How can I append arrays onto the same variable in a series of loops?

Hi, I am a beginner in MATLAB and was wondering if someone with more experience would find it easyto do what I am trying to achieve. I have a number of 2x6 matrices that I would like to "append" to one another in the z-dimension in order to make a single 3D variable. I would then like to output this variable as a .mat file. For example, lets say I have 3 different options, and each one has 3 sub-options which yield me a total of 9 different matrices. I had envisioned something like the following, but I can't get it to work. The goal would be to have a single variable that is 2x6x9 Any help is greatly appreciated. Thanks!

My bad example:

    x = zeros(2,6);
    counter=0;
    for i=1:3
      counter = counter+1;
      for j=1:3
        x(:,:,counter)= rand(2,6);
        save('test.mat','x','-append');
        counter = counter+1;
      end
    end

JH

Answer 1

[Stephen23]

Solution One: Vectorized

Your example uses rand, which lets you generate the whole 3D array directly by specifying the outputs size in a vector, no loops or concatenation is required:

out = rand([2,5,3]);

This will be much faster than using any loop and concatenating arrays, and is a good introduction to vectorized code! However if you really want to concatenate arrays, then read on...

Solution Two: with for-Loop

For a start there is no reason why you need to use the variable counter when you already have the for-loop variable. Why bother? And preallocating the array will make this faster. But the basic issue is that you are trying to concatenate and save the matrix in each loop iteration. Why not just create the whole array, and then save it?:

>> N = 3;
>> X = nan([2,6,N]);
>> for k = 1:N
      X(:,:,k) = rand(2,6);
end
>> save('test.mat','X')

Bonus Information on Concatenation

One of the easiest way of concatenating matrices is to use cat:

>> A = [1,2;3,4];
>> B = [10,20;30,40];
>> C = [0,NaN;Inf,-Inf];
>> out = cat(3,A,B,C)
out(:,:,1) =
     1     2
     3     4
out(:,:,2) =
    10    20
    30    40
out(:,:,3) =
     0   NaN
   Inf  -Inf

Note that this is even easier if you place the variables into a cell array:

>> Z = {A,B,C};
>> out = cat(3,Z{:})

[Jono Horne]

Thanks Stephen, I used "rand" as an example, but my problem is that the actual 2x6 arrays are uniquely generated depending on the path. If you can imagine a flowchart with 3 main headers, each branching into 3 sub-headers..(and on and on, but that shouldn;t change the "building block") I have it currently set up that the user would input their choices starting at the top and eventually getting to the bottom which will spit out a unique matrix. I want to modify the script so that it automates this and sequentially outputs all iterations. aM I making myself clear? I can attach a picture if that helps. Thanks again for your time, I really appreciate it. JH

[Stephen23]

Sure, attach some graphic / daigram, as this is not clear to me. Did Solution Two not work?

[Jono Horne]

I have attached a drawing of the current workflow. I have the script set up so that a user can choose the path via input prompts. They select a Modifier 1, and Modifier 2, and that gives them a unique 2x6 matrix. I instead would like to make Matlab run through each iteration and save the results into one 2x6x9 matrix. Any info is greatly appreciated. LEt mek now if you have any further questions

Answer 2

[Delladj Kamel]

 *It's easy, just type this code and you'll get what you are expecting*   
    x = zeros(2,6);
    for i=1:9
        x(:,:,i)= rand(2,6);
    end
    save('test.mat','x');

results

   x(:,:,1) =

    0.3724    0.4897    0.9516    0.0527    0.2691    0.5479
    0.1981    0.3395    0.9203    0.7379    0.4228    0.9427

   x(:,:,2) =

    0.4177    0.3015    0.6663    0.6981    0.1781    0.9991
    0.9831    0.7011    0.5391    0.6665    0.1280    0.1711

   x(:,:,3) =

    0.0326    0.8819    0.1904    0.4607    0.1564    0.6448
    0.5612    0.6692    0.3689    0.9816    0.8555    0.3763

   x(:,:,4) =

    0.1909    0.4820    0.5895    0.3846    0.2518    0.6171
    0.4283    0.1206    0.2262    0.5830    0.2904    0.2653

   x(:,:,5) =

    0.8244    0.7302    0.5841    0.9063    0.8178    0.5944
    0.9827    0.3439    0.1078    0.8797    0.2607    0.0225

   x(:,:,6) =

    0.4253    0.1615    0.4229    0.5985    0.6959    0.6385
    0.3127    0.1788    0.0942    0.4709    0.6999    0.0336

   x(:,:,7) =

    0.0688    0.5309    0.4076    0.7184    0.5313    0.1056
    0.3196    0.6544    0.8200    0.9686    0.3251    0.6110

   x(:,:,8) =

    0.7788    0.0908    0.1537    0.4401    0.4574    0.5181
    0.4235    0.2665    0.2810    0.5271    0.8754    0.9436

   x(:,:,9) =

    0.6377    0.2407    0.2891    0.6951    0.2548    0.6678
    0.9577    0.6761    0.6718    0.0680    0.2240    0.8444

[Stephen23]

This would be significantly improved by using array preallocation, simply by defining the final size of the matrix at the beginning:

x = zeros([2,6,9]);

[Delladj Kamel]

thanks, I find this comment very helpful and I accept it.