First find out indices where the items are not nan, and then pass the filtered versions of a and weights to numpy.average:

>>> import numpy as np
>>> a = np.array([1, 2, np.nan,4])
>>> weights = np.array([4, 3, 2, 1])
>>> indices = np.where(np.logical_not(np.isnan(a)))[0]
>>> np.average(a[indices], weights=weights[indices])
1.75

As suggested by @mtrw in comments, it would be cleaner to use masked array here instead of index array:

>>> indices = ~np.isnan(a)
>>> np.average(a[indices], weights=weights[indices])
1.75

Also, there is a function named nanmedian which ignores NaN values. Signature of that function is: numpy.nanmedian(a, axis=None, out=None, overwrite_input=False, keepdims=<no value>)

You can simply multiply the input array with the weights and sum along the specified axis ignoring NaNs with np.nansum. Thus, for your case, assuming the weights are to be used along axis = 1 on the input array sst_filt, the summations would be -

np.nansum(sst_filt*weights,axis=1)

Accounting for the NaNs while averaging, we will end up with :

def nanaverage(A,weights,axis):
    return np.nansum(A*weights,axis=axis)/((~np.isnan(A))*weights).sum(axis=axis)

Sample run -

In [200]: sst_filt  # 2D array case
Out[200]: 
array([[  0.,   1.],
       [ nan,   3.],
       [  4.,   5.]])

In [201]: weights
Out[201]: array([ 0.25,  0.75])

In [202]: nanaverage(sst_filt,weights=weights,axis=1)
Out[202]: array([0.75, 3.  , 4.75])