Most efficient solution?

Alex new_name at mit.edu
Mon Jul 16 11:09:49 EDT 2001


> map(A.remove, B)

That'll throw an error on elements of B that aren't in A, and if elt
occurs twice in A and once in B, map(A.remove, B) will still have a copy
of elt in it, even if no error is thrown.

Alex.



More information about the Python-list mailing list