Question

I have a list that contains 10**7 lists in the format:

big_list = [[1, 2, 3, 4, 5, 6], [2, 3, 4, 5, 6, 7], [2, 3, 4, 26, 33, 40], [10, 23, 33, 45, 46, 47]]

Every list contains 6 unique ints.

I need to compare every list to another list:

lst = [1, 3, 4, 10, 23, 46]

and return those where list item intersection is less than 3.
So newlist would be:

new_list = [[2, 3, 4, 5, 6, 7], [2, 3, 4, 26, 33, 40]]

At the moment I’m using set intersection, but it takes about 30 seconds to run

Answer 1

>>> big_list = [[1, 2, 3, 4, 5, 6], [2, 3, 4, 5, 6, 7], [2, 3, 4, 26, 33, 40], [10, 23, 33, 45, 46, 47]]
>>> normal = set([1, 3, 4, 10, 23, 46])
>>> [x for x in big_list if len(set(x).intersection(normal)) < 3]
[[2, 3, 4, 5, 6, 7], [2, 3, 4, 26, 33, 40]]