Fast check for NaN in NumPy

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 loopIn [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 loopIn [42]: %timeit np.isnan(np.sum(x))10000 loops, best of 3: 95.9 us per loopIn [43]: x[50000] = np.nanIn [44]: %timeit np.isnan(np.min(x))1000 loops, best of 3: 239 us per loopIn [45]: %timeit np.isnan(np.sum(x))10000 loops, best of 3: 95.8 us per loopIn [46]: x[0] = np.nanIn [47]: %timeit np.isnan(np.min(x))1000 loops, best of 3: 326 us per loopIn [48]: %timeit np.isnan(np.sum(x))10000 loops, best of 3: 95.9 us per loop

I think np.isnan(np.min(X)) should do what you want.

There are two general approaches here:

• Check each array item for nan and take any.
• Apply some cumulative operation that preserves nans (like sum) and check its result.

While the first approach is certainly the cleanest, the heavy optimization of some of the cumulative operations (particularly the ones that are executed in BLAS, like dot) can make those quite fast. Note that dot, like some other BLAS operations, are multithreaded under certain conditions. This explains the difference in speed between different machines.

import numpy as npimport perfplotdef min(a):    return np.isnan(np.min(a))def sum(a):    return np.isnan(np.sum(a))def dot(a):    return np.isnan(np.dot(a, a))def any(a):    return np.any(np.isnan(a))def einsum(a):    return np.isnan(np.einsum("i->", a))b = perfplot.bench(    setup=np.random.rand,    kernels=[min, sum, dot, any, einsum],    n_range=[2 ** k for k in range(25)],    xlabel="len(a)",)b.save("out.png")b.show()