# 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?

### Related

Do you consider `[1,2]` to intersect with `[1, ]`? That is, is it only the numbers you care about, or the list structure as well?

If only the numbers, investigate how to “flatten” the lists, then use the `set()` method.

You should flatten using this code ( taken from http://kogs-www.informatik.uni-hamburg.de/~meine/python_tricks ), the code is untested, but I’m pretty sure it works:

``````
def flatten(x):
"""flatten(sequence) -> list

Returns a single, flat list which contains all elements retrieved
from the sequence and all recursively contained sub-sequences
(iterables).

Examples:
>>> [1, 2, [3,4], (5,6)]
[1, 2, [3, 4], (5, 6)]
>>> flatten([[[1,2,3], (42,None)], [4,5], , 7, MyVector(8,9,10)])
[1, 2, 3, 42, None, 4, 5, 6, 7, 8, 9, 10]"""

result = []
for el in x:
#if isinstance(el, (list, tuple)):
if hasattr(el, "__iter__") and not isinstance(el, basestring):
result.extend(flatten(el))
else:
result.append(el)
return result
``````

After you had flattened the list, you perform the intersection in the usual way:

``````
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]]

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

print intersect(flatten(c1), flatten(c2))

``````

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.

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])
``````

### 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]]
``````

The functional approach:

``````input_list = [[1, 2, 3, 4, 5], [2, 3, 4, 5, 6], [3, 4, 5, 6, 7]]

result = reduce(set.intersection, map(set, input_list))
``````

and it can be applied to the more general case of 1+ lists

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)

Since `intersect` was defined, a basic list comprehension is enough:

``````>>> c3 = [intersect(c1, i) for i in c2]
>>> c3
[[32, 13], [28, 13, 7], [1, 6]]
``````

Improvement thanks to S. Lott’s remark and TM.’s associated remark:

``````>>> c3 = [list(set(c1).intersection(i)) for i in c2]
>>> c3
[[32, 13], [28, 13, 7], [1, 6]]
``````

I don’t know if I am late in answering your question. After reading your question I came up with a function intersect() that can work on both list and nested list. I used recursion to define this function, it is very intuitive. Hope it is what you are looking for:

``````def intersect(a, b):
result=[]
for i in b:
if isinstance(i,list):
result.append(intersect(a,i))
else:
if i in a:
result.append(i)
return result
``````

Example:

``````>>> 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]]
>>> print intersect(c1,c2)
[[13, 32], [7, 13, 28], [1, 6]]

>>> b1 = [1,2,3,4,5,9,11,15]
>>> b2 = [4,5,6,7,8]
>>> print intersect(b1,b2)
[4, 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]
``````
``````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 = [list(set(c2[i]).intersection(set(c1))) for i in xrange(len(c2))]

c3
->[[32, 13], [28, 13, 7], [1, 6]]
``````

We can use set methods for this:

``````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]]

result = []
for li in c2:
res = set(li) & set(c1)
result.append(list(res))

print result
``````

I was also looking for a way to do it, and eventually it ended up like this:

``````def compareLists(a,b):
removed = [x for x in a if x not in b]
added = [x for x in b if x not in a]
overlap = [x for x in a if x in b]
``````

The & operator takes the intersection of two sets.

``````{1, 2, 3} & {2, 3, 4}
Out: {2, 3}
``````

A pythonic way of taking the intersection of 2 lists is:

``````[x for x in list1 if x in list2]
``````

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]
``````
``````# Problem:  Given c1 and c2:
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]]
# how do you get c3 to be [[13, 32], [7, 13, 28], [1, 6]] ?
``````

Here’s one way to set `c3` that doesn’t involve sets:

``````c3 = []
for sublist in c2:
c3.append([val for val in c1 if val in sublist])
``````

But if you prefer to use just one line, you can do this:

``````c3 = [[val for val in c1 if val in sublist]  for sublist in c2]
``````

It’s a list comprehension inside a list comprehension, which is a little unusual, but I think you shouldn’t have too much trouble following it.

``````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 = [list(set(i) & set(c1)) for i in c2]
c3
[[32, 13], [28, 13, 7], [1, 6]]
``````

For me this is very elegant and quick way to to it 🙂

## 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.

# Simple way to find difference and intersection between iterables

Use this method if repetition matters

``````from collections import Counter

def intersection(a, b):
"""
Find the intersection of two iterables

>>> intersection((1,2,3), (2,3,4))
(2, 3)

>>> intersection((1,2,3,3), (2,3,3,4))
(2, 3, 3)

>>> intersection((1,2,3,3), (2,3,4,4))
(2, 3)

>>> intersection((1,2,3,3), (2,3,4,4))
(2, 3)
"""
return tuple(n for n, count in (Counter(a) & Counter(b)).items() for _ in range(count))

def difference(a, b):
"""
Find the symmetric difference of two iterables

>>> difference((1,2,3), (2,3,4))
(1, 4)

>>> difference((1,2,3,3), (2,3,4))
(1, 3, 4)

>>> difference((1,2,3,3), (2,3,4,4))
(1, 3, 4, 4)
"""
diff = lambda x, y: tuple(n for n, count in (Counter(x) - Counter(y)).items() for _ in range(count))
return diff(a, b) + diff(b, a)
``````
``````from random import *

a = sample(range(0, 1000), 100)
b = sample(range(0, 1000), 100)
print(a)
print(b)
print(set(a).intersection(set(b)))
``````
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