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

>>> 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.

Best I could come up with:

any([True for e in (1, 2) if e in a])

In python 3 we can start make use of the unpack asterisk. Given two lists:

bool(len({*a} & {*b}))

Edit: incorporate alkanen's suggestion

In some cases (e.g. unique list elements), set operations can be used.

>>> a=[2,3,4]
>>> set(a) - set([2,3]) != set(a)
True
>>> 

Or, using set.isdisjoint(),

>>> not set(a).isdisjoint(set([2,3]))
True
>>> not set(a).isdisjoint(set([5,6]))
False
>>> 

I collected several of the solutions mentioned in other answers and in comments, then ran a speed test. not set(a).isdisjoint(b) turned out the be the fastest, it also did not slowdown much when the result was False.

Each of the three runs tests a small sample of the possible configurations of a and b. The times are in microseconds.

Any with generator and max
        2.093 1.997 7.879
Any with generator
        0.907 0.692 2.337
Any with list
        1.294 1.452 2.137
True in list
        1.219 1.348 2.148
Set with &
        1.364 1.749 1.412
Set intersection explcit set(b)
        1.424 1.787 1.517
Set intersection implicit set(b)
        0.964 1.298 0.976
Set isdisjoint explicit set(b)
        1.062 1.094 1.241
Set isdisjoint implicit set(b)
        0.622 0.621 0.753

import timeit

def printtimes(t):
    print '{:.3f}'.format(t/10.0),

setup1 = 'a = range(10); b = range(9,15)'
setup2 = 'a = range(10); b = range(10)'
setup3 = 'a = range(10); b = range(10,20)'

print 'Any with generator and max\n\t',
printtimes(timeit.Timer('any(x in max(a,b,key=len) for x in min(b,a,key=len))',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('any(x in max(a,b,key=len) for x in min(b,a,key=len))',setup=setup2).timeit(10000000))
printtimes(timeit.Timer('any(x in max(a,b,key=len) for x in min(b,a,key=len))',setup=setup3).timeit(10000000))
print

print 'Any with generator\n\t',
printtimes(timeit.Timer('any(i in a for i in b)',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('any(i in a for i in b)',setup=setup2).timeit(10000000))
printtimes(timeit.Timer('any(i in a for i in b)',setup=setup3).timeit(10000000))
print

print 'Any with list\n\t',
printtimes(timeit.Timer('any([i in a for i in b])',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('any([i in a for i in b])',setup=setup2).timeit(10000000))
printtimes(timeit.Timer('any([i in a for i in b])',setup=setup3).timeit(10000000))
print

print 'True in list\n\t',
printtimes(timeit.Timer('True in [i in a for i in b]',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('True in [i in a for i in b]',setup=setup2).timeit(10000000))
printtimes(timeit.Timer('True in [i in a for i in b]',setup=setup3).timeit(10000000))
print

print 'Set with &\n\t',
printtimes(timeit.Timer('bool(set(a) & set(b))',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('bool(set(a) & set(b))',setup=setup2).timeit(10000000))
printtimes(timeit.Timer('bool(set(a) & set(b))',setup=setup3).timeit(10000000))
print

print 'Set intersection explcit set(b)\n\t',
printtimes(timeit.Timer('bool(set(a).intersection(set(b)))',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('bool(set(a).intersection(set(b)))',setup=setup2).timeit(10000000))
printtimes(timeit.Timer('bool(set(a).intersection(set(b)))',setup=setup3).timeit(10000000))
print

print 'Set intersection implicit set(b)\n\t',
printtimes(timeit.Timer('bool(set(a).intersection(b))',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('bool(set(a).intersection(b))',setup=setup2).timeit(10000000))
printtimes(timeit.Timer('bool(set(a).intersection(b))',setup=setup3).timeit(10000000))
print

print 'Set isdisjoint explicit set(b)\n\t',
printtimes(timeit.Timer('not set(a).isdisjoint(set(b))',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('not set(a).isdisjoint(set(b))',setup=setup2).timeit(10000000))
printtimes(timeit.Timer('not set(a).isdisjoint(set(b))',setup=setup3).timeit(10000000))
print

print 'Set isdisjoint implicit set(b)\n\t',
printtimes(timeit.Timer('not set(a).isdisjoint(b)',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('not set(a).isdisjoint(b)',setup=setup1).timeit(10000000))
printtimes(timeit.Timer('not set(a).isdisjoint(b)',setup=setup3).timeit(10000000))
print

This will do it in one line.

>>> a=[2,3,4]
>>> b=[1,2]
>>> bool(sum(map(lambda x: x in b, a)))
True

I have to say that my situation might not be what you are looking for, but it may provide an alternative to your thinking.

I have tried both the set() and any() method but still have problems with speed. So I remembered Raymond Hettinger said everything in python is a dictionary and use dict whenever you can. So that's what I tried.

I used a defaultdict with int to indicate negative results and used the item in the first list as the key for the second list (converted to defaultdict). Because you have instant lookup with dict, you know immediately whether that item exist in the defaultdict. I know you don't always get to change data structure for your second list, but if you are able to from the start, then it's much faster. You may have to convert list2 (larger list) to a defaultdict, where key is the potential value you want to check from small list, and value is either 1 (hit) or 0 (no hit, default).

from collections import defaultdict
already_indexed = defaultdict(int)

def check_exist(small_list, default_list):
    for item in small_list:
        if default_list[item] == 1:
            return True
    return False

if check_exist(small_list, already_indexed):
    continue
else:
    for x in small_list:
        already_indexed[x] = 1