# How do I fit a regression equation to find coefficients and exponents?

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I'm currently working with a wide dataset, where I'm using a machine learning technique to select key identifiers (SIMPLS partial least squares). Then I want to use the identifiers and my outcome to create a predictive equation. I've tried a bunch of linear regression tools but they only find the predictor's coefficents, where I am trying to find the coefficients and the exponents. To get around this I'm trying to use 'nlinfit' to force the final equation into the desired form. This is where I'm having an issue, when I run the code I get the following error:

"The function you provided as the MODELFUN input has returned Inf or NaN values."

I've also tried inputting the model in the following form:

modelfun = 'y~(b1 + b2*x1.^b3 + b4*x2.^b5 + b6*x3.^b7)';

For reference my current data set is a 13x16 matrix, at it's largest it will be 24x35 matrix where the last column represents the outcome. Once the variable selection is complete (works without issue) the matrix is reduced to an nx4 matrix

Here is my code:

clear

clc

% Imports data and removed first text column

data = readtable('PMHS PLS Practice.xlsx',"textType","string");

data.SpecimenID = [];

% Splits data into independant and dependant

% variables and normalizes values

X = data(:,1:15);

X = X{:,:};

Xnorm = normalize(X);

Y = data(:,16);

Y = table2array(Y);

Ynorm = normalize(Y);

% Performs Simpls PLS on normalized data returning

% the X scores and percent varience per variable

% for the first 5 latent variables

Cpart = cvpartition(13,"LeaveOut");

[~,~,SCR,~,~,PCTVAR,~,~] = plsregress(Xnorm,Ynorm,5,'cv',Cpart);

X_VAR = PCTVAR(1,:);

Y_VAR = PCTVAR(2,:);

% pareto(X_VAR)

% finds the dependant variable with the largest

% contribution to the first 3 latent variables

[Var1,id1] = max(abs(SCR(:,1)));

[Var2,id2] = max(abs(SCR(:,2)));

[Var3,id3] = max(abs(SCR(:,3)));

% creates a matrix containing the selected variables

X_reg = [data{:,id1} data{:,id2} data{:,id3}];

% Fits the data with a non-linear model with

% initial coefficient guesses of beta0

modelfun = @(b,x) (b(1)+b(2)*x(:,1).^b(3)+b(4)*x(:,2).^b(5)+b(6)*x(:,3).^b(7));

beta0 = ones(1,7);

[coeff] = nlinfit(X_reg,Y,modelfun,beta0);

##### 19 Comments

Matt J
on 17 Nov 2022

I can't provide my raw data because it has identifiable donor health information.

But why 2 .mat files instead of 1.

### Accepted Answer

Matt J
on 16 Nov 2022

Edited: Matt J
on 16 Nov 2022

fminspleas from the File Exchange (Download) fared better, but it looks like a highly ill-conditioned data set:

load('X_reg.mat')

load('Y.mat')

flist={1, @(b,x)x(:,1).^b(1), @(b,x)x(:,2).^b(2) , @(b,x)x(:,3).^b(3)};

[exps,coef]=fminspleas(flist,ones(1,3),X_reg,Y);

[b1,b2,b4,b6]=dealThem(coef)

[b3,b5,b7]=dealThem(exps)

function varargout=dealThem(z)

varargout=num2cell(z);

end

##### 3 Comments

Matt J
on 18 Nov 2022

Matt J
on 20 Nov 2022

Edited: Matt J
on 20 Nov 2022

If you pre-normalize the columns of X_reg, fminspleas gives the same results as Alex and my conditioning test shows a much better condition number on the solution for the coefficients:

load('X_reg.mat')

load('Y.mat')

flist={1, @(b,x)x(:,1).^b(1), @(b,x)x(:,2).^b(2) , @(b,x)x(:,3).^b(3)};

s=max(X_reg);

n=size(Y,1);

[exponents,coefficients]=fminspleas(flist,ones(1,3),X_reg./s,Y);

coefficients(2:end)=coefficients(2:end)./(s.^exponents)';

format longG

exponents,coefficients

A=(X_reg./s).^exponents; A=[ones(n,1),A];

cond(A)

### More Answers (1)

Alex Sha
on 17 Nov 2022

Although the results may seem strange, mathematically speaking, the result below is the best one:

Sum Squared Error (SSE): 875229.002284955

Root of Mean Square Error (RMSE): 259.47120816783

Correlation Coef. (R): 0.945217846153423

R-Square: 0.893436776686916

Parameter Best Estimate Std. Deviation Confidence Bounds[95%]

--------- ------------- -------------- --------------------------------

b1 12988.1117515585 6.29998367140824 [12972.6962468509, 13003.5272562661]

b2 1.533510681096E126 0.070612379218076 [1.533510681096E126, 1.533510681096E126]

b3 109.691046465045 0.167406667907439 [109.281417105381, 110.100675824708]

b4 -97.0000950427331 1048.74886499901 [-2663.19612168365, 2469.19593159819]

b5 -3.07105087006786 14192.7795542942 [-34731.5515429606, 34725.4094412205]

b6 -9.2988265393269E18 1.40462270054153E-15 [-9.2988265393269E18, -9.2988265393269E18]

b7 12.1425640359073 3.30305643552503E-124 [12.1425640359073, 12.1425640359073]

##### 6 Comments

Matt J
on 20 Nov 2022

If there are various solutions as you said, what are the objective function values (SSE)

Probably very similar to what you got. My solution may be local and your solution may be global, but that does not mean the global solution is unique.

Alex Sha
on 20 Nov 2022

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