Numpy mean of nonzero values

Get the count of non-zeros in each row and use that for averaging the summation along each row. Thus, the implementation would look something like this –

np.true_divide(matrix.sum(1),(matrix!=0).sum(1))

If you are on an older version of NumPy, you can use float conversion of the count to replace np.true_divide, like so –

matrix.sum(1)/(matrix!=0).sum(1).astype(float)

Sample run –

In [160]: matrix
Out[160]: 
array([[0, 0, 1, 0, 2],
       [1, 0, 0, 2, 0],
       [0, 1, 1, 0, 0],
       [0, 2, 2, 2, 2]])

In [161]: np.true_divide(matrix.sum(1),(matrix!=0).sum(1))
Out[161]: array([ 1.5,  1.5,  1. ,  2. ])

Another way to solve the problem would be to replace zeros with NaNs and then use np.nanmean, which would ignore those NaNs and in effect those original zeros, like so –

np.nanmean(np.where(matrix!=0,matrix,np.nan),1)

From performance point of view, I would recommend the first approach.