• Nov 19, 2022 •CodeCatch
0 likes • 1 view
def key_of_min(d): return min(d, key = d.get) key_of_min({'a':4, 'b':0, 'c':13}) # b
• May 31, 2023 •CodeCatch
0 likes • 6 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]
• Jun 26, 2025 •AustinLeath
0 likes • 2 views
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
• Aug 1, 2025 •AustinLeath
import re _proposal_regex = r'(?:(?:(IKE|ESP):)?[\w/]+(?:/NO_EXT_SEQ)?(?:, ?(IKE|ESP):[\w/]+(?:/NO_EXT_SEQ)?)*)?' _proposals_re = rf'(?P<proposals>{_proposal_regex}|)' pattern = rf'received proposals: {_proposals_re}' match = re.match(pattern, 'received proposals: ') print(match.group('proposals') if match else "No match") # Prints "No match"
• 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]
• Aug 12, 2024 •AustinLeath
0 likes • 5 views
magnitude = lambda bits: 1_000_000_000_000_000_000 % (2 ** bits) sign = lambda bits: -1 ** (1_000_000_000_000_000_000 // (2 ** bits)) print("64 bit sum:", magnitude(64) * sign(64)) print("32 bit sum:", magnitude(32) * sign(32)) print("16 bit sum:", magnitude(16) * sign(16))