Nonlinear data-fitting
조회 수: 2 (최근 30일)
이전 댓글 표시
I have the data below, where y=variables(:,2), x=variables(:,1) and t=length(x).
Do you have any idea how to get e better fitting in mdl2?
myfun=@(b,x) b(1)+b(2)*t+b(3)*t.^2;
InitGuess=[8.2075e+05 1 1];
mdl1=fitnlm(t,y,myfun,InitGuess);
YY=feval(mdl1,t);
figure,plot(t,1-YY./y'-b',t,0*t,'-r'), title('Relativ error ')
figure, plot(t,y,'-g',t,YY,'-b'), title('Data and model'), legend('Data','Model')
%%
Z=y-YY;
figure,plot(Z)
X=[t x];
f1=@(a,X) sin(a*X);
f2=@(a,X) cos(a*X);
myfun=@(b,X) b(1)*f1(2*pi/30.5,X(:,1)).*X(:,2) + b(2)*f2(2*pi/31.5,X(:,1)) + ...
+ b(3)*f2(pi/2.3,X(:,1))+ b(4)*f1(pi/2.3,X(:,2)).*X(:,1)+b(5)*X(:,2).*X(:,1) +b(6)*X(:,2)+b(7)*X(:,1);
InitGuess=[-7.6342e-08 1 1 1 1 1 1];
mdl2=fitnlm(X,Z,myfun,InitGuess)
ZZ=feval(mdl2,X);
댓글 수: 2
Walter Roberson
2020년 1월 29일
myfun=@(b,X) b(1)*f1(2*pi/30.5,X(:,1)).*X(:,2) + b(2)*f2(2*pi/31.5,X(:,1)) + ...
+ b(3)*f2(pi/31.5,X(:,1))+ b(4)*f1(pi/2.3,X(:,2)).*X(:,1)+b(5)*X(:,2).*X(:,1) +b(6)*X(:,2)+b(7)*X(:,1);
Has subexpression
b(2)*f2(2*pi/31.5,X(:,1)) + ...
+ b(3)*f2(pi/31.5,X(:,1))
The f2 parts are the same so that is (b(2)+b(3)) times the f2 part. You would then combine b(2)+b(3) in to a single parameter.
Or is there a mistake in the formula?
채택된 답변
Walter Roberson
2020년 1월 30일
The second system of equations has no constant terms, so the model fits perfectly if you set all of the parameters to 0.
This is not just an accident: you can do a calculus analysis of the sum of squares of the entries and show that all zeroes is the only critical point.
댓글 수: 6
Walter Roberson
2020년 2월 3일
In the function handle you have + 2*X(:,2) . In the line
A* [b1; b2; b3; b4; b5; b6; b7] ==2*X(:,2);
you have 2*X(:,2) on the right hand side: if you were to bring it to the left hand side to have an equation equal to 0, then it would correspond to - 2*X(:,2) rather than to + 2*X(:,2)
B = A \ 2*X(:,2);
That corresponds to - 2*X(:,2) not to + 2*X(:,2)
filestruct = load('Variables.mat');
variables = filestruct.Data1Rm;
y = variables(:,2);
x = variables(:,1);
t = length(x);
X=[(1:t).', x];
myfun2 =@(b,X) b(1)*f1(2*pi/30.5,X(:,1)).*X(:,2) + b(2)*f2(2*pi/31.5,X(:,1)) + ...
+ b(3)*f2(pi/2.3,X(:,1))+ b(4)*f1(pi/2.3,X(:,2)).*X(:,1)+b(5)*X(:,2).*X(:,1) +b(6)*X(:,2)+b(7)*X(:,1) + 2*X(:,2)
b = sym('b',[1 7]);
MF = myfun2(b,X);
[A,B] = equationsToMatrix(MF,b);
fit_coefficients = double(A)\double(B);
sum(myfun2(fit_coefficients,X).^2)
ans =
1.24829349642173e-26
That is a pretty good fit.
Note that your fitting function makes no reference to y. Your myfun functions are predicting x not y.
Predicting y might be something like
filestruct = load('Variables.mat');
variables = filestruct.Data1Rm;
y = variables(:,2);
x = variables(:,1);
t = length(x);
X = [(1:t).', x, y];
myfun2 = @(b,X) b(1)*f1(2*pi/30.5,X(:,1)).*X(:,2) + b(2)*f2(2*pi/31.5,X(:,1)) + ...
+ b(3)*f2(pi/2.3,X(:,1))+ b(4)*f1(pi/2.3,X(:,2)).*X(:,1)+b(5)*X(:,2).*X(:,1) +b(6)*X(:,2)+b(7)*X(:,1) - X(:,3)
b = sym('b',[1 7]);
MF = myfun2(b, X);
[A,B] = equationsToMatrix(MF, b);
fit_coefficients = double(A)\double(B);
fit_coefficients =
-1231.92429160698
20370.1016972963
-16.9180535390798
873.045987898082
-465.9789817208
54259.3096584326
6265.66604056945
ypred = myfun2(fit_coefficients,X) + X(:,3);
scatter(x, y, 'b');
hold on
scatter(x, ypred, 'r');
hold off
legend({'original', 'predicted'})
추가 답변 (1개)
Hiro Yoshino
2020년 1월 29일
I just wonder if this is a linear model?
(1) Is b a coefficient vector?
(2) Do you need to estimate "a" too?
if (1) yes, (2) no, then this is a linear model and you can solve this analytically.
댓글 수: 4
Hiro Yoshino
2020년 1월 30일
sorry for late.
Well, these are linear models, i.e., you don't need to run optimization to obtain parameters.
The solutions are analytically calculated.
As long as the parameters $$\mathbf{b}$$ are linear with respect to the given data $$X$$, the problem is called "linear problem". The solution can be given by
.
in matlab, fitlm is the one you should apply to this problem.
참고 항목
카테고리
Help Center 및 File Exchange에서 Descriptive Statistics에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!