numpy get 2d array where last dimension is indexed according to a 2d array

问题

I did read on numpy indexing but I didn't find what I was looking for.

I have a 288*384 image, where each pixel can have a labelling in [0,15]. It is stored in a 3d (288,384,16)-shaped numpy array im.

With im[:,:,1], I can for example get the image where all pixels have the label 1.

I have another 2d array labelling, (288*384)-shaped, containing a label for each pixel.

How do I get the image where each pixel has the corresponding pixel using some clever slicing?

Using loops, that would be:

result = np.zeros((288,384))
for x,y in zip(range(288), range(384)):
    result[x,y] = im[x,y,labelling[x,y]]

But this is of course inefficient.

回答1:

New Result

The short result is

np.choose(labelling,im.transpose(2,0,1))

Old Result

Try this

im[np.arange(288)[:,None],np.arange(384)[None,:],labelling]

It works for the following situation:

import numpy
import numpy.random
import itertools

a = numpy.random.randint(5,size=(2,3,4))
array([[[4, 4, 0, 0],
        [0, 4, 1, 1],
        [3, 4, 4, 2]],

      [[4, 0, 0, 2],
        [1, 4, 2, 2],
        [4, 2, 4, 4]]])

b = numpy.random.randint(4,size=(2,3))
array([[1, 1, 0],
       [1, 2, 2]])

res = a[np.arange(2)[:,None],np.arange(3)[None,:],b]
array([[4, 4, 3],
       [0, 2, 4]])

# note that zip is not doing what you expect it to do
result = np.zeros((2,3))
for x,y in itertools.product(range(2),range(3)):
    result[x,y] = a[x,y,b[x,y]]

array([[4., 4., 3.],
       [0., 2., 4.]])

Note that zip is not doing what you expect

zip(range(2),range(3))
[(0, 0), (1, 1)]

Probably you meant something like itertools.product

list(itertools.product(range(2),range(3)))
[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)]

The horribly looking [:,None] etc. can be avoided by using numpy.ix_

xx,yy = np.ix_( np.arange(2), np.arange(3) )

res = a[xx,yy,b]

来源：https://stackoverflow.com/questions/29098417/numpy-get-2d-array-where-last-dimension-is-indexed-according-to-a-2d-array