I have an array like this:

array = np.random.randint(1, 100, 10000).astype(object)
array[[1, 2, 6, 83, 102, 545]] = np.nan
array[[3, 8, 70]] = None

Now, I want to find the indices of the NaN items and ignore the None ones. In this example, I want to get the [1, 2, 6, 83, 102, 545] indices. I can get the NaN indices with np.equal and np.isnan:

np.isnan(array.astype(float)) & (~np.equal(array, None))

I checked the performance of this solution with %timeit and got the following result:

243 µs ± 1.32 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Is there faster solution?

>Solution :

array != array

The classic NaN test. Writing NaN tests like this is one of the reasons that motivated the NaN != NaN design decision, since the IEEE 754 designers couldn’t assume programmers would have access to an isnan routine.

This significantly outperforms the code in the question when I try it:

In [1]: import numpy as np

In [2]: array = np.random.randint(1, 100, 10000).astype(object)
   ...: array[[1, 2, 6, 83, 102, 545]] = np.nan
   ...: array[[3, 8, 70]] = None

In [3]: %timeit array != array
139 µs ± 46.6 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)

In [4]: %timeit np.isnan(array.astype(float)) & (~np.equal(array, None))
755 µs ± 123 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)

And of course, it does give the same output:

In [5]: result1 = array != array

In [6]: result2 = np.isnan(array.astype(float)) & (~np.equal(array, None))

In [7]: np.array_equal(result1, result2)
Out[7]: True