how to replace via regex with group enclosed in python

Question

Here's the scenario:

import re

if __name__ == '__main__':
    s = "s = \"456\";"
    ss = re.sub(r'(.*s\s+=\s+").*?(".*)', r"\1123\2", s)
    print ss

What I intend to do is to replace '456' with 123, but the result is 'J3";'. I try to print '\112', it turns out to be character 'J'. Thus, is there any method to specify that \1 is the group in regex, not something like a escape character in Python? Thanks in advance.

Avinash Raj · Accepted Answer · 2015-07-26 07:04:28Z

1

Just change \1 to \g<1>

>>> re.sub(r'(.*s\s+=\s+").*?(".*)', r"\g<1>123\2", s)
's = "123";'

If there was no numbers present next to the backreference (like \1,\2), you may use \1 or \2 but if you want to put a number next to \1 like \11, it would give you a garbage value . In-order to differntiate between the backreferences and the numbers, you should use \g<num> as backrefernce where num refers the capturing group index number.

edited Jul 26, 2015 at 7:04

answered Jul 26, 2015 at 6:45

Avinash Raj

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2 Comments

Judking Over a year ago

Thanks! it works. Could you plz paste the document URL relevant to this topic, so I could get an overall comprehension of it.

Wolph Over a year ago

@Judking: The Python regex howto (in the core docs) is probably your best start: docs.python.org/2/howto/regex.html

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