On Mar 20, 9:54 am, "thomasvang...@gmail.com" <thomasvang...@gmail.com> wrote: > You could use: > B=list(set(A)).sort() > Hope that helps. > T
That may hurt more than help, sort() only works in-place, and does *not* return the sorted list. For that you want the global built-in sorted: >>> data = map(int,"6 1 3 2 5 2 5 4 2 0".split()) >>> print sorted(list(set(data))) [0, 1, 2, 3, 4, 5, 6] To retain the original order, use the key argument, passing it a function - simplest is to pass the index of the value in the original list: >>> print sorted(list(set(data)), key=data.index) [6, 1, 3, 2, 5, 4, 0] If data is long, all of those calls to data.index may get expensive. You may want to build a lookup dict first: >>> lookup = dict((v,k) for k,v in list(enumerate(data))[::-1]) >>> print sorted(list(set(data)), key=lookup.__getitem__) [6, 1, 3, 2, 5, 4, 0] -- Paul -- http://mail.python.org/mailman/listinfo/python-list
