traffic light. Think about the information you need to generate to
# make your program mimic the real world.
a=1
b=7
c=3
if (a>b):
print("A is the greatest number")
elif(b>c):
print("B is the greatest number")
else:
print("C is the greatest number")
green = "green"
yellow = "yellow"
red = "red"
light = green
print(f"the light is {light} so you have to")
if(light == green):
print("go")
elif(light == yellow):
print("slow down")
else:
print("stop")
| true |
ff642a983c6b0b05b6ee01647ad0b27cba2a4032 | pinaxtech/hellocode | /alphabet.py | 201 | 4.15625 | 4 | n = input("Enter a character: ")
if(n=='A' or n=='a' or n=='E' or n =='e' or n=='I'
or n=='i' or n=='O' or n=='o' or n=='U' or n=='u'):
print(n, "is a Vowel")
else:
print(n, "is a Consonant") | false |
c6dd39f9a500b01dcb8afdebf50d7d2e78c242c2 | ethender/pythonlearn | /Lectures/Lecture3/lecture.py | 1,325 | 4.15625 | 4 | ##x = int(raw_input('Enter an integer'))
##ans = 0
##while ans*ans*ans < abs(x):
## ans = ans + 1
## #print 'Current guess =', ans
##
##if ans*ans*ans != abs(x):
## print x, 'is not a perfect cube'
##else:
## if x < 0:
## ans = -ans
## print 'Cube root of '+str(x)+' is '+str(ans)
###Find The cube root of a perfect cube
##
##x = int(raw_input('Enter an integer: '))
##for ans in range(0, abs(x)+1):
## if ans**3 == abs(x):
## break
##if ans**3 != abs(x):
## print x, 'is not perfect cube'
##else:
## if x < 0:
## ans = -ans
## print 'Cube root of '+str(x)+' is '+str(ans)
#Approximation
##x = 25
##epsilons = 0.01
##numGuesses = 0
##ans = 0.1
##while (abs(ans**2) = x) >= epsilons and ans <= x:
## ans += 0.00001
## numGuesses += 1
##print 'numGuesses =', numGuesses
##if abs(ans**2 = x) >= epsilons:
## print 'Failed on square root of ',x
##else:
## print ans, 'is close to square root of ',x
x = 12345
epsilon = 0.01
numGuesses = 0
low = 0.0
high = x
ans = (high + low )/2.0
while (abs(ans**2 = x) >= epsilon and ans <= x):
#print low , high, ans
numGuesses += 1
if ans**2 < x:
low = ans
else:
high = ans
ans = (high + low)/2.0
#print 'numguesses =', numGuesses
print ans, 'is close to square root of ',x
| false |
989b4ee75c7584e4b59fcb0ac42e51c2de51b532 | Stevemz09/Hello-World | /Number digits.py | 781 | 4.25 | 4 | '''
Find the number of digits in a given integer
'''
num = int(input("Enter the number :")) # Enter the number using keyboard
div = 1 # Start with 1 the first power of 10 in the integer division
while num// 10**div != 0: # Check the condition: is the quotient of the # integer division num divided by 10 to the power of div
# while loop it keeps executing the loop body as long as the condition is true.
# integer division num divided by 10 to the power of div
# not equal to zero?
div = div + 1 # If yes execute the loop body
print("the number :", num, "has ", div, "digits") # if no, get out of the loop and print
# the message
| true |
0c62653ae1900ca3497a3053b44cc220622aca9f | Stevemz09/Hello-World | /prime_num.py | 350 | 4.3125 | 4 | '''
Find whether a given number is PRIME
using a for loop
'''
num = int(input("Enter your number :"))
for divisor in range(2, num):
print(divisor)
# if num % divisor == 0:
# print(divisor )
# print( "The number is not prime, it has a divisor: ", divisor)
# exit()
#print("The number is prime")
#exit()
| true |
4e6a83486a7803b9d52cdd5d3872cb99c87095c9 | nmanley73/Tasks | /Task2_bmi.py | 268 | 4.4375 | 4 | # Program calculates somebody's BMI
# Create Variables and input values
Weight = float(input("Enter weight:"))
Height = float(input("Enter height:"))
# Calculate BMI
BMI = Weight / ((Height/100)**2)
# Print out result
print("BMI is", round(BMI,2), "metres squared") | true |
b5786eba9fefaea6b43d2389d93dba0f899c8802 | farukara/Automate-Boring-Stuff | /files/Ch7 regex version of strip.py | 751 | 4.4375 | 4 | #! Python3
# Regex version of strip.
import re
def regex_ver_strip(text, remove=''):
""" strips optional remove (defaults space)from text using regex"""
if remove:
extras = re.compile(r'^(%s*)'%remove)
mo = extras.sub('', text)
mo = mo[::-1]
mo = extras.sub('', mo)
mo = mo[::-1]
return mo
else:
spaces = re.compile(r'^(?: *)(.+)')
mo = spaces.findall(text)
mo = mo[0][::-1]
mo = spaces.findall(mo)
mo = mo[0][::-1]
return mo
if __name__ == '__main__':
text = ' $ Hello World 12 '
print(regex_ver_strip(text)) # strip spaces
text = 'tt$ Hello World 12tt'
print(regex_ver_strip(text, remove='tt')) # strip tt
| false |
fead3b2d65931a5e6a7bbb648436f0c053339f66 | gospodenkods/python_basic | /homeworks/les1/task1.py | 803 | 4.4375 | 4 | """
1. Поработайте с переменными, создайте несколько, выведите на экран, запросите у пользователя несколько чисел и строк
и сохраните в переменные, выведите на экран.
"""
firs_variable = "Sunday"
next_variable = "Moscow"
last_variable = 10
print(firs_variable, 'Day of week')
print(next_variable, 'City where you live')
print(last_variable, 'Your favorite number')
firs_variable = input('What day is today?\n')
next_variable = input('What city do you live in?\n')
last_variable = input('What you favorite number?\n')
summary_line = f"Day of week '{firs_variable}' city where you live '{next_variable}' \
Your favorite number '{last_variable}'"
print(summary_line)
| false |
bc845943c04f4a7572eae88f0a89163e7cafd854 | shinoherrera/ML_project | /Decisiontree.py | 2,742 | 4.375 | 4 |
#Here we call the essential libraries to help with our ML Algorithm
import matplotlib.pyplot as plt #necessary to help with our data visualization
import pandas as pd #help with working on multi-dimensional arrays
from sklearn.tree import DecisionTreeClassifier #the main libaries that will help with the learning/training
from sklearn import tree #using this to build a decision tree based on the code
data_set = pd.read_csv("test_data/train.csv") #read the data set by giving the function a path to where it is
clf =DecisionTreeClassifier() #this is our classifier that'll take in data and perform the
#training dataset
x_train = data_set.iloc[0:21000,1:].values #will use the first 21k rows, and will not use the first column because it includes the answer
train_label = data_set.iloc[0:21000,0].values #train label is what our x_train will be comparing itself to
clf.fit(x_train,train_label) #now we call the DTC function and fit the x_train and train_label to it
#testing data
x_test = data_set.iloc[21000:,1:] #Now we will use the rest of the rows of data to test our training model
actual_label = data_set.iloc[21000:,0].values #give it some information to check itself with
Pred=clf.predict(x_test) #Now we want to start the prediction by using x_test
count=0 #initializing a counter in order to be able to calculate accuracy
for i in range(0,21000): #we start the for-loop and tell it how many times to run
count +=1 if Pred[i] == actual_label[i] else 0 # will add 1 to the count if it guesses right,wont add if wrong
print("Accuracy= ",(count/21000)*100) #print statement that'll print after each iteration or about 21000 times
fig = plt.figure(figsize=(10,10)) #plt.fig will call on the matplotlib to construct a decision tree figure of the code above
F = tree.plot_tree(clf,filled=True) #clf is the classifier and the filled statement just assigns different colors to different nodes
fig.savefig("decision_tree.png") #save the figure as a png named decision_tree that is included in the report
text_representation = tree.export_text(clf) #just some extra way of visualize what each of the nodes say though it does get really messy and hard to follow
print(text_representation) | true |
ada91a70d3d157f9e4ac4b629a87e5200392629f | MEENUVIJAYAN/python | /python lab 3/logical.py | 236 | 4.34375 | 4 | #Program to perform logical operations
x =True
y =True
print(x and not y)
print(not x and y)
print(not x and not y)
print(x or not y)
print(not x or y)
print(not x or not y)
print("Precedence order is")
print("Not->And->Or")
| true |
274da6fa73ea02888de1175d2f706f46bbbf7e57 | William-cordingley/CP1404 | /Lectures/Lecture_04/lecture.py | 233 | 4.21875 | 4 | """
How many Vowels in a Name or String that has been inputted
"""
name = input("Name: ")
vowels = "AEIOUaeiou"
count = 0
for i in name:
if i in vowels:
count += 1
print("There is {} vowels in {} ".format(count, name))
| true |
8a3f4d95e5e7034409f7d519d6f166036a24838b | bakerb0473/cti110 | /P3HW2_SoftwareSales_Baker.py | 886 | 4.125 | 4 | # CTI-110
# P3H2 - Software Sales
# Bonnita Baker
# 6/24/2018
# This program will ask the user to enter the number of packages purchased
# it will display the amount of the discount
# and the total purchase cost with the discount applied
# If the quantity is less than 10 there is no discount
package = 99
quantity = float(input("How many packages are your ordering? "))
if quantity < 10:
print("Total: $ ", package)
if quantity >= 10 and quantity <= 19:
print("Total: $ ", (package * quantity) - (package * quantity) * .10)
if quantity >= 20 and quantity <= 49:
print("Total: $ ", (package * quantity) - (package * quantity) * .20)
if quantity >= 50 and quantity <= 99:
print("Total: $ ", (package * quantity) - (package * quantity) * .30)
if quantity >= 100:
print("Total: $ ", (package * quantity) - (package * quantity) * .40)
| true |
821499c9657944823af32525f42ec3e4a1387a77 | Coralma/python-learning | /basic/basicIterator.py | 1,552 | 4.375 | 4 | # 迭代器
# 迭代是Python最强大的功能之一，是访问集合元素的一种方式。。
# 迭代器是一个可以记住遍历的位置的对象。
# 迭代器对象从集合的第一个元素开始访问，直到所有的元素被访问完结束。迭代器只能往前不会后退。
# 迭代器有两个基本的方法：iter() 和 next()。
list=[1,2,3,4]
it = iter(list) # 创建迭代器对象
for x in it:
print (x, end=" ")
print();
# next() 函数：
import sys # 引入 sys 模块
list=[1,2,3,4]
it = iter(list) # 创建迭代器对象
while True:
try:
print(next(it))
except StopIteration:
sys.exit()
# 生成器
# 在 Python 中，使用了 yield 的函数被称为生成器（generator）。
# 跟普通函数不同的是，生成器是一个返回迭代器的函数，只能用于迭代操作，更简单点理解生成器就是一个迭代器。
# 在调用生成器运行的过程中，每次遇到 yield 时函数会暂停并保存当前所有的运行信息，返回yield的值。并在下一次执行 next()方法时从当前位置继续运行。
# 以下实例使用 yield 实现斐波那契数列：
def fibonacci(n): # 生成器函数 - 斐波那契
a, b, counter = 0, 1, 0
while True:
if (counter > n):
return
yield a
a, b = b, a + b
counter += 1
f = fibonacci(10) # f 是一个迭代器，由生成器返回生成
while True:
try:
print(next(f), end=" ")
except StopIteration:
sys.exit()
| false |
7e3eeba883995a42a99a18df17d87ba8d94a6799 | hkhurram147/ESC180 | /lab02.py | 2,272 | 4.34375 | 4 | # Lab 2: Calculator
current_value = 0
operations = []
# Lab 3
def get_current_value():
pass
# Problem 2:
def display_current_value():
global current_value
print("Current value: " + str(current_value))
def add_to_operations():
global current_value, operations
operations.append(current_value)
if len(operations) >= 3:
del operations[0]
# Problem 3: Addition
def add(to_add):
global current_value, operations
# global is used to update the current_value,
# if not used, the updates would only be local to this function
current_value += to_add
add_to_operations()
def subtract(to_subtract):
global current_value, operations
current_value -= to_subtract
add_to_operations()
# Problem 5: Multiplication and Division
def mult(to_multiply):
global current_value, operations
current_value *= to_multiply
add_to_operations()
def divide(to_divide):
global current_value, operations
if (to_divide != 0):
current_value /= to_divide
add_to_operations()
# Problems with the division function:
# It may lead to floats that are imprecise
saved = ""
# Problem 6: Memory and Recall
def memory():
# saves the current value
global current_value, saved
saved = str(current_value)
def recall():
# restores the saved value
global current_value, saved
current_value = int(saved)
# Problem 7: Undo
def undo():
"""restores the previous value that appeared on screen
before the current value"""
# pressing undo twice restores the original value
global current_value, operations
# solution: since there are only 2 values stored, swap
operations[0], operations[1] = operations[1], operations[0]
current_value = operations[-1]
# global variable representing the current file as main
if __name__ == "__main__":
print("Welcome to the calculator program")
display_current_value() # 0
add(5) # 5
subtract(2)
display_current_value() # 3
undo()
display_current_value() # 5
undo()
display_current_value() # 3
mult(10)
display_current_value() # 30
undo()
undo()
display_current_value() # 30
undo()
undo()
undo()
display_current_value() # 3
| true |
52fa8fa88a6728dd8c377c0074caee5f57861f59 | tcu93/py4e_exercises | /9dicts01.py | 760 | 4.1875 | 4 | '''Suppose you are given a string and you want
to count how many times each letter appears.'''
word1 = 'brontosaurus'
dict1 = dict()
# Create dict from string (word1)
for c in word1: # for c'th element of word1
dict1[c] = dict1.get(c, 0) + 1 # c'th elem of dict1 = value of that elem + 1
print(dict1)
print(dict1['r'])
'''
THIS ALSO WORKS
for x in word1:
y = x.split()
if x not in dict1: # confusing -- x acts as both letter & index at same time
dict1[x] = 1 # this is value, not key
else:
dict1[x] += 1
'''
# Check for presence of particular key value
if 'b' in dict1:
print('true')
# Create list of dict1 keys
k1 = list(dict1.keys())
print(k1)
# Create list of dict1 values
v1 = list(dict1.values())
print(v1) | true |
9cdbe833a1f8786cbc027acba3099a5b9338be08 | tcu93/py4e_exercises | /2vars3.py | 341 | 4.1875 | 4 | '''Exercise 3 Write a program to prompt the user for hours and rate per hour to compute gross pay.
