Fast check for NaN in NumPy

Question

I\'m looking for the fastest way to check for the occurrence of NaN (np.nan) in a NumPy array X. np.isnan(X) is out of the question, since

Answer 1

Even there exist an accepted answer, I'll like to demonstrate the following (with Python 2.7.2 and Numpy 1.6.0 on Vista):

In []: x= rand(1e5)
In []: %timeit isnan(x.min())
10000 loops, best of 3: 200 us per loop
In []: %timeit isnan(x.sum())
10000 loops, best of 3: 169 us per loop
In []: %timeit isnan(dot(x, x))
10000 loops, best of 3: 134 us per loop

In []: x[5e4]= NaN
In []: %timeit isnan(x.min())
100 loops, best of 3: 4.47 ms per loop
In []: %timeit isnan(x.sum())
100 loops, best of 3: 6.44 ms per loop
In []: %timeit isnan(dot(x, x))
10000 loops, best of 3: 138 us per loop

Thus, the really efficient way might be heavily dependent on the operating system. Anyway dot(.) based seems to be the most stable one.