np.count_nonzero(~np.isnan(data))

~ inverts the boolean matrix returned from np.isnan.

np.count_nonzero counts values that is not 0\false. .sum should give the same result. But maybe more clearly to use count_nonzero

Testing speed:

In [23]: data = np.random.random((10000,10000))

In [24]: data[[np.random.random_integers(0,10000, 100)],:][:, [np.random.random_integers(0,99, 100)]] = np.nan

In [25]: %timeit data.size - np.count_nonzero(np.isnan(data))
1 loops, best of 3: 309 ms per loop

In [26]: %timeit np.count_nonzero(~np.isnan(data))
1 loops, best of 3: 345 ms per loop

In [27]: %timeit data.size - np.isnan(data).sum()
1 loops, best of 3: 339 ms per loop

data.size - np.count_nonzero(np.isnan(data)) seems to barely be the fastest here. other data might give different relative speed results.

You can add axis=0 to count null per column:

number_nan_in_arr = np.count_nonzero(np.isnan(arr), axis=0)

Example:

import numpy as np 

a = np.array([[0, np.nan, 7, 0],
              [3, 0, 2, np.nan]])
np.count_nonzero(np.isnan(a), axis=0)

Output:

array([0, 1, 0, 1], dtype=int64)