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    <p>Or more concisely</p> <pre><code>if set(L) &amp; set(M): # there is an intersection else: # no intersection </code></pre> <p>If you really need <code>True</code> or <code>False</code></p> <pre><code>bool(set(L) &amp; set(M)) </code></pre> <p>After running some timings, this seems to be a good option to try too</p> <pre><code>m_set=set(M) any(x in m_set for x in L) </code></pre> <p>If the items in M or L are not hashable you have to use a less efficient approach like this</p> <pre><code>any(x in M for x in L) </code></pre> <p>Here are some timings for 100 item lists. Using sets is considerably faster when there is no intersection, and a bit slower when there is a considerable intersection.</p> <pre><code>M=range(100) L=range(100,200) timeit set(L) &amp; set(M) 10000 loops, best of 3: 32.3 µs per loop timeit any(x in M for x in L) 1000 loops, best of 3: 374 µs per loop timeit m_set=frozenset(M);any(x in m_set for x in L) 10000 loops, best of 3: 31 µs per loop L=range(50,150) timeit set(L) &amp; set(M) 10000 loops, best of 3: 18 µs per loop timeit any(x in M for x in L) 100000 loops, best of 3: 4.88 µs per loop timeit m_set=frozenset(M);any(x in m_set for x in L) 100000 loops, best of 3: 9.39 µs per loop # Now for some random lists import random L=[random.randrange(200000) for x in xrange(1000)] M=[random.randrange(200000) for x in xrange(1000)] timeit set(L) &amp; set(M) 1000 loops, best of 3: 420 µs per loop timeit any(x in M for x in L) 10 loops, best of 3: 21.2 ms per loop timeit m_set=set(M);any(x in m_set for x in L) 1000 loops, best of 3: 168 µs per loop timeit m_set=frozenset(M);any(x in m_set for x in L) 1000 loops, best of 3: 371 µs per loop </code></pre>
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    1. USJohn La Rooy
    1. USJohn La Rooy
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    1. CO@gnibbler - Is it provable that the `any()` version is less efficient? It seems like it would go through `M` only until it found an element in `L`, at which point `any` would return `True` and be done. This sounds more efficient than converting both `L` and `M` to sets beforehand. At least, on paper.
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      1. USChris Lutz
    2. COThis here, this is the answer.
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    3. CO@Chris, worst case is when when there is no intersection - O(l*m). With sets i believe it is O(l+m)
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