how to run a certain code loop for 'N' times and get 'N' number of output outside the loop.

I have certain code which I run for 'N' times using 'for' loop but after running the loop when I want to get all the N value of the parameters, it only gives the Last (Nth) value. for example this code generates the five matrices if I put A inside.

clc clear all N = 5; for i = 1:N A = rand(3,5); A end

% output is

A =

    0.4352    0.9375    0.5505    0.2475    0.3547
    0.1577    0.1078    0.4274    0.4474    0.7731
    0.6005    0.9000    0.1524    0.5328    0.8817

A =

    0.7341    0.6411    0.3105    0.4269    0.9250
    0.4064    0.1275    0.5786    0.0331    0.3583
    0.6042    0.4962    0.9436    0.9294    0.2600

A =

    0.7869    0.6848    0.9429    0.9094    0.6503
    0.5116    0.0924    0.0966    0.0113    0.3851
    0.5625    0.8726    0.8459    0.5237    0.6493

A =

    0.7629    0.2782    0.6316    0.4008    0.0904
    0.5757    0.8398    0.8335    0.5543    0.7444
    0.6319    0.4268    0.2702    0.4439    0.0326

A =

    0.4297    0.5223    0.8845    0.8946    0.5676
    0.0373    0.9096    0.2550    0.3985    0.8945
    0.9758    0.3832    0.9090    0.6250    0.2142

% If I put 'A' outside the loop it gives only the last (5th) value.

clc clear all N = 5; for i = 1:N A = rand(3,5); end A

A =

    0.4297    0.5223    0.8845    0.8946    0.5676
    0.0373    0.9096    0.2550    0.3985    0.8945
    0.9758    0.3832    0.9090    0.6250    0.2142

I want all the five matrices outside the loop as A1, A2, A3,A4 and A5. Please help me.

Answer 1

[KSSV]

Make A a 3D matrix.

N = 5; 
A = zeros(3,5,N) ;
for i = 1:N 
    A(:,:,i) = rand(3,5); 
end

You can access A using A(:,:,i) where i = 1,2,3,4,5.

Answer 2

[Thorsten]

You don't need the loop:

A = rand(3, 5, N);

[MANISH KUMAR]

Sir, actually the Matrices I need have some constraints. I give you the code than suggest me how to get 'N' matrices using this code.

the code is

clc

clear all

N = 5; % Population size

P = 5; % The number of competitive projects

T = 10; % The number of time periods

% The Expected benefit in each time period

B = [25 23 20 17 16 14 10 0 0 0; 30 28 25 21 19 17 0 0 0 0; 20 19 17 15 14 12 10 9 0 0; 15 13 12 11 11 10 7 5 3 0; 25 24 21 19 17 14 8 0 0 0];

% Duration of projects

D = [4; 5; 3; 2; 4];

% Resource requirement of a project in each period

% Res Req (type 1) = [3; 5; 1; 2; 4]; Max Availability = 6

R1 = [3*ones(1,10);5*ones(1,10);ones(1,10);2*ones(1,10);4*ones(1,10)];

% Res Req (type 2) = [2; 4; 1; 2; 3]; Max Availability = 4

R2 = [2*ones(1,10);4*ones(1,10);ones(1,10);2*ones(1,10);3*ones(1,10)];

excludedcount = D-1;

X = zeros(P,T);

for i = 1:N

    while(1)

        for row = 1:P

            rv = [1, zeros(1, T - excludedcount(row) - 1)];

            X(row, 1 : (T - excludedcount(row))) = rv(randperm(numel(rv)));

        end

        X;

       if any(X(1,:))
            X([4 5],:)=0;
        end
       if any(X(2,:))
            X(4,:)=0;
        end
       if any(X(3,:))
            X(5,:)=0;
        end
       if any(X(4,:))
            X(2,:)=0;
        end
       if any(X(5,:))
            X([1 3],:)=0;
       end

        X;
        Q = X.*B;
        TF = sum(sum(Q));
        Y=X;
        for row = 1:P
            ic=find(Y(row,:));
            if ~isempty(ic)
              Y(row,ic:ic+D(row)-1)=ones(1,D(row));
            end
        end
        Y;
        RR1 = R1.*Y;
        CRR1 = sum(RR1);
        RR2 = R2.*Y;
        CRR2 = sum(RR2);
        if ((max(CRR1)>6)||(max(CRR2)>4))
            continue;
        else
            break;
        end
        CRR1;
        CRR2;
    end
    X
    TF
    Y
    RR1
    RR2
    CRR1
    CRR2
end

Sir i want this matrix X which gives N=5 different matrices inside the loop but outside the loop it gives the last on only.

Please help.

[Jan]

@MANISH KUMAR: Please use the "{} Code" button to post readable code. Do not hide an important detail of the question in the comment of an answer, but all required details shsould be found inside the original question. Therefore use the possibility to edit the question. Thanks.

You got several suggestion already. All you have to do is storing the matrix in each iteration.

Result = cell(1, N);
for i = 1:N
  ...
  ResultC{i} = X;
end
...
% Perhaps:
ResultM = cat(3, ResultC{:});

[MANISH KUMAR]

@Jan Simon: Sir thanks for the kind support and sorry for the mistake. I will keep this in mind next time. Thank you very much again.

Answer 3

[Stefano Gianoli]

You might allocate the memory required for A so it can hold 3 x 5 x N elements instead of just 3 x 5:

>>N = 5; for i = 1:N A(:,:,i) = rand(3,5); end, A

A(:,:,1) =

    0.8147    0.9134    0.2785    0.9649    0.9572
    0.9058    0.6324    0.5469    0.1576    0.4854
    0.1270    0.0975    0.9575    0.9706    0.8003

A(:,:,2) =

    0.1419    0.7922    0.0357    0.6787    0.3922
    0.4218    0.9595    0.8491    0.7577    0.6555
    0.9157    0.6557    0.9340    0.7431    0.1712

A(:,:,3) =

    0.7060    0.0462    0.6948    0.0344    0.7655
    0.0318    0.0971    0.3171    0.4387    0.7952
    0.2769    0.8235    0.9502    0.3816    0.1869

A(:,:,4) =

    0.4898    0.7094    0.6797    0.1190    0.3404
    0.4456    0.7547    0.6551    0.4984    0.5853
    0.6463    0.2760    0.1626    0.9597    0.2238

A(:,:,5) =

    0.7513    0.6991    0.5472    0.2575    0.8143
    0.2551    0.8909    0.1386    0.8407    0.2435
    0.5060    0.9593    0.1493    0.2543    0.9293

you might also preallocate A for speed:

>> A = zeros(3,5,N);

[KSSV]

N = 5; for i = 1:N, eval(sprintf('A%i = rand(3,5);A%i',i,i)), end

This is not suggested.

[Stephen23]

@Stefano Gianoli: is there a reason why you are advising other beginners to use buggy and slow eval for such a trivial bit of code? eval is totally unnecessary in this code, and is best avoided (especially by beginners).

http://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval