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

1 次查看(过去 30 天)
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
  2 个评论
Stephen23
Stephen23 2015-1-23
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 2015-1-23
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;

请先登录,再进行评论。

回答(1 个)

Matt J
Matt J 2015-1-23
编辑:Matt J 2015-1-23
Your pA1 and pB1 are 4D arrays. However linsolve(A,B) expects A and B to be 2D matrices, and similarly for '\'.
  4 个评论
seprienna
seprienna 2015-1-23
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 2015-1-23
Nope, no tips. We have no way of understanding what your code is trying to do unless you explain it to us.

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 MATLAB 的更多信息

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by