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From: Ian Kelly <ian.g.kelly@gmail.com>
Date: Fri, 1 Apr 2011 16:42:51 -0600
Subject: Re: Extracting repeated words
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On Fri, Apr 1, 2011 at 2:54 PM, candide <candide@free.invalid> wrote:
> Another question relative to regular expressions.
>
> How to extract all word duplicates in a given text by use of regular
> expression methods ? =A0To make the question concrete, if the text is
>
> ------------------
> Now is better than never.
> Although never is often better than *right* now.
> ------------------
>
> duplicates are :
>
> ------------------------
> better is now than never
> ------------------------
>
> Some code can solve the question, for instance
>
> # ------------------
> import re
>
> regexp=3Dr"\w+"
>
> c=3Dre.compile(regexp, re.IGNORECASE)
>
> text=3D"""
> Now is better than never.
> Although never is often better than *right* now."""
>
> z=3D[s.lower() for s in c.findall(text)]
>
> for d in set([s for s in z if z.count(s)>1]):
> =A0 =A0print d,
> # ------------------
>
> but I'm in search of "plain" re code.

You could use a look-ahead assertion with a captured group:

>>> regexp =3D r'\b(?P<dup>\w+)\b(?=3D.+\b(?P=3Ddup)\b)'
>>> c =3D re.compile(regexp, re.IGNORECASE | re.DOTALL)
>>> c.findall(text)

But note that this is computationally expensive.  The regex that you
posted is probably more efficient if you use a collections.Counter
object instead of z.count.

Cheers,
Ian