Azzera filtri
Azzera filtri

How to index with a changing index

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xiaojuezi
xiaojuezi il 8 Dic 2021
Risposto: DGM il 8 Dic 2021
Hi, I would like to construct a matrix L of size nxn that looks like below:
L = [a,b,c,0,0,0,0,...
0,a,b,c,0,0,0,...
0,0,a,b,c,0,0,...
0,0,0,a,b,c,0,...
...
0,0,0,0,...a,b,c];
I'm trying to do this without the for loop.
i = 2:n-1;
L(:,i-1:i+1) = repmat([a,b,c],n,1);
However, this doesn't give what I want, and I was warned by the complier that i is a non-scalar operator.
Is there a way to fix this?
Thank you very much!

Risposte (1)

DGM
DGM il 8 Dic 2021
You can try something like this:
a = 1;
b = 2;
c = 3;
A = toeplitz([a 0 0 0 0 0 0],[a b c 0 0 0 0]);
A = A(1:end-2,:)
A = 5×7
1 2 3 0 0 0 0 0 1 2 3 0 0 0 0 0 1 2 3 0 0 0 0 0 1 2 3 0 0 0 0 0 1 2 3

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R2021b

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