Please help with For Loop

2017 11 6
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更新时间：2017 11 7
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u1 = v(:,1);
a2 = v(:,1)\v(:,2);
u2 = v(:,2) - v(:,1)*a2;
a3 = v(:,[1 2])\v(:,3);
u3 = v(:,3) - v(:,[1 2])*a3;
a4 = v(:,[1 2 3])\v(:,4);
u4 = v(:,4) - v(:,[1 2 3])*a4;
% Structural residuals and covariance
b02 = [ 1 0 0 0;
-a2(1) 1 0 0;
-a3(1) -a3(2) 1 0;
-a4(1) -a4(2) -a4(3) 1 ];
u = [u1 u2 u3 u4 ];
Please advise how can I estimate a , u and b with for loop...in case I have many more variables then just four.

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Walter Roberson
Walter Roberson 2017-11-6

0 个投票

Index your variables instead of using fixed variable name. Use cell arrays.
a{K} = v(:,K-1)\v(:,K);
u{K} = v(:,K) - v(:,K-1)*a{K};

6 个评论
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Raisul Islam
Raisul Islam 2017-11-6
Are you sure Walter? Each v is a 5000 by 1 vector. And how will I create the lower triangular matrix with loop?
Raisul Islam
Raisul Islam 2017-11-6
OK walter. for a{k} its fine, because a returns scaler. for 32 column that's fine. But for u , its not fine. The loop returns scaler. Where the loop is supposed to return vectors of size m by n. Note, v is sized m by n.
Walter Roberson
Walter Roberson 2017-11-6
u(:,1) = v(:,1);
for K = 2 : size(v,2)
a(K) = v(:,K-1)\v(:,K);
u(:,K) = v(:,K) - v(:,K-1)*a(K);
end
Raisul Islam
Raisul Islam 2017-11-6
Thank you so much. That worked easily. Can you help me with the final part. How can I make the lower triangular matrix with a loop? Thnx
Walter Roberson
Walter Roberson 2017-11-6
Correction:
ncol = size(v,2);
u(:,1) = v(:,1);
a = zeros(ncol, ncol);
for K = 2 : size(v,2)
a(1:K-1,K) = v(:,1:K-1)\v(:,K);
u(:,K) = v(:,K) - v(:,1:K-1)*a(1:K-1,K);
end
b = diag(ones(1,ncol)) - tril(a.');
Raisul Islam
Raisul Islam 2017-11-7
thank you so much

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