For loop with horzcat

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Ryosuke Saito
Ryosuke Saito 2020-4-12
评论： Ryosuke Saito 2020-4-12
Why does this not success?? Would someone help me correct the code?
A=[1 5; 3 3; 8 6; 4 8; 9 14; 11 11; 13 12; 15 7; 17 15; 1 20]
B=[]
for i=1:5
for j=1:5
B(i,:)=horzcat(B(i,:),A(i+j-1,:));
end
end
I'd like to get this
B=[1 5 3 3 8 6 4 8 9 14;3 3 8 6 4 8 9 14 11 11; 8 6 4 8 9 14 11 11 13 12; 4 8 9 14 11 11 13 12 15 7;9 14 11 11 13 12 15 7 17 15;11 11 13 12 15 7 17 15 1 20]

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采纳的回答

Matt J
Matt J 2020-4-12
编辑：Matt J 2020-4-12
Another way,
AA=reshape(A.',[],1).';
B=cell(numel(AA),1);
for i=1:2:11
B{i}=AA(i:i+9);
end
B=cell2mat(B)
  1 个评论
Ryosuke Saito
Ryosuke Saito 2020-4-12
Thanks so much! This is exactly what I'd like to do and I found this works!!

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更多回答（1 个）

Matt J
Matt J 2020-4-12
编辑：Matt J 2020-4-12
horzcating in for-loops is bad practice. Better to do,
AA=reshape(A.',[],1).';
B=AA( (1:10)+(0:2:10).' )
  2 个评论
Ryosuke Saito
Ryosuke Saito 2020-4-12
Thanks for your comment. I am sorry that I set up a simple table as the contents of A here. Actual contents of A in the computation I am trying to do is much complicated. So the proposed solution is not useful.
Matt J
Matt J 2020-4-12
My solution is independent of the contents of A.

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