> Both solutions seem to be equivalent in that concerns the number of needed > loop runs, but this two-step operation might require one less loop over list1. > The set&set solution, in contrary, might require one loop while transforming > to a set and another one for the & operation.
python -m timeit -s "l1 = range(1, 10000); l2 = range(5000, 15000)" "l3 = list(set(l1) & set(l2))" 100 loops, best of 3: 2.19 msec per loop python -m timeit -s "l1 = range(1, 10000); l2 = range(5000, 15000)" "s=set(l2); l3 = [i for i in l1 if i in s]" 100 loops, best of 3: 2.45 msec per loop python -m timeit -s "l1 = range(1, 100000); l2 = range(50000, 150000)" "l3 = list(set(l1) & set(l2))" 10 loops, best of 3: 28 msec per loop python -m timeit -s "l1 = range(1, 100000); l2 = range(50000, 150000)" "s=set(l2); l3 = [i for i in l1 if i in s]" 10 loops, best of 3: 28.1 msec per loop So even with conversion back into list set&set is still marginally faster. -- With best regards, Daniel Kluev -- http://mail.python.org/mailman/listinfo/python-list
