a = np.zeros(1_000_000) a[100] = 1
%timeit np.any(a) 814 µs ± 17.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each) %timeit np.any(a == 1) 488 µs ± 5.68 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each) Haven't investigated further since your times are much longer than mine and secondly the equality check for 1 implies that perhaps a short circuit actually exists somewhere -- Sent from: http://numpy-discussion.10968.n7.nabble.com/ _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@python.org https://mail.python.org/mailman/listinfo/numpy-discussion
