python 'x days ago' to datetime

Question

I have strings that show a date in the following format:

x minutes/hours/days/months/years ago

I need to parse that to a datetime using pyt

Answer 1

Custom function to convert x hours ago to datetime, x hour, y mins ago to datetime, etc in Python.

Function takes single parameter of type string which is parsed using RegExp. RegExp can be customized to match function input.

For usage see examples below.

import re
from datetime import datetime, timedelta


def convert_datetime(datetime_ago):
    matches = re.search(r"(\d+ weeks?,? )?(\d+ days?,? )?(\d+ hours?,? )?(\d+ mins?,? )?(\d+ secs? )?ago", datetime_ago)

    if not matches:
        return None

    date_pieces = {'week': 0, 'day': 0, 'hour': 0, 'min': 0, 'sec': 0}

    for i in range(1, len(date_pieces) + 1):
        if matches.group(i):
            value_unit = matches.group(i).rstrip(', ')
            if len(value_unit.split()) == 2:
                value, unit = value_unit.split()
                date_pieces[unit.rstrip('s')] = int(value)

    d = datetime.today() - timedelta(
        weeks=date_pieces['week'],
        days=date_pieces['day'],
        hours=date_pieces['hour'],
        minutes=date_pieces['min'],
        seconds=date_pieces['sec']
    )

    return d

Example usage:

dates = [
    '1 week, 6 days, 11 hours, 20 mins, 13 secs ago',
    '1 week, 10 hours ago',
    '1 week, 1 day ago',
    '6 days, 11 hours, 20 mins ago',
    '1 hour ago',
    '11 hours, 20 mins ago',
    '20 mins 10 secs ago',
    '10 secs ago',
    '1 sec ago',
]    

for date in dates:
    print(convert_datetime(date))

Output:

2019-05-10 06:26:40.937027
2019-05-16 07:46:53.937027
2019-05-15 17:46:53.937027
2019-05-17 06:26:53.937027
2019-05-23 16:46:53.937027
2019-05-23 06:26:53.937027
2019-05-23 17:26:43.937027
2019-05-23 17:46:43.937027
2019-05-23 17:46:52.937027