Matlab last dimension access on ndimensions matrix

Question

I have as entry a matrix which can have an several number of dimensions : n × m or n × m × p or n × m × p × q or ...

What I want to do is to access to the last dimension, something like:

data = input(:,:,1)

The problem is that the number of : can change.

Is the possible number of dimensions finite (and small)? If so, you could query the size of the matrix using the `size` function and have various cases using a `switch` statement. — am304, Nov 13 '13 at 14:07
Note that it is not recommended to use a variable called `input` as this is the name of a standard function. — Dennis Jaheruddin, Nov 13 '13 at 14:12
possible duplicate of [Indexing of unknown dimensional matrix](http://stackoverflow.com/questions/10146082/indexing-of-unknown-dimensional-matrix) — Eitan T, Nov 14 '13 at 12:16
@EitanT Though the solution to that question should be applicable here, this one is more specific. Hence 'shortcuts' may be taken here that are not viable solutions for the linked question. — Dennis Jaheruddin, Dec 24 '13 at 16:26
@DennisJaheruddin then perhaps it's a duplicate of [Using a colon for indexing in matrices of unknown dimensions](http://stackoverflow.com/questions/16924273/using-a-colon-for-indexing-in-matrices-of-unknown-dimensions). I go by the theory that most of the new questions are recycled... :) — Eitan T, Dec 24 '13 at 16:49
@EitanT I agree that with some minor rewording that question would make this one obsolete. Have suggested a merge as that question seems to be worded better, whilst this one has a larger variety of answers. — Dennis Jaheruddin, Dec 27 '13 at 08:34

score 14 · Accepted Answer · answered Nov 13 '13 at 15:36

14

You should make use of the fact that indices into an array can be strings containing ':':

>> data = rand(2, 2, 2, 5);
>> otherdims = repmat({':'},1,ndims(data)-1);
>> data(otherdims{:}, 1)
ans(:,:,1) =
    7.819319665880019e-001    2.940663337586285e-001
    1.006063223624215e-001    2.373730197055792e-001
ans(:,:,2) =
    5.308722570279284e-001    4.053154198805913e-001
    9.149873133941222e-002    1.048462471157565e-001

See the documentation on subsref for details.

answered Nov 13 '13 at 15:36

Rody Oldenhuis

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Very nice, I started out like this, but didn't find the `otherdims{:}` step! – Dennis Jaheruddin Nov 13 '13 at 15:42
You can use strings? Wow. +1 – Dan Nov 14 '13 at 15:37

score 5 · Answer 2 · answered Nov 13 '13 at 14:08

5

It is a bit of a hack, but here is how you can do it:

data = rand(2,2,3);

eval(['data(' repmat(':,',1,ndims(data)-1) '1)'])

This will give (depending on the randon numbers):

ans =

      0.19255      0.56236
      0.62524      0.90487

answered Nov 13 '13 at 14:08

Dennis Jaheruddin

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Well seen. Let's clean up `eval`'s image. It's hackish but useful – Luis Mendo Nov 13 '13 at 15:18
No `eval` needed; MATLAB's default `subsref` allows string arguments for indices, provided those strings are `':'`; see my answer. Still, +1 :) – Rody Oldenhuis Nov 13 '13 at 15:40

Mohsen Nosratinia · Answer 3 · 2013-11-13T14:53:54.367

2

You can use shiftdim to bring the last dimension to the first and index that one and reshape it back

x = rand(2,3,4,5,6);

sz = size(x);
A = shiftdim(x, numel(sz)-1);
B = reshape(A(1,:), sz(1:end-1));

and

>> isequal(B, x(:,:,:,:,1))
ans =
     1

or alternatively you can use subsref to index it:

B = subsref(x, substruct('()', [num2cell(repmat(':', 1, ndims(x)-1)) 1]))

edited Nov 13 '13 at 14:53

answered Nov 13 '13 at 14:46

Mohsen Nosratinia

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+1, but that last step is a bit of an awkward way of doing it...see my answer. – Rody Oldenhuis Nov 13 '13 at 15:45

Luis Mendo · Answer 4 · 2013-11-13T15:36:30.837

2

Nice question!

Another possibility is to use linear indexing in blocks and then reshape:

x = rand(2,3,4,5); % example data
n = 2; % we want x(:,:,...,:,n)

siz = size(x);
b = numel(x)/siz(end); % block size
result = reshape(x(b*(n-1)+1:b*n),siz(1:end-1));

This seems to be the fastest approach in my computer (but do try yourself; that may depend on Matlab version and system)

edited Nov 13 '13 at 15:36

answered Nov 13 '13 at 15:13

Luis Mendo

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+1 This is almost _exactly_ what I was typing up. It's questions like this that make me want to stay glued to SO, but alas, I was away. :( – chappjc Nov 14 '13 at 00:48
@chappjc I knew you would like this question :-D – Luis Mendo Nov 14 '13 at 07:50

score 1 · Answer 5 · answered Nov 13 '13 at 14:40

I think this does it:

a = rand(2,2,2,3)

s = size(a)
r = reshape(a, [], s(end))
reshape(r(:,1), s(1:end-1)) %// reshape(r(:,2), s(1:end-1)) gives you a(:,:,:,...,2) etc...

I benchmarked against Dennis's (correct) answer and it gives the same result but doesn't need eval which is always worth avoiding if possible.

score 0 · Answer 6 · answered Nov 10 '16 at 02:07

Just in case GNU Octave readers reach here.

@Rody Oldenhuis's answer can be written as an one-liner in Octave:

> data = reshape(1:8, 2, 2, 2);
> data({':'}(ones(1,ndims(data)-1)){:}, 1)
ans =

   1   3
   2   4

Here:

{':'}(ones(1,ndims(data)-1)){:}

means:

tmp = {':'};
tmp = tmp(ones(1,ndims(data)-1));
tmp{:}

Of couse, repmat({':'},1,ndims(data)-1){:} works also.

Matlab last dimension access on ndimensions matrix

6 Answers6