Enter Hours: 35
Enter Rate: 2.75
Pay: 96.25
'''
crt_product = lambda arg1, arg2: arg1 * arg2
hours = int(input('Enter Hours: '))
rate = int(input('Rate: '))
#pay = hours * rate
#print(f'Pay: {pay}')
print('Pay:', crt_product(hours, rate)) | true |
11cbeb799bdf0b595f369bc1db8096dae310fb2a | Jemmy-Hong/myPython | /pyPro/cn/py/chap3/cars.py | 842 | 4.40625 | 4 | #sort排序,永久改变列表
# cars = ['bmw', 'audi', 'toyota', 'subaru']
# print(cars)
# cars.sort()
# print(cars)
# cars.sort(reverse=True)
# print(cars)
#sorted函数，临时排序，不改变列表元素位置，同样有reverse=True参数
# cars = ['bmw', 'audi', 'toyota', 'subaru' ]
# print("Here is the original list:")
# print(cars)
#
# print("\nHere is the sorted list:")
# print(sorted(cars))
#
# print("\nHere is the original list again:")
# print(cars)
#倒着打印列表(永久性改变列表)
# cars = ['bmw', 'audi', 'toyota', 'subaru']
# print(cars)
#
# cars.reverse()
# print(cars)
#确定列表的长度
# cars = ['bmw', 'audi', 'toyota', 'subaru']
# print(len(cars))
#获取列表的最后一个元素
motorcycles = ['honda', 'yamaha', 'suzuki' ]
print(motorcycles[-1])
motorcycles = []
print(motorcycles[-1])
| false |
01269ea0a69f1fd48fefc05c429edb7641d95e61 | Shadowztorm/python-practice | /listprac.py | 1,567 | 4.21875 | 4 | # 1. Let us say your expense for every month are listed below,
# 1. January - 2200
# 2. February - 2350
# 3. March - 2600
# 4. April - 2130
# 5. May - 2190
#
# Create a list to store these monthly expenses and using that find out,
#
# 1. In Feb, how many dollars you spent extra compare to January?
# 2. Find out your total expense in first quarter (first three months) of the year.
# 3. Find out if you spent exactly 2000 dollars in any month
# 4. June month just finished and your expense is 1980 dollar. Add this item to our monthly expense list
# 5. You returned an item that you bought in a month of April and
# got a refund of 200$. Make a correction to your monthly expense list
# based on this
exp=[2200,2350,2600,2130,2000]
# 1. In Feb, how many dollars you spent extra compare to January?
print("In Feb money spent compared to january is ", exp[1]-exp[0],"$")
# 2. Find out your total expense in first quarter (first three months) of the year.
print("Total expense in first quarter is ", sum(exp[0:3]))
# 3. Find out if you spent exactly 2000 dollars in any month
print("Did I spent 2000$ in any month? ", 2000 in exp)
# 4. June month just finished and your expense is 1980 dollar. Add this item to our monthly expense list
exp.append(1980)
print("June month expense",exp)
# 5. You returned an item that you bought in a month of April and
# got a refund of 200$. Make a correction to your monthly expense list
# based on this
exp[3]=exp[3]-200
print("Returned money of april and got discount",exp) | true |
f4ff2a3508fc767eac67126559a03fd239c82b63 | barawalojas/Python-tutorial | /Season 02 - Python Fundamentals/Episode 16 - Functions.py | 1,032 | 4.59375 | 5 | # Functions
""" We Studied functions like zip, print, enumerate, etc.., but here we'll study how to define your own function. """
""" def is a python keyword used for defining functions.
# def greet():
here, greet() is the function, we've defined and (:) colon is to be followed as of syntax.. """
def greet():
name = input('Enter your name : ')
print(f'Hello, {name}')
""" Now, if we run the code, python will run from the first line of code, and recognize that we've defined a function
greet() only, and comes out of the body, overall if we run the above chunk of code, it will not run the body of the
function, so we need to call the function..., """
greet()
""" -- OUTPUT --
Enter your name : Pythobit
Hello, Pythobit
"""
# we can only call the function, after defining the function only, if we call the function before defining the function,
# it will give an Traceback error..
""" Variables created inside function, die at the end of the function.., """
# So, if you
print(name) # OUTPUT - Traceback error..!
| true |
07bbbbe40e1c56bb0b03697cb7bdec93394161b4 | barawalojas/Python-tutorial | /Season 02 - Python Fundamentals/Episode 01 - if statements.py | 1,979 | 4.34375 | 4 | # if Statements
"""if statements allow our programs to make decision depending on the boolean"""
friend = 'Rolf'
username = input('Enter your name : ')
if True:
print('Hello friend..!')
# Above code will print 'Hello friend..!' only if the user input name as Rolf, otherwise nothing.
""" !! -- as true will be always true, so the print statement will be printed always. -- !! """
if username == friend:
print('Hello friend..!') # Output - Enter your name : bob
""" -- What will happen here is that python checks if username is equal to friend,
if yes, then it will run the print statement inside it. else nothing will be printed -- """
""" SYNTAX
if username == friend }- Boolean Comparison, : }- Colon depicts the start of the body.
body }- indentation is necessary, so python knows that these statements are inside if statements."""
"""as soon as you terminate the space in front of print statements, python will give an error.."""
""" if [Bool]: -- if true, returns print statment, otherwise else
print()
else:
print()
using if and else at same time also known as "Compound if-Statement" -- """
# Things you put inside if condition, doesn't needs to be a bool comparison.
friends = ['Charlie','Smith']
family = ['Johny','Bob']
username = input('Enter your name : ')
if username in friends:
print('Hello friend..!')
elif username in family:
print('Hello Family..!')
else:
print('I Don\'t know you..!')
""" !! -- !! IMPORTANT THINGS TO NOTICE !! -- !! """
# 01. Take care of INDENTATION, mostly beginners make a lot mistake in this
""" 02. AN Example - of not giving indentation properly will give an error.. """
"""
if username = friend:
print('Hello friend ..!')
print('hello bello') -- Right here this statement will give an error,
python interprets this as an error.
"""
# 03. from now onwards, i won't be giving the outputs, you have to try yourself..
# Any editor you like, i prefer to PyCharm..
| true |
54e664190c974dd17397667d9f91935f6c1b132a | barawalojas/Python-tutorial | /Season 02 - Python Fundamentals/Episode 11 - List Comprehension.py | 2,181 | 4.625 | 5 | # List Comprehension
# Suppose we have list of numbers as given below :
numbers = [0,1,2,3,4]
# and we want to double all the number in the list, i.e: each number to be multiplied by
doubled_numbers = []
for number in numbers:
doubled_numbers.append(number * 2)
print(doubled_numbers) # OUTPUT - [0, 2, 4, 6, 8]
# Above method is long and time consuming to run,
# so, we'll use list comprehension instead
doubled_numbers = [number * 2 for number in numbers]
print(doubled_numbers) # OUTPUT - [0, 2, 4, 6, 8]
# Above method is way easier than that 5 line code above.
""" Also we can use range, instead of using a list.. """
doubled_numbers = [number * 2 for number in range(10)]
print(doubled_numbers) # OUTPUT - [0, 2, 4, 6, 8, 10, 12, 14, 16, 18]
# Examples
# 01.
friend_ages = [22,31,35,37]
ages = [f'My friend is {age} years old' for age in friend_ages]
print(friend_ages) # OUTPUT - [22, 31, 35, 37]
# 02.
names = ['Rolf', 'Bob' ,'Jen']
lower = [name.lower() for name in names]
print(lower) # OUTPUT - ['rolf', 'bob', 'jen']
# Above method will give print the letters in lowercase
# Above method comes best use in user_input..tasks..
#friend = input('Enter your name : ')
friends = ['Rolf','Anne','Jen','Bob','Charlie']
# friends_lower = [name.lower() for name in friends]
#if friend in friends:
# print(f'{friend} is one of your friend.') # OUTPUT - Enter your name : Rolf
# Rolf is one of your friend.
# for the best use
friend = input('Enter your name : ')
friends = ['Rolf','Anne','Jen','Bob','Charlie']
friends_lower = [name.lower() for name in friends]
#if friend.lower() in friends_lower:
# print(f'{friend} is one of your friend.') # OUTPUT - Enter your name : jen
# jen is one of your friend.
# we can also do title casing, meaning that the first letter will be capital and rest letters as lowercase.
print(f'{friend.title()} is one of your friend.') # OUTPUT - Enter your name : jen
# Jen is one of your friend.
| true |
c430e45aabc74b8fad06b15ab7a05a8a2f11cad3 | Nadeemk07/Rock-Paper-Scissors | /Rock Paper Scissor.py | 1,286 | 4.21875 | 4 | import random
rock = '''
_______
---' ____)
(_____)
(_____)
(____)
---.__(___)
'''
paper = '''
_______
---' ____)____
______)
_______)
_______)
---.__________)
'''
scissors = '''
_______
---' ____)____
______)
__________)
(____)
---.__(___)
'''
choice=int(input("What do you want to Choose 0 for rock, 1 for paper, 2 for scissors\n"))
print("You choose\n")
if(choice==0):
print(rock)
elif(choice==1):
print(paper)
elif(choice==2):
print(scissors)
list=[rock,paper,scissors]
computer_choice=random.choice(list)
print("Computer choose")
print(computer_choice)
if(choice==0 and computer_choice==rock):
print("No one wins")
if(choice==0 and computer_choice==paper):
print("You Lose")
if(choice==0 and computer_choice==scissors):
print("You won")
if(choice==1 and computer_choice==rock):
print("You won")
if(choice==1 and computer_choice==paper):
print("Tie")
if(choice==1 and computer_choice==scissors):
print("You lose")
if(choice==2 and computer_choice==rock):
print("You lose")
if(choice==2 and computer_choice==paper):
print("You won")
if(choice==2 and computer_choice==scissors):
print("Tie") | false |
3db329e28cc4475ab59303785dcb2c78a6b7a8c6 | dentalmisorder/PythonUdemy | /basics (not actual projects)/logicalNbool.py | 427 | 4.21875 | 4 | #boolean
print(2 > 1) #true 2 is greater then 1
print(2 <= 1) #false 2 is not lower or equal then 1
print(2 != 1) #true 2 is not equal 1
print(2 == 2) #true 2 is equal 2
#logical
print(2 == 2 and 2 > 1) #true, 2 is equal to 2 and 2 greater then 1
print(2 == 2 or 2 < 1) #true, 2 is equal to 2 (cause OR operator needs only 1 true)
print(not 2 == 2) #false cause NOT operator gets the oposite of what expected (expected true) | true |
b4cdbf06e668f46be3f18b1e7c914d73d75064ce | frclasso/2nd_Step_Python_Fabio_Classo | /Cap02_Classes_e_Programacao_Orientada_a_Objetos/class-updated/script7_heranca.py | 2,685 | 4.78125 | 5 | #!/usr/bin/env python3
# heranca
"""Nos inicializamos nossa variavel last_name com a string "Fish"
porque sabemos que a maioria dos peixes tera isso como seu sobrenome."""
# Parent Class
# class Fish:
# def __init__(self, first_name, last_name="Fish"):
# self.first_name = first_name
# self.last_name = last_name
#
#
# # adicinando metodos
# def swim(sef):
# print("The Fish is swimming")
#
# def swim_backwards(self):
# print("The Fish can swim in backwards")
"""Adicionamos os metodos swim () e swim_backwards () a classe Fish,
para que cada subclasse tambem possa usar esses metodos.
