• Jul 8, 2025 •AustinLeath
0 likes • 5 views
from datetime import datetime epoch_time = 1753823646 # Example epoch time (March 15, 2023 00:00:00 UTC) # Convert epoch time to a UTC datetime object utc_datetime = datetime.utcfromtimestamp(epoch_time) print(f"Epoch time: {epoch_time}") print(f"UTC datetime: {utc_datetime}") # You can also format the output string formatted_utc_time = utc_datetime.strftime('%m-%d-%Y %H:%M:%S UTC') print(f"Formatted UTC datetime: {formatted_utc_time}")
• Nov 19, 2022 •CodeCatch
0 likes • 2 views
def byte_size(s): return len(s.encode('utf-8')) byte_size('😀') # 4 byte_size('Hello World') # 11
• May 31, 2023 •CodeCatch
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# Function to check Armstrong number def is_armstrong_number(number): # Convert number to string to iterate over its digits num_str = str(number) # Calculate the sum of the cubes of each digit digit_sum = sum(int(digit) ** len(num_str) for digit in num_str) # Compare the sum with the original number if digit_sum == number: return True else: return False # Prompt user for a number number = int(input("Enter a number: ")) # Check if the number is an Armstrong number if is_armstrong_number(number): print(number, "is an Armstrong number.") else: print(number, "is not an Armstrong number.")
• Oct 7, 2022 •KETRICK
x[cat_var].isnull().sum().sort_values(ascending=False)
• Sep 6, 2020 •C S
0 likes • 3 views
def Fibonacci(n): if n<0: print("Incorrect input") # First Fibonacci number is 0 elif n==1: return 0 # Second Fibonacci number is 1 elif n==2: return 1 else: return Fibonacci(n-1)+Fibonacci(n-2) # Driver Program print(Fibonacci(9))
• Apr 15, 2021 •NoahEaton
import anytree as at import random as rm # Generate a tree with node_count many nodes. Each has a number key that shows when it was made and a randomly selected color, red or white. def random_tree(node_count): # Generates the list of nodes nodes = [] for i in range(node_count): test = rm.randint(1,2) if test == 1: nodes.append(at.Node(str(i),color="white")) else: nodes.append(at.Node(str(i),color="red")) #Creates the various main branches for i in range(node_count): for j in range(i, len(nodes)): test = rm.randint(1,len(nodes)) if test == 1 and nodes[j].parent == None and (not nodes[i] == nodes[j]): nodes[j].parent = nodes[i] #Collects all the main branches into a single tree with the first node being the root for i in range(1, node_count): if nodes[i].parent == None and (not nodes[i] == nodes[0]): nodes[i].parent = nodes[0] return nodes[0]