what is the most efficient way to find the position of the first np.nan value?

Question

consider the array a

a = np.array([3, 3, np.nan, 3, 3, np.nan])

I could do

np.isnan(a).argmax()

Answer 1

Here is a pythonic approach using itertools.takewhile():

from itertools import takewhile
sum(1 for _ in takewhile(np.isfinite, a))

Benchmark with generator_expression_within_next approach: ¹

In [118]: a = np.repeat(a, 10000)

In [120]: %timeit next(i for i, j in enumerate(a) if np.isnan(j))
100 loops, best of 3: 12.4 ms per loop

In [121]: %timeit sum(1 for _ in takewhile(np.isfinite, a))
100 loops, best of 3: 11.5 ms per loop

But still (by far) slower than numpy approach:

In [119]: %timeit np.isnan(a).argmax()
100000 loops, best of 3: 16.8 µs per loop

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_{1. The problem with this approach is using enumerate function. Which returns an enumerate object from the numpy array first (which is an iterator like object) and calling the generator function and next attribute of the iterator will take time.}