Alex Martelli wrote: > is the "one obvious way to do it" (the set(...) is just a simple and > powerful optimization -- checking membership in a set is roughly O(1), > while checking membership in a list of N items is O(N)...).
Depending on a how a set is stored, I'd estimate any membership check in a set to be O(log N). That's if it's stored in a tree of some kind, which you'd need to fast finding. Say a balanced binary tree. Worst case, you'd have to search half of the elements to find what you were looking for. > > > Alex -- http://mail.python.org/mailman/listinfo/python-list