non linear regression problem. fitnlm gives error

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sepideh
sepideh 2023-7-18
评论： sepideh 2023-7-18
Hello
I need to do curve fitting with 2 set of data and I know the model function and initial guess. I used "fitnlm " but it gives error.
would you please check my code and tell me whether I made a mistake or it is the nature of these data.
load FITDATAK
load FITDATAB
tb1= table (FITDATA1,FITDATA)
modelfun= @(b,x) b(1)*x(:,1)^b(2)
beta0= [0.22 -0.5]
md1= fitnlm(tb1,modelfun,beta)
it gives me this errors :
Error using nlinfit>checkFunVals
The function you provided as the MODELFUN input has returned Inf or NaN values.
Error in nlinfit>LMfit (line 596)
if funValCheck && ~isfinite(sse), checkFunVals(r); end
Error in nlinfit (line 284)
[beta,J,~,cause,fullr] = LMfit(X,yw, modelw,beta,options,verbose,maxiter);
Error in NonLinearModel/fitter (line 1153)
nlinfit(X,y,F,b0,opts,wtargs{:},errormodelargs{:});

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采纳的回答

Torsten
Torsten 2023-7-18
编辑：Torsten 2023-7-18
Ran in:
Look at the values of K and B. Do you really want to approximate B by a*K^b ?
format long
K = load("FITDATAK.mat");
K = K.FITDATA1
K = 526×1
1.0e+00 * 0.000000000165736 0.000000000146554 0.000000000138203 0.000000000104830 0.000000000067225 0.000000000053905 0.000000000043810 0.000000000022345 0.000000000019131 0.000000000009961
nnz(isnan(K))
ans =
0
nnz(isinf(K))
ans =
0
B = load("FITDATAB.mat");
B = B.FITDATA
B = 526×1
308500 439800 410200 749100 721800 900400 1072400 1304600 3641600 4397100
nnz(isnan(B))
ans =
0
nnz(isinf(B))
ans =
0
modelfun = @(b) b(1)*K.^b(2)-B;
beta0= [1e17 1];
sol = lsqnonlin(modelfun,beta0,[],[],optimset('MaxFunEvals',100000,'MaxIter',100000))
Local minimum possible. lsqnonlin stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance.
sol = 1×2
1.0e+17 * 0.999999999999999 0.000000000000000
norm(modelfun(sol))
ans =
6.834566371856458e+10
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Torsten
Torsten 2023-7-18
编辑：Torsten 2023-7-18
My code from above gives the result of the linearized and the nonlinear problem in one plot.
The result for the nonlinear problem formulation looks far off, but remember that the plot is in loglog scale where the result of the linearized problem is the optimal one.

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更多回答（1 个）

Matt J
Matt J 2023-7-18
编辑：Matt J 2023-7-18
Ran in:
Your model function can easily generate NaN's and Infs because with fitnlm there is nothing to bound the b parameters (see below).. It would be better to use fit() with the 'power1' model, and with appropriate bounds on b(2).
load FITDATAK
modelfun= @(b,x) b(1)*x^b(2);
modelfun([1,-100],FITDATA1(50))
ans = Inf
modelfun([0,-100],FITDATA1(50))
ans = NaN

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