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import timeit init = 'temp1 = list(range(100)); temp2 = [i * 2 for i in range(50)]' print timeit.timeit('list(set(temp1) - set(temp2))', init, number = 100000) print timeit.timeit('s = set(temp2);[x for x in temp1 if x not in s]', init, number = 100000) print timeit.timeit('[item for item in temp1 if item not in temp2]', init, number = 100000) The method I presented as well as preserving order is also (slightly) faster than the set subtraction because it doesn't require construction of an unnecessary set.The performance difference would be more noticable if the first list is considerably longer than the second and if hashing is expensive.That the memory of their virtues, which still survives among us, would. Nay, she said, I am persuaded that all men do all things.

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