Answers for "numpy count the number of 1s in array"

2

numpy count the number of 1s in array

a = numpy.array([0, 3, 0, 1, 0, 1, 2, 1, 0, 0, 0, 0, 1, 3, 4])
unique, counts = numpy.unique(a, return_counts=True)
dict(zip(unique, counts))
{0: 7, 1: 4, 2: 1, 3: 2, 4: 1}
Posted by: Guest on March-25-2020
1

numpy count where

print(np.sum(a % 2 == 1))
Posted by: Guest on March-09-2020
2

numpy count the number of 1s in array

a = numpy.array([0, 3, 0, 1, 0, 1, 2, 1, 0, 0, 0, 0, 1, 3, 4])
unique, counts = numpy.unique(a, return_counts=True)
dict(zip(unique, counts))
{0: 7, 1: 4, 2: 1, 3: 2, 4: 1}
Posted by: Guest on March-25-2020
1

numpy count where

print(np.sum(a % 2 == 1))
Posted by: Guest on March-09-2020

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