Nico Grubert wrote: >> May be not so much pythonic, but works >> >> for i in range(len(q)): >> for x in q[i:]: >> if x.startswith(q[i]) and x!=q[i]: >> q.remove(x) > > ...but works fine. Thanks, Eugene. > Also thanks to Andrew. Your example works fine, too. Thanks to remind me > of the 'yield' statement! ;-)
in case it's not clear, the method above is O(n^2) while mine was O(n). that will not matter for small lists, but for long ones mine will be much faster. andrew -- http://mail.python.org/mailman/listinfo/python-list
