On Fri, Dec 11, 2009 at 9:49 PM, Ed Keith <e_...@yahoo.com> 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. > > Does anyone have a better solution?
Regexes seem like the proverbial sledgehammer to crack a nut here. Note that '*' if it is present, is always 1 character, so we can write: n = [x[1:-1] for x in l if x.startswith("*") and x.endswith("*")] -- André Engels, andreeng...@gmail.com -- http://mail.python.org/mailman/listinfo/python-list