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

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Answer 1

flat list can be made through reduce easily.

All you need to use initializer - third argument in the reduce function.

reduce(
   lambda result, _list: result.append(
       list(set(_list)&set(c1)) 
     ) or result, 
   c2, 
   [])

Above code works for both python2 and python3, but you need to import reduce module as from functools import reduce. Refer below link for details.

• for python2

• for python3

Answer 2

For people just looking to find the intersection of two lists, the Asker provided two methods:

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]

and

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

print intersect(b1, b2)

But there is a hybrid method that is more efficient, because you only have to do one conversion between list/set, as opposed to three:

b1 = [1,2,3,4,5]
b2 = [3,4,5,6]
s2 = set(b2)
b3 = [val for val in b1 if val in s2]

This will run in O(n), whereas his original method involving list comprehension will run in O(n^2)

Answer 3

The & operator takes the intersection of two sets.

{1, 2, 3} & {2, 3, 4}
Out[1]: {2, 3}

Answer 4

To define intersection that correctly takes into account the cardinality of the elements use Counter:

from collections import Counter

>>> c1 = [1, 2, 2, 3, 4, 4, 4]
>>> c2 = [1, 2, 4, 4, 4, 4, 5]
>>> list((Counter(c1) & Counter(c2)).elements())
[1, 2, 4, 4, 4]

Answer 5

Given:

> 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]]

I find the following code works well and maybe more concise if using set operation:

> c3 = [list(set(f)&set(c1)) for f in c2] 

It got:

> [[32, 13], [28, 13, 7], [1, 6]]

If order needed:

> c3 = [sorted(list(set(f)&set(c1))) for f in c2] 

we got:

> [[13, 32], [7, 13, 28], [1, 6]]

By the way, for a more python style, this one is fine too:

> c3 = [ [i for i in set(f) if i in c1] for f in c2]

Answer 6

Pure list comprehension version

>>> 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]]
>>> c1set = frozenset(c1)

Flatten variant:

>>> [n for lst in c2 for n in lst if n in c1set]
[13, 32, 7, 13, 28, 1, 6]

Nested variant:

>>> [[n for n in lst if n in c1set] for lst in c2]
[[13, 32], [7, 13, 28], [1, 6]]