• Nov 18, 2022 •AustinLeath
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# List lst = [1, 2, 3, 'Alice', 'Alice'] # One-Liner indices = [i for i in range(len(lst)) if lst[i]=='Alice'] # Result print(indices) # [3, 4]
• 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
• 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]
• May 5, 2026 •CodeCatch
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class Rectangle: pass class Square(Rectangle): pass rectangle = Rectangle() square = Square() print(isinstance(rectangle, Rectangle)) # True print(isinstance(square, Rectangle)) # True print(isinstance(square, Square)) # True print(isinstance(rectangle, Square)) # False
• 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 ?