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}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}Efficiently count zero elements in numpy array?
In [3]: arr
Out[3]:
array([[1, 2, 0, 3],
[3, 9, 0, 4]])
In [4]: np.count_nonzero(arr==0)
Out[4]: 2
In [5]:def func_cnt():
for arr in arrs:
zero_els = np.count_nonzero(arr==0)
# here, it counts the frequency of zeroes actuallyCopyright © 2021 Codeinu
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