2D data fitting - Surface

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Patrick 2018-10-12

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https://ww2.mathworks.cn/matlabcentral/answers/423673-2d-data-fitting-surface

评论： Patrick 2018-10-12

fit2d.m

在 MATLAB Online 中打开

Dear all,

I wanted to adapt the post 2D data fitting - Surface that uses lsqcurvefit to fit data defined on a 2D grid and use instead nlinfit and fitnlm.

For nlinfit replacing the following

B = lsqcurvefit(surfit, [0.5 -0.5 -0.5], XY, z, [0 -10 -10], [1 10 10])

with

flatten = @(X) X(:);
Surfit = @(B,XY)  flatten(surfit(B,XY));
B = nlinfit(XY,flatten(z),Surfit,[0.5 -0.5 -0.5],statset('Display','final'))

seems to works fine even though results differ slightly but I can't figure out how to do the same with fitnlm. I tried

flatten = @(X) X(:);
Surfit = @(B,XY)  flatten(surfit(B,XY));
B = fitnlm(XY,flatten(z),Surfit,[0.5 -0.5 -0.5],'Options',statset('Display','final','Robust','On'))

but I get the following error

Error using classreg.regr.FitObject/assignData (line 140)
All predictor and response variables must be vectors or matrices.
Error in NonLinearModel.fit (line 1417)
                model =
                assignData(model,X,y,weights,[],model.Formula.VariableNames,exclude);
Error in fitnlm (line 99)
model = NonLinearModel.fit(X,varargin{:});
Error in fit2d (line 69)
B = fitnlm(XY,flatten(z),Surfit,[0.5 -0.5 -0.5],...

Any suggestion on how to proceed most welcome!

Many thanks, Patrick

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Patrick 2018-10-12

在 MATLAB Online 中打开

fit2d.m

I think I found a solution using a table

tbl = table(flatten(XY(:,:,1)),flatten(XY(:,:,2)),flatten(z));
Surfit = @(B,XY)  B(1)*exp(B(2).*XY(:,1)) + (1 - exp(B(3).*XY(:,2)));
nlm = fitnlm(tbl,Surfit,[0.5 -0.5 -0.5],'Options',statset('Display','final','Robust','On'));
nlm.Coefficients{:,1}'

But they seem to provide quite different answers with these data as seen in the output of the updated code attached

      Local minimum possible.
      lsqcurvefit stopped because the final change in the sum of squares relative to 
      its initial value is less than the default value of the function tolerance.
      B =
          0.6343   -0.4053   -0.2029
      Iterations terminated: relative change in SSE less than OPTIONS.TolFun
      B =
          0.6340   -0.4044   -0.2026
      Iterations terminated: relative norm of the current step is less than OPTIONS.TolX
      ans =
          0.1564   -0.2730   -0.2366