Test if lists share any items in python

def lists_overlap3(a, b):
    return bool(set(a) & set(b))

Note: the above assumes that you want a boolean as the answer. If all you need is an expression to use in an if statement, just use if set(a) & set(b):

def lists_overlap(a, b):
  sb = set(b)
  return any(el in sb for el in a)

This is asymptotically optimal (worst case O(n + m)), and might be better than the intersection approach due to any's short-circuiting.

E.g.:

lists_overlap([3,4,5], [1,2,3])

will return True as soon as it gets to 3 in sb

EDIT: Another variation (with thanks to Dave Kirby):

def lists_overlap(a, b):
  sb = set(b)
  return any(itertools.imap(sb.__contains__, a))

This relies on imap's iterator, which is implemented in C, rather than a generator comprehension. It also uses sb.__contains__ as the mapping function. I don't know how much performance difference this makes. It will still short-circuit.

You could also use any with list comprehension:

any([item in a for item in b])

In python 2.6 or later you can do:

return not frozenset(a).isdisjoint(frozenset(b))