Finding if any element in a list is in another list and return the first element found

Question

It is easy to check if an element of a list is in another list using any():

any(elem in list2 for elem in list1)

but is there

Answer 1

This answer is similar to an answer on a similar question, where @jamylak goes into more detail of timing the results compared to other algorithms.

If you just want the first element that matches, use next:

>>> a = [1, 2, 3, 4, 5]
>>> b = [14, 17, 9, 3, 8]
>>> next(element for element in a if element in b)
3

This isn't too efficient as it performs a linear search of b for each element. You could create a set from b which has better lookup performance:

>>> b_set = set(b)
>>> next(element for element in a if element in b_set)

If next doesn't find anything it raises an exception:

>>> a = [4, 5]
>>> next(element for element in a if element in b_set)
Traceback (most recent call last):
StopIteration

You can give it a default to return instead, e.g. None. However this changes the syntax of how the function parameters are parsed and you have to explicitly create a generator expression:

>>> None is next((element for element in a if element in b_set), None)
True