Bengt Richter wrote: > Perhaps newbies should be advised that > > [x for x in l1 if x in set(l2)] But the resulting list is a representative of bag not a set ( contains multiple occurrences of elements ): >>> [x for x in [3, 3] if s in Set([3])] [3,3] Same with Raymonds solution: >>> filter(Set([3]).__contains__, [3,3]) [3, 3] Kay