creating a large matrix with a pattern

2019 5 16
2 个回答
回答已采纳
更新时间：2019 5 16
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0 个投票

Hi,
I would like to create a 800x200 matrix which looks like:
1 0 0 0 0 0 0 0 0 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 0 0 0 0 0 0 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 1 0 0 0 0 0 0 0 0 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 0 0 0 0 0 0 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 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
To put in in words, every new column should start with the column [1 1 0 1].' when the other is finished.
Tried to do this with toeplitz. However this results in only 1 step change of columns e.g.
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0
This is not what I desire. Does anyone know another way to do this?

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convert_to_metric
convert_to_metric 2019-5-16
编辑：convert_to_metric 2019-5-16

2 个投票

Hi bus14,
Try this:
m=zeros(804,200);
m([1,2,4],:)=1;
m2=reshape(m,800,[]);
m2(:,201)=[];

更多回答（1 个）

Andrei Bobrov
Andrei Bobrov 2019-5-16
编辑：Andrei Bobrov 2019-5-16

1 个投票

out = kron(eye(200),[1;1;0;1]);
or
a = [1;1;0;1;zeros(796,1)];
out = a(mod((1:800)'-(1:4:800),800)+1);

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