"""
"""Construir uma classe pai segue a mesma metodologia que construir qualquer
outra classe, exceto que estamos pensando sobre quais metodos as classes
filhas poderao usar uma vez que as criarmos."""
# Child Class
# class Trout(Fish):
# pass
# 2-------------------------------------------
class Fish:
def __init__(self, first_name, last_name="Fish",
skeleton='Bone', eyelids=False):
self.first_name = first_name
self.last_name = last_name
self.skeleton = skeleton
self.eyelids = eyelids
def swim(sef):
print("The Fish is swimming")
def swim_backwards(self):
print("The Fish can swim in backwards")
class Trout(Fish):
pass
# vamos criar um objeto da classe Trout que ainda nao possui nenhum metodo proprio
terry = Trout("Terry")
print(terry.first_name + " " + terry.last_name)
terry.swim()
terry.swim_backwards()
print(terry.eyelids)
print(terry.skeleton)
"""Nos criamos um objeto Trout terry que faz uso de cada um dos metodos da classe
Fish, embora nao tenhamos definido esses metodos na classe filha Trout. Nos so
precisavamos passar o valor de "Terry" para a variavel first_name porque todas as
outras variaveis foram inicializadas.
"""
# vamos criar outra subclasse Clownfish
class Clownfish(Fish):
def live_with_anemone(self):
print("The clownfish is coexisting with sea anemone.")
print()
# vamos instanciar o objeto clownfish
casey = Clownfish("Casey")
print(casey.first_name + " " + casey.last_name)
casey.swim()
casey.swim_backwards()
casey.live_with_anemone()
"""A saida mostra que o objeto Clownfish casey e capaz de usar os metodos Fish __init __ ()
e swim (), assim como seu metodo de classe filho de live_with_anemone ()."""
"""As classes filhas herdam os metodos da classe pai a que pertencem, portanto, cada classe
filha pode fazer uso desses metodos dentro dos programas."""
"""
Referencia:
https://www.digitalocean.com/community/tutorials/
understanding-class-inheritance-in-python-3""" | false |
3cdba55cf89594113473cdd16ef2fcf4c6be9e28 | frclasso/2nd_Step_Python_Fabio_Classo | /Cap01-Elementos-de-sintaxe-especifica/02_list_comprehension/extras/comprehension2.py | 294 | 4.15625 | 4 | """In fact, lists, sets, dictionaries, and generators can all
be built with comprehensions """
print([ord(x)for x in 'spam']) # List of character ordinals
print({ord(x)for x in 'spam'}) # Set
print({x : ord(x)for x in 'spam'}) # Dictionary
print((ord(x)for x in 'spam')) # Generator of values
| true |
4e9e683720ec7e5386244e90a89e720bbea2d514 | frclasso/2nd_Step_Python_Fabio_Classo | /Cap02_Classes_e_Programacao_Orientada_a_Objetos/empregados.py | 1,593 | 4.34375 | 4 | #!/usr/bin/env python3
# -*-coding:utf-8 -*-
class Employee:
'Classe Base para todos os empregados/Commom base class for all employees.'
empCount = 0
def __init__(self, name, salary):
self.name = name
self.salary = salary
Employee.empCount += 1
def displayCount(self):
print(f"Total de empregados: {Employee.empCount}")
def displayEmployee(self):
print("Nome: ", self.name, "-", "Salario: R$", self.salary)
#This would create the first object of Eployee class
emp1 = Employee('Zara', 2000)
#This would create the second object of Eployee class
emp2 = Employee('Manni', 5000)
# Acessing Attributes
emp1.displayEmployee()
emp2.displayEmployee()
print("Total Employee {}".format(Employee.empCount))
# Adicionando, modificando e deletando atributos
# emp1.salary = 3000 # altera o valor do salario
# emp1.name = 'Joseph' # altera o nome do empregado 1
#emp1.displayEmployee()
#del emp1.salary
# Verificar atributos dos objetos
# print(hasattr(emp1, 'salary')) # Retorna True se o atributo 'salary' existir
# print(getattr(emp1, 'salary')) # Retorna o valor do atributo 'salary'
# setattr(emp1, 'salary', 4000) # Define o novo valor
# print(getattr(emp1, 'salary'))
# delattr(emp1,'salary') # Deleta o atributo 'salary'
# print(hasattr(emp1, 'salary'))
# Verificar atributos de classe
print("Employee.__doc__:", Employee.__doc__)
print("Employee.__name__:", Employee.__name__)
print("Emplyee.__module: ", Employee.__module__)
print("Employee.__bases__: ", Employee.__bases__)
print("Employee.__dict__: ", Employee.__dict__)
| false |
936b135496ab68554a661f1fd381fabb2dee8c08 | frclasso/2nd_Step_Python_Fabio_Classo | /Cap02_Classes_e_Programacao_Orientada_a_Objetos/class-updated/script8.py | 2,728 | 4.9375 | 5 | #!/usr/bin/env python3
# Sobreescrevendo metodos da classe Pai
"""Ate agora, examinamos a classe filha Trout que utilizou a palavra-chave pass
para herdar todos os comportamentos Fish da classe pai e outra classe filha,
Clownfish, que herdou todos os comportamentos da classe pai e tambem criou seu proprio
metodo exclusivo. especifico para a classe filho. As vezes, no entanto, queremos usar
alguns dos comportamentos da classe pai, mas nao todos eles. Quando mudamos os metodos
de classe pai, nos os substituimos."""
"""Ao construir classes pai e filho, e importante manter o design do programa em mente
para que a substituicao nao produza codigo desnecessario ou redundante."""
"""Diante disso, substituiremos o metodo construtor __init __ () e o metodo swim_backwards ().
Nao precisamos modificar o metodo swim (), pois os tubaroes sao peixes que podem nadar.
Vamos dar uma olhada nesta classe infantil:
"""
class Fish:
def __init__(self, first_name, last_name="Fish",
skeleton='Bone', eyelids=False):
self.first_name = first_name
self.last_name = last_name
self.skeleton = skeleton
self.eyelids = eyelids
def swim(sef):
print("The Fish is swimming")
def swim_backwards(self):
print("The Fish can swim in backwards")
class Shark(Fish):
def __init__(self, first_name, last_name="Shark",
skeleton='Cartilage', eyelids=True):
self.first_name = first_name
self.last_name = last_name
self.skeleton = skeleton
self.eyelids = eyelids
def swim_backwards(self):
print("The Fish can not swim in backwards, but can sink backwards")
"""Nos substituimos os parametros inicializados no metodo __init __ (), para que a variavel
last_name seja agora igual a string "Shark", a variavel esqueleto agora e igual a string
"cartilage", e a variavel eyelids agora esta definida para o valor booleano True.
Cada instancia da classe tambem pode substituir esses parametros.
O metodo swim_backwards () agora imprime uma string diferente da que esta na classe pai
Fish, porque os tubaroes nao sao capazes de nadar para tras da maneira que os peixes com ossos
conseguem.
"""
# instanciando um objeto shark
sammy = Shark("Sammy")
print(sammy.first_name+ " "+sammy.last_name)
sammy.swim()
sammy.swim_backwards()
print(sammy.eyelids)
print(sammy.skeleton)
"""A classe secundaria Shark substituiu com exito os metodos __init __ () e swim_backwards ()
da classe pai Fish, enquanto tambem herdava o metodo swim () da classe pai.
Quando houver um numero limitado de classes filhas que sao mais exclusivas do que outras,
a substituicao dos metodos de classe pai pode ser util"""
| false |
c28c4ab306be61d3067b915c907685ef11bd43d5 | frclasso/2nd_Step_Python_Fabio_Classo | /Cap01-Elementos-de-sintaxe-especifica/03_dict_comprehension/extras/01_exemplo1.py | 232 | 4.125 | 4 | #!/usr/bin/env python3
# dict comprehension criando um dicionario apartir do for i converte int em str como key
d = {str(i):i for i in [1,2,3,4,5]}
print(d)
# http://cmdlinetips.com/2018/01/5-examples-using-dict-comprehension/ | false |
ec301abb2c01e8acdaabc92785acfaa2a249504e | frclasso/2nd_Step_Python_Fabio_Classo | /Cap02_Classes_e_Programacao_Orientada_a_Objetos/class-updated/script10.py | 2,842 | 4.25 | 4 | #!/usr/bin/env python3
# Heranca multipla
"""Herança múltipla é quando uma classe pode herdar atributos e métodos de mais de uma
classe pai. Isso pode permitir que os programas reduzam a redundância, mas também
pode introduzir uma certa complexidade e ambigüidade, portanto, isso deve ser feito
pensando no projeto geral do programa.
Para mostrar como funciona a herança múltipla, vamos criar uma classe-filha Coral_reef
que herda de uma classe Coral e uma classe Sea_anemone. Podemos criar um método em cada
um e, em seguida, usar a palavra-chave pass na classe-filha Coral_reef:
"""
class Coral:
def community(self):
print("Coral lives in community")
class Anenome:
def protect_clowfish(self):
print("The anenome is protecting the clownfish.")
class CoralReef(Coral, Anenome):
pass
"""A classe Coral tem um método chamado community () que imprime uma linha, e a classe
Anemone possui um método chamado protect_clownfish () que imprime outra linha.
Então nós chamamos ambas as classes na tupla de herança. Isso significa que Coral está
herdando de duas classes pai."""
# instanciando um objeto CoralReef()
great_barrier = CoralReef()
great_barrier.community()
great_barrier.protect_clowfish()
"""A saída mostra que os métodos de ambas as classes pai foram efetivamente usados na
classe filha.
A herança múltipla nos permite usar o código de mais de uma classe pai em uma classe
filha. Se o mesmo método for definido em vários métodos pai, a classe filha usará o
método do primeiro pai declarado em sua lista de tupla.
Embora possa ser usado com eficácia, a herança múltipla deve ser feita com cuidado para
que nossos programas não se tornem ambíguos e difíceis para outros programadores entenderem.
"""
"""Conclusão
Este tutorial passou pela construção de classes pai e de classes filhas, substituindo
métodos e atributos pai em classes filhas, usando a função super () e permitindo que
classes filhas herdassem de várias classes pai.
A herança na codificação orientada a objetos pode permitir a adesão ao princípio DRY
(não se repita) de desenvolvimento de software, permitindo que mais seja feito com menos
código e repetição. A herança também obriga os programadores a pensar em como estão
projetando os programas que estão criando para garantir que o código seja eficaz e claro."""
"""A programação orientada a objetos (OOP) se concentra na criação de padrões reutilizáveis
de código, em contraste com a programação procedural, que se concentra em instruções
explícitas em seqüência. Ao trabalhar em programas complexos em particular, a programação
orientada a objetos permite reutilizar códigos e escrever códigos mais legíveis, o que,
por sua vez, os torna mais fáceis de manter."""
| false |
c07f6f5d58d0bda201befe51f092eb47a2732940 | frclasso/2nd_Step_Python_Fabio_Classo | /Cap02_Classes_e_Programacao_Orientada_a_Objetos/representation.py | 556 | 4.25 | 4 | #!/usr/bin/env python3
class bunny:
def __init__(self, n):
self._n = n
def __repr__(self):
return f'repr: the number of bunnies is {self._n}'
def __str__(self):
return f'str: the number of bunnies id {self._n}'
x = bunny(47)
#print(repr(x))
print((x)) # RETIRANDO REPR, DA PREFERENCIA PARA STRING METHOD
# print(chr(128406) * 10)
# print(chr(128527) * 10) # Emojis http://graphemica.com/
"""Para mais informacoes e exemplos de Funcoes Built-in acessem:
https://docs.python.org/3/library/functions.html
""" | false |
d27c8f4a3a48ac6ffed4ae610676c24a64736f72 | frclasso/2nd_Step_Python_Fabio_Classo | /Cap02_Classes_e_Programacao_Orientada_a_Objetos/class-updated/script5_metodos_construtores.py | 2,742 | 4.8125 | 5 | #!/usr/bin/env python3
"""O método construtor
O método construtor é usado para inicializar dados. É executado assim que um objeto
de uma classe é instanciado. Também conhecido como o método __init__, será a primeira
definição de uma classe e se parece com isso:
"""
# class Shark:
#
# def __init__(self):
# print("This is the constructor method")
#
# def swim(self):
# print("The shark is swimming")
#
# def be_awsome(self):
# print("The shark is being awesome.")
#
# sammy = Shark()
# sammy.swim()
# sammy.be_awsome()
"""Se você adicionou o método __init__ acima à classe Shark no programa acima, o programa
produziria o seguinte sem modificar nada dentro da instanciação sammy:"""
# output
"""This is the constructor method
The shark is swimming
The shark is being awesome."""
"""
Isso ocorre porque o método construtor é inicializado automaticamente. Você deve usar este
método para realizar qualquer inicialização que você gostaria de fazer com seus objetos de classe.
