Time Duration to Numeric Minutes using Python Library or Regex

Question

I have a free form time duration strings containing hour and minute values one of then could be optional

1 hour
12 hours 3 mins
47 mins
10 hours
1 min

I have to convert them to number of minutes. First searched for Python library which converts times and duration but the string format does not allow to use such approach

Then I tried with regex to extract the number groups:

re.search("(\d+)?.*(\d+\w)", string).group(1)
re.search("(\d+)?.*(\d+\w)", string).group(2)

which worked for most cases when hour and minute values are present or when only the minute value is present (since I made the first group optional) This regex fails when the hour is single digit (1 hour). Also because I am extracting only digits groups without the descriptive text (hour(s) and/or min(s) the calculation is wrong when there is only the hour value (with two digits) - like 10 hours and it wrongly is extracted as the 2nd group as minutes.

You can extract time data from a string using datetime.strptime() from python library datetime — Mael Galliffet
– Mael Galliffet, Commented Oct 24, 2018 at 11:14

blhsing · Accepted Answer · 2018-10-24 11:28:58Z

1

You can use re.findall with the following regex:

import re
s = '''1 hour
12 hours 3 mins
47 mins
10 hours
1 min'''
for h, m in re.findall(r'(?=\d+ *hours?| *\d+ *min(?:ute)?s?)(?:(\d+) *hours?)?(?: *(\d+) *min(?:ute)?s?\b)?', s, flags=re.IGNORECASE):
    print(int(h or 0) * 60 + int(m or 0))

This outputs:

answered Oct 24, 2018 at 11:28

blhsing

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Comments

Arman Ordookhani · Accepted Answer · 2018-10-24 11:29:32Z

I wrote this simple snippet that parses all your cases. Ask if you had any problem.

Output:

1 hour -> 1:00:00
12 hours 3 mins -> 12:03:00
47 mins -> 0:47:00
10 hours -> 10:00:00
1 min -> 0:01:00
random text -> 0:00:00

Code:

import re
from datetime import timedelta


number_word_regex = re.compile(r'(\d+) (\w+)')


def parse_fuzzy_duration(s):
    ret = timedelta(0)

    for number, word in number_word_regex.findall(s):
        number = int(number)

        if word in ['minute', 'min', 'minutes', 'mins']:
            ret += timedelta(minutes=number)
        elif word in ['hour', 'hours']:
            ret += timedelta(hours=number)

    return ret


for s in ['1 hour', '12 hours 3 mins', '47 mins', '10 hours', '1 min', 'random text']:
    print(s, '->', parse_fuzzy_duration(s))

Rakesh · Accepted Answer · 2018-10-24 11:39:47Z

1

You can try using dateutil and Regex

Demo:

import dateutil.parser as dparser
import re

s = """1 hour
12 hours 3 mins
47 mins
10 hours
1 min"""

for line in s.splitlines():
    print(dparser.parse(re.sub(r"(mins?)", "minutes", line), fuzzy=True).strftime("%H:%M:%S") )

Output:

01:00:00
12:03:00
00:47:00
10:00:00
00:01:00

answered Oct 24, 2018 at 11:39

Rakesh

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Comments

Slam · Accepted Answer · 2018-10-24 11:42:05Z

0

Other answers is fine, another way to do this with regex (if you really want to):

match = re.match(
    r'((?P<hours>\d+) hours?)? ?((?P<mins>\d+) mins?)?',
    '12 hours 3 mins'
)

match.groupdicts()

This way it may make more sense you you (this is the main problem with a lot of regex engineering. I'd suggest to try out any regex you choose at some resource like https://regex101.com/ to have some testing and description.

answered Oct 24, 2018 at 11:42

Slam

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1 Comment

ch-28 Over a year ago

blhsing's answer was more fool proof engineered so even if I put arbitrary number of white-spaces between the number and the hour/min pattern it selects the digits. However your's was shorter and thus easier to understand so I was able to change it to the same behavior : ((?P<hours>\d+) *hours?)? ?((?P<mins>\d+) *mins?)? - thanks

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