Alexzive wrote: > I'd like to get the same result of set() but getting an indexable > object. > How to get this in an efficient way? > > Example using set > > A = [1, 2, 2 ,2 , 3 ,4] > B= set(A) > B = ([1, 2, 3, 4]) > > B[2] > TypeError: unindexable object
If the initial list is ordered or at least equal items are neighbours you can use groubpy(): >>> from itertools import groupby >>> a = [1,1,1,2,2,3,4,4,4] >>> [key for key, group in groupby(a)] [1, 2, 3, 4] Here's what happens if there are equal items that are not neigbours: >>> b = [1,1,1,2,2,2,3,3,2,1,1,1,1] >>> [key for key, group in groupby(b)] [1, 2, 3, 2, 1] Peter -- http://mail.python.org/mailman/listinfo/python-list
