A[A==NDV]=numpy.nan

A==NDV will produce a boolean array that can be used as an index for A

Ray's solution is good. However, on my machine it is about 2.5x faster to use numpy.sum in place of numpy.min:

In [13]: %timeit np.isnan(np.min(x))
1000 loops, best of 3: 244 us per loop

In [14]: %timeit np.isnan(np.sum(x))
10000 loops, best of 3: 97.3 us per loop

Unlike min, sum doesn't require branching, which on modern hardware tends to be pretty expensive. This is probably the reason why sum is faster.

edit The above test was performed with a single NaN right in the middle of the array.

It is interesting to note that min is slower in the presence of NaNs than in their absence. It also seems to get slower as NaNs get closer to the start of the array. On the other hand, sum's throughput seems constant regardless of whether there are NaNs and where they're located:

In [40]: x = np.random.rand(100000)

In [41]: %timeit np.isnan(np.min(x))
10000 loops, best of 3: 153 us per loop

In [42]: %timeit np.isnan(np.sum(x))
10000 loops, best of 3: 95.9 us per loop

In [43]: x[50000] = np.nan

In [44]: %timeit np.isnan(np.min(x))
1000 loops, best of 3: 239 us per loop

In [45]: %timeit np.isnan(np.sum(x))
10000 loops, best of 3: 95.8 us per loop

In [46]: x[0] = np.nan

In [47]: %timeit np.isnan(np.min(x))
1000 loops, best of 3: 326 us per loop

In [48]: %timeit np.isnan(np.sum(x))
10000 loops, best of 3: 95.9 us per loop