Em vez de usar o método do construtor acima, vamos criar um que use uma variável de nome que
possamos usar para atribuir nomes a objetos. Vamos passar o nome como parâmetro e definir self.name
igual ao nome:
"""
# class Shark:
#
# def __init__(self, name):
# self.name = name
#
#
# def swim(self):
# #referenciando a variavel name
# print(self.name + "is swimming")
#
# def be_awsome(self):
# print(self.name + "is being awesome.")
"""Finalmente, podemos definir o nome do sammy do objeto Shark como igual a "Sammy", passando-o
como um parâmetro da classe Shark:"""
# def main():
# sammy = Shark("Sammy")
# sammy.swim()
# sammy.be_awsome()
#
# if __name__=="__main__":
# main()
"""Vemos que o nome que passamos para o objeto está sendo impresso. Definimos o método __init__
com o nome do parâmetro (juntamente com a palavra-chave self) e definimos uma variável dentro
do método.
Como o método construtor é automaticamente inicializado, não precisamos chamá-lo explicitamente,
apenas passar os argumentos entre parênteses após o nome da classe quando criamos uma nova
instância da classe.
Se quiséssemos adicionar outro parâmetro, como a idade, poderíamos fazer isso passando-o para o
método __init__:"""
class Shark:
def __init__(self, name, age):
self.name = name
self.age = age
def swim(self):
#referenciando a variavel name
print(self.name + "is swimming")
def be_awsome(self):
print(self.name + "is being awesome.")
def main():
sammy = Shark("Sammy", 34)
sammy.swim()
sammy.be_awsome()
if __name__=="__main__":
main() | false |
60fe114c0e0981b625a2c5f2bcf60053b87c2828 | frclasso/2nd_Step_Python_Fabio_Classo | /Cap02_Classes_e_Programacao_Orientada_a_Objetos/class-updated/script3_juntando_tudo.py | 1,916 | 4.8125 | 5 | #!/usr/bin/env python3
"""Trabalhando com variáveis de classe e instância juntas
Variáveis de classe e variáveis de instância costumam ser utilizadas ao mesmo tempo,
portanto, vamos ver um exemplo disso usando a classe Shark que criamos. Os comentários
no programa descrevem cada etapa do processo."""
class Shark:
# variaveis de classe
animal_type = 'fish'
location = 'ocean'
# metodo construtor com as variaves de instancia name e age
def __init__(self, name, age):
self.name = name
self.age = age
# metodo para definir a variavel followers
def set_follwers(self, followers):
print("This user has " + str(followers) + " followers")
def main():
sammy = Shark('Sammy', 32)
print(sammy.name)
print(sammy.age)
print(sammy.location)
print()
# segundo objeto
stevie = Shark('Stevie', 34)
print(stevie.name)
print(stevie.age)
print(stevie.location)
print(stevie.animal_type)
if __name__=="__main__":
main()
"""Aqui, usamos as variáveis class e instance em dois objetos da classe Shark, sammy
e stevie."""
"""
Conclusão
Na programação orientada a objetos, as variáveis no nível da classe são referidas como
variáveis de classe, enquanto as variáveis no nível do objeto são chamadas de variáveis
de instância.
Essa diferenciação nos permite usar variáveis de classe para inicializar objetos com um
valor específico atribuído a variáveis e usar variáveis diferentes para cada objeto com
variáveis de instância.
O uso de variáveis específicas de classe e de instância pode garantir que nosso código
siga o princípio DRY para reduzir a repetição no código."""
"""referencias:
https://www.digitalocean.com/community/tutorials/
understanding-class-and-instance-variables-in-python-3""" | false |
9c4251d34eff84b1451c32763c238b5ebbd45762 | frclasso/2nd_Step_Python_Fabio_Classo | /Cap01-Elementos-de-sintaxe-especifica/06_itertools2.py | 462 | 4.125 | 4 | #!/usr/bin/env python3
# Itertools Part 2
import itertools
## Permutations, order matters
election = {1:'Barb', 2:'Karen', 3:'Erin'}
# for p in itertools.permutations(election):
# print(p)
# for p1 in itertools.permutations(election.values()):
# print(p1)
## Combinations: Order does not matter
colorsForPainting = ["Red", "Blue", "Purple", "Orange", "Yellow", "Pink"]
for c in itertools.combinations(colorsForPainting, 2): # 2 elements
print(c)
| true |
d0110beec760bf30f9c8cef83d025d67872b1b80 | Belindalinx/Time_to_Practice | /Complex Datastructures/Complex Datastructures.py | 1,361 | 4.28125 | 4 | #Create a list of “person” dictionaries with a name, age and list of hobbies for each person. Fill in any data you want.
person = [{"name": "Alice", "age": "25", "hobbies": ["music", "art", "coloring"]},
{"name": "Bob", "age": "27", "hobbies": ["science", "gaming", "sleep"]},
{"name": "Corgi", "age": "3", "hobbies": ["ball", "run", "eat"]}
]
#Use a list comprehension to convert this list of persons into a list of names (of the persons).
"""
name = []
for i in person:
for k, v in i.items():
if k == "name":
name.append(v)
"""
name=[i["name"] for i in person]
print(name)
#Use a list comprehension to check whether all persons are older than 20.
"""
age = []
for i in person:
for k, v in i.items():
if k == "age":
age.append(int(v))
if all(age) > 20:
print("All persons are older than 20.")
else:
print("All persons are not older than 20.")
"""
age=all([int(i["age"]) >20 for i in person])
print(age)
#Copy the person list such that you can safely edit the name of the first person (without changing the original list).
c_person = [i.copy() for i in person]
c_person[0]["name"] = "Gina"
print(person)
print(c_person)
#Unpack the persons of the original list into different variables and output these variables.
a, b, c, = person
print(a)
print(b)
print(c)
| true |
c14ae5ba10c5384fda85b575af34916f9b25924a | chen13545260986/hello_python | /6并发编程/2线程/1线程.py | 962 | 4.125 | 4 | """
1进程
进程是最小的内存分配单位
进程中至少含有一个线程
进程中可以开启多个线程
2线程
线程是进程中的执行单位
线程被CPU执行的最小单位
线程之间资源共享
开启一个线程所需要的时间要远远小于开启一个进程
python 与 2线程
Cpython解释器在解释代码过程中容易产生数据不安全的问题
GIL 全局解释器锁 锁的是线程
"""
from threading import Thread
import time
# def func(n):
# time.sleep(1)
# print(n)
#
# # 线程并发执行
# for i in range(10):
# # 将这个函数注册到一个线程
# t = Thread(target=func,args=(i,))
# t.start()
"""
多线程中数据的共享
"""
def func(n):
global a
a += n
print(a)
a = 100
lst = []
for i in range(10):
# 注册线程并启动
t = Thread(target=func,args=(i,))
t.start()
lst.append(t)
for t in lst:
t.join()
print(a) | false |
040a30a9bf29b237ccb4ef75489a5d307d03fe61 | naveen-nayan/pyproto | /Hello.py | 2,562 | 4.75 | 5 | print("Hello World\n")
# Creating a list
List = []
# Creating a List with the use of string
List1 = ["Hello"]
# Creating a List with the use of multiple string
List2 = ["Good", "Morning"]
# Creating a Multi-Dimensional List
# (By Nesting a list inside a List)
List3 = [["Good", "Morning"], ["Hello"]]
# Print nested list
print(List3)
# Creating a List with
# the use of Numbers
# (Having duplicate values)
List4 = [1, 2, 4, 4, 3, 3, 3, 6, 5]
# Creating a List with
# mixed type of values
# (Having numbers and strings)
List5 = [1, 2, 'Good', 4, 'Morning', 6, 'Hello']
# Addition of Elements
# in the List
List.append(1)
List.append(2)
List.append(4)
# Adding elements to the List
# using Iterator
for i in range(1, 4):
List.append(i)
# Inserting element to list insert(index, value)
List.insert(2, 13)
print(List)
# Addition of multiple elements
# to the List at the end
# (using Extend Method)
List.extend([8, 'Hello', 'World'])
# accessing a element using
# negative indexing
# print the last element of list
print(List[-1])
# print the third last element of list
print(List[-3])
# Removing elements from List
# using Remove() method
List.remove(8)
# Creating a List
List7 = ['G', 'E', 'E', 'K', 'S', 'F',
'O', 'R', 'G', 'E', 'E', 'K', 'S']
print("Intial List: ")
print(List7)
# Print elements of a range
# using Slice operation
Sliced_List = List7[3:8]
print("\nSlicing elements in a range 3-8: ")
print(Sliced_List)
# Print elements from a
# pre-defined point to end
Sliced_List = List7[5:]
print("\nElements sliced from 5th "
"element till the end: ")
print(Sliced_List)
# Printing elements from
# beginning till end
Sliced_List = List7[:]
print("\nPrinting all elements using slice operation: ")
print(Sliced_List)
print(List7)
# Negative Index list slicing
# Creating a List
List8 = ['G', 'E', 'E', 'K', 'S', 'F',
'O', 'R', 'G', 'E', 'E', 'K', 'S']
print("Initial List: ")
print(List8)
# Print elements from beginning
# to a pre-defined point using Slice
Sliced_List = List8[:-6]
print("\nElements sliced till 6th element from last: ")
print(Sliced_List)
# Print elements of a range
# using negative index List slicing
Sliced_List = List8[-6:-1]
print("\nElements sliced from index -6 to -1")
print(Sliced_List)
# Printing elements in reverse
# using Slice operation
Sliced_List = List[::-1]
print("\nPrinting List in reverse: ")
print(Sliced_List)
for item in list:
print(item) | true |
509df17b2dc3eea4823eb50d9c548c53ae3261a7 | playerlove1/python_practice | /basic/Prime_number.py | 599 | 4.125 | 4 | #這是python求質數的範例
#求質數所用的方式:小於該整數一半以下的數 都無法整除該整數則為質數
#接收輸入的參數
input_number=int(input('輸入數字:'))
#該整數除以2的商
half_input_number=input_number//2
#迴圈(由 i=2 跑到一半+1)
for i in range(2,half_input_number+1):
#如果該數被一半以下的數整除(即取餘數為0)
if(input_number%i)==0:
print(input_number,"不是質數,可被",i,"整除")
break
#此為for迴圈的else敘述 當迴圈沒有被任何break中斷時會執行的敘述
else:
print(input_number,"是質數") | false |
72b0a8b9551ee1f989b8bd262d9890988e542461 | playerlove1/python_practice | /basic/object.py | 2,504 | 4.15625 | 4 | #Python的所有東西都是Object,每個object都有 id(),type(),print()的函式
#list(串列)
simple_list=[1,"two",True]
print(simple_list)
# 即在串列的最後接上"新元素"字串
simple_list.append("新元素")
print(simple_list)
# 即移除串列中最後一個元素
simple_list.pop()
print(simple_list)
#即為將"two"元素由串列中移除
simple_list.remove("two")
print(simple_list)
#即為在list index為0的位置插入 "zero"的元素
simple_list.insert(0,"zero")
print(simple_list)
#透過len函數取得串列長度
len(simple_list)
#也可在[]中指定index值 直接進行修改
simple_list[len(simple_list)-1]=len(simple_list)-1
print(simple_list)
#刪除index 1以後的元素(包含index=1) 若輸入為 del simple_list[:] 即為清空該串列
del simple_list[1:]
print(simple_list)
# "[:]"對串列使用等於建立一個新串列，每個索引值都會參考到舊串列中每個所印位置的元素
#也就是所謂的淺層複製 (Shallow copy)
copy_list=simple_list[:]
print(copy_list)
#將list使用"*"運算 將會對list內容進行 淺層複製 (Shallow copy)
double_list=simple_list*2
print(double_list)
#將list使用"+"運算 將會建立一個新的物件 長度等於兩個串列的和 並合併兩個串列
add_list=copy_list+double_list
print(add_list)
#set (集合)
admins_set={"root","fong","s"}
users_set={"user1","user2","s"}
#交集
print("交集:",admins_set&users_set)
#聯集
print("聯集:",admins_set|users_set)
#差集
print("差集:",admins_set-users_set)
#XOR :排除共同元素
print("XOR:",admins_set^users_set)
#父集>子集 子集<父集 (> 與 < 回傳布林值 主要判斷兩個集合是否有父集與子集的關係)
#dictionary 即為key value 的模式 類似Java的map
passwords_dict={"user1":123,"user2":456} #同義於 passwords=dict(user1=123,user2=456)
#可以直接使用passwords["user1"]取得該變數的值 但若名稱不存在 則會拋出KeyError
#因此可以透過get取得 若不存在則回傳None(deault) 可在後面加入字串修改default回傳的結果
print(passwords_dict.get("user3","變數不存在"))
# 使用update() 增加元素
passwords_dict.update({"user3":789})
print(passwords_dict)
# 使用pop() 刪除元素
passwords_dict.pop("user3")
print(passwords_dict)
#可以使用passwords_dict.items()取得tuple
#可以使用passwords_dict.keys()取得key
#可以使用passwords_dict.value()取得value
#Tuple 像list一樣 但list是可變動 tuple是不可變動 | false |
0158a3c83488fbc7bcd3a27a2cdb781c1fb836bf | tbremm/Hacker_Rank_Python | /map_and_lambda_function.py | 1,775 | 4.6875 | 5 | # Let's learn some new Python concepts! You have to generate a list of the first fibonacci numbers, being the first
