# Question on creating dynamic matrix variables

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Tsz Tsun il 12 Apr 2024
Commentato: Tsz Tsun il 12 Apr 2024
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?
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Stephen23 il 12 Apr 2024
Modificato: Stephen23 il 12 Apr 2024

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### Risposta accettata

Stephen23 il 12 Apr 2024
Modificato: Stephen23 il 12 Apr 2024
"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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##### 5 CommentiMostra 3 commenti meno recentiNascondi 3 commenti meno recenti
Stephen23 il 12 Apr 2024
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 il 12 Apr 2024
Thank you very much for your help!

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### Più risposte (1)

Taylor il 12 Apr 2024
May be worth looking into structures instead of matrices. https://www.mathworks.com/help/matlab/matlab_prog/generate-field-names-from-variables.html
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