Creating a matrix having repeating elements

2019 9 4
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更新时间：2019 9 4
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% Dear users,
% Assume that I have a A matrix having the size of 5x5 and in the end;
% I want to get A=[1 -1 0 0 0; 1 -1 0 0 0; 0 1 -1 0 0; 0 1 -1 0 0; 0 0 1 -1 0]
% What is the easiest way to do it?
% Thanks in advance

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Fabio Freschi
Fabio Freschi 2019-9-4
Is this matrix what you want? I don't understand the structure
A =
1 -1 0 0 0
1 -1 0 0 0
0 1 -1 0 0
0 1 -1 0 0
0 0 1 -1 0
Rengin
Rengin 2019-9-4
编辑：Rengin 2019-9-4
Yes, that's right! the element of (1,-1) should shift one column to the right in every odd row. If the row has an even number, it should repeat itself.
Bruno Luong
Bruno Luong 2019-9-4
The easiest way is just type you matrix.
A=[1 -1 0 0 0; 1 -1 0 0 0; 0 1 -1 0 0; 0 1 -1 0 0; 0 0 1 -1 0]
Rengin
Rengin 2019-9-4
编辑：Rengin 2019-9-4
Since I can do it by myself, I am asking obviously something different.
Bruno Luong
Bruno Luong 2019-9-4
OK here is different and still simple:
A=[1 -1 0 0 0; 1 -1 0 0 0; 0 1 -1 0 0; 0 1 -1 0 0; 0 0 1 -1 0]+0
The point is your question is not well formulated
Rengin
Rengin 2019-9-4
Thanks but I was still be able to get my answer. It works perfectly fine.

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 采纳的回答

Andrei Bobrov
Andrei Bobrov 2019-9-4
编辑：Andrei Bobrov 2019-9-4

1 个投票

A = kron(eye(5),[1;1])+kron(diag(-ones(4,1),1),[1;1]);
out = A(1:5,:);
or
n = 1:5;
A = repmat([1,-1,0,0,0],5,1);
out = A(mod(repmat(n,5,1) - ceil(n(:)/2),5)*5 + n(:));

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