# number. Then, apply the map function and a lambda expression to cube each fibonacci number and print the list.
#
# Concept
# The map() function applies a function to every member of an iterable and returns the result. It takes two parameters:
# first, the function that is to be applied and secondly, the iterables.
# Let's say you are given a list of names, and you have to print a list that contains the length of each name.
#
# >> print (list(map(len, ['Tina', 'Raj', 'Tom'])))
# [4, 3, 3]
#
# Lambda is a single expression anonymous function often used as an inline function. In simple words, it is a function
# that has only one line in its body. It proves very handy in functional and GUI programming.
#
# >> sum = lambda a, b, c: a + b + c
# >> sum(1, 2, 3)
# 6
#
# Note:
# Lambda functions cannot use the return statement and can only have a single expression. Unlike def, which creates a
# function and assigns it a name, lambda creates a function and returns the function itself. Lambda can be used inside
# lists and dictionaries.
#
# Input Format
# One line of input: an integer N.
#
# Constraints
# 0 < N < 15
#
# Output Format
# A list on a single line containing the cubes of the first N fibonacci numbers.
cube = lambda x: x * x * x # complete the lambda function
def fibonacci(num_fib_terms):
# return a list of fibonacci numbers
if num_fib_terms == 0:
return []
if num_fib_terms == 1:
return [0]
fibs = [0, 1] # Fib series always starts with [0, 1]
for _ in range(2, num_fib_terms):
fibs.append(fibs[-1] + fibs[-2])
return fibs
n = int(input())
print(list(map(cube, fibonacci(n))))
| true |
c84bf6274eb980f1db1fb2816fdddaeba36909e6 | nmelgar/python_for_students | /loops.py | 372 | 4.1875 | 4 | pizza_toppings = ["cheese","pepperoni","mushrooms"]
#for loop after the FOR it goes whatever i want to call it
for topping in pizza_toppings:
message = f"I would like {topping} on my pizza"
print(message)
my_numbers = [1, 2, 3]
my_square_numbers = []
for number in my_numbers:
my_square_numbers.append(number**2)
print(my_square_numbers)
| true |
e1f4534a4ffcd110bb7f4c6929831e2ecc749f92 | Chiedu99/Unit10 | /Unit 10.py | 1,233 | 4.1875 | 4 | # Chiedu Nduka-Eze 11/22/16 Unit 10
# This program creates a pyramid with 5 rows of different colored bricks
import pygame, sys
from pygame.locals import *
import brick
pygame.init()
mainSurface = pygame.display.set_mode((500, 400), 0, 32)
BLUE = (0, 0, 255)
RED = (255, 0, 0)
BLACK = (0, 0, 0)
GREEN = (0, 255, 0)
GOLDROD = (184, 131, 11)
WHITE = (255, 255, 255)
mainSurface.fill(WHITE)
listColors = [BLACK, GREEN, GOLDROD, BLUE, RED]
SPACES = 5
height = mainSurface.get_height()
numberOfBricks = 9
widthOfBrick = (mainSurface.get_width() - ((numberOfBricks + 1) * SPACES ))/ 9
xPos = SPACES
yPos = height - 25
rowNumber = 0
# This loop creates a brick, places them, and changes the color for all 5 rows
for color in listColors:
# This indents the first brick of each row
xPos = rowNumber * (widthOfBrick + SPACES)
for x in range(numberOfBricks):
pie = brick.Brick(mainSurface, 25, widthOfBrick, color)
pie.draw(xPos, yPos)
xPos += (widthOfBrick + SPACES)
yPos -= 30
xPos = SPACES
numberOfBricks -= 2
rowNumber += 1
pygame.display.update()
while True:
for event in pygame.event.get():
if event.type == QUIT:
pygame.quit()
sys.exit()
| true |
c8a9544cf61cbc8a85148ee53fd579dca64a51db | nar0se/NCI | /Challenges/challenge6.py | 2,561 | 4.28125 | 4 | # Write an application for Cody's Car Care Shop that shows a user a list of available services:
# - Oil Change
# - Tire Rotation
# - Battery Check
# - Brake Inspection
#
# Allow the user to enter a string that corresponds to one of the options and display the option
# and its price as $25, $22, $15, or $5, accordingly. Display an error message if the user enters
# an invalid item.
#
# It might not be reasonable to expect users to type long entries such as 'Oil Change' accurately.
# as long as the user enters the first three characters of service, the choice is considered valid.
#
# Create a list with all the available services and another with the list of prices in another
# file and import them to the main file.
#
# Create a list of lists and print all elements of every list
#
# Create a list containing lists of tuple, an array, two dictionaries, a set, two strings, two floats, two integers, two
# booleans. Print each element of the list
import json, os
from json import dumps, loads
from os import system
# from prices import *
# from services import *
system('cls')
counter = 0
path = './Challenges/services.txt'
services = ['Oil Change', 'Tire Rotation', 'Battery Check', 'Brake Inspection']
with open (path, 'w') as file:
file.write('This is the list of services we provide')
file.close()
with open(path, 'a') as file:
file.write(f'\n{services}')
file.close()
path = './Challenges/prices.txt'
prices = ['25', '22', '15', '5']
with open(path, 'w') as file:
file.write('This is the list of prices for our services')
file.close()
with open(path, 'a') as file:
file.write(f'\n{prices}')
file.close()
cselect = str(input('Please select one of our services >> '))
dselect = cselect.lower()
string = dselect[0:3]
if (string) == 'oil':
string = services[0]
print(string)
elif (string) == 'tir':
string = services[1]
elif (string) == 'bat':
string = services[2]
elif (string) == 'bra':
string = services[3]
listofservices = dumps(services)
print(listofservices)
listofprices = dumps(prices)
print(listofprices)
listoflists = [
['Oil Change', 'Tire Rotation', 'Battery Check', 'Brake Inspection']
['25', '22', '15', '5']
]
# while True:
# content = input('What content do you want to add to the file? >> ')
# with open(path, 'a') as file:
# file.write(f'\n{content}')
# file.close()
# response = input('Do you want to add more content [Y/N]? >> ')
# if(str.upper(response) == 'N'):
# break
| true |
e149e67b55d68e45e070ffd9c164b52afcbde777 | kuranyukie/Projects | /Python/2020/Python3-Cookbook-1.2.py | 1,119 | 4.125 | 4 | # -*- coding: UTF-8 -*-
# 1.2 解压可迭代对象赋值给多个变量
a = (1,2,3,4,5)
x1, *x, x2 = a
print(f'{x1 = }') # x1 = 1
print(f'{x = }') # x = [2, 3, 4]
print('*x = ', *x) # *x = 2 3 4
print(f'{x2 = }') # x2 = 5
# 星号表达式在迭代元素为可变长元组的序列时是很有用的。
records = [
('foo', 1, 2),
('bar', 'hello'),
('foo', 3, 4),
]
def do_foo(x, y):
print('foo', x, y)
def do_bar(s):
print('bar', s)
for tag, *args in records:
if tag == 'foo':
do_foo(*args)
elif tag == 'bar':
do_bar(*args)
# 理解 x / *x / **x 的区别
def fun1(x, y = 'hello') : return (str(x) + str(y))
def fun2(**x) : return fun1(*x) # 此时的*x = (a, b) 是两个参数
def fun3(**x) : return fun1(x) # 此时的x = {'a': 1, 'b': 2} 是一个参数
def fun4(*x) : return fun1(x) # 此时的x = (1, 2) 是一个参数
def fun5(*x) : return fun1(*x) # 此时的*x = (1, 2) 是两个参数
print(fun2(a=1,b=2)) # ab
print(fun3(a=1,b=2)) # {'a': 1, 'b': 2}hello
print(fun4(1, 2)) # (1, 2)hello
print(fun5(1, 2)) # 12
| false |
95e01befa693e7805497c80c00eeb2fc166d4a04 | kuranyukie/Projects | /Python/2018/exercise/Ch5.py | 1,097 | 4.34375 | 4 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Tue Jan 30 19:24:35 2018
@author: Yukie
"""
"""
Assume the availability of a function is_prime.
Assume a variable n has been associated with positive integer.
Write the statements needed to find out how many prime numbers
(starting with 2 and going in increasing order with successively
higher primes [2,3,5,7,11,13,17,19,23...]) can be added before exceeding n.
Associate this number with the variable k.
"""
def is_prime(a):
if a == 0 or a == 1:
return False
for i in range(2,a):
if a % i == 0:
return False
return True
def test(n):
sum = 0
k = -1
num = 2
while sum <= n:
if is_prime(num):
sum += num
k += 1
num += 1
return(k)
# print(test(1))
for i in range(20):
print("test:", i, "result:", test(i))
"""
interest rate
"""
p = float(input("Enter current bank balance:"))
i = float(input("Enter interest rate:"))
t = float(input("Enter the amount of time that passes:"))
def calc(p,i,t):
f = p * (1 + i) ** t
return f
print(calc(p,i,t)) | true |
aad1b6a40b920be33aae31e7ebe184409efcea55 | santosh241/python-exercise | /function.py | 1,445 | 4.1875 | 4 | def main():
print("hello world!")
main()
print("print mee")
#
#
# #built_in_function
# a=9
# b=8
# c=sum((a,b))
# print(c)
# #user_define_function
# a=9
# b=8
# c=a+b
# print(c)
def function(a,b):
print("your are in function one" , a+b)
function(6,7)
#
def function1(a,b):
""" this method use for add two number"""
average=(a+b)/2
return average
v=function1(6,8)
print(v)
print(function1.__doc__)
#
#
def wish():
return "hello good morning"
print(wish())
#
#
def sum(a,b):
print("sum of two number",a+b)
sum(5,10)
a=int(input("enter your first value"))
b=int(input("enter your second value"))
def sum(c,d):
print("sum of two numbers",a+b)
sum(a,b)
#
def sum(c,a=10,b=20):
print("sum of two number",a+b+c)
sum(a=20,c=10)
#
class Circle:
pi = 3.14
##########Circle gets instantiated with a radius (default is 1)
def __init__(self, radius=1):
self.radius = radius
self.area = radius * radius * Circle.pi
####Method for resetting Radius
def setRadius(self, new_radius):
self.radius = new_radius
self.area = new_radius * new_radius * self.pi
### Method for getting Circumference
def getCircumference(self):
return self.radius * self.pi * 2
c = Circle()
print('Radius is: ',c.radius)
print('Area is: ',c.area)
print('Circumference is: ',c.getCircumference()) | true |
39c13556ed9b929cac6f24576fdbe0ce596a7e7c | killerCult/Basic-Python-Codes | /guess.py | 1,092 | 4.1875 | 4 | '''
You, the user, will have in your head a number between 0 and 100.
The program will guess a number, and you, the user, will say whether it is too high, too low, or your number.
At the end of this exchange, this program should print out how many guesses it took to get your number.
'''
# Killer_Cult
begin = 50 #Start with middle number between 1 and 100
flag = 0 #Indicate if correct guess or not
high = 100 #Default high, subject to change
low = 0 #Default low, subject to change
count = 0 #Count the number of guesses
while flag==0:
count+=1
print "Is the number", begin,"? (yes/no):"
guess = raw_input()
if guess=="yes":
print "Correct Guess"
print "Took", count, "tries"
flag = 1
elif guess=="no":
print "Wrong Guess"
print "Was the guess higher or lower? (higher/lower)"
hl = raw_input()
if hl=="higher":
high = begin
begin = begin - ((high-low)/2)
elif hl=="lower":
low = begin
begin = begin + ((high-low)/2)
else:
print "Invalid Input"
else:
print "Invalid Input"
| true |
6ee12f8281d9f8af9bfd1ba3d40cee89bebf577b | Ariz23746/MyPrograms-python | /array-usingClass.py | 1,647 | 4.21875 | 4 | class Array:
# Constructing a Constructor
def __init__(self):
self.length = 0
self.array = {}
# --------------------------------------------------------------------#
# Function to return our array:-
def ourArray(self, index):
return self.array[index]
# --------------------------------------------------------------------#
# Function to add element at the last index of an array :-
def push(self, item):
self.array[self.length] = item
self.length = self.length + 1
return self.array
# --------------------------------------------------------------------#
# Function to remove element from the last index of an array :-
def pop(self):
del(self.array[self.length - 1])
self.length = self.length - 1
return self.array
# --------------------------------------------------------------------#
# Function to delete an element from desire index of an array
def delete(self, index):
item = self.array[index] # O(1)
while(index < self.length - 1): # O(n)
self.array[index] = self.array[index + 1]
index = index + 1
self.length = self.length - 1 # O(1)
self.array[self.length] = item # O(1)
del(self.array[self.length]) # O(1)
return self.array # O(1)
# time complexity will be O(5 + n) = O(n)
# --------------------------------------------------------------------#
arr = Array()
arr.push('hi')
arr.push('my')
arr.push('name')
arr.push('is')
arr.push('Ariz')
print(arr.push('khan'))
print(arr.delete(5))
| true |
84eebe7fad08c17ad92b04069eb631cd9016459b | natwilkinson/advent-of-code-2020 | /day_02/day_2.py | 1,312 | 4.15625 | 4 | # Advent of Code - Day 2
def get_valid_passwords(filename):
"""
Finds valid password count using part 2 rules.
