Python: Determine if an unsorted list is contained in a 'list of lists', regardless of the order to the elements

Question

I have a similar question to this question: Determine if 2 lists have the same elements, regardless of order?

What is the best/quickest way to determine whether an unsor

Answer 1

This algorithm appears to be slightly faster:

l1 = [3, 4, 1, 2, 3]
l2 = [4, 2, 3, 3, 1]
same = True
for i in l1:
    if i not in l2:
        same = False
        break

For 1000000 loops, this takes 1.25399184227 sec on my computer, whilst

same = sorted(l1) == sorted(l2)

takes 1.9238319397 sec.