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Unexpected singleton dimension removal

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When I create a three dimensional array and assign it to an undefined array, the assignment does not behave as I expect if the first two dimensions are singletons:
>> it = rand(1,1,3)
it(:,:,1) =
it(:,:,2) =
it(:,:,3) =
>> vom(1,:,:) = it
vom =
0.9157 0.7922 0.9595
>> vom(:,1,:) = it
vom =
I know there are a million ways around this, but this behavior seems unnatural to me. Is this singleton dimension removal documented anywhere?
Sara on 8 Jul 2014
But what are you trying to achieve? If you want to assign it to vom, you don't need any indexing. vom1(1,1,1:3) = it will do not remove any dimension.
Thomas on 8 Jul 2014
As I said there are plenty of ways around this, in my case size(it,2) is not fixed, and I was doing the assigment
vom(m,:,:) = it;
in a loop. Alternatives are
1) initialize vom by
vom = zeros(0, size(it,2),3)
2) do the assignment as
vom(m, 1:size(it,2), :) = it
3) Initialize vom as empty and assign via concatenation
vom = [vom;it]
Doubtless there are other ways to do this.

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Accepted Answer

Matt J
Matt J on 11 Nov 2014
Edited: Matt J on 11 Nov 2014
Is this singleton dimension removal documented anywhere?
I think the best documentation you'll find will be this excerpt from the SUBSASGN documentation, note the portion in bold
"For multidimensional arrays, a(I,J,K,...) = b assigns b to the specified elements of a. b must be length(I)-by-length(J)-by-length(K)-... or be shiftable to that size by adding or removing singleton dimensions. "
So, in other words, it is applying shiftdim() to the right hand side "b", until it reaches a shape that will work. The question is, when vom is undefined, how does it choose the direction to shift? What accounts for differences in the 3 tests below? I believe the answer is, it will start with a right-shift of 1 and continue to right shift until it finds a shape that will work.
>> clear vom; vom(1,:,:) = it
vom =
0.9134 0.6324 0.0975
>> clear vom; vom(1,:,:) = it(:)
vom =
0.9134 0.6324 0.0975
>> clear vom; vom(1,:,:) = it(:)'
vom(:,:,1) =
vom(:,:,2) =
vom(:,:,3) =
So, in the first test above, "it" is right shiftdimmed by 1 making it into a column vector, but a column vector is not a shape compatible with the indexing vom(1,:,:). The target shape has to be either a row vector or 1x1x3. So, "it" is right shifted again, making it a row vector, which works.
Similarly, in the second test, the right hand side starts as a column vector, gets right shifted by 1, making it a row vector which is a legal target shape, and so that's the shape that gets chosen
Finally, the 3rd test is the most interesting. The given right hand side it(:)' starts as a row vector, and the sensible thing would be to stop there, since this is a legal target shape. However, as I mentioned, the rule is that the code will always attempt at least one right shift and so it converts the right hand side to 1x1x3. But this is also a legal target shape and so that's the shape that gets selected.

More Answers (1)

John D'Errico
John D'Errico on 8 Jul 2014
Clearly, "vom" stands for Volatile Organic Matter, and "it" always means Information Technology. But put them together and you get ... something that makes MATLAB upset.
Seriously, I was a bit surprised myself at first. Apparently MATLAB sees a vector as a vector, as just a vector and no more.
A = rand(1,3);
clear B
B(1,:) = A
B =
0.98946 0.86134 0.70694
clear B
B(:,1) = A
B =
An array is a different thing though.
A = rand(2,2);
clear B
B(1,:) = A
Subscripted assignment dimension mismatch.
I'd prefer there was consistency, but I'll bet this is a legacy thing, harking back to the days of MATLAV Version 1.
  1 Comment
Matt J
Matt J on 11 Nov 2014
An array is a different thing though.
Not as different as you might think:
>> A=rand(2,2)
A =
0.8003 0.4218
0.1419 0.9157
>> clear B; B(1,:,:)=A
B(:,:,1) =
0.8003 0.1419
B(:,:,2) =
0.4218 0.9157

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