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
<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
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 & operator takes the intersection of two sets.
{1, 2, 3} & {2, 3, 4}
Out[1]: {2, 3}
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]
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]]
>>> 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]]