regex group reference error

Question

p = r'([\,|\.]\d{1}$)'
re.sub(p, r"\1", v)

works, but I want to add a zero to the capture group, not replace with capture group '10', how can I do this?

re.sub(p, r"\10", v)

fails:

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/re.py", line 151, in sub
    return _compile(pattern, flags).sub(repl, string, count)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/re.py", line 275, in filter
    return sre_parse.expand_template(template, match)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/sre_parse.py", line 802, in expand_template
    raise error, "invalid group reference"
sre_constants.error: invalid group reference

As I said in my answer: re.sub(p, r"\g<1>0", v) – qstebom Dec 09 '13 at 10:39 — qstebom, Dec 09 '13 at 10:39
`[\,|\.]` doesn't look right. Do you mean `[.,]`? – georg Dec 09 '13 at 11:18 — georg, Dec 09 '13 at 11:18

qstebom · Accepted Answer · 2013-12-09T21:48:52.630

6

Just wrap the group reference in '\g<#>':

import re
pattern = r'([\,|\.]\d{1}$)'
string = 'Some string .1\n'
rep = r'\g<1>0'
re.sub(pattern, rep, string)
> 'Some string .10\n'

Source: http://docs.python.org/2/library/re.html#re.sub

edited Dec 09 '13 at 21:48

answered Dec 09 '13 at 10:35

qstebom

589
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Hedde van der Heide · Answer 2 · 2013-12-09T10:35:40.377

1

Use a named capture group:

p = r'(?P<var>[\,|\.]\d{1})$'
re.sub(p, r"\g<var>0", v)

e.g.

>>> p = r'(?P<var>[\,|\.]\d{1})$'
>>> v = '235,5'
>>> re.sub(p, r"\g<var>0", v)
'235,50'

edited Dec 09 '13 at 10:35

answered Dec 09 '13 at 10:29

Hedde van der Heide

19,278
13
58
93

score 0 · Answer 3 · answered Dec 09 '13 at 10:30

The simplest method (which might also be the only method, I'm not actually sure) is to name the capturing group, then refer back to it by name:

>>> re.sub(p, r'\10', '1.2')
Traceback (most recent call last):
   ...
sre_constants.error: invalid group reference
>>> p = r'(?P<frac>[\,|\.]\d{1}$)'
>>> re.sub(p, r'\g<frac>0', '1.2')
'1.20'

Pick some name better than just "frac" (which I pulled out my ... er, ear, yes, let's go with "ear" :-) ).

Chris

regex group reference error

3 Answers3