Why does this regex not match my groups in python?

Question

I have the following complete code example

import re

examples = [
    "D1",       # expected: ('1')
    "D1sjdgf",  # ('1')
    "D1.2",     # ('1', '2')
    "D1.2.3",   # ('1', '2', '3')
    "D3.10.3x", # ('3', '10', '3')
    "D3.10.11"  # ('3', '10', '11')
]

for s in examples:
    result = re.search(r'^D(\d+)(?:\.(\d+)(?:\.(\d+)))', s)
    print(s, result.groups())

where I want to match the 1, 2 or 3 numbers in the expression always starting with the letter "D". It could be 1 of them, or 2, or three. I am not interested in anything after the last digit.

I would expect that my regex would match e.g. D3.10.3x and return ('3','10','3'), but instead returns only ('3',). I do not understand why.

^D(\d+\)(?:\.(\d+)(?:\.(\d+)))

^D matches "D" at the start
\d matches the first one-digit number inside a group.
(?: starts a non-matching group. I do not want to get this group back.
\. A literal point
(\d+) A group of one or more numbers I want to "catch"

I also do not know what a "non-capturing" group means in that context as for this answer.

[\.(\d+)+] is a character class that has no group, [\.(\d+)+] = [.)(+\d]. It should be (?:\.(\d+))?, i.e. you must use an optional non-capturing group instead of a character class. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Aug 11, 2022 at 15:21
Ok I removed the rectangular brackets. Still does not work. Even with non-capturing groups. Should I create a new question what will get closed without valid reason? — Alex
– Alex, Commented Aug 11, 2022 at 15:29

anubhava · Accepted Answer · 2022-08-11 15:41:24Z

1

You may use this regex solution with a start anchor and 2 capture groups inside the nested optional capture groups:

^D(\d+)(?:\.(\d+)(?:\.(\d+))?)?

RegEx Demo

Explanation:

^: Start
D: Match letter D
(\d+): Match 1+ digits in capture group #1
(?:: Start outer non-capture group
- \.: Match a dot
- (\d+): Match 1+ digits in capture group #2
- (?:: Start inner non-capture group
  - \.: Match a dot
  - (\d+): Match 1+ digits in capture group #3
- )?: End inner optional non-capture group
)?: End outer optional non-capture group

Code Demo:

import re

examples = [
    "D1",       # expected: ('1')
    "D1sjdgf",  # ('1')
    "D1.2",     # ('1', '2')
    "D1.2.3",   # ('1', '2', '3')
    "D3.10.3x", # ('3', '10', '3')
    "D3.10.11"  # ('3', '10', '11')
]

rx = re.compile(r'^D(\d+)(?:\.(\d+)(?:\.(\d+))?)?')

for s in examples:
    result = rx.search(s)
    print(s, result.groups())

Output:

D1 ('1', None, None)
D1sjdgf ('1', None, None)
D1.2 ('1', '2', None)
D1.2.3 ('1', '2', '3')
D3.10.3x ('3', '10', '3')
D3.10.11 ('3', '10', '11')

edited Aug 11, 2022 at 15:41

answered Aug 11, 2022 at 15:31

anubhava

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1 Comment

Alex Over a year ago

Sorry, I used your suggestion in some other regex. Seems to work. And maybe I even understand it a bit how it works. Thanks!

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