How to expend 1 dim for array ?
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il 12 Lug 2023
Commentato: Walter Roberson
il 17 Lug 2023
I already have a vector s , which size is ![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1432318/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1432318/image.png)
Now want to change its size to
for another function that normally has an input size of 4x4x4xn (n >=1). I tried this command:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1432323/image.png)
s_expend(:,:,:,1) = s
but I found that the size of s_expend is still
(Use command: size(s_expend))
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1432328/image.png)
It seems that MATLAB ignores dimensions of size 1 that are the last dimension? How do I implement something like expend_dim in numpy or unsqueeze in Pytorch using Matlab?
1 Commento
Stephen23
il 12 Lug 2023
"It seems that MATLAB ignores dimensions of size 1 that are the last dimension?"
Not at all. If you check that dimension, you will find that it has size 1 (as do all infinite trailing dimensions):
A = rand(4,4,4);
size(A,4)
Risposta accettata
Walter Roberson
il 12 Lug 2023
Spostato: Matt J
il 12 Lug 2023
MATLAB does not exactly ignore trailing dimension of 1, but thinking of it that way is "close enough" for most work.
Every array in MATLAB is treated as having an indefinite number of trailing dimensions of size 1. A 4 x 4 x 4 array is treated the same as a 4 x 4 x 4 x 1 array or a 4 x 4 x 4 x 1 x 1 x 1 x 1 x 1 x 1 array.
When MATLAB is storing the dimensions of an array internally, all singular (length 1) trailing dimensions are omitted from being explicitly stored, and the number of stored dimensions (ndims) for any array is always the minimum of 2 and "the last dimension number that is not 1". Trailing dimensions that are size 1 are "collapsed".
If you look at an array and it is 4 x 4 x 4, ndims 3, and you need to check whether it is compatible with a 4 x 4 x 4 x n array, then when it makes sense to do so, MATLAB automatically internally replicates data to match. If you were, for example, adding a 4 x 4 x 4 to a 4 x 4 x 4 x 7 then MATLAB would do the internal equivalent of repmat(The3DArray, 1, 1, 1, 7) and add that to the 4 x 4 x 4 x 7. If you were doing cat(4, The3DArray, The4DArray) then that would work without difficulty and would give you a 4D array .
There are cases where you do need exactly the same size of array; in such a case you can explicitly test isequal(size(TheFirstArray), size(TheSecondArray)) . But if you are checking for arithematic compatibility then ndims(TheFirstArray) <= ndims(TheSecondArray)
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Più risposte (2)
Kanishk Singhal
il 12 Lug 2023
Yeah, as you said MATLAB ignnores dimension of size 1, but if you do,
A = [1 2;3 4];
size(A,3)
You'll get 1 which I think is what you want. If you want to loop in the function you can use size to find the dimension you need.
Hope it helps.
1 Commento
Steven Lord
il 12 Lug 2023
In release R2019b we enhanced the size function to accept a vector of values for the dimension input. So if that's all the data you need, no loop is required.
A = ones(4, 5, 6);
sz = size(A, 1:10) % Size of A in dimensions 1 through 10
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il 17 Lug 2023
1 Commento
Walter Roberson
il 17 Lug 2023
isequal(size(Tensor, 1:3), [4 4 4]) & ndims(Tensor) <= 4
should work for the original arrangement
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