Unexpected singleton dimension removal

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) =
    0.9157
it(:,:,2) =
    0.7922
it(:,:,3) =
    0.9595

>> vom(1,:,:) = it

vom =
    0.9157    0.7922    0.9595

>> vom(:,1,:) = it

vom =

      0.2769
      0.0462
      0.0971

I know there are a million ways around this, but this behavior seems unnatural to me. Is this singleton dimension removal documented anywhere?

[Sara]

May squeeze do what you need?

[Thomas]

'Squeeze' is what I don't want it to do, I would be fine if MATLAB just performed the assignment so the shape of the LHS is the same as the shape of the RHS.

[John D'Errico]

I don't think squeeze is the issue, but rather an understanding of why something unexpected happens. Predictability is of utmost importance in programming.

[Sara]

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]

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.

Answer 1

[Matt J]

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) =

        0.9134

    vom(:,:,2) =

        0.6324

    vom(:,:,3) =

        0.0975

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.

Answer 2

[John D'Errico]

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 =
      0.98946
      0.86134
      0.70694

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.

[Matt J]

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