12/12/09 @ 08:16 (-0800), thus spake Keith Goodman: > If a and b are as short as in your example, which I doubt, here's a faster > way: > > >> timeit np.nonzero(reduce(np.logical_or, [a == i for i in b])) > 100000 loops, best of 3: 14 µs per loop > >> timeit [i for i, z in enumerate(a) if z in b] > 100000 loops, best of 3: 3.43 µs per loop > > Looping over a instead of b is faster if len(a) is much less than len(b): > > >> a = np.random.randint(0,100,10000) > >> b = tuple(set(a[:50].tolist())) > >> len(b) > 41 > >> timeit np.nonzero(reduce(np.logical_or, [a == i for i in b])) > 100 loops, best of 3: 2.65 ms per loop > >> timeit [i for i, z in enumerate(a) if z in b] > 10 loops, best of 3: 37.7 ms per loop
Nice. Well speed was not critical in this case, sometimes is good to know alternative ways of doing things. Thanks for your suggestions. Ernest _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
