slicing matrix in efficient way

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ARN
ARN 2020 年 6 月 5 日
回答済み: James Tursa 2020 年 6 月 5 日
I have a 120*1 matrix and i want to divide into 3 matrix x,y,z where first 10 elements goes to x, next 10 to y and next 10 to z and then next 10 elements to x as the second column. I did that
x = [1:10; 31:40; 61:70; 91:100] %=[10x4]
y = [11:20;41:50;71:80;101:.110] % =[10x4]
z= [21:30; 51:60; 81:90;111:120] %=[10x4]
Is there any efficient way to do that?

採用された回答

Rik
Rik 2020 年 6 月 5 日
Something like this should do the trick:
a=1:120;
b=reshape(a,10,3,[]);
x=squeeze(b(:,1,:));
y=squeeze(b(:,2,:));
z=squeeze(b(:,3,:));

その他の回答 (2 件)

madhan ravi
madhan ravi 2020 年 6 月 5 日
Matrix = reshape(1:120,10,[]);
x = Matrix(:,1:3:end)
y = Matrix(:,2:3:end)
z = Matrix(:,3:3:end)

James Tursa
James Tursa 2020 年 6 月 5 日
Another way:
a = 1:120;
r = reshape(a,30,[]);
x = r( 1:10,:);
y = r(11:20,:);
z = r(21:30,:);

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