fast numpy addnan

Here is one possibility:

>>> x = np.array([1, 2, np.nan, 3, np.nan, 4])... y = np.array([1, np.nan, 2, 5, np.nan, 8])>>> x = np.ma.masked_array(np.nan_to_num(x), mask=np.isnan(x) & np.isnan(y))>>> y = np.ma.masked_array(np.nan_to_num(y), mask=x.mask)>>> (x+y).filled(np.nan)array([  2.,   2.,   2.,   8.,  nan,  12.])

The real difficulty is that you seem to want nan to be interpreted as zero unless all values at a particular position are nan. This means that you must look at both x and y to determine which nans to replace. If you are okay with having all nan values replaced, then you can simply do np.nan_to_num(x) + np.nan_to_num(y).

You could do something like:

arr1 = np.array([1.0, 1.0, np.nan, 1.0, 1.0, np.nan])arr2 = np.array([1.0, 1.0, 1.0, 1.0, 1.0, np.nan])flags = np.isnan(arr1) & np.isnan(arr2)copy1 = arr1.copy()copy2 = arr2.copy()copy1[np.isnan(copy1)] = 0.0copy2[np.isnan(copy2)] = 0.0out = copy1 + copy2out[flags] = np.NaNprint outarray([  2.,   2.,   1.,   2.,   2.,  NaN])

to find the locations in the arrays where both have a NaN in that index. Then, do essentially what @mgilson suggested, as in make copies and replace the NaNs with 0.0, add the two arrays together, and then replace the flagged indices above with np.NaN.

import numpy as npz=np.nansum([X,Y],axis=0)