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<time>\d\d:\d\d:\d\d.\d\d\d)\s{1}',
           '(?P<date>\d\d/\d\d/\d\d)\s',
           '(?P<pr>(?:.*)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')

Answer 1

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<time>\d\d:\d\d:\d\d.\d\d\d)\s{1}(?P<date>\d\d/\d\d/\d\d)\s.*?(?P<pr>Pt\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<time>\d\d:\d\d:\d\d.\d{3})\s(?P<date>\d\d/\d\d/\d\d)\s.*?(?P<pr>Pt\s{3}\d+(?:\.\d+)?\s[a-z]+)

• (?P<time> Group time
• \d\d:\d\d:\d\d.\d{3} Match a time like format
• )\s Close group and match a whitespace char
• (?P<date> 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<pr> 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