efficiently knowing if intersection of two list is empty or not, in python

Question

Hi!

Suppose I have two lists, L and M. Now I want to know if they share an element. Which would be the fastest way of asking (in python) if they share an element? I don't care which elements they share, or how many, just if they share or not.

For example, in this case

L = [1,2,3,4,5,6]
M = [8,9,10]

I should get False, and here:

L = [1,2,3,4,5,6]
M = [5,6,7]

I should get True.

I hope the question's clear. Thanks!

Manuel

Answer 1

First of all, if you do not need them ordered, then switch to the set type.

If you still need the list type, then do it this way: 0 == False

len(set.intersection(set(L), set(M)))

Answer 2

Or more concisely

if set(L) & set(M):
    # there is an intersection
else:
    # no intersection

If you really need True or False

bool(set(L) & set(M))

After running some timings, this seems to be a good option to try too

m_set=set(M)
any(x in m_set  for x in L)

If the items in M or L are not hashable you have to use a less efficient approach like this

any(x in M for x in L)

Here are some timings for 100 item lists. Using sets is considerably faster when there is no intersection, and a bit slower when there is a considerable intersection.

M=range(100)
L=range(100,200)

timeit set(L) & set(M)
10000 loops, best of 3: 32.3 µs per loop

timeit any(x in M for x in L)
1000 loops, best of 3: 374 µs per loop

timeit m_set=frozenset(M);any(x in m_set  for x in L)
10000 loops, best of 3: 31 µs per loop

L=range(50,150)

timeit set(L) & set(M)
10000 loops, best of 3: 18 µs per loop

timeit any(x in M for x in L)
100000 loops, best of 3: 4.88 µs per loop

timeit m_set=frozenset(M);any(x in m_set  for x in L)
100000 loops, best of 3: 9.39 µs per loop


# Now for some random lists
import random
L=[random.randrange(200000) for x in xrange(1000)]
M=[random.randrange(200000) for x in xrange(1000)]

timeit set(L) & set(M)
1000 loops, best of 3: 420 µs per loop

timeit any(x in M for x in L)
10 loops, best of 3: 21.2 ms per loop

timeit m_set=set(M);any(x in m_set  for x in L)
1000 loops, best of 3: 168 µs per loop

timeit m_set=frozenset(M);any(x in m_set  for x in L)
1000 loops, best of 3: 371 µs per loop

Answer 3

To avoid the work of constructing the intersection, and produce an answer as soon as we know that they intersect:

m_set = frozenset(M)
return any(x in m_set for x in L)

Update: gnibbler tried this out and found it to run faster with set() in place of frozenset(). Whaddayaknow.

Answer 4

That's the most generic and efficient in a balanced way I could come up with (comments should make the code easy to understand):

import itertools, operator

def _compare_product(list1, list2):
    "Return if any item in list1 equals any item in list2 exhaustively"
    return any(
        itertools.starmap(
            operator.eq,
            itertools.product(list1, list2)))

def do_they_intersect(list1, list2):
    "Return if any item is common between list1 and list2"

    # do not try to optimize for small list sizes
    if len(list1) * len(list2) <= 100: # pick a small number
        return _compare_product(list1, list2)

    # first try to make a set from one of the lists
    try: a_set= set(list1)
    except TypeError:
        try: a_set= set(list2)
        except TypeError:
            a_set= None
        else:
            a_list= list1
    else:
        a_list= list2

    # here either a_set is None, or we have a_set and a_list

    if a_set:
        return any(itertools.imap(a_set.__contains__, a_list))

    # try to sort the lists
    try:
        a_list1= sorted(list1)
        a_list2= sorted(list2)
    except TypeError: # sorry, not sortable
        return _compare_product(list1, list2)

    # they could be sorted, so let's take the N+M road,
    # not the N*M

    iter1= iter(a_list1)
    iter2= iter(a_list2)
    try:
        item1= next(iter1)
        item2= next(iter2)
    except StopIteration: # one of the lists is empty
        return False # ie no common items

    while 1:
        if item1 == item2:
            return True
        while item1 < item2:
            try: item1= next(iter1)
            except StopIteration: return False
        while item2 < item1:
            try: item2= next(iter2)
            except StopIteration: return False

HTH.