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hi everyone , how can I built a matrix of circularly shifted array in efficient way

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fatema hamodi
fatema hamodi 2018-2-19
编辑： Stephen23 2018-2-19
I have a given initial array of length M , I want to built a matrix of MxM , by circularly shifting this array, example
for array=[2,0,0] the matrix should be matrix=[2,0,0;0,2,0;0,0,2]
my code works but it not efficient for large M I would be happy if someone can gives me an efficient suggestion
my code:
array=[2,0,0]; % given Initial array
M=length(array);
mat=zeros(M,M);%matrix
mat(1,:)=array;
for i=2:M
array=circshift(array(1,1:M),[0 1]);
mat(i,:)=array;
end

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Stephen23
Stephen23 2018-2-19
编辑：Stephen23 2018-2-19
Multiple circshift calls in a loop is not required. One call to toeplitz is simpler:
>> V = [2,0,0];
>> toeplitz(V([1,end:-1:2]),V)
ans =
2 0 0
0 2 0
0 0 2
Or a clearer example:
>> V = [4,3,2,1,0];
>> toeplitz(V([1,end:-1:2]),V)
ans =
4 3 2 1 0
0 4 3 2 1
1 0 4 3 2
2 1 0 4 3
3 2 1 0 4

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Guillaume
Guillaume 2018-2-19
编辑：Guillaume 2018-2-19
To be compared to a loop version, it's very possible that a loop is faster:
array = [2, 0, 0];
shiftidx = hankel(1:numel(array), circshift(1:numel(array), 1))
M = array(shiftidx)
Or, for a shift in the other direction:
array = [2, 0, 0];
shiftidx = toeplitz(1:numel(array), [1, numel(array):-1:2])
M = array(shiftidx.')

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