Finding If Any Element In A List Is In Another List And Return The First Element Found

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 anyway to idiomatic way to return the first element f

Solution 1:

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

Solution 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:

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

Solution 3:

do like this:

[e for e in a if e in b]

Solution 4:

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']

Solution 5:

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]