Hello Fred, > duplicates = filter((lambda x:list.count(x)>1), list) If I understood the previous posters here, the "list.count" part is the problem, because it goes thru the whole list every time, which is expensive. Or doesn't this apply here? Y, Scot -- Scot W. Stevenson -- scot@possum.in-berlin.de -- Zepernick, Germany