re.sub(): replace longest match instead of leftmost match?

[MRAB]

On 16/12/2011 16:49, John Gordon wrote:
> According to the documentation on re.sub(), it replaces the leftmost
> matching pattern.
>
> However, I want to replace the *longest* matching pattern, which is
> not necessarily the leftmost match.  Any suggestions?
>
> I'm working with IPv6 CIDR strings, and I want to replace the longest
> match of "(0000:|0000$)+" with ":".  But when I use re.sub() it replaces
> the leftmost match, even if there is a longer match later in the string.
>
> I'm also looking for a regexp that will remove leading zeroes in each
> four-digit group, but will leave a single zero if the group was all
> zeroes.
>
How about this:

result = re.sub(r"\b0+(\d)\b", r"\1", string)