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From: Devin Jeanpierre <jeanpierreda@gmail.com>
Date: Fri, 16 Dec 2011 11:56:32 -0500
Subject: Re: re.sub(): replace longest match instead of leftmost match?
To: John Gordon <gordon@panix.com>
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Xref: x330-a1.tempe.blueboxinc.net comp.lang.python:17363

You could use re.finditer to find the longest match, and then replace
it manually by hand (via string slicing).

(a match is the longest if (m.end() - m.start()) is the largest --
so, max(re.finditer(...), key=3Dlambda m: (m.end() =3D m.start()))

-- Devin

P.S. does anyone else get bothered by how it's slice.start and
slice.stop, but match.start() and match.end() ?

On Fri, Dec 16, 2011 at 11:49 AM, John Gordon <gordon@panix.com> 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. =C2=A0Any suggestions?
>
> I'm working with IPv6 CIDR strings, and I want to replace the longest
> match of "(0000:|0000$)+" with ":". =C2=A0But when I use re.sub() it repl=
aces
> 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.
>
> Thanks!
>
> --
> John Gordon =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=
=A0 A is for Amy, who fell down the stairs
> gordon@panix.com =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0B is for=
 Basil, assaulted by bears
> =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=
=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0-- Edward Gorey, "The Gashlycrumb Tin=
ies"
>
> --
> http://mail.python.org/mailman/listinfo/python-list