This question already has an answer here:
I have an (2n-1)-by-1 vector with certain values and I want to obtain an n-n matrix with the diagonals filled using the same value.
Eg. if I have
a = [1; 2; 3; 4; 5];
I want to obtain
A = [[3 4 5];[2 3 4];[1 2 3]]
= 3 4 5
2 3 4
1 2 3
My matrix dimensions are a lot bigger so I'd want this as efficient as possible. I already found following solutions:
n = 3;
A = toeplitz(a);
A = A(1:n,end-n+1:end)
and
A = a(n)*eye(n);
for j=1:n-1
A(1+j:n+1:end-j*n) = a(n-j);
A(j*n+1:n+1:end) = a(n+j);
end
I wonder if there are more efficient ways to obtain this result, keeping in mind that I am working with huge matrices and really need the speed.
Go with:
or
here is what
hankel
does internally (at least in ML R2013a), adapted to this problem:I guess the bsxfun solution and what
thewaywewalk
supposed is the fastest (it's basically the same)