bambam wrote: > > In this case it doesn't matter - my lists don't contain > duplicate elements this time - but I have worked with lists in > money market and in inventory, and finding the intersection > and difference for matching off and netting out are standard > operations.
I would use a list comprehension for that case: A = ['a','b','c','a','c','d'] U = ['a','b','e'] B = [x for x in A if x in U] The result would be B==['a','b','a'] /MiO -- http://mail.python.org/mailman/listinfo/python-list