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Simple Matrix operation using for loop

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Nikhil
Nikhil el 3 de Oct. de 2013
Comentada: Azzi Abdelmalek el 3 de Oct. de 2013
Hello all, I want to do following matrix operation, I have two matrices y and errorsample as: y=y=[0.53, 0.6, 0.58, 0.68, 0.85, 0.85];1X6 matrix errorsample=normrnd(0,stdev,1,1e4);1X1e4 matrix
Now I want to create new matrix Y (1e4X6) such that Y(i)(j)=y(j)-errorsample(i)
For that I have written following code: y=[0.53, 0.6, 0.58, 0.68, 0.85, 0.85]; stdev=0.0523; errorsample=normrnd(0,stdev,1,1e4); for i=1:1:1e4; for j=1:1:6; Y(i)(j)=y(j)-errorsample(i); end end
but after running this, I am getting following error: ??? Error: File: Q4.m Line: 23 Column: 5 ()-indexing must appear last in an index expression.
Can anybody please tell me how to debug this problem, It's urgent Thanks in advance,
Nikhil

Respuesta aceptada

Azzi Abdelmalek
Azzi Abdelmalek el 3 de Oct. de 2013
Editada: Azzi Abdelmalek el 3 de Oct. de 2013
Maybe you want
y=[0.53, 0.6, 0.58, 0.68, 0.85, 0.85];
stdev=0.0523;
errorsample=normrnd(0,stdev,1,1e4);
out=zeros(1e4,6);
for i=1:1:1e4;
for j=1:1:6;
out(i,j)=y(j)-errorsample(i);
end
end
disp(out)
  2 comentarios
Nikhil
Nikhil el 3 de Oct. de 2013
Editada: Azzi Abdelmalek el 3 de Oct. de 2013
Hey, Thanks a lot for your quick reply. My next stage is I want to perform regression analysis on each row matrix Y. And I want to store the answers of each regression analysis on matrix form. generally output of regress command is two vectors my code for this is as follows:
x=[0,0.2, 0.4, 0.6, 0.8, 1.0];
errorsample=normrnd(0,stdev,1,1e4);
for i=1:1:1e4;
for j=1:1:6;
Y(i,j)=y(j)-errorsample(i);
end
end
xx=[ones(6,1),x'];
for i=1:1:1e4;
b=regress(Y(i,:)',xx);
Z(i)=(b);
end
The error which I am getting is as follows: ??? In an assignment A(I) = B, the number of elements in B and I must be the same.
Error in ==> Q4 at 29 Z(i)=(b);
kindly help me to debug this, Thanks in advance,
Nikhil
Azzi Abdelmalek
Azzi Abdelmalek el 3 de Oct. de 2013
y=[0.53, 0.6, 0.58, 0.68, 0.85, 0.85];
stdev=0.0523;
errorsample=normrnd(0,stdev,1,1e4);
Y=zeros(1e4,6);
for i=1:1:1e4;
for j=1:1:6;
Y(i,j)=y(j)-errorsample(i);
end
end
x=[0,0.2, 0.4, 0.6, 0.8, 1.0];
xx= [ones(6,1),x'];
Z=zeros(1e4,2);
for i=1:1:1e4;
b=regress(Y(i,:)',xx);
Z(i,:)=b';
end

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