Question

Suppose I have the following numpy arrays:

>>a
array([[0, 0, 2],
       [2, 0, 1],
       [2, 2, 1]])
>>b
array([[2, 2, 0],
       [2, 0, 2],
       [1, 1, 2]])

that I then vertically stack

c=np.dstack((a,b))

resulting in:

>>c
array([[[0, 2],
        [0, 2],
        [2, 0]],

       [[2, 2],
        [0, 0],
        [1, 2]],

       [[2, 1],
        [2, 1],
        [1, 2]]])

From this I wish to, for each 3rd dimension of c, check which combination is present in this subarray, and then number it accordingingly with the index of the list-match. I’ve tried the following, but it is not working. The algorithm is simple enough with double for-loops, but because c is very large, it is prohibitively slow.

classes=[(0,0),(2,1),(2,2)]
out=np.select( [h==c for h in classes], range(len(classes)), default=-1)

My desired output would be

out = [[-1,-1,-1],
       [3,  1,-1],
       [2,  2,-1]]

Answer 1

How about this:

(np.array([np.array(h)[...,:] == c for h in classes]).all(axis = -1) *
         (2 + np.arange(len(classes)))[:, None, None]).max(axis=0) - 1

It returns, what you actually need

array([[-1, -1, -1],
       [ 3,  1, -1],
       [ 2,  2, -1]])