• Nov 19, 2022 •CodeCatch
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def clamp_number(num, a, b): return max(min(num, max(a, b)), min(a, b)) clamp_number(2, 3, 5) # 3 clamp_number(1, -1, -5) # -1
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from functools import partial def curry(fn, *args): return partial(fn, *args) add = lambda x, y: x + y add10 = curry(add, 10) add10(20) # 30
• May 31, 2023 •CodeCatch
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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]
• Aug 1, 2025 •AustinLeath
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from typing import Optional from datetime import datetime def convert_timestamp_string_to_epoch(timestamp: str) -> Optional[int]: epoch_time = None time_obj = datetime.strptime(timestamp, "%Y-%m-%d %H:%M:%S.%f") epoch_time = int((time_obj - datetime(1970, 1, 1)).total_seconds() * 1000) return epoch_time print(int(convert_timestamp_string_to_epoch("2025-08-01 13:11:47.171"))) #above outputs 1754053907171.0 #how to I remove the .0 ?
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# function which return reverse of a string def isPalindrome(s): return s == s[::-1] # Driver code s = "malayalam" ans = isPalindrome(s) if ans: print("Yes") else: print("No")
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import itertools def compute_permutations(string): # Generate all permutations of the string permutations = itertools.permutations(string) # Convert each permutation tuple to a string permutations = [''.join(permutation) for permutation in permutations] return permutations # Prompt the user for a string string = input("Enter a string: ") # Compute permutations permutations = compute_permutations(string) # Display the permutations print("Permutations:") for permutation in permutations: print(permutation)