"""
f = open(filename, "r")
valid_passwords = 0
for line in f:
data = line.split()
[min_num, max_num] = [int(x) - 1 for x in data[0].split("-")]
if has_valid_password_part_2(min_num, max_num, data[1][0], data[2]):
valid_passwords += 1
return valid_passwords
def has_valid_password_part_1(min_num, max_num, target, password):
"""
Determines the validity of a password for Part 1. A password is valid if the target
character is in the password >= min_num and <= max_num.
"""
target_count = 0
for char in password:
if char == target:
target_count += 1
if target_count > max_num:
return False
if target_count < min_num:
return False
return True
def has_valid_password_part_2(index_1, index_2, target, password):
"""
Determines the validity of a password for Part 2. A password is valid if the target
character is found at index_1 or index_2, but not both (exclusive or).
"""
if (password[index_1] == target) is not (password[index_2] == target):
return True
return False
print(get_valid_passwords("input_day_2.txt"))
| true |
16e145e6deef9b5ccbc72ab067d4c684fab50890 | younes-alouani/MITx_6.00.1x_edx | /FInal_Exam/Problem4.py | 1,366 | 4.125 | 4 | """
Date : 26/02/2018
Program COntext : 6.00.1 EDX course
"""
def primes_list(N):
'''
N: an integer
Returns a list of prime numbers
'''
####
"""
Private function that returns :
True if a number is Prime
False if not
"""
def isPrime(number) :
import math
if number == 0 or number == 1 :
return False
elif number == 2 :
return True
else :
maxDenom = int(math.sqrt(number))
for denom in range(2,maxDenom+1):
if number % denom == 0 :
return False
return True
#####
primesList = []
for n in range(N+1) :
if isPrime(n) :
primesList += [n]
return primesList
"""
Tests
"""
#Test 1 :
N = 0
print("Test 1 result is : ",primes_list(N))
if primes_list(N) == [] :
test1 = "Test 1 passed"
else :
test1 = "Test 1 failed"
#Test 2 :
N = 3
print("Test 2 result is : ", primes_list(N))
if primes_list(N) == [2,3] :
test2 = "Test 2 passed"
else :
test2 = "Test 2 failed"
#Test 3 :
N = 11
print("Test 3 result is : ", primes_list(N))
if primes_list(N) == [2,3,5,7,11] :
test3 = "Test 3 passed"
else :
test3 = "Test 3 failed"
print(test1)
print(test2)
print(test3)
#All tests done - 10/10 full MARK
| true |
f3929ace00207f0fb2396de21a57d6cdd3d6ad13 | Aakash-Rajbhar/Library-Management-System | /library_manage_system.py | 2,351 | 4.40625 | 4 | #Implement a student library system using oops where student can borrow a book from the list of books.
#Create a seperate library & student class. Your program must be menu driver.You are free to choose methods and atributes of your choice to implement the functionality.
class Library:
def __init__(self, listofBooks):
self.books = listofBooks #This is for displaying the Books
def displayAvailableBooks(self):
print("Books present in the Library are : ")
for book in self.books:
print("* "+book)
def borrowBook(self,bookName):
if bookName in self.books:
print(f"You have been isued {bookName}. Please keep it safe and return it within 30 days.")
self.books.remove(bookName)
return True
def returnBook(self, bookName):
self.books.append(bookName)
print("Thanx for returning this book I hope you enjoyed it. Have a great day Ahead!!")
#----------------------Student------------------------------------------------------
class Student:
def requestBook(self):
self.book = input("Enter the name of book you want to borrow : ")
return self.book
def returnBook(self):
self.book = input("Enter the name of the book you want to return : ")
return self.book
if __name__ == "__main__" :
centralLibrary = Library(["Jungle Book", "Arabian Nights", "Avengers", "Honeycomb", "Inception", "Matrix", "King Kong", "Vistas", "Flamingo", "RD Sharma"])
student = Student()
while (True):
try:
Wlcm_msg = "************************ WELCOME TO THE CENTRAL LIBRARY **************************\n* Please Choose an option\n1. Show the List of books.\n2. Request for a book.\n3. Add/Return a book.\n4. Exit"
print(Wlcm_msg)
a = int(input("Enter a choice : "))
if a == 1:
centralLibrary.displayAvailableBooks()
elif a == 2 :
centralLibrary.borrowBook(student.requestBook())
elif a == 3 :
centralLibrary.returnBook(student.returnBook())
elif a == 4 :
print("Thanx for coming to the CENTRAL LIBRARY....")
exit()
else :
print("Invalid Choice.")
except :
print("Sorry Something wrong") | true |
ef6b1aca554648b945706f3398a8d1178da33bd7 | pallavijoshi92/Python_General_codes | /Palindrome.py | 450 | 4.28125 | 4 | #program to check if a given number/string is a palindrome
import sys
def isPalindrome(given):
given="".join(given.split())
if len(given)==0:
print "No string entered"
sys.exit()
for i in range(len(given)//2):
if given[i]!=given[-1-i]:
return False
return True
if __name__=="__main__":
given=raw_input("Enter string/number:")
print isPalindrome(given)
| true |
2d9b398818845fb013e96c97dae162326c492778 | carolinecav/test | /card.py | 251 | 4.125 | 4 | suits = ["hearts","clubs","diamonds","spades"]
values = ["2","3","4","5","6","7","8","9","10","jack","queen","king","ace"]
count = 0
for value in values:
for suit in suits:
print(value + " of "+ suit)
count = count +1
print(count)
| false |
acb9996d891d2f6955b56ed3d95a5a501d65542e | aditiabhang/my-python-stuff | /concepts_practice/problem_solving/fancy_number.py | 1,817 | 4.125 | 4 | '''
Given a mobile number and some conditions for a fancy number,
find if the given number is fancy. A 10 digit mobile number is called fancy
if it satisfies any of the following three conditions.
1. A single number occurs three consecutive times. Like 777.
2. Three consecutive digits are either in increasing or decreasing fashion. Like 456 or 987.
3. A single digit occurs four or more times in the number. Like 9859009976 – here the digit 9 occurs 4 times.
'''
import collections
def cond1(number_str):
for i in range(len(number_str)-2):
if (number_str[i] == number_str[i+1] and number_str[i+1] == number_str[i+2]):
return True
else:
return False
def cond2(number_str):
for i in range(len(number_str)-2):
if (number_str[i] < number_str[i + 1] and number_str[i + 1] < number_str[i + 2] or number_str[i] > number_str[i + 1] and number_str[i + 1] > number_str[i + 2]):
return True
else:
return False
# def cond3(number_str):
# countdict = collections.Counter(number_str)
# for keys in countdict:
# if countdict[keys] > 4:
# return True
# else:
# return False
def isFancy(num_str):
if (cond1(num_str) or cond2(num_str)): #or cond3(num_str)
return True
else:
return False
# Driver program
mobile_number = '1746294766666'
if (isFancy(mobile_number)):
print("Yes. You got a fancy number!")
else:
print(" Neahh, thats not a fancy number.")
# s = '90999904835545'
# # l2 = []
# # for i in s:
# # l2.append(i)
# # print(l2)
# listy = [n for n in s]
# my_dict = collections.Counter(listy)
# print(my_dict)
# for keys in my_dict:
# if my_dict[keys] > 4:
# print("YES")
# else:
# print("NO")
| true |
ef978c8a5467f6394fc5ce375b075c5afae369d9 | tarushsinha/pyLearn | /1.7 RegExp.py | 311 | 4.25 | 4 | ## Regular Expressions
## --> Powerful tool for string manipulation
## Useful for pattern matching, and string substitutions
## Regular expressions in Python can be accessed with the `re` module
import re
pattern = r"spam"
if re.match(pattern, "spamspamspam"):
print("match")
else:
print("no match") | true |
0bd4bf89f8afb934d4f092034aaebe120dd0805c | Mims20/turtle_race | /main.py | 1,241 | 4.3125 | 4 | from turtle import Turtle, Screen
import random
screen = Screen()
screen.setup(width=500, height=400)
user_bet = screen.textinput(title='Make a Bet', prompt='Which turtle will win? Enter a color: ')
colors = ['yellow', 'orange', 'red', 'green', 'blue', 'purple']
y_coordinate = [-100, -60, -20, 20, 60, 100]
all_turtles = []
for turtle_index in range(0, 6):
new_turtle = Turtle(shape="turtle")
new_turtle.penup()
new_turtle.color(colors[turtle_index])
new_turtle.goto(x=-240, y=y_coordinate[turtle_index])
all_turtles.append(new_turtle)
is_race_on = False
if user_bet:
is_race_on = True
while is_race_on:
for turtle in all_turtles:
if turtle.xcor() > 230: # checks to see if a turtle has reached the finish line
is_race_on = False
winner_turtle = turtle.pencolor() # get the color of the winner
if user_bet == winner_turtle:
print(f"You've Won! {winner_turtle} turtle won the race")
else:
print(f"You've Lost! {winner_turtle} turtle won the race")
# randomizing the forward distance each turtle takes
forward_distance = random.randint(0, 10)
turtle.forward(forward_distance)
screen.exitonclick()
| true |
4bf9285fbecc7434a559b8a0c38461850cd9cf8b | orangeshushu/LeetCode | /206. Reverse Linked List.py | 634 | 4.125 | 4 | '''
Reverse a singly linked list.
click to show more hints.
Hint:
A linked list can be reversed either iteratively or recursively. Could you implement both?
'''
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def reverseList(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
p =head
res = []
while p:
res.insert(0, p.val)
p = p.next
p = head
for i in res:
p.val = i
p = p.next
return head | true |
d6eb2b2a0ef3f34e8c872d6a69b4d91211c60e1b | sgriffin10/BIDA | /Class Activity #1/PyBasics__2__Variables_Types.py | 708 | 4.28125 | 4 | #Variables
x = 4 # x is of type int
x = "Sally" # x is now of type str
print(x)
#To combine both text and a variable, Python uses the + character:
x = "awesome"
print("Python is " + x)
#If you try to combine a string and a number, Python will give you an error:
# Uncomment the 3 lines below to test this out.
# Select the lines you want to uncomment and click Control-/ (i.e. control + forward slash)
# x = 5
# y = "John"
# print(x + y)
#To verify the type of any object in Python, use the type() function:
x = 1 # int
y = 2.8 # float
z = "Hello" # complex
print(type(x))
print(type(y))
print(type(z))
#Casting: Specify a Variable Type
y = int(2.8) # y will be 2
print(y) | true |
42f3f3eef47765bca4caa3d6f32a334bd815093a | sgriffin10/BIDA | /Class Activity #1/PyBasics__3__Conditional.py | 1,432 | 4.28125 | 4 | #Conditional Statements
#IF..ELSE
a = 200
b = 33
if b > a:
print("b is greater than a")
elif a == b:
print("a and b are equal")
else:
print("a is greater than b")
#shorthand
print("b is greater than a") if b > a else print("a and b are equal") if a == b else print("a is greater than b")
# #WHILE Loop
# i = 1
# while i < 6:
# print(i)
# # i += 1
# #With the break statement we can stop the loop even if the while condition is true
# i = 1
# while i < 6:
# print(i)
# if i == 3:
# break
# i += 1
# #With the continue statement we can stop the current iteration, and continue with the next
# i = 0
# while i < 6:
# i += 1
# if i == 3:
# continue
# print(i)
# # FOR Loops
# #A for loop is used for iterating over a sequence (that is either a list, a tuple, a dictionary, a set, or a string)
# fruits = ["apple", "banana", "cherry"]
# for x in fruits:
# print(x)
# for x in "banana":
# print(x)
# for x in range(6):
# print(x)
# else:
# print("Finally finished!")
# adj = ["red", "big", "tasty"]
# fruits = ["apple", "banana", "cherry"]
# for x in adj:
# for y in fruits:
# print(x, y)
for x in "banana":
print(x)
for x in range(6):
print(x)
else:
print("Finally finished!")
