In Python 3.x (and Python 2.7, when it's released), you can use collections.Counter for this:
>>> from collections import Counter
>>> list((Counter([2,2,1,1]) & Counter([1,3,3,1])).elements())
[1, 1]
Here's an alternative using collections.defaultdict (available in Python 2.5 and later). It has the nice property that the order of the result is deterministic (it essentially corresponds to the order of the second list).
from collections import defaultdict
def list_intersection(list1, list2):
bag = defaultdict(int)
for elt in list1:
bag[elt] += 1
result = []
for elt in list2:
if elt in bag:
# remove elt from bag, making sure
# that bag counts are kept positive
if bag[elt] == 1:
del bag[elt]
else:
bag[elt] -= 1
result.append(elt)
return result
For both these solutions, the number of occurrences of any given element x
in the output list is the minimum of the numbers of occurrences of x
in the two input lists. It's not clear from your question whether this is the behavior that you want.