# Find intersection of two nested lists?

### Question

I know how to get an intersection of two flat lists:

``````b1 = [1,2,3,4,5,9,11,15]
b2 = [4,5,6,7,8]
b3 = [val for val in b1 if val in b2]
``````

or

``````def intersect(a, b):
return list(set(a) & set(b))

print intersect(b1, b2)
``````

But when I have to find intersection for nested lists then my problems starts:

``````c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]
c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
``````

In the end I would like to receive:

``````c3 = [[13,32],[7,13,28],[1,6]]
``````

Can you guys give me a hand with this?

1
466
1/17/2018 7:11:28 AM

If you want:

``````c1 = [1, 6, 7, 10, 13, 28, 32, 41, 58, 63]
c2 = [[13, 17, 18, 21, 32], [7, 11, 13, 14, 28], [1, 5, 6, 8, 15, 16]]
c3 = [[13, 32], [7, 13, 28], [1,6]]
``````

Then here is your solution for Python 2:

``````c3 = [filter(lambda x: x in c1, sublist) for sublist in c2]
``````

In Python 3 `filter` returns an iterable instead of `list`, so you need to wrap `filter` calls with `list()`:

``````c3 = [list(filter(lambda x: x in c1, sublist)) for sublist in c2]
``````

Explanation:

The filter part takes each sublist's item and checks to see if it is in the source list c1. The list comprehension is executed for each sublist in c2.

176
5/29/2016 5:00:30 AM

You don't need to define intersection. It's already a first-class part of set.

``````>>> b1 = [1,2,3,4,5,9,11,15]
>>> b2 = [4,5,6,7,8]
>>> set(b1).intersection(b2)
set([4, 5])
``````