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

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it easy check if element of list in list using any():

any(elem in list2 elem in list1)

but there anyway idiomatic way return first element found?

i'd prefer one-line solution rather than:

for elem in list1:    if elem in list2:        return elem

this answer similar an answer on a similar question, @jamylak goes more detail of timing results compared other algorithms.

if want first element matches, use next:

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

this isn't efficient performs linear search of b each element. create set b has better lookup performance:

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

if next doesn't find raises exception:

>>> = [4, 5] >>> next(element element in if element in b_set) traceback (most recent call last): stopiteration

you can give default return instead, e.g. none. changes syntax of how function parameters parsed , have explicitly create generator expression:

>>> none next((element element in if element in b_set), none) true

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