Hi, I liked the twist at the end when you state that only the first two 2's count. It reminded me of my maths O'level revision where you always had to read the question thoroughly.
Here is what I came up with: >>> ref [2, 2, 4, 1, 1] >>> lst [2, 2, 5, 2, 4] >>> tmp = [ [val]*min(lst.count(val), ref.count(val)) for val in set(ref)] >>> tmp [[], [2, 2], [4]] >>> answer = [x for y in tmp for x in y] >>> answer [2, 2, 4] >>> I took a lot from Peter Ottens reply to generate tmp then flattened the inner lists. After a bit more thought, the intermediate calculation of tmp can be removed with a little loss in clarity though, to give: >>> answer = [ val for val in set(ref) for x in range(min(lst.count(val), >>> ref.count(val)))] >>> answer [2, 2, 4] - Cheers, Paddy. -- http://mail.python.org/mailman/listinfo/python-list
