## The Question :

*471 people think this question is useful*

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

*The Question Comments :*

## The Answer 1

*178 people think this answer is useful*

**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.

## The Answer 2

*894 people think this answer is useful*

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])

## The Answer 3

*61 people think this answer is useful*

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)

## The Answer 4

*29 people think this answer is useful*

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

## The Answer 5

*27 people think this answer is useful*

### 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 Answer 6

*21 people think this answer is useful*

The & operator takes the intersection of two sets.

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

## The Answer 7

*13 people think this answer is useful*

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

[x for x in list1 if x in list2]

## The Answer 8

*8 people think this answer is useful*

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], [6], 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))

## The Answer 9

*8 people think this answer is useful*

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]]

## The Answer 10

*5 people think this answer is useful*

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]

## The Answer 11

*3 people think this answer is useful*

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]

## The Answer 12

*2 people think this answer is useful*

Do you consider `[1,2]`

to intersect with `[1, [2]]`

? 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.

## The Answer 13

*1 people think this answer is useful*

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]
return [removed,added,overlap]

## The Answer 14

*0 people think this answer is useful*

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]]

## The Answer 15

*0 people think this answer is useful*

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

## The Answer 16

*0 people think this answer is useful*

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]

## The Answer 17

*0 people think this answer is useful*

# 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.

## The Answer 18

*0 people think this answer is useful*

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 ðŸ™‚

## The Answer 19

*0 people think this answer is useful*

## 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.

## The Answer 20

*-1 people think this answer is useful*

# 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)