Error Using Accumarray, Third input SZ must be a full row vector with one element for each column of SUBS

Hi All, I am having trouble using accumarray. I am getting the following error and I am having trouble figuring out why. Thanks in advance.

Error using accumarray Third input SZ must be a full row vector with one element for each column of SUBS.

Error in approach2 (line 13) p = accumarray(subs, val, sz);

And the following is the code:

d = [3 3 4 4 4 5 6 5 5 4 3 2 3 4 3 2 5 6 6 5 5 4 3];
precision = 1;
f = (round(d/precision)).*precision;
[c, ~, ic] = unique(f);
N = numel(c);
subs = [ic(1:end - 3), ic(2:end-2), ic(3:end-1), ic(4:end)]; val = 1; sz = [N^2, N^2];
p = accumarray(subs, val, sz);
p = bsxfun(@rdivide, p, sum(p, 2));

Answer 1

[Star Strider]

If you don’t want to use the third argument, you don’t need to include it.

See if substituting this for your accumarray call does what you want:

p = accumarray(subs, val);

[Ellie]

Well, i think I found the problem by doing that. p becomes a 4-D array, which is not what I want. Any idea how to fix that? or is that a new question?

[Star Strider]

What do you want?

I would not consider it a new Question.

Answer 2

[Walter Roberson]

When you use accumarray, each row of subs is used as an index into array, as if

idx = num2cell(subs(K,:));
result(idx{:}) = val(K);

Your subs is 4 wide, so you are asking to create a 4 dimensional array.

My guess is that you want sz = [N, N, N, N]

You could always reshape() the result to N^2 x N^2

[Walter Roberson]

   subs_sq = [(subs(:,2) - 1)*N + subs(:,1), (subs(:,4) - 1)*N + subs(:,3)];
   p = accumarray(subs_sq, val);
   p = bsxfun(@rdivide, p, max(1,sum(p, 2)));

Or alternately,

p = reshape(accumarray(subs, val, [N, N, N, N]), [N^2, N^2]);
p = bsxfun(@rdivide, p, max(1,sum(p, 2)));

Yes this does create a temporary 4-dimensional array, and then immediately reshapes it to 2D. It also has exactly the same effect as the subs_sq version that builds the 2D array directly by assigning a number to each possible 2D vector.