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least square method.getting error using '\' or linsolve. pls help

조회 수: 1 (최근 30일)
seprienna
seprienna 2015년 1월 23일
댓글: Matt J 2015년 1월 23일
any kind folks who can help me out. i'm trying to implement least square method, but i'm getting some error using linsolve and '\' function. please help, and million thanks in advance!
for a =1:400
for P=1:49
o=(P-1)*243;
for r=1:35
if r==35
t=-2;
pA1(1:5,1,r,P)=A(7*r-6+o:7*r+t+o,a);
pB1(1:5,1,r,P)=B(7*r-6+o:7*r+t+o,a);
beta(r,P)=linsolve(pA1,pB1); <--- error
% beta(r,P)=pA1\pB1;
else
t=0;
pA(1:7+t,1,r,P)=A(7*r-6+o:7*r+t+o,a);
pB(1:7+t,1,r,P)=B(7*r-6+o:7*r+t+o,a);
beta(r,P)=linsolve(pA,pB);; <---- error
% beta(r,P)=pA\pB;
end
end
end
  댓글 수: 2
Stephen23
Stephen23 2015년 1월 23일
Can you please tell us what the complete error message is. We cannot read minds, and we also cannot read what is on your computer screen.
seprienna
seprienna 2015년 1월 23일
many thanks for willing to help me out.
The complete message for linsolve is as follows:
Error using linsolve
First and second arguments must be 2D.
Error in modifiedregression (line 34)
beta(r,P)=linsolve(pA,pB);
whereas for \ is as follows:
Error using \
Inputs must be 2-D, or at least one input must be scalar.
To compute elementwise LDIVIDE, use LDIVIDE (.\) instead.
Error in modifiedregression (line 36)
beta(r,P)=pA\pB;

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답변 (1개)

Matt J
Matt J 2015년 1월 23일
편집: Matt J 2015년 1월 23일
Your pA1 and pB1 are 4D arrays. However linsolve(A,B) expects A and B to be 2D matrices, and similarly for '\'.
  댓글 수: 4
seprienna
seprienna 2015년 1월 23일
i'm still trying to figure out a method to overcome the arrays to matrices. any tips from you? your kind reply is greatly appreciated
Matt J
Matt J 2015년 1월 23일
Nope, no tips. We have no way of understanding what your code is trying to do unless you explain it to us.

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