Ed Keith wrote:
> I have a problem and I am trying to find a solution to it that is both
> efficient and elegant.
>
> I have a list call it 'l':
>
> l = ['asc', '*nbh*', 'jlsdjfdk', 'ikjh', '*jkjsdfjasd*', 'rewr']
>
> Notice that some of the items in the list start and end with an '*'. I
> wish to construct a new list, call it 'n' which is all the members of l
> that start and end with '*', with the '*'s removed.
>
> So in the case above n would be ['nbh', 'jkjsdfjasd']
>
> the following works:
>
> r = re.compile('\*(.+)\*')
>
> def f(s):
> m = r.match(s)
> if m:
> return m.group(1)
> else:
> return ''
>
> n = [f(x) for x in l if r.match(x)]
>
>
>
> But it is inefficient, because it is matching the regex twice for each
> item, and it is a bit ugly.
>
> I could use:
>
>
> n = []
> for x in keys:
> m = r.match(x)
> if m:
> n.append(m.group(1))
>
>
> It is more efficient, but much uglier.
It's efficient and easy to understand; maybe you have to readjust your
taste.
> Does anyone have a better solution?
In this case an approach based on string slicing is probably best. When the
regular expression gets more complex you can use a nested a generator
expression:
>>> items = ['asc', '*nbh*', 'jlsdjfdk', 'ikjh', '*jkjsdfjasd*', 'rewr']
>>> match = re.compile(r"\*(.+)\*").match
>>> [m.group(1) for m in (match(s) for s in items) if m is not None]
['nbh', 'jkjsdfjasd']
Peter
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