re.sub replace with matched content

Solution 1:

Simply use \1 instead of $1:

In [1]: import re

In [2]: method = 'images/:id/huge'

In [3]: re.sub(r'(:[a-z]+)', r'<span>\1</span>', method)
Out[3]: 'images/<span>:id</span>/huge'

Also note the use of raw strings (r'...') for regular expressions. It is not mandatory but removes the need to escape backslashes, arguably making the code slightly more readable.

Solution 2:

Use \1 instead of $1.

\number Matches the contents of the group of the same number.

http://docs.python.org/library/re.html#regular-expression-syntax

Solution 3:

A backreference to the whole match value is \g<0>, see re.sub documentation:

The backreference \g<0> substitutes in the entire substring matched by the RE.

See the Python demo:

import re
method = 'images/:id/huge'
print(re.sub(r':[a-z]+', r'<span>\g<0></span>', method))
# => images/<span>:id</span>/huge

If you need to perform a case insensitive search, add flag=re.I:

re.sub(r':[a-z]+', r'<span>\g<0></span>', method, flags=re.I)

Solution 4:

For the replacement portion, Python uses \1 the way sed and vi do, not $1 the way Perl, Java, and Javascript (amongst others) do. Furthermore, because \1 interpolates in regular strings as the character U+0001, you need to use a raw string or \escape it.

Python 3.2 (r32:88445, Jul 27 2011, 13:41:33) 
[GCC 4.0.1 (Apple Inc. build 5465)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> method = 'images/:id/huge'
>>> import re
>>> re.sub(':([a-z]+)', r'<span>\1</span>', method)
'images/<span>id</span>/huge'
>>>