2009/12/11 Ed Keith <e_...@yahoo.com>:
> 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.
>
> Does anyone have a better solution?
>
> Thank,
>
> -EdK
>
>
> Ed Keith
> e_...@yahoo.com
>
> Blog: edkeith.blogspot.com
>
>
>
> --
> http://mail.python.org/mailman/listinfo/python-list
>
Hi,
maybe you could use a list comprehension or the equivalent loop just
using the string methods and slicing?
>>> lst = ['asc', '*nbh*', 'jlsdjfdk', 'ikjh', '*jkjsdfjasd*', 'rewr']
>>> [item[1:-1] for item in lst if (item.startswith("*") and
>>> item.endswith("*"))]
['nbh', 'jkjsdfjasd']
>>>
hth,
vbr
--
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