Singleton dimention as last dimension in matrix

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Bernard
Bernard 2011-11-22
编辑： Stephen23 2024-3-28
Hello,
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!
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Erik Newton
Erik Newton 2024-3-28
In my use case, I'm encoding to Json and sending to a remote (3rd-party non-matlab) system which has to parse it. It is expecting 3 dimensions. and so as an example of my problem:
>> jsonencode(zeros(3,2,2))
ans =
'[[[0,0],[0,0]],[[0,0],[0,0]],[[0,0],[0,0]]]'
>> jsonencode(zeros(3,2,1))
ans =
'[[0,0],[0,0],[0,0]]'
I wish/need to keep a consistent level of square brackets, independent of whether I just happen to only have a single case in the 3rd dimension.
Stephen23
Stephen23 2024-3-28
编辑：Stephen23 2024-3-28
Ran in:
@Erik Newton: here is a workaround (does not work for scalar nor empty inputs). Assumes you want a 3D array.
A = randi(9,3,2,1);
[R,C,~] = size(A);
T = jsonencode(cat(3,A,nan(R,C,1)));
T = strrep(T,',null]',']')
T = '[[[8],[5]],[[3],[4]],[[1],[1]]]'

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采纳的回答

the cyclist
the cyclist 2011-11-22
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);
>> x(2,3,1,1,1,1,1,1,1,1,1,1,1)
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?

更多回答（3 个）

Fangjun Jiang
Fangjun Jiang 2011-11-22
Unless you have further process need, there is really no need to do that.
>> size(rand(10))
ans =
10 10
>> size(rand(10),3)
ans =
1
>> size(rand(10),5)
ans =
1
David
David 2013-10-25
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
%filter function.
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');
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David
David 2013-10-25
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');

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Hin Kwan Wong
Hin Kwan Wong 2011-11-22
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=[5] has dimension 1 as well as 1x1 and 1x1x1 and 1x1x1x1...

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