`parser.parse()` - how to specify allowed formats (and it's order)

Question

I want to create my own date parser. I like parser.parse() function but if I have:

date1 = '2022/11/10'
date2 = '10/11/2022'

parser.parse(date1)
parser.parse(date2)

Then the results are:

datetime.datetime(2022, 11, 10, 0, 0)
datetime.datetime(2022, 10, 11, 0, 0)

I want to always first try to parse the day with month between day and year. So the parser.parse(date2) should return November, not October. What is more I want to parse only the dates which includes days, month, and years, so parser.parse("20") should return an error and not datetime.datetime(2023, 1, 20, 0, 0). I would simply write my own function to parse dates (couple of if statements) but I'm afraid that it might works very slow with big data.

You could try different formats with a `try/except` around each, until it succeeds. — Barmar, Jan 07 '23 at 21:03

score 0 · Accepted Answer · answered Jan 09 '23 at 07:17

Do you know datetime.datetime.strptime ? It looks like what you want : just call each time with a different format you want to try for the input.

import datetime as dt
from typing import Optional


def my_date_parser(some_date_string: str) -> Optional[dt.date]:
    expected_date_formats = [
        # see https://docs.python.org/3/library/datetime.html#strftime-strptime-behavior
        "%d/%m/%Y",
        "%Y/%m/%d",
        # ...
    ]
    for expected_date_format in expected_date_formats:
        try:
            return dt.datetime.strptime(some_date_string, expected_date_format).date()
        except ValueError:
            continue  # ignore this one
    else:  # if the for-loop went through, it means that nothing matches
        return None


date1 = '2022/11/10'
date2 = '10/11/2022'
print(date1, my_date_parser(date1))  # 2022/11/10 2022-11-10
print(date2, my_date_parser(date2))  # 10/11/2022 2022-11-10

`parser.parse()` - how to specify allowed formats (and it's order)

1 Answers1