Error Using fitnlm Function

Question

Ricardo Gutierrez 2021-3-19

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/777957-error-using-fitnlm-function

回答： Star Strider 2021-3-19

function regresion_parametros_secador
clc, clear, clear all
TR  =[0.422103409 0.42981426 0.437540563 0.445266867 0.45299317 0.460719473 0.468445777 0.47617208 0.483898384 0.491624687 0.499350991 0.507077294 0.514803597 0.522529901 0.530256204 0.537982508 0.545708811 0.553435115 0.561161418 0.568887721 0.576614025 0.592066632 0.607519238 0.622971845 0.638424452 0.653877059 0.669329666 0.684782273 0.70023488 0.715687486 0.731140093 0.7465927 0.762045307 0.777497914 0.792950521 0.808403128 0.823855734 0.839308341 0.854760948 0.870213555 0.885666162 0.901118769 0.916571376 0.932023982 0.947476589 0.962929196 0.978381803 0.99383441 0.99993819];
HFG =[2500.9 2489.1 2477.2 2465.3 2453.5 2441.7 2429.8 2417.9 2406.0 2394.0 2382.0 2369.8 2357.6 2345.4 2333.0 2320.6 2308.0 2295.3 2282.5 2269.5 2256.4 2229.7 2202.1 2173.7 2144.2 2113.7 2081.9 2048.8 2014.2 1977.9 1939.7 1899.7 1857.3 1812.7 1765.4 1715.1 1661.6 1604.4 1543.0 1476.7 1404.6 1325.7 1238.4 1140.1 1027.3 892.7 719.8 443.8 0.0];
modelfun = @(p,HFG) (exp((p(1)+p(2)*(log(1./TR))^0.1+p(3)/TR.^2+p(4)/TR.^3+p(5)/TR.^4)^0.5));
beta0    = [0 53.63746882 0.01601773 0.031311254 0.071298912];
mdl      = fitnlm(TR,HFG,modelfun,beta0)
end

%ERROR MESSAGE%

Error using nlinfit (line 205)

Error evaluating model function

'@(p,HFG)(exp((p(1)+p(2)*(log(1./TR))^0.1+p(3)/TR.^2+p(4)/TR.^3+p(5)/TR.^4)^0.5))'.

Error in NonLinearModel/fitter (line 1123)

nlinfit(X,y,F,b0,opts,wtargs{:},errormodelargs{:});

Error in classreg.regr.FitObject/doFit (line 94)

model = fitter(model);

Error in NonLinearModel.fit (line 1430)

model = doFit(model);

Error in fitnlm (line 94)

model = NonLinearModel.fit(X,varargin{:});

Error in regresion_parametros_secador (line 14)

mdl = fitnlm(TR,HFG,modelfun,beta0)

Caused by:

Error using ^

One argument must be a square matrix and the other must be a scalar. Use POWER (.^) for elementwise

power.

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Answer 1

Star Strider 2021-3-19

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/777957-error-using-fitnlm-function#answer_652932

在 MATLAB Online 中打开

Use element-wise operations on every applicable operation in ‘modelfun’ and transpose ‘modelfun’ to return a column vector, and it works:

TR  =[0.422103409 0.42981426 0.437540563 0.445266867 0.45299317 0.460719473 0.468445777 0.47617208 0.483898384 0.491624687 0.499350991 0.507077294 0.514803597 0.522529901 0.530256204 0.537982508 0.545708811 0.553435115 0.561161418 0.568887721 0.576614025 0.592066632 0.607519238 0.622971845 0.638424452 0.653877059 0.669329666 0.684782273 0.70023488 0.715687486 0.731140093 0.7465927 0.762045307 0.777497914 0.792950521 0.808403128 0.823855734 0.839308341 0.854760948 0.870213555 0.885666162 0.901118769 0.916571376 0.932023982 0.947476589 0.962929196 0.978381803 0.99383441 0.99993819];
HFG =[2500.9 2489.1 2477.2 2465.3 2453.5 2441.7 2429.8 2417.9 2406.0 2394.0 2382.0 2369.8 2357.6 2345.4 2333.0 2320.6 2308.0 2295.3 2282.5 2269.5 2256.4 2229.7 2202.1 2173.7 2144.2 2113.7 2081.9 2048.8 2014.2 1977.9 1939.7 1899.7 1857.3 1812.7 1765.4 1715.1 1661.6 1604.4 1543.0 1476.7 1404.6 1325.7 1238.4 1140.1 1027.3 892.7 719.8 443.8 0.0];
modelfun = @(p,HFG) (exp(sqrt(p(1)+p(2).*(log(1./TR)).^0.1+p(3)./TR.^2+p(4)./TR.^3+p(5)./TR.^4))).';
beta0    = [0 53.63746882 0.01601773 0.031311254 0.071298912];
mdl      = fitnlm(TR,HFG,modelfun,beta0)

producing:

mdl = 
Nonlinear regression model:
    y ~ F(p,HFG)
Estimated Coefficients:
          Estimate      SE        tStat       pValue  
          ________    _______    _______    __________
    b1    -5.3764     0.53639    -10.023    6.2155e-13
    b2     80.604      1.2895      62.51    1.2803e-44
    b3    -12.018      1.5173    -7.9205    5.2486e-10
    b4     6.8052      1.0612      6.413    8.3294e-08
    b5    -1.1397      0.2096    -5.4375    2.2468e-06
Number of observations: 49, Error degrees of freedom: 44
Root Mean Squared Error: 12.4
R-Squared: 1,  Adjusted R-Squared 1
F-statistic vs. zero model: 2.6e+05, p-value = 3.84e-97