元素相同的两个列表中的公共元素

Question

I have two lists like thw following:

a=['not','not','not','not']
b=['not','not']

and I have to find the len of the list containing the intesection of the two above list, so that the result is:

intersection=['not','not']
len(intersection)
2

Now the problem is that I have tried filter(lambda x: x in a,b) and filter (lambda x: x in b,a) but when one of two list in longer than the other I do not get an intersection but just a membership checking. In the example above, since all the members of a are in b I get a len of common elements of 4; what I instead want is the intersection, which is len 2. Using set().intersection(set()) would instead create a set, which is not what I want since all the elements are the same. Can you suggest me any valuable and compact solution to the problem?

Answer 1

If you don't mind using collections.Counter , then you could have a solution like

>>> import collections
>>> a=['not','not','not','not']
>>> b=['not','not']

>>> c1 = collections.Counter(a)
>>> c2 = collections.Counter(b)

and then index by 'not'

>>> c1['not'] + c2['not']
6

For the intersection, you need to

>>> (c1 & c2) ['not']
2

Answer 2

I don't see any particularly compact way to compute this. Let's just go for a solution first.

The intersection is some sublist of the shorter list (eg b ). Now, for better performance when the shorter list is not extremely short, make the longer list a set (eg set(a) ). The intersection can then be expressed as a list comprehension of those items in the shorter list which are also in the longer set:

def common_elements(a, b):
    shorter, longer = (a, b) if len(a)<len(b) else (b, a)
    longer = set(longer)
    intersection = [item for item in shorter if item in longer]
    return intersection

a = ['not','not','not','not']
b = ['not','not']
print(common_elements(a,b))

Answer 3

Do it by set . First make those lists to sets and then take their intersection. Now there might be repetitions in the intersection. So for each elements in intersection take the minimum repetitions in a and b .

>>> a=['not','not','not','not']
>>> b=['not','not']
>>> def myIntersection(A,B):
...     setIn = set(A).intersection(set(B))
...     rt = []
...     for i in setIn:
...         for j in range(min(A.count(i),B.count(i))):
...             rt.append(i)
...     return rt
...
>>> myIntersection(a,b)
['not', 'not']

Answer 4

Have you considered the following approach?

a = ['not','not','not','not']
b = ['not','not']

min(len(a), len(b))
# 2

Since all the elements are the same, the number of common elements is just the minimum of the lengths of both lists.