Looping through 3D matrix

Hi, I'm trying to populate a 3D matrix of zeros (3x14x12), nextvoy with a subset of rows in another matrix, trade, according to a lookup table, desigcty. The matrices look like this: trade =

    682         
    682         
    682         
    566         
    566         
    566         
    579        
    579       
    579        

desigcty =

            682         
            579        
            566         
   >> tmp_x

tmp_x =

       1
       1
       1
       2
       2
       2
       3
       3
       3
       4
       4
       4 
For example, nextvoy(:,:,1) should have all the entries of 682, then nextvoy(:,:,2) has 579, and then 566 for nextvoy(:,:,3). Then it should repeat this for the next 3 i's in nextvoy and so on.

The code I have overwrites each array in nextvoy with 566 because it's the last j (it's not performing according to expectations):

for i = 1:(size(desigcty,1)*nx); %for all load ports and ships: (:,:,i)
   for j = 1:size(desigcty,1); %loop through each load port (j,:,:)- 3 for now
    nextvoy(:,1,i)= tmp_x(i,1); %ship ID 
    [r1,~,~]= find(trade(:,1)==desigcty(j,1),1, 'first');
    [r2,~,~]= find(trade(:,1)==desigcty(j,1),1, 'last');
    nextvoy(:,2:9,i)= trade(r1:r2,1:8); %copies subset of trade matrix according to load code
    end;
end;

For example, part of the sequence this code produces is >> nextvoy(:,1:2,1:6)

ans(:,:,1) =

     1   566
     1   566
     1   566

ans(:,:,2) =

     1   566
     1   566
     1   566

ans(:,:,3) =

     1   566
     1   566
     1   566

ans(:,:,4) =

     2   566
     2   566
     2   566

ans(:,:,5) =

     2   566
     2   566
     2   566

ans(:,:,6) =

     2   566
     2   566
     2   566

But I need: >> nextvoy(:,1:2,1:3)

    ans(:,:,1) =

         1   682
         1   682
         1   682

    ans(:,:,2) =

        1  579
        1  579
        1  579

    ans(:,:,3) =

       1   566
       1   566
       1   566

    ans(:,:,4) =

         2   682
         2   682
         2   682

    ans(:,:,5) =

         2   579
         2   579
         2   579

    ans(:,:,6) =

         2   566
         2   566
         2   566

And so on...

[Jan]

Please format your code correctly as explained in the "Markup help" link on this page. Currently the question is not readable.

In addition you forgot to ask a specific question.

[Jan]

@Sophia: again, please format the code in your original message. You only have to press the "edit" button, mark the text with the mouse and press the "{} code" button. It is really worth to spend the necessary 4 seconds to make it easier for the readers.

I still do not get the point iof the question. Does the code do what you want? Do you get an error message? Does the result differ from your expectations - if so, how?

It would be much easier to answer, if you post running code. Currently I cannot copy&paster the definitions of trade and desigcty. But I want to concentrate on producing an answer, not on finding out, what the question could be.

[Sophia]

Does that help? Solution doesn't have to be in order of desigcty.

Answer 1

[Andrei Bobrov]

in your case

a1 = sortrows(tride);
nextvoy = permute(reshape(a1,3,3,[]),[1 3 2]);

variant 2

a1 = sortrows(tride);
a2 = sort(desigcty(:,1));
[b,n] = histc(a1(:,1),a2);
nextvoy = zeros(max(b),14,numel(b));
m = size(tride,2);
for i1 = 1:numel(b)
    nextvoy(1:b(i1),1:m,i1) = a1(n==i1,:);
end

EDIT variant

[c,idx] = sort(desigcty(:,1));
[b,n] = histc(tride(:,1),c);
nextvoy = [permute(ones(max(b),1)*tmp_x(:)',[1 3 2]),zeros(max(b),13,numel(tmp_x))];
k = size(tride,2);
p0 = numel(unique(tmp_x));
p = numel(tmp_x)/p0;
for i1 = 1:numel(b)
    i2 = idx(i1);
    nextvoy(1:b(i2),2:k+1,i1:p:end) = repmat(tride(i2==n,:),[1,1,p0]);
end

[Sophia]

Thanks, but that's not exactly what I'm looking for. The point of having it search for first and last occurrence of codes is to allow for flexibility since there may be more than 3 occurrences in later iterations. Also, I need it to populate a 3D matrix size = 3 14 12 4 times (3 of same subset, 4 times). Thanks for the help.

[Sophia]

Brilliant, that worked, thank you very much!