How to get the indices list of all NaN value in numpy array?
Question:
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?
Answers:
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]])
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)]
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.
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?
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]])
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)]
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.