find and delete from more-dimensional numpy array

Question

I have two numpy-arrays:

p_a_colors=np.array([[0,0,0],
                     [0,2,0],
                     [119,103,82],
                     [122,122,122],
                     [122,122,122],
                     [3,2,4]])

p_rem = np.array([[119,103,82],
                     [122,122,122]])

I want to delete all the columns from p_a_colors that are in p_rem, so I get:

p_r_colors=np.array([[0,0,0],
                    [0,2,0],
                    [3,2,4]])

I think, something should work like

p_r_colors= np.delete(p_a_colors, np.where(np.all(p_a_colors==p_rem, axis=0)),0)

but I just don't get the axis or [:] right.

I know, that

p_r_colors=copy.deepcopy(p_a_colors)
for i in range(len(p_rem)):
    p_r_colors= np.delete(p_r_colors, np.where(np.all(p_r_colors==p_rem[i], axis=-1)),0)

would work, but I am trying to avoid (python)loops, because I also want the performance right.

Answer 1

This is how I would do it:

dtype = np.dtype((np.void, (p_a_colors.shape[1] * 
                            p_a_colors.dtype.itemsize)))
mask = np.in1d(p_a_colors.view(dtype), p_rem.view(dtype))
p_r_colors = p_a_colors[~mask]

>>> p_r_colors
array([[0, 0, 0],
       [0, 2, 0],
       [3, 2, 4]])

You need to do the void dtype thing so that numpy compares rows as a whole. After that using the built-in set routines seems like the obvious way to go.

Answer 2

It's ugly, but

tmp = reduce(lambda x, y: x |  np.all(p_a_colors == y, axis=-1), p_rem, np.zeros(p_a_colors.shape[:1], dtype=np.bool))

indices = np.where(tmp)[0]

np.delete(p_a_colors, indices, axis=0)

(edit: corrected)

>>> tmp = reduce(lambda x, y: x |  np.all(p_a_colors == y, axis=-1), p_rem, np.zeros(p_a_colors.shape[:1], dtype=np.bool))
>>> 
>>> indices = np.where(tmp)[0]
>>> 
>>> np.delete(p_a_colors, indices, axis=0)
array([[0, 0, 0],
       [0, 2, 0],
       [3, 2, 4]])
>>>