Parsing string to time value in python

Question

I need to convert the following strings into either datetimes or strings in a different format. My input looks like these examples:

196h 26m 13s
95h 19m
45m 28s

My desired string output would be (But my final goal is to convert these strings into datetime objects):

196:26:13
95:19:00
00:45:28

In Excel, it would be [h]:mm:ss.

Note: As you can see, hours can be higher than 24.

I have tried to parse this string with the time.strptime() method, but it doesn't work with hours higher than 24. I have a solution with regular expressions, but I want to know if there is a more straightforward way to do this. What is the best way to solve this?

Those aren't points in time; those are durations of time. Thus, datetime, date, and time are entirely the wrong types to use; the correct type is timedelta, which does not have a built-in string-parsing method. — jwodder
– jwodder, Commented Jan 26, 2017 at 21:50
Thanks guys for the help! I would stay by the good old regex way — patex1987
– patex1987, Commented Jan 26, 2017 at 21:58

Mike Müller · Accepted Answer · 2017-01-26 22:06:14Z

2

This would give you time deltas:

from datetime import timedelta

def to_timedelta(time_string):
    units = {'h': 'hours', 'd': 'days', 'm': 'minutes', 's': 'seconds'}
    return timedelta(**{units[x[-1]]: int(x[:-1]) for x in time_string.split()})

Test:

times = ['196h 26m 13s', '95h 19m', '45m 28s']
for t in times:
    print(to_timedelta(t))

Output:

8 days, 4:26:13
3 days, 23:19:00
0:45:28

timedelta takes these arguments;

datetime.timedelta(days=0, seconds=0, microseconds=0, milliseconds=0, minutes=0, hours=0, weeks=0)

Using this mapping:

units = {'h': 'hours', 'd': 'days', 'm': 'minutes', 's': 'seconds'}

allows the short units in the string to be mapped to the corresponding names of the arguments. Using Pythons ** syntax, the resulting dictionary can be used as single argument that will be converted in to the matching keyword arguments.

edited Jan 26, 2017 at 22:06

answered Jan 26, 2017 at 21:57

Mike Müller

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Comments

eyllanesc · Accepted Answer · 2017-01-26 21:57:59Z

1

The first thing we should do is use a regular expression and use timedelta instead of datetime.

import datetime
import re

regex = re.compile(r'((?P<hours>\d+?)h)?((?P<minutes>\d+?)m)?((?P<seconds>\d+?)s)?')

def parse_time(time_str):
    parts = regex.match(time_str)
    if not parts:
        return
    parts = parts.groupdict()
    time_params = {}
    for (name, param) in parts.items():
        if param:
            time_params[name] = int(param)
    return datetime.timedelta(**time_params)


L = ["196h 26m 13s", "95h 19m", "45m 28s"]

for l in L:
    print(parse_time(l))

Output:

8 days, 4:00:00
3 days, 23:00:00
0:45:00

answered Jan 26, 2017 at 21:57

eyllanesc

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1 Comment

patex1987 Over a year ago

Thanks! I had a very similar solution based on regex

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