• Apr 15, 2021 •NoahEaton
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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]
• Jun 26, 2025 •AustinLeath
def format_timestamp(timestamp_epoch): """ Convert epoch timestamp to formatted datetime string without using datetime package. Args: timestamp_epoch (int/float): Unix epoch timestamp (seconds since 1970-01-01 00:00:00 UTC) Returns: str: Formatted datetime string in 'YYYY-MM-DD HH:MM:SS' format """ # Constants for time calculations SECONDS_PER_DAY = 86400 SECONDS_PER_HOUR = 3600 SECONDS_PER_MINUTE = 60 # Handle negative timestamps and convert to integer timestamp = int(timestamp_epoch) # Calculate days since epoch and remaining seconds days_since_epoch = timestamp // SECONDS_PER_DAY remaining_seconds = timestamp % SECONDS_PER_DAY # Calculate hours, minutes, seconds hours = remaining_seconds // SECONDS_PER_HOUR remaining_seconds %= SECONDS_PER_HOUR minutes = remaining_seconds // SECONDS_PER_MINUTE seconds = remaining_seconds % SECONDS_PER_MINUTE # Calculate date (simplified, ignoring leap seconds) year = 1970 days = days_since_epoch while days >= 365: is_leap = (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0) days_in_year = 366 if is_leap else 365 if days >= days_in_year: days -= days_in_year year += 1 # Month lengths (non-leap year for simplicity, adjusted later for leap years) month_lengths = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] if (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0): month_lengths[1] = 29 month = 0 while days >= month_lengths[month]: days -= month_lengths[month] month += 1 # Convert to 1-based indexing for month and day month += 1 day = days + 1 # Format the output string return f"{year:04d}-{month:02d}-{day:02d} {hours:02d}:{minutes:02d}:{seconds:02d}" # Example timestamp (Unix epoch seconds) timestamp = 1697054700 formatted_date = format_timestamp(timestamp) print(formatted_date + " UTC") # Output: 2023-10-11 18:45:00
• Nov 19, 2022 •CodeCatch
0 likes • 4 views
def max_n(lst, n = 1): return sorted(lst, reverse = True)[:n] max_n([1, 2, 3]) # [3] max_n([1, 2, 3], 2) # [3, 2]
• Aug 1, 2025 •AustinLeath
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 ?
• Mar 12, 2021 •mo_ak
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prime_lists=[] # a list to store the prime numbers def prime(n): # define prime numbers if n <= 1: return False # divide n by 2... up to n-1 for i in range(2, n): if n % i == 0: # the remainder should'nt be a 0 return False else: prime_lists.append(n) return True for n in range(30,1000): # calling function and passing starting point =30 coz we need primes >30 prime(n) check=0 # a var to limit the output to 10 only for n in prime_lists: for x in prime_lists: val= n *x if (val > 1000 ): check=check +1 if (check <10) : print("the num is:", val , "=",n , "* ", x ) break
• Oct 7, 2022 •KETRICK
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x[cat_var].isnull().sum().sort_values(ascending=False)