regexp substitution - a lot of work!

Fredrik Lundh fredrik at pythonware.com
Fri Jun 18 19:05:39 CEST 2004


Lukas Holcik quoted someone writing:

> > I don't understand your point. The Python equivalent is:
> >
> >   re.sub('(..)', r'x\1x', s)
> >
> > or using a precompiled pattern:
> >
> >  pat.sub(r'x\1x', s)

footnote: you can use a callback instead of the replacement pattern.
callbacks are often faster, and can lead more readable code:

    http://effbot.org/zone/re-sub.htm#callbacks

(as the other examples on that page show, you can do a lot of weird
stuff with re.sub callbacks...)

</F>







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