How to check if one of the following items is in a list?

Question

>>> L1 = [2,3,4]
>>> L2 = [1,2]
>>> [i for i in L1 if i in L2]
[2]


>>> S1 = set(L1)
>>> S2 = set(L2)
>>> S1.intersection(S2)
set([2])

Both empty lists and empty sets are False, so you can use the value directly as a truth value.

Ah, Tobias you beat me to it. I was thinking of this slight variation on your solution:

>>> a = [1,2,3,4]
>>> b = [2,7]
>>> any(x in a for x in b)
True

Maybe a bit more lazy:

a = [1,2,3,4]
b = [2,7]

print any((True for x in a if x in b))

Think about what the code actually says!

>>> (1 or 2)
1
>>> (2 or 1)
2

That should probably explain it. :) Python apparently implements "lazy or", which should come as no surprise. It performs it something like this:

def or(x, y):
    if x: return x
    if y: return y
    return False

In the first example, x == 1 and y == 2. In the second example, it's vice versa. That's why it returns different values depending on the order of them.

a = {2,3,4}
if {1,2} & a:
    pass

Code golf version. Consider using a set if it makes sense to do so. I find this more readable than a list comprehension.