Thanks for the replies... the matrix is fed to another function which expects a 3D matrix - The function takes the size of the matrix but processes it incorrectly if the matrix is 2D (I'm fairly confident it will work if it sees a 64x64x1 matrix, for example). I rather avoid changing the latter function since it is part of a suite of programs written by another party and has been extensively used previously.
Singleton dimention as last dimension in matrix
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Bernard on 22 Nov 2011
Commented: the cyclist on 29 May 2023 at 18:33
How do I create a singleton dimension as a last dimension in matlab, for example, so that size = 64 64 1.
I've tried reshape(x,[64 64 1]), but the resultant matrix is 64x64, not 64x64x1. Similarly with permute.
Thanks for any help!
the cyclist on 29 May 2023 at 18:33
I nominate @Stephen23's comment for the Best Comment Posted More Than 10 Years After the Original Question Award.
the cyclist on 22 Nov 2011
Arrays in MATLAB have an implied infinitely long series of trailing singleton dimensions. You can index into them with no problem. For example
>> x = rand(3,3);
is a valid indexing into x.
What is it that you are trying to do, that you need to emphasize that the array is 64x64x1?
More Answers (3)
Fangjun Jiang on 22 Nov 2011
Unless you have further process need, there is really no need to do that.
Trailing singleton dimensions ARE useful, and I want them too. This is useful for passing arguments to functions like convn. I would like to calculate discrete derivatives of a 3D data set as shown here. The calculation fails because reshape passes the simple partial derivative filter as 3x1 instead of 3x1x1. I can get around this by making the filter 3x3x3 and padding zeros, but its less clean:
%Bthree is 121,121,161 3D array
%Now we need to get the derivatives. I will try to convolve a simple linear
dBdX = convn(reshape([-5 0 5],3,1,1),Bthree,'same');
dBdY = convn(reshape([-5 0 5],1,3,1),Bthree,'same');
dBdZ = convn(reshape([-5 0 5],1,1,3),Bthree,'same');
Nevermind... convn actually does add extra singleton dimensions as needed- the reason this wasn't working is that the 'same' option matches to the first matrix, not the second. So this modification works:
dBdX = convn(Bthree,reshape([-5 0 5],3,1,1),'same');
dBdY = convn(Bthree,reshape([-5 0 5],1,3,1),'same');
dBdZ = convn(Bthree,reshape([-5 0 5],1,1,3),'same');
Hin Kwan Wong on 22 Nov 2011
64 x 64 x 1 with all due consideration is identical to 64 x 64...
You with see why with zeros(5,5,1), zeros(5,5,2) and zeros(5,5)
It's same as saying data= has dimension 1 as well as 1x1 and 1x1x1 and 1x1x1x1...
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