Help for linear regression

Question

b 2012-5-20

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/38871-help-for-linear-regression

Hi friends, I have a problem in finding the regression by gradient descent. I coded it.

My code is as shown; 
x=[x1 x2 x3 x4 x5];
  y=realoutput;
  alpha=0.001;
  m=length(y);
  n=5;
  theta_vec=zeros(5,1);
  error=zeros(5,1);
  itr=100;
  for max_iter=1:itr
      for j=1:n
          sum_theta_vec=0;
          for i=1:m
              h_theta_v=0;
              for inner_j=1:n
              h_theta=(x.^i)*theta_vec;
              h_theta_v=(h_theta * ones(1,n)).';
              y_v= (y * ones(1,n)).';
              end
              sum_theta_vec= (sum_theta_vec + ((h_theta_v - y_v(i)) * x(:,inner_j).^i));          
          end
          theta_vec=theta_vec(j) - (alpha* 1/m * sum_theta_vec);
      end
      theta_vec
end

But theta_vec occurs as NaN. Why? Can anybody help me?

Thank you very much.

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b 2012-5-20

"inner_j=1:n" means that x values in the last equation take all rows and 1 column in for loop.

Walter Roberson 2012-5-21

You did not code an assignment for "inner_j": you coded a "for" loop. The "for" loop ends after the calculation of y_v, and everything in that loop will be done "n" times. After the loop is finished, the loop index variable will have the value it was last assigned, so inner_j will be "n" not 1:n .

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Answer 1

Walter Roberson 2012-5-20

0
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/38871-help-for-linear-regression#answer_48414

You have m=length(y) . If y is empty then m would be 0. Then when you calculate theta_vec you have 1/m * sum_theta_vec and sum_theta_vec would be 0 because the "for i=1:m" loop would execute 0 times when m is 0. 1/0 * 0 will give you NaN.