On Fri, Mar 19, 2010 at 8:17 AM, Joe Kington <jking...@wisc.edu> wrote: > See itertools.permutations (python standard library) > e.g. > In [3]: list(itertools.permutations([1,1,0,0])) > Out[3]: > [(1, 1, 0, 0), > (1, 1, 0, 0), > (1, 0, 1, 0), > (1, 0, 0, 1), > (1, 0, 1, 0), > (1, 0, 0, 1), > (1, 1, 0, 0), > (1, 1, 0, 0), > (1, 0, 1, 0), > (1, 0, 0, 1), > (1, 0, 1, 0), > (1, 0, 0, 1), > (0, 1, 1, 0), > (0, 1, 0, 1), > (0, 1, 1, 0), > (0, 1, 0, 1), > (0, 0, 1, 1), > (0, 0, 1, 1), > (0, 1, 1, 0), > (0, 1, 0, 1), > (0, 1, 1, 0), > (0, 1, 0, 1), > > > > (0, 0, 1, 1), > (0, 0, 1, 1)] > Hope that helps, > -Joe
That treats each 1 as distinct. set() solves that: >> list(set(itertools.permutations([1,1,0,0]))) [(1, 0, 1, 0), (1, 1, 0, 0), (0, 0, 1, 1), (1, 0, 0, 1), (0, 1, 1, 0), (0, 1, 0, 1)] _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
