I have an n * x numpy matrix, which could look like this:
a = np.array([[1, 0], [0, 1]])
and I have another n * n numpy matrix, which look like this:
b = np.array([[2, 2], [2, 2]])
I would like to replace zero elements of a with the corresponding element of b so I would get:
[[1, 2],
[2, 1]]
How can I do that?
You can just use a boolean mask:
mask = (a == 0)
a[mask] = b[mask]
This is efficient if you want to update the original array a since it only assigns to the zero elements, not the whole array.
numpy.where is the elegant way to do it
numpy.where for that task is going to be inefficient if only a few number of elements are zero as it needs to assign the whole array back to a.
where. np.copyto with a where parameter is a bit faster than this, np.copyto(a, b, where=a==0) . Exact time advantages may very with the proportion of 0s in a.np.copyto can loop all parts together and doesn't need a temporary array. Very good!You can use np.where:
np.where(a!=0, a, b)
array([[1, 2],
[2, 1]])
