How to get the indices list of all NaN value in numpy array?

PythonNumpyScipy

Python Problem Overview


Say now I have a numpy array which is defined as,

[[1,2,3,4],
[2,3,NaN,5],
[NaN,5,2,3]]

Now I want to have a list that contains all the indices of the missing values, which is [(1,2),(2,0)] at this case.

Is there any way I can do that?

Python Solutions


Solution 1 - Python

np.isnan combined with np.argwhere

x = np.array([[1,2,3,4],
              [2,3,np.nan,5],
              [np.nan,5,2,3]])
np.argwhere(np.isnan(x))

output:

array([[1, 2],
       [2, 0]])

Solution 2 - Python

You can use np.where to match the boolean conditions corresponding to Nan values of the array and map each outcome to generate a list of tuples.

>>>list(map(tuple, np.where(np.isnan(x))))
[(1, 2), (2, 0)]

Solution 3 - Python

Since x!=x returns the same boolean array with np.isnan(x) (because np.nan!=np.nan would return True), you could also write:

np.argwhere(x!=x)

However, I still recommend writing np.argwhere(np.isnan(x)) since it is more readable. I just try to provide another way to write the code in this answer.

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Questionxxx222View Question on Stackoverflow
Solution 1 - Pythonmichael_j_wardView Answer on Stackoverflow
Solution 2 - PythonNickil MaveliView Answer on Stackoverflow
Solution 3 - Pythonjohnnyasd12View Answer on Stackoverflow