adj = ["red", "big", "tasty"]
fruits = ["apple", "banana", "cherry"]
for x in adj:
for y in fruits:
print(x, y)
| false |
359112b336e60bbb88aefb54a313e9a4612382a3 | AngusBangus/Python | /python/caesar.py | 2,473 | 4.4375 | 4 | # Exercise 1: Convert a string of characters to a list of numbers by taking the ASCII value of the numbers
def chars_to_nums(c):
char = []
sizeofList = len(c)
i = 0
while i < sizeofList:
char.append(ord(c[i]))
i += 1
return char
# Provide the correct implementation of this function
return "Not the correct result 1 "
print chars_to_nums("HELLO WORLD")
# Exercise 2: Convert a list of numbers to a string of characters by converting the ASCII values to characters
def nums_to_chars(n):
# Provide the correct implementation of this function
i = 0
num = []
sizeofList = len(n)
while i < sizeofList:
num.append(chr(n[i]))
i += 1
return num
return "Not the correct result 2"
print nums_to_chars([40, 37, 44, 44, 47, 0, 55, 47, 50, 44, 36])
# Exercise 3: Implement an encode procedure for the Caesar cipher. It should take in the plaintext as a string, the key as an integer, and a modulus as an integer, which corresponds to the size of the alphabet, and produce the appropriate ciphertext. In most cases the modulus will be 95, since there are 95 printable ASCII characters that we can use. The pair (key, mod) makes up the encryption key.
def caesar_encode(plaintext, key, mod):
# Provide the correct implementation of this function
#grab ascii values and change according to the key,
i = 0
string = ""
sizeofList = len(plaintext)
while i < sizeofList:
string+= (chr((ord(plaintext[i]) + key)%mod))
i += 1
return string
return "Not the correct result 3"
print caesar_encode("HELLO", 3, 95)
# Exercise 4: Implement the decode procedure for the Caesar cipher. It should take in the ciphertext and the encryption key, and produce the plaintext.
def caesar_decode(ciphertext, key, mod):
i = 0
string = ""
sizeofList = len(ciphertext)
while i < sizeofList:
string+= (chr((ord(ciphertext[i]) - key)%mod))
i += 1
return string
# Provide the correct implementation of this function
return "Not the correct result 4"
print caesar_decode("KHOOR", 3, 95)
# Exercise 5: Assuming the following string was encoded with the Caesar cipher, using some unknown encryption key, recover the plaintext. Type your answer in the string 'ex_5_plaintext' below, do not include any characters in the string, other than the recovered plaintext:
ex_5_ciphertext = "-HIIJSYCMY=IIF"
print caesar_decode(ex_5_ciphertext, 89, 95)
ex_5_plaintext = "3NOOPYISCOOL"
print "The message is:", ex_5_plaintext
| true |
1180e2154deb033234fe9d300b002974ec568a8d | VincentGFL/python_challenge | /programs/twob.py | 1,079 | 4.46875 | 4 | '''
Given a string containing lowercase letters (a-z) and numbers,
determine whether the string contains more numbers, more
letters, or an equal number of both. If the string contains an
invalid character (not a-z or 0-9) it should be ignored. If no
numbers or letters are present, the default response should be
“equal”.
countstring("abc123") -> "equal"
countstring("hello123") -> "letters"
countstring("sdf12345") -> "numbers"
countstring("1£2$aAbBcC") -> "letters"
countstring("ASD£!((") -> "equal"
'''
def countstring(string):
letter = 0
number = 0
for index in string:
if index.isalpha():
letter += 1
elif index.isalnum():
number += 1
continue
if letter == 0 or number == 0:
return "equal"
elif letter > number:
return "letters"
elif number > letter:
return "number"
else:
return "equal"
print(countstring("abc123"))
print(countstring("hello123"))
print(countstring("sdf12345"))
print(countstring("1£2$aAbBcC"))
print(countstring("ASD£!(("))
| true |
30ca13e3529c756573a0b0e7867bc5bfd763b50c | gitdxb/learning-python | /free-python/udemy-free-python/course-2/control-flow.py | 1,359 | 4.15625 | 4 | #if statement
#Turn light on and off
light = False
if light:
light = False
else:
light = True
print(light)
##The above code the same with 'light = not light'. output: false
# Odd or even number
a = int(input("Please enter a number: ")) #Use int() to convert input to interger
if a % 2 == 0:
print("You just entered an even number!")
else:
print("That is an odd number!")
#FOR STATEMENT
for number in range(1,11):
print(number)
_list = [True, 123,1212,345,1212,42, "tomorrow"]
for items in _list:
print(items)
#Or print only first 5 items
#for items in _list[:5]
# print(items)
#find top 3 value items
_list = [1,2,3,5,7,87,12,4323,653,999,343,88888,32,0,23]
max1 = max2 = max3 = min(_list)
for x in _list:
if x > max3:
max1 = max2
max2 = max3
max3 = x
elif x > max2:
max1 = max2
max2 = x
elif x > max1:
max1 = x
print(max1)
print(max2)
print(max3)
#write a program that checks if a list contains a sublist
a = [7,11,23,34,56,89,1]
b = [23,34]
is_sub_list = False
for i in range(len(a)):
if(a[i] == b[0]):
n = 1
while(n < len(b)) and a[i + n] == b[n]:
n += 1
if(n == len(b)):
is_sub_list = True
print(is_sub_list)
#Continue condition
for i in rang(1,11):
if i % 2 == 0:
continue
print(i) | true |
91d29bae42fd0a9193987d204c9ca3c8126ff616 | kamilzazi/Python_Bootcamp_Krakow | /1_podstawy_pgg/zad_11.py | 1,333 | 4.375 | 4 | """
Napisz program, który na podstawie pozycji gracza (x, y) na
planszy w przedziale od 0 do 100 wyświetli jego przybliżone
położenie (centrum, prawy górny róg, górna krawędź, . . . ) lub
informację o pozycji poza planszą. Przyjmij wartość 10 jako
margines krawędzi.
Przykładowy komunikat programu:
Podaj pozycję gracza X: 95
Podaj pozycję gracza Y: 95
Gracz znajduje się w prawym górnym rogu.
"""
x = int(input("Podaj pozycję gracza X: "))
y = int(input("Podaj pozycję gracza Y: "))
# if x >= 0 and x <= 10: -> if 0 <= x <= 10:
# elif x > 10 and x <= 90: -> elif 10 < x <= 90:
if 0 <= x <= 10:
if 0 <= y <= 10:
print("lewy dolny róg")
elif 10 < y <= 90:
print("lewy krawędź")
elif 90 < y <= 100:
print("lewy górny róg")
else:
print("jesteś poza planszą")
elif 10 < x <= 90:
if 0 <= y <= 10:
print("lewa krawędź")
elif 10 < y <= 90:
print("centrum")
elif 90 < y <= 100:
print("góna krawędź")
else:
print("jesteś poza planszą")
elif 90 < x <= 100:
if 0 <= y <= 10:
print("prawy dolny róg")
elif 10 < y <= 90:
print("prawa krawędź")
elif 90 < y <= 100:
print("prawy górny róg")
else:
print("jesteś poza planszą")
else:
print("poza planszą") | false |
5e519eced0ecafc616024ba968a4db8d7a7c4fb4 | cdong5/Dog-Walking-Excerciser | /GUI.py | 2,255 | 4.28125 | 4 | # Created by Calvin Dong - 12/30/2018
# Learning tkinter and random Library
from random import *
from tkinter import *
def exercise():
# Reads a txt file named exercises
# Places each line of text into a list
# Uses the list to generate exercises
exerciselist = []
file = open('exercises.txt', 'r')
for line in file:
ex = line.strip()
exerciselist.append(ex)
len_exlist = int(len(exerciselist))
ran_ex = exerciselist[randint(0,len_exlist-1)]
return ran_ex
def num_sec_ex():
# Creates a random number of reps to do for exercise
ex_time = randint(10, 20)
return ex_time
def gen_ex():
# Creates a Label of the exercise and number of reps
# Places into the application
textframe = Frame(root)
textframe.pack()
e1 = Label(textframe, text=(f'{exercise()} for {num_sec_ex()} reps'))
e1.pack()
def num_sec_walk():
# Creates a random variable to tell how long to walk
walk_time = randint(120,240)
return walk_time
def gen_walking():
# Generates a label to tell how long to walk for
# Places into the application
textframe = Frame(root)
textframe.pack()
e2 = Label(textframe, text=(f'Walk for {round(num_sec_walk()//60, 2)} minutes'))
e2.pack()
return e2
def create_list():
# Creates a random list of exercises and walking time
# Gets the time you have, and creates a list of exercises to fit in the time
time = entry_box.get()
current_time = 0
total_time = int(time)*60
while current_time < total_time:
gen_walking()
gen_ex()
current_time += num_sec_ex()
current_time += num_sec_walk()
# Formating and organization of the GUI
root = Tk()
root.title('Dog Walk Exerciser')
labelframe = Frame(root)
labelframe.pack(side=TOP)
top_label = Label(labelframe, text='Dog Walk Exerciser').pack(side=TOP)
Entry_frame = Frame(root)
Entry_frame.pack(side=TOP)
entry_label = Label(Entry_frame, text='How Long?')
entry_label.pack(side="left", padx=2)
entry_box = Entry(Entry_frame)
entry_box.pack(side="left", padx=2, pady=2)
gen_frame = Frame(root)
gen_frame.pack(side=TOP)
gen_button = Button(gen_frame, text='Generate', command= create_list)
gen_button.pack(pady=2)
root.mainloop()
| true |
cf7c96d4d93b347ef3df853f46725ef048951754 | enanibus/biopython | /curso_python/ejemplos/banco.py | 988 | 4.125 | 4 | #! /usr/bin/env python
class Banco():
""" Simula las transacciones de un banco"""
__clientes = {}
def __init__(self, saldo, interes = 2):
self.__interes = interes
self.__saldo = saldo
def Prestamo(self, cliente, dinero):
if self.__saldo <= dinero:
print "Andamos escasos de dinero :)."
else:
if cliente in self.__clientes:
self.__clientes[cliente] += dinero
else:
self.__clientes[cliente] = dinero
self.__saldo -= dinero
def Devolucion(self, cliente):
if not cliente in self.__clientes:
return "No eres cliente nuestro."
else:
self.__saldo += (self.__clientes[cliente] + (self.__clientes[cliente] * self.__interes))
del self.__clientes[cliente]
def setSaldo(self, saldo):
self.__saldo = saldo
def getSaldo(self):
return self.__saldo
def setInteres(self, interes):
self.__interes = interes
def getInteres(self):
return self.__interes
saldo = property(getSaldo, setSaldo)
interes = property(getInteres, setInteres)
| false |
5952c4e0a05fdabbc468cf305d8edbc210cfeb2a | Dim-Nike/Py_start | /lessons_2/variables.py | 2,417 | 4.125 | 4 | # ------TRIGGER__PY_START------
word = "Some" # String
number = 7 # Integer
num_2 = 5.68132 # Float
bool = False # Boolean
word = "Результат:"
number -= 5
x = 8
res = number % x
a = z = q = 10
w, r, b = 12, "Word", True
print(word, res, w, r, b)
age = input("Введите ваш возраст: ")
print("Ваш возраст", age)
# -------HOMEWORK--------
# TODO Нахождение числа Создайте программу, которая будет принимать число (n), введенное пользователем, и выдавать
# результат в виде (n + n * 2).
# TODO Работа с переменными
# Создайте переменную со значением 46 и переменную со значением "string". Последнюю переменную умножьте на 5.
# Выведите на экран обе переменные.
# TODO Простые переменные
# Создайте переменные со значениями: 5, F, Привет, 90.2.
# Создайте переменную, которую нельзя будет изменить и установите ей значение 67.
# Выведите переменную со значением "Привет" на экран.
# TODO Разделение числа на символы
# Напишите программу, которая будет получать от пользователя число с 4 числами.
# Реализуйте разделение этого числа на отдельные цифры.
# Пример:
# Число 5934
# Результат 5, 9, 3, 4
# TODO Получение данных от пользователя
# Создайте программу, что будет запрашивать данные пользователя: имя, фамилию, возраст и выводить их на экран.
# TODO Математические операции
# Создайте программу, которая будет запрашивать три переменные у пользователя и после их получения выводить на экран математически операции над ними:
# сумма
# вычитание
# деление
# умножение
# остаток при делении | false |
86759b830fd23e3b4ce3e7b72f6285678b184784 | floydchenchen/leetcode | /116-populating-next-right-pointers-in-each-node.py | 2,965 | 4.25 | 4 | # 116. Populating Next Right Pointers in Each Node
# Given a binary tree
#
# struct TreeLinkNode {
# TreeLinkNode *left;
# TreeLinkNode *right;
# TreeLinkNode *next;
# }
# Populate each next pointer to point to its next right node. If there is no next right node,
# the next pointer should be set to NULL.
#
# Initially, all next pointers are set to NULL.
#
# Note:
#
# You may only use constant extra space.
# Recursive approach is fine, implicit stack space does not count as extra space for this problem.
# You may assume that it is a perfect binary tree (ie, all leaves are at the same level,
# and every parent has two children).
# Example:
#
# Given the following perfect binary tree,
#
# 1
# / \
# 2 3
# / \ / \
# 4 5 6 7
# After calling your function, the tree should look like:
#
# 1 -> NULL
# / \
# 2 -> 3 -> NULL
# / \ / \
# 4->5->6->7 -> NULL
# Definition for binary tree with next pointer.
class TreeLinkNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
self.next = None
class Solution:
# recursive solution
def connect(self, root):
if not root or not root.left:
return
root.left.next = root.right
if root.next:
root.right.next = root.next.left
self.connect(root.left)
self.connect(root.right)
# iterative solution
def connect1(self, root):
while root and root.left:
next = root.left
while root:
root.left.next = root.right
root.right.next = root.next and root.next.left
root = root.next
root = next
# universal iterative solution for 116 and 117
def connect(self, root):
cur = root
# loopcur nodeƶͨdown_ptr¼
while cur:
# create two dummy pointer nodes
right_ptr, down_ptr = Node(-1), Node(-1)
# ֻdown_ptrָһߵnodeҵԺͲ
down_ptr_found = False
# loopcur nodeƶͨright_ptr¼
while cur:
# Ϊdown_ptrright_ptrʼһnodedown_ptrָparentһߵnode
if cur.left:
if not down_ptr_found:
down_ptr.next = cur.left
down_ptr_found = True
right_ptr.next = cur.left
right_ptr = right_ptr.next
if cur.right:
if not down_ptr_found:
down_ptr.next = cur.right
down_ptr_found = True
right_ptr.next = cur.right
right_ptr = right_ptr.next
cur = cur.next
cur = down_ptr.next
return root
| true |
0ffda0479b8b905b89affa0271a628933b5fd92f | floydchenchen/leetcode | /752-open-the-lock.py | 2,913 | 4.1875 | 4 | # 752. Open the Lock
# You have a lock in front of you with 4 circular wheels. Each wheel has 10 slots: '0', '1', '2', '3', '4', '5', '6', '7', '8', '9'. The wheels can rotate freely and wrap around: for example we can turn '9' to be '0', or '0' to be '9'. Each move consists of turning one wheel one slot.
