On Feb 15, 8:31 pm, "Gabriel Genellina" <[EMAIL PROTECTED]>
wrote:
> En Fri, 15 Feb 2008 19:25:59 -0200, Jonathan Lukens
> <[EMAIL PROTECTED]> escribió:
>
>
>
> >> What would you like to see instead?
>
> > I had mostly just expected that there was some method that would
> > return each entire match as an item on a list. I have this pattern:
>
> >>>> import re
> >>>> corporate_names =
> >>>> re.compile(u'(?u)\\b([А-Я]{2,}\\s+)([<<"][а-яА-Я]+)(\\s*-?[а-яА-Я]+)*([>>"])')
> >>>> terms = corporate_names.findall(sourcetext)
>
> > Which matches a specific way that Russian company names are
> > formatted. I was expecting a method that would return this:
>
> >>>> terms
> > [u'string one', u'string two', u'string three']
>
> > ...mostly because I was working it this way in Java and haven't
> > learned to do things the Python way yet. At the suggestion from
> > someone on the list, I just used list() on all the tuples like so:
>
> The group() method of match objects does what you want:
>
> terms = [match.group() for match in corporate_names.finditer(sourcetext)]
>
> Seehttp://docs.python.org/lib/match-objects.html
>
> >>>> detupled_terms = [list(term_tuple) for term_tuple in terms]
> >>>> delisted_terms = [''.join(term_list) for term_list in detupled_terms]
>
> > which achieves the desired result, but I am not a programmer and so I
> > would still be interested to know if there is a more elegant way of
> > doing this.
>
> That ''.join(...) works equally well on tuples; you don't have to convert
> tuples to lists first:
>
> delisted_terms = [''.join(term_list) for term in terms]
>
> --
> Gabriel Genellina
Thanks Gabriel,
That is just what I was looking for.
Jonathan
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