- the only variable with an index in position 2 is U. Apparently U only has 2 columns, and n is a value >2, hence the error.
Index in position 2 exceeds array bounds (must not exceed 2
3 次查看(过去 30 天)
显示 更早的评论
Background : I am writing a MATLAB algorithm LinearReg that takes as an input the vectors x and y and solves the linear regression problem associated with the data stored in x and y using a version decomposition that I wrote and then plots the data and the obtained line using plot(x,y,’bo’,x,ysol,’g-’). It should output ysol, the vector of y coordinates of the obtained line at the x points.
- First, I wrote the modified algorithm as follow :
function x=QRQ(A,b,n)
[Q1,R1]=qr(A);
c1=Q1'*b;
n=length(c1);
x=backward(R1,c1,n);
end
function x=backward(U,y,n)
x=zeros(n,1);
x(n)=y(n)/U(n,n);
for i=n -1 : -1 : 1
x(i)=(y(i)-U(i,i+1 : n)*x(i+1 : n))/U(i,i);
end
end
- The second and vital part is writing the MATLAB function LinearReg that should solve and plot the graph of the linear function :
function ysol = LinearReg(x,y)
A=[x ones(21,1)];
z=QRQ(A,y,2);
ysol=z(1)*x+z(2);
plot(x,y,'bo',x,ysol,'g-');
end
Running this code using the following data points :
x=[0;0.25;0.5;0.75;1;1.25;1.5;1.75;2;2.25;2.5;2.75;3;3.25;3.5;3.75;4;4.25;4.5;4.75;5];
y=[4;3;7;7;1;4;4;6;7;7;2;6;6;1;1;4;9;3;5;2;7];
have resulted in an error stating that Index in position 2 exceeds array bounds (must not exceed 2). I am afraid that my mistake is something trivial or stupid. I hope someone assists me in functioning this code. Thank you very much!
0 个评论
采纳的回答
Cris LaPierre
2021-2-12
编辑:Cris LaPierre
2021-2-12
The full error message is
Index in position 2 exceeds array bounds (must not exceed 2).
Error in untitled>backward (line 12)
x(n)=y(n)/U(n,n);
Error in untitled>QRQ (line 8)
x=backward(R1,c1,n);
Error in untitled>LinearReg (line 20)
z=QRQ(A,y,2);
The line causing the error is x(n)=y(n)/U(n,n);
Using the debugger, I see that U is a 21x2 array, and n has a value of 21.
U = rand(21,2);
U(3,3)
更多回答(0 个)
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Matrix Indexing 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!