#
# The lock initially starts at '0000', a string representing the state of the 4 wheels.
# You are given a list of deadends dead ends, meaning if the lock displays any of these codes, the wheels of the lock will stop turning and you will be unable to open it.
# Given a target representing the value of the wheels that will unlock the lock, return the minimum total number of turns required to open the lock, or -1 if it is impossible.
# Example 1:
# Input: deadends = ["0201","0101","0102","1212","2002"], target = "0202"
# Output: 6
# Explanation:
# A sequence of valid moves would be "0000" -> "1000" -> "1100" -> "1200" -> "1201" -> "1202" -> "0202".
# Note that a sequence like "0000" -> "0001" -> "0002" -> "0102" -> "0202" would be invalid,
# because the wheels of the lock become stuck after the display becomes the dead end "0102".
# Example 2:
# Input: deadends = ["8888"], target = "0009"
# Output: 1
# Explanation:
# We can turn the last wheel in reverse to move from "0000" -> "0009".
# Example 3:
# Input: deadends = ["8887","8889","8878","8898","8788","8988","7888","9888"], target = "8888"
# Output: -1
# Explanation:
# We can't reach the target without getting stuck.
# Example 4:
# Input: deadends = ["0000"], target = "8888"
# Output: -1
class Solution:
# two-way BFS
def openLock(self, deadends: List[str], target: str) -> int:
# helper function to help move up and down
def move(s: str, pos: int, up: bool) -> str:
s = list(s)
if up and s[pos] == '9':
s[pos] = '0'
if not up and s[pos] == '0':
s[pos] = '9'
else:
s[pos] = str(int(s[pos]) + 1) if up else str(int(s[pos]) - 1)
return ''.join(s)
deads = set(deadends)
q1, q2, visited = set(), set(), set()
step = 0
q1.add('0000')
q2.add(target)
while q1 and q2:
temp_q1 = set()
for s in q1:
if s in deads:
continue
if s in q2:
return step
visited.add(s)
# move one digit at each position up and down
for i in range(4):
up = move(s, i, True)
if up not in visited:
temp_q1.add(up)
down = move(s, i, False)
if down not in visited:
temp_q1.add(down)
step += 1
# q1 and q2
q1 = q2
q2 = temp_q1
return -1 | true |
0e8182fc17ff332c6930b1e85c6615707661d1a2 | floydchenchen/leetcode | /150-evaluate-reverse-polish-notation.py | 1,703 | 4.125 | 4 | # 150. Evaluate Reverse Polish Notation
# Evaluate the value of an arithmetic expression in Reverse Polish Notation.
#
# Valid operators are +, -, *, /. Each operand may be an integer or another expression.
#
# Note:
#
# Division between two integers should truncate toward zero.
# The given RPN expression is always valid. That means the expression would always evaluate to a
# result and there won't be any divide by zero operation.
# Example 1:
#
# Input: ["2", "1", "+", "3", "*"]
# Output: 9
# Explanation: ((2 + 1) * 3) = 9
# Example 2:
#
# Input: ["4", "13", "5", "/", "+"]
# Output: 6
# Explanation: (4 + (13 / 5)) = 6
# Example 3:
#
# Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
# Output: 22
# Explanation:
# ((10 * (6 / ((9 + 3) * -11))) + 17) + 5
# = ((10 * (6 / (12 * -11))) + 17) + 5
# = ((10 * (6 / -132)) + 17) + 5
# = ((10 * 0) + 17) + 5
# = (0 + 17) + 5
# = 17 + 5
# = 22
class Solution:
def evalRPN(self, tokens):
"""
:type tokens: 1List[str]
:rtype: int
"""
def operation(op, second, first):
if op == "*":
return first * second
elif op == "/":
# since python // is different than java /: python always does floor division
# https://stackoverflow.com/questions/5535206/negative-integer-division-surprising-result
return int(first / second)
elif op == "+":
return first + second
else:
return first - second
result = []
for token in tokens:
if token.lstrip("-").isdigit():
result.append(int(token))
else:
result.append(operation(token, result.pop(), result.pop()))
return result.pop()
sol = Solution()
print(sol.evalRPN(["10","6","9","3","+","-11","*","/","*","17","+","5","+"]))
| true |
27fe6d0149fa51fd5e9ff87d321bbc575238f416 | floydchenchen/leetcode | /366-find-leaves-of-binary-tree.py | 1,493 | 4.25 | 4 | # 366. Find Leaves of Binary Tree
# Given a binary tree, collect a tree's nodes as if you were doing this: Collect and remove all leaves,
# repeat until the tree is empty.
#
#
#
# Example:
#
# Input: [1,2,3,4,5]
#
# 1
# / \
# 2 3
# / \
# 4 5
#
# Output: [[4,5,3],[2],[1]]
#
#
# Explanation:
#
# 1. Removing the leaves [4,5,3] would result in this tree:
#
# 1
# /
# 2
#
#
# 2. Now removing the leaf [2] would result in this tree:
#
# 1
#
#
# 3. Now removing the leaf [1] would result in the empty tree:
#
# []
# Definition for a binary tree node.
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
from collections import defaultdict
class Solution:
# recursive divide and conquer + dictionary
def findLeaves(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
# 和depth是反的，leaf时是order为0，root的order是max(left, right) + 1
def order(root, dic):
if not root:
return 0
left, right = order(root.left, dic), order(root.right, dic)
level = max(left, right) + 1
dic[level].append(root.val)
return level
dic, result = defaultdict(list), []
order(root, dic)
for i in range(1, len(dic) + 1):
result.append(list(dic[i]))
return result
| true |
51378900f7fedfe0974f3da45c274d008581d4ce | floydchenchen/leetcode | /588-design-in-memory-file-system.py | 2,328 | 4.28125 | 4 | # 588. Design In-Memory File System
# Design an in-memory file system to simulate the following functions:
# ls: Given a path in string format. If it is a file path, return a list that only contains this file's name. If it is a directory path, return the list of file and directory names in this directory. Your output (file and directory names together) should in lexicographic order.
# mkdir: Given a directory path that does not exist, you should make a new directory according to the path. If the middle directories in the path don't exist either, you should create them as well. This function has void return type.
# addContentToFile: Given a file path and file content in string format. If the file doesn't exist, you need to create that file containing given content. If the file already exists, you need to append given content to original content. This function has void return type.
# readContentFromFile: Given a file path, return its content in string format.
# Example:
# Input:
# ["FileSystem","ls","mkdir","addContentToFile","ls","readContentFromFile"]
# [[],["/"],["/a/b/c"],["/a/b/c/d","hello"],["/"],["/a/b/c/d"]]
# Output:
# [null,[],null,null,["a"],"hello"]
# https://assets.leetcode.com/uploads/2018/10/12/filesystem.png
from collections import defaultdict
class TrieNode:
def __init__(self):
self.children = defaultdict(TrieNode)
self.content = ""
class FileSystem(object):
def __init__(self):
self.root = TrieNode()
#find and return TrieNode at path.
def find(self, path):
curr = self.root
# no nested path
if len(path) == 1:
return self.root
for word in path.split("/")[1:]:
curr = curr.children[word]
return curr
def ls(self, path):
curr = self.find(path)
# file path, return file name if curr is a file
if curr.content:
return [path.split('/')[-1]]
# if directory, return children keys
return sorted(curr.children.keys())
def mkdir(self, path):
self.find(path)
def addContentToFile(self, filePath, content):
curr = self.find(filePath)
curr.content += content
def readContentFromFile(self, filePath):
curr = self.find(filePath)
return curr.content | true |
25d0869cffcf30ed5612478ba699701be0b6396c | floydchenchen/leetcode | /043-mutiply-strings.py | 1,929 | 4.21875 | 4 | # 43. Multiply Strings
# Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2,
# also represented as a string.
#
# Example 1:
#
# Input: num1 = "2", num2 = "3"
# Output: "6"
# Example 2:
#
# Input: num1 = "123", num2 = "456"
# Output: "56088"
# Note:
#
# The length of both num1 and num2 is < 110.
# Both num1 and num2 contain only digits 0-9.
# Both num1 and num2 do not contain any leading zero, except the number 0 itself.
# You must not use any built-in BigInteger library or convert the inputs to integer directly.
# 思路： product和sum做法不同，不像sum一样两个pointer从右到左扫过，而是每一位都可能过很多遍（下面的例子）
# 所以注意在每次操作之后 product[tempPos] %= 10
# 1. 先建一个len(num1) + len(num2) 大的product list
# 2. 两个数从后往前，两个for-loop做乘法（如下图）
# 3. 为了避免2*3=6而不是06这种情况，将leading zeros去掉
# _ _ _ 1 2 3
# x _ _ _ 4 5 6
# -------------
# 7 3 8
# 6 1 5
# 4 9 2
# -------------
# 5 6 0 8 8
class Solution:
def multiply(self, num1, num2):
# 1. 先建一个len(num1) + len(num2) 大的product list
product = [0] * (len(num1) + len(num2))
# 2. 两个数从后往前，两个for-loop做乘法
for i in range(len(num1) - 1, -1, -1):
for j in range(len(num2) - 1, -1, -1):
product[j + i + 1] += int(num1[i]) * int(num2[j])
product[j + i] += product[j + i + 1] // 10
product[j + i + 1] %= 10
# print(product)
# 3. 为了避免2*3=6而不是06这种情况，将leading zeros去掉
start_index = 0
while start_index < len(product) - 1 and product[start_index] == 0:
start_index += 1
return ''.join(map(str, product[start_index:]))
print(Solution().multiply("2","3")) | false |
ce0e25ce4367119a43a29bf4c85711ac93d8deb7 | floydchenchen/leetcode | /226-invert-binary-tree.py | 1,583 | 4.15625 | 4 | # 226. Invert Binary Tree
# Invert a binary tree.
#
# Example:
#
# Input:
#
# 4
# / \
# 2 7
# / \ / \
# 1 3 6 9
# Output:
#
# 4
# / \
# 7 2
# / \ / \
# 9 6 3 1
# Definition for a binary tree node.
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
# recursive, divide and conquer
def invertTree(self, root):
"""
:type root: TreeNode
:rtype: TreeNode
"""
if not root:
return root
# divide and conquer
# 这里一定要写成1行，不然需要temp去存root.right的值（因为它会改变）
root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)
return root
# BFS, level order traversal
def invertTree1(self, root):
if not root:
return root
q = [root]
while q:
node = q.pop(0)
node.left, node.right = node.right, node.left
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
return root
# iterative DFS, stack
def invertTree2(self, root):
if not root:
return root
stack = [root]
while stack:
node = stack.pop()
node.left, node.right = node.right, node.left
if node.left:
stack.append(node.left)
if node.right:
stack.append(node.right)
return root
| true |
4273cd45078991b43b59bb232b6206161bed0798 | geiyer/python | /Module01/math-operators.py | 383 | 4.125 | 4 | print('Welcome to Python')
print('Addition: (2+7): ' + str(2+7))
print('subsctraction: (2-7) : ' + str(2-7))
print('Multiplication: (2*7):' + str(2*7))
print('Division: (7/2): ' + str(7/2))
print('Modulus: (7%2): ' + str(7%2))
print('Floor division: (8//3): ' + str(7//2))
print('Exponent: (2**3)' + str(2**3))
# Comments
# Line 2
# Line 3
"""
Comments are written here.
""" | false |
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