Finding intersection of two lists of strings in python

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?

Solution 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']

Solution 2

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))

Solution 3

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'}

Solution 4

In case you have to fit it on a fortune cookie:

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

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