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

def intersect(a, b):
  return list(set(a) & set(b))

print intersect(b1, b2)

Answer 2

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

Answer 3

do like this:

[e for e in a if e in b]

Answer 4

Use sets: https://docs.python.org/2/library/sets.html

result = set(list1) & set(list2)

if you want to make it a conditional like any:

if (set(list1) & set(list2)):
    do something

Answer 5

Yes, it is possible, by using a filter when doing your list comprehension :

list1 = ['a', 'b', 'c', 'd', 'e']
list2 = ['g', 'z', 'b', 'd', '33']
[elem for elem in list1 if elem in list2]
# ['b', 'd']

Answer 6

Use a list comprehension:

[elem for elem in list1 if elem in list2]

Example:

list1 = [1, 2, 3, 4, 5]
list2 = [1, 10, 2, 20]

c = [elem for elem in list1 if elem in list2]
print(c)

Output

[1, 2]