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

If you're comfortable with numba it allows to create a fast short-circuit (stops as soon as a NaN is found) function:

import numba as nb
import math

@nb.njit
def anynan(array):
    array = array.ravel()
    for i in range(array.size):
        if math.isnan(array[i]):
            return True
    return False

If there is no NaN the function might actually be slower than np.min, I think that's because np.min uses multiprocessing for large arrays:

import numpy as np
array = np.random.random(2000000)

%timeit anynan(array)          # 100 loops, best of 3: 2.21 ms per loop
%timeit np.isnan(array.sum())  # 100 loops, best of 3: 4.45 ms per loop
%timeit np.isnan(array.min())  # 1000 loops, best of 3: 1.64 ms per loop

But in case there is a NaN in the array, especially if it's position is at low indices, then it's much faster:

array = np.random.random(2000000)
array[100] = np.nan

%timeit anynan(array)          # 1000000 loops, best of 3: 1.93 µs per loop
%timeit np.isnan(array.sum())  # 100 loops, best of 3: 4.57 ms per loop
%timeit np.isnan(array.min())  # 1000 loops, best of 3: 1.65 ms per loop

Similar results may be achieved with Cython or a C extension, these are a bit more complicated (or easily avaiable as bottleneck.anynan) but ultimatly do the same as my anynan function.