How to deal with the structure?

Dear All,

I have a struct which saves a number of records. I have the following codes which takes more time than expected.

Resstd2 = [];
resstd20 = [];
treestruct2 = [];
nnzV = find(v);
m1 = length(TreeStruct2);
temp = [];
for i = 1 : m1
    vblockid = TreeStruct2(i).treestruct.vblockID;
    solvedvid = v(vblockid);
    resstd2 = TreeStruct2(i).Resstd;
    if nnz(solvedvid) == 0
        if resstd2(3) < voltagelimit
            treestruct2 = [treestruct2; TreeStruct2(i).treestruct];
            Resstd2 = [Resstd2; resstd2];
        else
            resstd20 = [resstd20; resstd2];
        end
    end
    temp = [temp; resstd2(3)];
end

It seems to me that using structures takes long time. Does anyone have the experience in cell or table? Do you think using cells or tables will be faster than using structures?

Thanks.

Benson

[Rik]

It seems to me your style of dynamically growing the array is probably the real culprit, not the data type used.

The code in the loop will only run once, as max(size([])) is 1. It will error at the first line, as [] is not a struct array with at least 1 element.

Have a read here and here. It will greatly improve your chances of getting an answer.

[Rik]

Your edit doesn't seem to address the issue I raised in my comment. Without a good description of what you want, the only thing we can say is not to use a dynamic expansion.

If you want help with your code specifically, you should make sure we can run it.

Resstd2 = [];
resstd20 = [];
treestruct2 = [];
nnzV = find(v);

Unrecognized function or variable 'v'.

[Benson Gou]

Hi, Rik,

Thanks a lot for your reply.

The data is big and is hard to display the data in my question. Is it possible to attached the mat file in my question?

Thanks.

Benson

Answer 1

[Jan]

Do not let arrays grow iteratively because this causes exponentially growing computing times. Example:

x = [];
for k = 1:1e6
   x(k) = k;
end

Although the final array has only 8MB (8 bytes per double), Matlab has to allocate a new array in each iteration and copy the old values. In total this allocates sum(1:1e6)*8 = 4TB!

m1          = length(TreeStruct2);
Resstd2     = zeros(m1, 1);
Resstd2_i   = 0;
resstd20    = zeros(m1, 1);
resstd20_i  = 0;
treestruct2_m = false(m1, 1);
% nnzV = find(v);  % Not used
temp = zeros(m1, 1);
for i = 1 : m1
    vblockid  = TreeStruct2(i).treestruct.vblockID;
    solvedvid = v(vblockid);
    resstd2   = TreeStruct2(i).Resstd;
    if ~any(solvedvid)   % Faster than nnz==0
        if resstd2(3) < voltagelimit
            treestruct2_m = true;
            Resstd2_i     = Resstd2_i + 1;
            Resstd2(Resstd2_i) = resstd2;
        else
            resstd20_i  = resstd20_i + 1;
            resstd20(resstd20_i) = resstd2;
        end
    end
    
    temp(i) = resstd2(3);
end

Resstd2  = Resstd2(1:Resstd2_i);    % Crop unused elements
resstd20 = resstd20(1:resstd20_i);  % Crop unused elements
treestruct2 = cat(1, TreeStruct2(treestruct2_m).treestruct);

Maybe Resstd2 and resstd20 need more dimension. Without provided input data, e.g. created by RAND() I cannot guess this. Adjust the dimensions on demand.

[Benson Gou]

Thanks a lot, Jan. Your code works very well.

Benson