• Nov 19, 2022 •CodeCatch
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def print_x_pattern(size): i,j = 0,size - 1 while j >= 0 and i < size: initial_spaces = ' '*min(i,j) middle_spaces = ' '*(abs(i - j) - 1) final_spaces = ' '*(size - 1 - max(i,j)) if j == i: print(initial_spaces + '*' + final_spaces) else: print(initial_spaces + '*' + middle_spaces + '*' + final_spaces) i += 1 j -= 1 print_x_pattern(7)
• Oct 15, 2022 •CodeCatch
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my_list = ["blue", "red", "green"] #1- Using sort or srted directly or with specifc keys my_list.sort() #sorts alphabetically or in an ascending order for numeric data my_list = sorted(my_list, key=len) #sorts the list based on the length of the strings from shortest to longest. # You can use reverse=True to flip the order #2- Using locale and functools import locale from functools import cmp_to_key my_list = sorted(my_list, key=cmp_to_key(locale.strcoll))
def key_of_min(d): return min(d, key = d.get) key_of_min({'a':4, 'b':0, 'c':13}) # b
• Jun 26, 2025 •AustinLeath
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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
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)
• Oct 7, 2022 •KETRICK
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x[cat_var].isnull().sum().sort_values(ascending=False)