在 Python 中计算 numpy ndarray 中非 NaN 元素的数量

Question

I need to calculate the number of non-NaN elements in a numpy ndarray matrix. How would one efficiently do this in Python? Here is my simple code for achieving this:

import numpy as np

def numberOfNonNans(data):
    count = 0
    for i in data:
        if not np.isnan(i):
            count += 1
    return count 

Is there a built-in function for this in numpy? Efficiency is important because I'm doing Big Data analysis.

Thnx for any help!

Answer 1

np.count_nonzero(~np.isnan(data))

~ inverts the boolean matrix returned from np.isnan .

np.count_nonzero counts values that is not 0\false. .sum should give the same result. But maybe more clearly to use count_nonzero

Testing speed:

In [23]: data = np.random.random((10000,10000))

In [24]: data[[np.random.random_integers(0,10000, 100)],:][:, [np.random.random_integers(0,99, 100)]] = np.nan

In [25]: %timeit data.size - np.count_nonzero(np.isnan(data))
1 loops, best of 3: 309 ms per loop

In [26]: %timeit np.count_nonzero(~np.isnan(data))
1 loops, best of 3: 345 ms per loop

In [27]: %timeit data.size - np.isnan(data).sum()
1 loops, best of 3: 339 ms per loop

data.size - np.count_nonzero(np.isnan(data)) seems to barely be the fastest here. other data might give different relative speed results.

Answer 2

Quick-to-write alterantive

Even though is not the fastest choice, if performance is not an issue you can use:

sum(~np.isnan(data)) .

Performance:

In [7]: %timeit data.size - np.count_nonzero(np.isnan(data))
10 loops, best of 3: 67.5 ms per loop

In [8]: %timeit sum(~np.isnan(data))
10 loops, best of 3: 154 ms per loop

In [9]: %timeit np.sum(~np.isnan(data))
10 loops, best of 3: 140 ms per loop

Answer 3

To determine if the array is sparse, it may help to get a proportion of nan values

np.isnan(ndarr).sum() / ndarr.size

If that proportion exceeds a threshold, then use a sparse array, eg - https://sparse.pydata.org/en/latest/

Answer 4

An alternative, but a bit slower alternative is to do it over indexing.

np.isnan(data)[np.isnan(data) == False].size

In [30]: %timeit np.isnan(data)[np.isnan(data) == False].size
1 loops, best of 3: 498 ms per loop 

The double use of np.isnan(data) and the == operator might be a bit overkill and so I posted the answer only for completeness.

Answer 5

len([i for i in data if np.isnan(i) == True])