• Nov 18, 2022 •AustinLeath
0 likes • 11 views
#Python 3: Fibonacci series up to n def fib(n): a, b = 0, 1 while a < n: print(a, end=' ') a, b = b, a+b print() fib(1000)
• May 31, 2023 •CodeCatch
0 likes • 7 views
my_list = [1, 2, 3, 4, 5] removed_element = my_list.pop(2) # Remove and return element at index 2 print(removed_element) # 3 print(my_list) # [1, 2, 4, 5] last_element = my_list.pop() # Remove and return the last element print(last_element) # 5 print(my_list) # [1, 2, 4]
• Nov 19, 2022 •CodeCatch
0 likes • 6 views
from collections import Counter def find_parity_outliers(nums): return [ x for x in nums if x % 2 != Counter([n % 2 for n in nums]).most_common()[0][0] ] find_parity_outliers([1, 2, 3, 4, 6]) # [1, 3]
0 likes • 3 views
# Python code to find the URL from an input string # Using the regular expression import re def Find(string): # findall() has been used # with valid conditions for urls in string regex = r"(?i)\b((?:https?://|www\d{0,3}[.]|[a-z0-9.\-]+[.][a-z]{2,4}/)(?:[^\s()<>]+|\(([^\s()<>]+|(\([^\s()<>]+\)))*\))+(?:\(([^\s()<>]+|(\([^\s()<>]+\)))*\)|[^\s`!()\[\]{};:'\".,<>?«»“”‘’]))" url = re.findall(regex,string) return [x[0] for x in url] # Driver Code string = 'My Profile: https://codecatch.net' print("Urls: ", Find(string))
• Jun 16, 2024 •lagiath
0 likes • 2 views
print('hello, world')
• May 5, 2026 •CodeCatch
0 likes • 1 view