Extract diagonal element from each frontal slice of tensor

Question

I have a p-by-p-by-n tensor. I want to extract diagonal element for each p-by-p slice. Are there anyone know how to do this without looping?

Thank you.

Answer 1

Behold the ever mighty and ever powerful bsxfun for vectorizing MATLAB problems to do this task very efficiently using MATLAB's linear indexing -

diags = A(bsxfun(@plus,[1:p+1:p*p]',[0:n-1]*p*p))

Sample run with 4 x 4 x 3 sized input array -

A(:,:,1) =
    0.7094    0.6551    0.9597    0.7513
    0.7547    0.1626    0.3404    0.2551
    0.2760    0.1190    0.5853    0.5060
    0.6797    0.4984    0.2238    0.6991
A(:,:,2) =
    0.8909    0.1493    0.8143    0.1966
    0.9593    0.2575    0.2435    0.2511
    0.5472    0.8407    0.9293    0.6160
    0.1386    0.2543    0.3500    0.4733
A(:,:,3) =
    0.3517    0.9172    0.3804    0.5308
    0.8308    0.2858    0.5678    0.7792
    0.5853    0.7572    0.0759    0.9340
    0.5497    0.7537    0.0540    0.1299
diags =
    0.7094    0.8909    0.3517
    0.1626    0.2575    0.2858
    0.5853    0.9293    0.0759
    0.6991    0.4733    0.1299

Benchmarking

Here's few runtime tests comparing this bsxfun based approach against repmat + eye based approach for big datasizes -

***** Datasize: 500 x 500 x 500 *****
----------------------- With BSXFUN
Elapsed time is 0.008383 seconds.
----------------------- With REPMAT + EYE
Elapsed time is 0.163341 seconds.

***** Datasize: 800 x 800 x 500 *****
----------------------- With BSXFUN
Elapsed time is 0.012977 seconds.
----------------------- With REPMAT + EYE
Elapsed time is 0.402111 seconds.

***** Datasize: 1000 x 1000 x 500 *****
----------------------- With BSXFUN
Elapsed time is 0.017058 seconds.
----------------------- With REPMAT + EYE
Elapsed time is 0.690199 seconds.

Answer 2

Just reading Divakar's answer and trying to understand why he again is roughly 10 times faster than my idea I put together code mixing both, and ended up with code which is faster:

A=reshape(A,[],n);
diags2 = A(1:p+1:p*p,:);

For a 500x500x500 tensor I get 0.008s for Divakar's solution and 0.005s for my solution using Matlab 2013a. Probably plain indexing is the only way to beat bsxfun.

Answer 3

One suggestion I have is to create a p x p logical identity matrix, replicate this n times in the third dimension, and then use this matrix to access your tensor. Something like this, supposing that your tensor was stored in A:

ind = repmat(logical(eye(p)), [1 1 n]);
out = A(ind);

Example use:

>> p = 5; n = 3;
>> A = reshape(1:75, p, p, n)

A(:,:,1) =

     1     6    11    16    21
     2     7    12    17    22
     3     8    13    18    23
     4     9    14    19    24
     5    10    15    20    25


A(:,:,2) =

    26    31    36    41    46
    27    32    37    42    47
    28    33    38    43    48
    29    34    39    44    49
    30    35    40    45    50


A(:,:,3) =

    51    56    61    66    71
    52    57    62    67    72
    53    58    63    68    73
    54    59    64    69    74
    55    60    65    70    75

>> ind = repmat(logical(eye(p)), [1 1 n]);
>> out = A(ind)

out =

     1
     7
    13
    19
    25
    26
    32
    38
    44
    50
    51
    57
    63
    69
    75

You'll notice that we grab the diagonals of the first slice, followed by the diagonals of the second slice, etc. up until the last slice. These values are all concatenated into a single vector.