That looks good, and perhaps a difference operator would be too simple to be useful anyway.
Steve. "Mikael Olofsson" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > > > 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
