How To Convert Standard Timedelta String To Timedelta Object

What is the simplest way to convert standard timedelta string to timedelta object? I have printed several timedelta objects and got these strings: '1157 days, 9:46:39' '12:00:01.82

Solution 1:

I cannot find a better way other than writing a parser myself. The code looks bulky but it is essentially parsing string into a dictionary which is useful not only to creating a timedelta object.

import re

defparse(s):
    if'day'in s:
        m = re.match(r'(?P<days>[-\d]+) day[s]*, (?P<hours>\d+):(?P<minutes>\d+):(?P<seconds>\d[\.\d+]*)', s)
    else:
        m = re.match(r'(?P<hours>\d+):(?P<minutes>\d+):(?P<seconds>\d[\.\d+]*)', s)
    return {key: float(val) for key, val in m.groupdict().iteritems()}

Test:

from datetime import timedelta

s1 = '1157 days, 9:46:39'
s2 = '12:00:01.824952'
s3 = '-1 day, 23:59:31.859767'
t1 = parse(s1)
t2 = parse(s2)
t3 = parse(s3)

timedelta(**t1) # datetime.timedelta(1157, 35199)
timedelta(**t2) # datetime.timedelta(0, 43201, 824952)
timedelta(**t3) # datetime.timedelta(-1, 86371, 859767)

Hope this can suit your purpose.

Solution 2:

This is an update of Ray's answer that works for Python 3. It will return an empty string if the regex is fed a bad string for parsing, otherwise a timedelta object is returned.

from datetime import timedelta
import re

defparse_timedelta(stamp):
    if'day'in stamp:
        m = re.match(r'(?P<d>[-\d]+) day[s]*, (?P<h>\d+):'r'(?P<m>\d+):(?P<s>\d[\.\d+]*)', stamp)
    else:
        m = re.match(r'(?P<h>\d+):(?P<m>\d+):'r'(?P<s>\d[\.\d+]*)', stamp)
    ifnot m:
        return''

    time_dict = {key: float(val) for key, val in m.groupdict().items()}
    if'd'in time_dict:
        return timedelta(days=time_dict['d'], hours=time_dict['h'],
                         minutes=time_dict['m'], seconds=time_dict['s'])
    else:
        return timedelta(hours=time_dict['h'],
                         minutes=time_dict['m'], seconds=time_dict['s'])