Matlab trailing singleton dimension

Question

I have the following code

o = ones(4,3,2)
c = cellfun(@squeeze,num2cell(o,[2 3]), 'UniformOutput', false)

which gives as expected the 4 cells, each containing 3x2 matrixes.

But if I reduce the last dimension of o to one, the behavior is totally not as expected:

o = ones(4,3,1)
c = cellfun(@squeeze,num2cell(o,[2 3]), 'UniformOutput', false)

The output is:

[1x3 double]
[1x3 double]
[1x3 double]
[1x3 double]

Whereas I would expect:

[3x1 double]
[3x1 double]
[3x1 double]
[3x1 double]

Any possiblity how to obtain the correct result?

score 3 · Accepted Answer · edited May 23 '17 at 12:08

3

When the dimension of o is 4 x 3 x 1, num2cell(o, [2 3]) gives a 1 x 3 vector. As stated in the documentation, squeeze has no effect on a 2D array, so it will remain a row or column vector:

Two-dimensional arrays are unaffected by squeeze; if A is a row or column vector or a scalar (1-by-1) value, then B = A.

size(squeeze(rand(1, 3)))
%   1   3

size(squeeze(rand(3, 1)))
%   3   1

As pointed out by @Luis, you could replace squeeze with permute to get the dimensions you'd expect.

c = cellfun(@(x)permute(x, [2 3 1]), num2cell(o,[2 3]), 'UniformOutput', false)

edited May 23 '17 at 12:08

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answered Sep 19 '16 at 15:53

Suever

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... so the solution would be to replace `@squeeze` by `@(x)permute(x,[2 3 1])` – Luis Mendo Sep 19 '16 at 15:55
@LuisMendo Thanks for that! – Suever Sep 19 '16 at 15:57

Matlab trailing singleton dimension

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