Numpy mask to count number of elements satisfying a condition

np.count_nonzero should be a bit faster than the sum:

np.count_nonzero(arr1 > 0.6)

In fact, it is three times as fast

>>> from timeit import repeat
>>> kwds = dict(globals=globals(), number=10000)
>>> 
>>> arr1 = np.random.rand(184,184)
>>> 
>>> repeat('np.count_nonzero(arr1 > 0.6)', **kwds)
[0.15281831508036703, 0.1485864429268986, 0.1477385900216177]
>>> repeat('(arr1 > 0.6).sum()', **kwds)
[0.5286932559683919, 0.5260644309455529, 0.5260107989888638]

Get a boolean mask and just count the "True"s:

(arr1 > 0.6).sum()