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)]
See http://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
--
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