在python中查找两个字符串列表的交集

Question

I have gone through Find intersection of two lists? , Intersection of Two Lists Of Strings , Getting intersection of two lists in python . However, I could not solve this problem of finding intersection between two string lists using Python.

I have two variables.

A = [['11@N3'], ['23@N0'], ['62@N0'], ['99@N0'], ['47@N7']]

B  = [['23@N0'], ['12@N1']]

How to find that '23@N0' is a part of both A and B?

I tried using intersect(a,b) as mentioned in http://www.saltycrane.com/blog/2008/01/how-to-find-intersection-and-union-of/ But, when I try to convert A into set, it throws an error:

File "<stdin>", line 1, in <module> TypeError: unhashable type: 'list'

To convert this into a set, I used the method in TypeError: unhashable type: 'list' when using built-in set function where the list can be converted using

result = sorted(set(map(tuple, A)), reverse=True)

into a tuple and then the tuple can be converted into a set. However, this returns a null set as the intersection.

Can you help me find the intersection?

Answer 1

You can use flatten function of compiler.ast module to flatten your sub-list and then apply set intersection like this

from compiler.ast import flatten

A=[['11@N3'], ['23@N0'], ['62@N0'], ['99@N0'], ['47@N7']]
B=[['23@N0'], ['12@N1']]

a = flatten(A)
b = flatten(B)
common_elements = list(set(a).intersection(set(b)))
common_elements
['23@N0']

Answer 2

Your datastructure is a bit strange, as it is a list of one-element lists of strings; you'd want to reduce it to a list of strings, then you can apply the previous solutions:

Thus a list like:

B = [['23@N0'], ['12@N1']]

can be converted to iterator that iterates over '23@N0', '12@N1'

with itertools.chain(*) , thus we have simple oneliner:

>>> set(chain(*A)).intersection(chain(*B))
{'23@N0'}

Answer 3

The problem is that your lists contain sublists so they cannot be converted to sets. Try this:

A=[['11@N3'], ['23@N0'], ['62@N0'], ['99@N0'], ['47@N7']]
B=[['23@N0'], ['12@N1']]

C = [item for sublist in A for item in sublist]
D = [item for sublist in B for item in sublist]

print set(C).intersection(set(D))

Answer 4

如果你必须把它装在幸运饼干上：

set(i[0] for i in A).intersection(set(i[0] for i in B))

Answer 5

You have two lists of lists with one item each. In order to convert that to a set you have to make it a list of strings:

set_a = set([i[0] for i in A])
set_b = set([i[0] for i in B])

Now you can get the intersection:

set_a.intersection(set_b)

Answer 6

A=[['11@N3'], ['23@N0'], ['62@N0'], ['99@N0'], ['47@N7']]
A=[a[0] for a in A]
B=[['23@N0'], ['12@N1']]
B=[b[0] for b in B]
print set.intersection(set(A),set(B))

Output: set(['23@N0'])

If each of your list has sublists of only 1 element you can try this.

Answer 7

My preference is to use itertools.chain from the standard library:

from itertools import chain

A = [['11@N3'], ['23@N0'], ['62@N0'], ['99@N0'], ['47@N7']]

B = [['23@N0'], ['12@N1']]

set(chain(*A)) & set(chain(*B))

# {'23@N0'}