RegEx with multiple groups?

Question

I’m getting confused returning multiple groups in Python. My RegEx is this:

lun_q = 'Lun:s*(d+s?)*'

And my string is

s = '''Lun:                     0 1 2 3 295 296 297 298'''`

I return a matched object, and then want to look at the groups, but all it shows it the last number (258):

r.groups()  
(u'298',)

Why isn’t it returning groups of 0,1,2,3,4 etc.?

Asked By: joslinm

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Source

Answer 1

Your regex only contains a single pair of parentheses (one capturing group), so you only get one group in your match. If you use a repetition operator on a capturing group (+ or *), the group gets “overwritten” each time the group is repeated, meaning that only the last match is captured.

In your example here, you’re probably better off using .split(), in combination with a regex:

lun_q = 'Lun:s*(d+(?:s+d+)*)'
s = '''Lun: 0 1 2 3 295 296 297 298'''

r = re.search(lun_q, s)

if r:
    luns = r.group(1).split()

    # optionally, also convert luns from strings to integers
    luns = [int(lun) for lun in luns]

Answered By: Ben Blank

Answer 2

Another approach would be to use the regex you have to validate your data and then use a more specific regex that targets each item you wish to extract using a match iterator.

import re
s = '''Lun: 0 1 2 3 295 296 297 298'''
lun_validate_regex = re.compile(r'Lun:s*((d+)(sd+)*)')
match = lun_validate_regex.match(s)
if match:
    token_regex = re.compile(r"d{1,3}")
    match_iterator = token_regex.finditer(match.group(1))
    for token_match in match_iterator:
        #do something brilliant

Answered By: pokstad

Answer 3

Sometimes, its easier without regex.

>>> s = '''Lun: 0 1 2 3 295 296 297 298'''
>>> if "Lun: " in s:
...     items = s.replace("Lun: ","").split()
...     for n in items:
...        if n.isdigit():
...           print n
...
0
1
2
3
295
296
297
298

Answered By: kurumi

Answer 4

If you are looking for an output such as 0,1,2,3,4 etc.:

print re.findall('d', s)

Answered By: Rakesh kumar

RegEx with multiple groups?

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