python re.sub group: number after \number

The answer is:

re.sub(r'(foo)', r'\g<1>123', 'foobar')

Relevant excerpt from the docs:

In addition to character escapes and backreferences as described above, \g<name> will use the substring matched by the group named name, as defined by the (?P<name>...) syntax. \g<number> uses the corresponding group number; \g<2> is therefore equivalent to \2, but isn’t ambiguous in a replacement such as \g<2>0. \20 would be interpreted as a reference to group 20, not a reference to group 2 followed by the literal character '0'. The backreference \g<0> substitutes in the entire substring matched by the RE.

For this problem I would prefer to match but not capture, by employing the following.

re.sub(r'(?<=foo)', r'123', 'foobar')
  #=> 'foo123bar'

which replaces the zero-width string after 'foo' (think between 'foo' and 'bar') with '123'. (?<=foo) is a positive lookbehind.

Demo

There are of course situations where a capture group is needed, such as

re.sub(r'(f\w*o)', r'\g<1>123', 'foobar')

Here

re.sub(r'(?<=f\w*o)', r'123', 'foobar')

does not work because Python's default regex engine does not support variable-length lookbehinds (the alternative PyPI regex module does, however).