Question on creating dynamic matrix variables

Hi all, I would like to know how to write dynamic matrix variables. For example, I have the following:

clear all;
n=11

n = 11

for i=1:n-1
    A1(i,i+1) = 1;
    A1(i+1,i) = 1;
end

A1

A1 = 11x11

0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1

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for i=1:n-2
    A2(i,i+2) = 1;
    A2(i+2,i) = 1;
end

A2

A2 = 11x11

0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0

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for i=1:n-3
    A3(i,i+3) = 1;
    A3(i+3,i) = 1;
end

A3

A3 = 11x11

0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0

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And I would like to make something like this:

for k=1:1:3
  for i=1:n-1

    A[k](i,i+1) = 1;

Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.

    A[k](i+1,i) = 1; 
  end
end

Or alternatively, is there a neat way to do so?

[Stephen23]

https://www.mathworks.com/matlabcentral/discussions/tips/849901-tutorial-why-variables-should-not-be-named-dynamically-eval

Answer 1

[Stephen23]

"Or alternatively, is there a neat way to do so?"

Of course there is: indexing. Either into a numeric array or into a container array (e.g. a cell array).

Indexing is neat, simple, and efficient. Unlike what you are trying to do.

n = 11; % size of each matrix
D = 2:4; % vector of indices of the 1's
C = cell(size(D));
for k = 1:numel(D)
    V = zeros(1,n);
    V(D(k)) = 1;
    C{k} = toeplitz(V);
end

Checking:

C{1}

ans = 11x11

0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1

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C{2}

ans = 11x11

0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0

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C{3}

ans = 11x11

0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0

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[Tsz Tsun]

Thanks a lot! May you explain why D=2:4 instead of D=1:3? I am not too familiar with using cell.

[Stephen23]

"May you explain why D=2:4 instead of D=1:3?"

Because your question shows the 1's starting in row/column 2, 3, and 4 in each matrix respectively.

"I am not too familiar with using cell."

https://www.mathworks.com/help/matlab/cell-arrays.html

https://www.mathworks.com/help/matlab/matlab_prog/access-data-in-a-cell-array.html

[Tsz Tsun]

Sorry, I would like to have half of the 1-entries only, how can I change the code accordingly?

for i=1:n-1
    A1(i,i+1) = 1;
end

[Stephen23]

Either modify the TOEPLITZ call with two vector inputs, or call DIAG:

n = 11; % size of each matrix
D = 1:3; % offset of the diagonal
C = cell(size(D));
for k = 1:numel(D)
    C{k} = diag(ones(1,n-D(k)),D(k));
end
C{1}

ans = 11x11

0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1

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C{2}

ans = 11x11

0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0

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C{3}

ans = 11x11

0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

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[Tsz Tsun]

Thank you very much for your help!

Answer 2

[Taylor]

May be worth looking into structures instead of matrices. https://www.mathworks.com/help/matlab/matlab_prog/generate-field-names-from-variables.html