sum on empty arrays

Can someone explain the logic of third statement (Run under R2021b) :

sum(zeros(0,0))

ans = 0

sum(zeros(1,0))

ans = 0

sum(zeros(2,0)) % This is returns an odd result, or not coherent with the second statement

ans =

  1×0 empty double row vector

sum(zeros(2,0),'all')

ans = 0

[jessupj]

a hint is to look at the type of objects. zeros(1,0) is a 'row vector' object, and its sum is a scalar operating along the only dimension. zeros(2,0) is a matrix object, so its sum operates across one of the two dimensions

i agree that its unexpected, but not counterintuitive looked at this way.

[Bruno Luong]

so zeros(0,0) is not a vector and not a matrix?

since if it's matrix IMO

sum(zeros(0,0)) % should return zeros(1,0) and not 0 

ans = 0

[jessupj]

(someone more expert than myself can answer this, but i think matrices adn arrays in matlab may just be, at a very base level, 1dimensional with accounting to provide multidimenional represnetations).

i'll play devil's advocate, because i don't know the answer.

zeros(0,0) has neither rows nor columns initialized, so seems like it should be scalar; initialized as null in both dimensions, there are no 'axes' to sum over so you get a scalar zero.

However, that doesn't explain why the nd-array case sum(zeros(0,0,0)) does give you want you want: a 1x0x0 empty array. I suspect this has to do with special rules for how empty objects are initialized. This is an interesting question.

[AndresVar]

It's weird but kinda helpful in some cases. Since you usually want to sum a nonempty matrix, this way your result tells you it was empty and not just 0. So actually it would be nice if the other cases resulted in empty array rather than 0.

[Bruno Luong]

I accept Steve's answer but surely jessupj comment is also spot on. Thanks.

Answer 1

[Steven Lord]

sum(zeros(0,0))

ans = 0

This is a special case, see the definition of the input argument A on the documentation page for the sum function.

case2 = sum(zeros(1,0)) % Sums along first nonsingleton dimension, dimension 2

case2 = 0

case3 = sum(zeros(2,0)) % Sums along first nonsingleton dimension, dimension 2

case3 =

  1×0 empty double row vector

If the dimension is not specified (as in case2 and case3) we sum along the first dimension whose size is not 1. The sizes of the result and A will be the same except that the size of the result in the first nonsingleton dimension of A is 1. In case2 the size of A is [1 0] and so the size of the result is [1 1]. In case3 the size of A is [2 0] and so the size of the result is [1 0].

sum(zeros(2,0),'all') % Sums along all dimensions, first dimension 1 then dimension 2

ans = 0

You can think of this as equivalent to reshaping the input into a vector then summing that vector. If you make zeros(2, 0) a vector what size is it?

A = zeros(2, 0);
v1 = reshape(A, 1, []) % case2 or

v1 =

  1×0 empty double row vector

v2 = reshape(A, [], 1) % the transpose of case2

v2 =

  0×1 empty double column vector

which when summed will give you the 1-by-1 value 0.

[Bruno Luong]

"This (BLU note: zeros(0,0) or []) is a special case, see the definition of the input argument A on the documentation page for the sum function."

Thanks, this is what we come up and I want to point out from the discussion.

It's quite odd that we use very often

sum([])

it's very convient to get 0, but without questioning the exeptional rule that leads to such result.

MATLAB is sometime tricky to correctly deal with.