On Mon, Mar 5, 2012 at 11:36 AM, <josef.p...@gmail.com> wrote: > How about numpy.ptp, to follow this line? I would expect it's single > pass, but wouldn't short circuit compared to cython of Keith
I[1] a = np.ones(100000) I[2] timeit (a == a[0]).all() 1000 loops, best of 3: 203 us per loop I[3] timeit a.min() == a.max() 10000 loops, best of 3: 106 us per loop I[4] timeit np.ptp(a) 10000 loops, best of 3: 106 us per loop I[5] a[1] = 9 I[6] timeit (a == a[0]).all() 10000 loops, best of 3: 89.7 us per loop I[7] timeit a.min() == a.max() 10000 loops, best of 3: 102 us per loop I[8] timeit np.ptp(a) 10000 loops, best of 3: 103 us per loop _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
