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?
This question is related to
python
list
intersection
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 :)
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]
>>> 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]]
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])
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 & operator takes the intersection of two sets.
{1, 2, 3} & {2, 3, 4}
Out[1]: {2, 3}
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.
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))
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]]
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)
# 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.
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]
A pythonic way of taking the intersection of 2 lists is:
[x for x in list1 if x in list2]
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 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
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.
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)
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 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]
Source: Stackoverflow.com