Oscar wrote: >>> def uniquecombinations(h, m): >>> for ha in submultisets(h, len(h)//2): >>> hb = list(h) >>> for c in ha: >>> hb.remove(c) >>> yield [m[0] + a for a in ha] + [m[1] + b for b in hb] >>> >>> h = ['A', 'A', 'B', 'B'] >>> m = ['a', 'b'] >>> >>> for x in uniquecombinations(h, m): >>> print(x) >>> ''' >>> >>> Output: >>> ['aB', 'aB', 'bA', 'bA'] >>> ['aA', 'aB', 'bA', 'bB'] >>> ['aA', 'aA', 'bB', 'bB']
On 3 October 2012 21:15, Steen Lysgaard <boxeakast...@gmail.com> wrote: > Hi, > > thanks for your interest. Sorry for not being completely clear, yes > the length of m will always be half of the length of h. > > /Steen Then you can make the uniquecombinations function recursive. First find the elements that go with 'a' then from the remaining elements find those that go with 'b', then 'c' and so on. Oscar -- http://mail.python.org/mailman/listinfo/python-list
