Removing NaNs in numpy arrays

Question

I have two numpy arrays that contains NaNs:

A = np.array([np.nan,   2,   np.nan,   3,   4])
B = np.array([   1  ,   2,     3   ,   4,  np.nan])

are there any smart way using numpy to remove the NaNs in both arrays, and also remove whats on the corresponding index in the other list? Making it look like this:

A = array([  2,   3, ])
B = array([  2,   4, ])

Answer 1

What you could do is add the 2 arrays together this will overwrite with NaN values where they are none, then use this to generate a boolean mask index and then use the index to index into your original numpy arrays:

In [193]:

A = np.array([np.nan,   2,   np.nan,   3,   4])
B = np.array([   1  ,   2,     3   ,   4,  np.nan])
idx = np.where(~np.isnan(A+B))
idx
print(A[idx])
print(B[idx])
[ 2.  3.]
[ 2.  4.]

output from A+B:

In [194]:

A+B
Out[194]:
array([ nan,   4.,  nan,   7.,  nan])

EDIT

As @Oliver W. has correctly pointed out, the np.where is unnecessary as np.isnan will produce a boolean index that you can use to index into the arrays:

In [199]:

A = np.array([np.nan,   2,   np.nan,   3,   4])
B = np.array([   1  ,   2,     3   ,   4,  np.nan])
idx = (~np.isnan(A+B))
print(A[idx])
print(B[idx])
[ 2.  3.]
[ 2.  4.]

Answer 2

A[~(np.isnan(A) | np.isnan(B))]

B[~(np.isnan(A) | np.isnan(B))]