Python regex with optional Groups, one match at least

Question

So I thought I could just quickly do a re.match() with my given String, but I'm stuck.

With a given list of Strings

The Time is 12H 3M 12S
The Time is 3M 12S
It is 12H 3M
Ready in 12S
The Time is 6H

I would like to extract into 3 groups H, M and S, somehow like

(?: (\d{1,2})H)?

(?: (\d{1,2})M)?

(?: (\d{1,2})S)?

Easyliy I could then access the H, M and S components by group(1-3). I just would like to restrict the match to fulfill the creteria, that at least one of the optionl groups has to be triggered or it's no match. Else this expression is optionally empty and matches everything, I guess.

Here's a link to the example: https://regex101.com/r/LKAKbx/5

How can I get the numbers only as groups from match, eg:

The Time is 12H 3M 12S

group(1) = 12, group(2) = 3, group(3) = 12

Or

Ready in 12S

group(1) = None, group(2) = None, group(3) = 12

What is your question?

mkrieger1
– mkrieger1

2020-01-03 16:27:07 +00:00
Commented Jan 3, 2020 at 16:27 — mkrieger1
– mkrieger1, Commented Jan 3, 2020 at 16:27

Toto · Accepted Answer · 2020-01-03 17:27:09Z

3

Use a positive lookahead to make sure we have at least one of H, M or S.

import re

strings = [
    'The Time is 12H 3M 12S',
    'The Time is 3M 12S',
    'It is 12H 3M',
    'Ready in 12S',
    'The Time is 6H',
]

for s in strings:
    res = re.search(r'(?= \d{1,2}[HMS])(?: (\d{1,2})H)?(?: (\d{1,2})M)?(?: (\d{1,2})S)?', s)
    #          here __^^^^^^^^^^^^^^^^^
    print(res.groups())

Output:

('12', '3', '12')
(None, '3', '12')
('12', '3', None)
(None, None, '12')
('6', None, None)

answered Jan 3, 2020 at 17:27

Toto

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1 Comment

engelant Over a year ago

Exactly what I was looking for!

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