Python: Non capturing group is not working in Regex

Question

I'm using non-capturing group in regex i.e., (?:.*) but it's not working.

I can still able to see it in the result. How to ignore it/not capture in the result?

Code:

import re

text = '12:37:25.790 08/05/20 Something   P  LR    0.156462 sccm   Pt   25.341343 psig something-else'

pattern = ['(?P\d\d:\d\d:\d\d.\d\d\d)\s{1}',
           '(?P\d\d/\d\d/\d\d)\s',
           '(?P(?:.*)Pt\s{3}\d*[.]?\d*\s[a-z]+)'
          ]

result = re.search(r''.join(pattern), text)

Output:

>>> result.group('pr')
            
'Something   P  LR    0.156462 sccm   Pt   25.341343 psig'

Expected output:

'Pt   25.341343 psig'

More info:

>>> result.groups()
            
('12:37:25.790', '08/05/20', 'Something   P  LR    0.156462 sccm   Pt   25.341343 psig')

user_550094 · Answer

Remove the non-capturing group from your named group. Using a non-capturing group means that no new group will be created in the match, not that that part of the string will be removed from any including group.

import re

text = 'Something   P  LR    0.156462 sccm   Pt   25.341343 psig something-else'

pattern = r'(?:.*)(?PPt\s{3}\d*[.]?\d*\s[a-z]+)'

result = re.search(pattern, text)
print(result.group('pr'))

Output:

Pt   25.341343 psig

Note that the specific non-capturing group you used can be excluded completely, as it basically means that you want your regex to be preceded by anything, and that's what search will do anyway.

user_5424988 · Answer

The quantifier is inside the named group, you have to place it outside and possibly make it non greedy to.

The updated pattern could look like:

(?P\d\d:\d\d:\d\d.\d\d\d)\s{1}(?P\d\d/\d\d/\d\d)\s.*?(?PPt\s{3}\d*[.]?\d*\s[a-z]+)

Note that with he current pattern, the number is optional as all the quantifiers are optional. You can omit {1} as well.

If the number after Pt can not be empty, you can update the pattern using \d+(?:\.\d+)? matching at least a single digit:

(?P\d\d:\d\d:\d\d.\d{3})\s(?P\d\d/\d\d/\d\d)\s.*?(?PPt\s{3}\d+(?:\.\d+)?\s[a-z]+)

(?P Group time
\d\d:\d\d:\d\d.\d{3} Match a time like format
)\s Close group and match a whitespace char
(?P Group date
- \d\d/\d\d/\d\d Match a date like pattern
)\s Close group and match a whitespace char
.*? Match any char except a newline, as least as possible
(?P Group pr
- Pt\s{3} Match Pt and 3 whitespace chars
- \d+(?:\.\d+)? Match 1+ digits with an optional decimal part
\s[a-z]+ Match a whitespace char an 1+ times a char a-z
) Close group

Regex demo

soulmerge_44562 · Answer

I think there is a confusion regarding the meaning of "non-capturing" here: It does not mean that the result omits this part, but that no match group is created in the result.

Example where the same regex is executed with a capturing, and a non-capturing group:

>>> import re
>>> match = re.search(r'(?Pfoo(.*))', 'foobar')
>>> match.groups()
('foobar', 'bar')
>>> match = re.search(r'(?Pfoo(?:.*))', 'foobar')
>>> match.groups()
('foobar',)

Note that match.group(0) is the same in both cases (group 0 contains the matching part of the string in full).

Python: Non capturing group is not working in Regex

3 Answers