Am 09.03.2010 13:02, schrieb Peter Otten: >>>> [sum(a for a,b in zip(x,y) if b==c)/y.count(c)for c in y] > [1.5, 1.5, 8.0, 4.0, 4.0, 4.0] > Peter ... pwned. Should be the fastest and shortest way to do it. I tried to do something like this, but my brain hurt while trying to visualize list comprehension evaluation orders ;) Regards, Michael