在MATLAB中将一维索引及其对应的第二维索引列表切成3D矩阵？

Question

How do I slice a 3D matrix with a list of first-dimension indices and its corresponding second-dimension indices?

For example, given

>> A = cat(3, [1 2 3; 4 5 6; 7 8 9], [10 20 30; 40 50 60; 70 80 90], [100 200 300; 400 500 600; 700 800 900])

A(:,:,1) =

     1     2     3
     4     5     6
     7     8     9


A(:,:,2) =

    10    20    30
    40    50    60
    70    80    90


A(:,:,3) =

   100   200   300
   400   500   600
   700   800   900

I want to slice out A(2, 3, :) and A(1, 2, :) to get [6 60 600; 2 20 200] [6 60 600; 2 20 200] .

I failed with

>> A([2, 1], [3, 2], :)

ans(:,:,1) =

     6     5
     3     2


ans(:,:,2) =

    60    50
    30    20


ans(:,:,3) =

   600   500
   300   200

I believe there's a one-liner/elegant solution.

Answer 1

To extract the desired elements subscripts should be converted to indices ( sub2ind ), but before it a 3D transpose ( permute ) should be applied so the third dimension becomes the first.

idx = [2 3; 1 2];
[m n z]= size(A);
B=permute(A,[3 1 2]);
result = B(:,sub2ind([m,n],idx(:,1),idx(:,2)))