Daniel Kluev wrote: >> 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.
If you are looking for speed, consider s = set(l1) s.intersection_update(l2) l3 = list(s) $ python -m timeit -s "l1 = range(1, 10000); l2 = range(5000, 15000)" "list(set(l1) & set(l2))" 100 loops, best of 3: 4 msec per loop $ python -m timeit -s "l1 = range(1, 10000); l2 = range(5000, 15000)" "s = set(l1); s.intersection_update(l2); list(s)" 100 loops, best of 3: 1.99 msec per loop -- http://mail.python.org/mailman/listinfo/python-list
