how to save an array in each iteration?

I have a parameter L and the following array SR:

L = 3;

SR = [ 0.0941 0.1129 0.0471 0 0 0

0.0941 0 0 0.0471 0.0941 0

0 0 0 0.0471 0 0.0471

0 0.1129 0 0 0.0941 0

0 0 0.0471 0 0 0.0471

0 0 0 0 0 0 ]

The code is:

for i = 1:L

A = - (1 / 2) * ( ( 2 * i - 1 ) / L) * SR

end

the result is. For each iteration a block.

A{1} = [ - 0.0157 - 0.0188 - 0.0078 0 0 0

- 0.0157 0 0 -0.0078 -0.0157 0

0 0 0 -0.0078 0 -0.0078

0 - 0.0188 0 0 -0.0157 0

0 0 - 0.0078 0 0 -0.0078

0 0 0 0 0 0 ]

A{2} = [ -0.0471 -0.0565 -0.0235 0 0 0

-0.0471 0 0 -0.0235 -0.0471 0

0 0 0 -0.0235 0 -0.0235

0 -0.0565 0 0 -0.0471 0

0 0 -0.0235 0 0 -0.0235

0 0 0 0 0 0 ]

A{3} = [ -0.0784 -0.0941 -0.0392 0 0 0

-0.0784 0 0 -0.0392 -0.0784 0

0 0 0 -0.0392 0 -0.0392

0 -0.0941 0 0 -0.0784 0

0 0 -0.0392 0 0 -0.0392

0 0 0 0 0 0]

The question is, how can I save a total array with all those three blocks?..

As the following configuration.

A = [ A{1} A{2} A{3} ]

Answer 1

[Torsten K]

Here a variant with a 3d-matrix:

clearvars;
A = zeros(6,6,3)
L = 3;
SR = [ 0.0941    0.1129    0.0471         0               0                0
          0.0941         0             0          0.0471     0.0941            0
                0            0             0           0.0471         0            0.0471
                0        0.1129         0              0          0.0941           0
                0            0         0.0471          0              0            0.0471
                0            0              0              0              0                 0 ]

for i = 1:L
    A(:,:,i) =  - (1 / 2) * ( ( 2 * i  -  1 ) / L) * SR 
end

Answer 2

[Torsten K]

Maybe a cell-array is what you are looking for?

A = cell(3,1);
L = 3;
SR = [ 0.0941    0.1129    0.0471         0               0                0
          0.0941         0             0          0.0471     0.0941            0
                0            0             0           0.0471         0            0.0471
                0        0.1129         0              0          0.0941           0
                0            0         0.0471          0              0            0.0471
                0            0              0              0              0                 0 ]

for i = 1:L
    A{i,1} =  - (1 / 2) * ( ( 2 * i  -  1 ) / L) * SR 
end

[Dany]

It could be, yes! .. I am only wanting to obtain the matrix, with all its iterations stored in another matrix variable.

[Dany]

I have the following warning. ("Unable to perform assignment because brace indexing is not supported for variables of this type")

Applying this code:

for i = 1:L

A{i,1} = - (1 / 2) * ( ( 2 * i - 1 ) / L) * SR

end

[madhan ravi]

A = cell(3, 1); % this is missing from your copied code, COPY IT properly

[Dany]

in other words, I want to get this total matrix !!

A = [

-0.0157 -0.0188 -0.0078 0 0 0 -0.0471 -0.0565 -0.0235 0 0 0 -0.0784 -0.0941 -0.0392 0 0 0

-0.0157 0 0 -0.0078 -0.0157 0 -0.0471 0 0 -0.0235 -0.0471 0 -0.0784 0 0 -0.0392 -0.0784 0

0 0 0 -0.0078 0 -0.0078 0 0 0 -0.0235 0 -0.0235 0 0 0 -0.0392 0 -0.0392

0 -0.0188 0 0 -0.0157 0 0 -0.0565 0 0 -0.0471 0 0 -0.0941 0 0 -0.0784 0

0 0 -0.0078 0 0 -0.0078 0 0 -0.0235 0 0 -0.0235 0 0 -0.0392 0 0 -0.0392

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ]

Answer 3

[Dany]

Thank you, so much!!!

[Torsten K]

You are welcome! :)