numpy count the number of 1s in array

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Luciana
26 Oct 2019
1a = numpy.array([0, 3, 0, 1, 0, 1, 2, 1, 0, 0, 0, 0, 1, 3, 4])
2unique, counts = numpy.unique(a, return_counts=True)
3dict(zip(unique, counts))
4{0: 7, 1: 4, 2: 1, 3: 2, 4: 1}
Karl
28 Nov 2018
1print(np.sum(a % 2 == 1))
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