![]() pd.todatetime ('2018 - 07 - 07 04 - PM', format'Y - m - d I - p') Timestamp (' 16:00:00') This behaviour is documented in the docs: When used with the strptime () function, the p directive only. Print('dateutil parse:\n', timeit.timeit(lambda: parser.parse(' 12:23:44.234'). Since you're parsing a 12-hour time format, you will need I instead of H, otherwise the p specifier has no effect. Numbers = ''.join(re.findall(r'\d+', format_date))ĭ = datetime.datetime(int(numbers), int(numbers), int(numbers))ĭ = datetime.datetime(int(numbers), int(numbers), int(numbers), int(numbers), int(numbers), int(numbers))ĭ = datetime.datetime(int(numbers), int(numbers), int(numbers), int(numbers), int(numbers), int(numbers), microsecond=1000*int(numbers)) import reĭef date2timestamp_anyformat(format_date): ![]() Using regex to cut string into then unpack it and fill into datetime class datetime.datetime(year, month, day, hour=0, minute=0, second=0, microsecond=0, tzinfo=None, *, fold=0), this is the fastest way I tested so far.
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