Singleton dimention as last dimension in matrix

2011 11 22
4 个回答
回答已采纳
更新时间：2024 3 28
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3 个投票

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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Bernard
Bernard 2011-11-23
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.
Jan
Jan 2011-11-23
You cannot create a [64x64x1] array in Matlab, as Fangjun, cyclist and Wong have explained (+1 for all).
If the function you want to use does not catch a trailing singelton dimension, and you do not want to modify this function, you cannot use it.
Fangjun Jiang
Fangjun Jiang 2011-11-23
If the function expects a 64x64x1 matrix, then a 64x64 matrix is right for it and it shouldn't cause syntax error or functionality error. For example, data(i,j,1) is the same as data(i,j). size(data,3) returns 1 for both. You have not presented a use case that proves otherwise. Take another look at your data, see if there is other reason that causes the incorrect result. For example, might it be the first dimension and second dimension are reversed?
Jan
Jan 2011-11-23
The called function can fail, if it uses "if ndim(In)~=3" or "s = size(In); dim3 = s(3)".
Bernard
Bernard 2011-11-28
Thanks for your help everyone (and sorry for the slow update), I've just written my own function instead of using the pre-existing one - the purpose of the function is to write >3 dimension data in a certain file format: it expects a 3D matrix as input, adds dimensions to it, then writes the file. In the old function, if a 2D matrix is input, then the number and order of the dimension written to the output file is wrong. Anyway, in the end I used my own function.
Thanks for you help, just wanted to absolutely make sure there's no way to have a singleton dimension at the end of a matrix!
Jan
Jan 2011-11-28
编辑：Jan 2013-10-25
There is a way to create a dimension as [64 x 64 x 1]: In a Mex-function you can write in the dimensions vector. The official functions like mxSetDimensions care for the removing of trailing ones, but you can access the dimensions directly using undocumented methods. Anyhow, I strongly suggest not to do this. A bug in this code might have desasterous effects.
Stephen23
Stephen23 2023-5-24
编辑：Stephen23 2023-6-27
As Jan pointed out here, the function the OP was given probably used V = SIZE(X) and wrongly assumed something about the dimensions of interest based on the number of elements in V. The robust approach would be to call either
  • N = size(X,D)
  • [~,~,..,N,~] = size(X)
for each dimension of interest, both of which also work without error for trailing singleton dimensions.
the cyclist
the cyclist 2023-5-29
I nominate @Stephen23's comment for the Best Comment Posted More Than 10 Years After the Original Question Award.
Fangjun Jiang
Fangjun Jiang 2023-6-27
Or the best mind reader? I agree that is a good use case.
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:
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

3 个投票

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

4 个投票

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

4 个投票

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

3 个投票

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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