Python - Intersection Of Two Lists Of Lists

These are my two lists; k = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4], [5,9]] kDash = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4], [5,6], [1,2]] My output should be the following; [

Solution 1:

You will have to convert the lists to list of tuples, and then use the intersection. Note that below solution may have elements in a different order, and duplicates will obviously not be there, since I'm using set.

In [1]: l1 = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4], [5,9]]

In [2]: l2 = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4], [5,6], [1,2]]

In [3]: [list(x) for x in set(tuple(x) for x in l1).intersection(set(tuple(x) for x in l2))]
Out[3]: [[1, 2], [5, 6, 2], [3], [4]]

You can alternatively save the intersection in a variable and get the final list, if order, duplicates are necessary:

In [4]: intersection = set(tuple(x) for x in l1).intersection(set(tuple(x) for x in l2))

In [5]: [x for x in l1 if tuple(x) in intersection]
Out[5]: [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4]]

And the intersection, just in case if you are interested.

In [6]: print intersection
set([(1, 2), (5, 6, 2), (3,), (4,)])

This will work pretty well for large lists, but if the lists are small, do explore the other solution by @timegb (whose solution will be highly unoptimal for longer lists)

Solution 2:

Since your output list has duplicate elements, you don't really seem to want a classical intersection. A basic list comprehension will do everything.

>>> k = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4], [5,9]]
>>> kDash = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4], [5,6], [1,2]]
>>> [x for x in k if x in kDash]
[[1, 2], [4], [5, 6, 2], [1, 2], [3], [4]]

For large lists, we want to get the time it takes to call __contains__ to O(1) instead of O(n):

>>>stuff_in_kDash = set(map(tuple, kDash))>>>[x for x in k iftuple(x) in stuff_in_kDash]
[[1, 2], [4], [5, 6, 2], [1, 2], [3], [4]]

Solution 3:

A cleaner way to write the intersection is

{tuple(x) for x in l1} & {tuple(x) for x in l2}

A good alternative is

{tuple(x) for x in l1}.intersection(map(tuple, l2))

Solution 4:

Even though what was written here is much more elegant solutions, here's another one

def foo(L1,L2):
    res=[]
    for lst in L1:
        if lst in L2:
            res.append(lst)
    return res