On Mon, 19 Dec 2005, Brian van den Broek wrote: > [EMAIL PROTECTED] said unto the world upon 2005-12-19 02:27: >> Steve Holden wrote: >> >>> Kevin Yuan wrote: >>> >>>> How to remove duplicated elements in a list? eg. >>>> [1,2,3,1,2,3,1,2,1,2,1,3] -> [1,2,3]? >>>> Thanks!! >>> >>> >>> list(set([1,2,3,1,2,3,1,2,1,2,1,3])) >>> [1, 2, 3] >> >> Would this have the chance of changing the order ? Don't know if he >> wants to maintain the order or don't care though. > > For that worry: > >>>> orig_list = [3,1,2,3,1,2,3,1,2,1,2,1,3] >>>> new_list = list(set(orig_list)) >>>> new_list.sort(cmp= lambda x,y: cmp(orig_list.index(x), > orig_list.index(y))) >>>> new_list > [3, 1, 2] >>>>
Ah, that gives me an idea: >>> import operator >>> orig_list = [3,1,2,3,1,2,3,1,2,1,2,1,3] >>> new_list = map(operator.itemgetter(1), ... filter(lambda (i, x): i == orig_list.index(x), ... enumerate(orig_list))) >>> new_list [3, 1, 2] This is a sort of decorate-fondle-undecorate, where the fondling is filtering on whether this is the first occurrance of the the value. This is, IMHO, a clearer expression of the original intent - "how can i remove such-and-such elements from a list" is begging for filter(), i'd say. My code is O(N**2), a bit better than your O(N**2 log N), but we can get down to O(N f(N)), where f(N) is the complexity of set.__in__ and set.add, using a lookaside set sort of gizmo: >>> orig_list = [3,1,2,3,1,2,3,1,2,1,2,1,3] >>> seen = set() >>> def unseen(x): ... if (x in seen): ... return False ... else: ... seen.add(x) ... return True ... >>> new_list = filter(unseen, orig_list) >>> new_list [3, 1, 2] Slightly tidier like this, i'd say: >>> orig_list = [3,1,2,3,1,2,3,1,2,1,2,1,3] >>> class seeingset(set): ... def see(self, x): ... if (x in self): ... return False ... else: ... self.add(x) ... return True ... >>> new_list = filter(seeingset().see, orig_list) >>> new_list [3, 1, 2] tom -- Hit to death in the future head -- http://mail.python.org/mailman/listinfo/python-list
