You can also use np.where to replace a number with NaN.

arr = np.where(arr==NDV, np.nan, arr)

For example, the following result can be obtained via

arr = np.array([[1, 1, 2], [2, 0, 1]])
arr = np.where(arr==1, np.nan, arr)

This creates a new copy (unlike A[A==NDV]=np.nan) but in some cases that could be useful. For example, if the array was initially an int dtype, it will have to converted into a float array anyway (because replacing values with NaN won't work otherwise) and np.where can handle that.

This should work:

from numpy import *

a = array([[1, 2, 3], [0, 3, NaN]])
where_are_NaNs = isnan(a)
a[where_are_NaNs] = 0

In the above case where_are_NaNs is:

In [12]: where_are_NaNs
Out[12]: 
array([[False, False, False],
       [False, False,  True]], dtype=bool)

A complement about efficiency. The examples below were run with numpy 1.21.2

>>> aa = np.random.random(1_000_000)
>>> a = np.where(aa < 0.15, np.nan, aa)
>>> %timeit a[np.isnan(a)] = 0
536 µs ± 8.11 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)
>>> a = np.where(aa < 0.15, np.nan, aa)
>>> %timeit np.where(np.isnan(a), 0, a)
2.38 ms ± 27.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
>>> a = np.where(aa < 0.15, np.nan, aa)
>>> %timeit np.nan_to_num(a, copy=True)
8.11 ms ± 401 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
>>> a = np.where(aa < 0.15, np.nan, aa)
>>> %timeit np.nan_to_num(a, copy=False)
3.8 ms ± 70.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In consequence a[np.isnan(a)] = 0 is faster.