Problems using Linear Regression and syntax

Question

Dylan 2023-2-24

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https://ww2.mathworks.cn/matlabcentral/answers/1918540-problems-using-linear-regression-and-syntax

评论： Torsten 2023-2-25

Can I get some help getting this code to run?

I am trying to use linear regression and the golden search method to find the constants in the Antoine equation. I think my logic is correct -- I just have issues with MATLAB's syntax. Any help is appreciated.

*************************************

function sumSquares = residuals(T,P,C)
Y = log(P);
X = [ones(size(T)), 1./(T+C)]; 
AB = X\Y; % complete the code
A = AB(1);
B = AB(2);
sumSquares = sum((log(P) - A - (B./(T+C))).^2);
end
********************************************
function ABC = fitAntoine(T,P)
T = [273.15; 300.00;   310.00;   320.00;   330.00;  340.00;  350.00;    360.00;    370.00;    380.00;    390.00;   400.00];
P = [728.56; 3911.92; 6820.02; 11205.84; 18049.22; 28158.07; 42655.42; 63062.41; 91129.51; 129027.61; 179180.44; 244541.42];
Ca = -10;
Cb = 10;
Cbeta = Cb - 0.8*(Cb-Ca);
Calpha = Ca + 0.8 * (Cb-Ca);
fCalpha = residuals(T, P, Calpha);
fCbeta = residuals(T, P, Cbeta);
while abs(fCalpha-fCbeta) >sqrt(eps('double'))
    
    Cbeta = Cb - 0.8*(Cb-Ca);
    Calpha = Ca + 0.8 * (Cb-Ca);
    fCalpha = residuals(T, P, Calpha);
    
    fCbeta = residuals(T, P, Cbeta);
    
    if fCbeta >= fCalpha
        Ca = Cbeta;
        C = min(Ca, Cb);
    else 
        Cb = Calpha;
        C = min(Ca, Cb);
    end
    
end
Y = log(P);
X = [ones(size(T)), 1./(T+C)]; 
AB = X\Y; % complete the code
A = AB(1);
B = AB(2);
ABC = [A, B, C]; % complete the code
end

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John D'Errico 2023-2-24

You are not doing something overtly bad, but it is a poor idea to write your own code to do that which is already done better using existing tools. For example, use polyfit to perform the linear regression, combined with a tool like fminbnd to minimize the sum of squares, and therefore to solve for C.

Essentially, you could write much of what you did here, in just a few short lines.

Dylan 2023-2-24

I appreciate the suggegtions, Ill try fiddling around with those functions. Currently I'm a student so I'm working off the professors example, and many of the students have less experience with MatLAB than myself, so I'm sure he wants us to understand the workflow on a more fundamental level. As it stands though I cannot get my output to match the professors. That being

ABC =

1.0e+03 *

0.0236

-4.1789

-0.0281

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Answer 1

Askic V 2023-2-24

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https://ww2.mathworks.cn/matlabcentral/answers/1918540-problems-using-linear-regression-and-syntax#answer_1179485

在 MATLAB Online 中打开

Can you clarify what exactly is the problem with Matlab syntax?

Here is what is obtained if your code is executed:

clear

clc

T = [273.15; 300.00; 310.00; 320.00; 330.00; 340.00; 350.00; 360.00; 370.00; 380.00; 390.00; 400.00];

P = [728.56; 3911.92; 6820.02; 11205.84; 18049.22; 28158.07; 42655.42; 63062.41; 91129.51; 129027.61; 179180.44; 244541.42];

Ca = -10;

Cb = 10;

Cbeta = Cb - 0.8*(Cb-Ca);

Calpha = Ca + 0.8 * (Cb-Ca);

fCalpha = residuals(T, P, Calpha);

fCbeta = residuals(T, P, Cbeta);

while abs(fCalpha-fCbeta) >sqrt(eps('double'))

Cbeta = Cb - 0.8*(Cb-Ca);

Calpha = Ca + 0.8 * (Cb-Ca);

fCalpha = residuals(T, P, Calpha);

fCbeta = residuals(T, P, Cbeta);

if fCbeta >= fCalpha

Ca = Cbeta;

C = min(Ca, Cb);

else

Cb = Calpha;

C = min(Ca, Cb);

end

Y = log(P);

X = [ones(size(T)), 1./(T+C)];

AB = X\Y; % complete the code

A = AB(1);

B = AB(2);

ABC = [A, B, C] % complete the code

ABC = 1×3

1.0e+03 * 0.0245 -4.6973 -0.0100

p1 = exp(A+B./(C+T)); % calculate new p values

plot(T,p1)

hold on

plot(T,P,'o')

legend('Calculated coefficients', 'Original data')

function sumSquares = residuals(T,P,C)

Y = log(P);

X = [ones(size(T)), 1./(T+C)];

AB = X\Y; % complete the code

A = AB(1);

B = AB(2);

sumSquares = sum((log(P) - A - (B./(T+C))).^2);

end

What issues are you experiencing?

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Dylan 2023-2-24

Well it looks like C is just approaching the bounds initially set by Cb whenever I change it. The syntax issues I'm having are related to the examples I'm using to build this code. Specifically naming conventions when dont you have to have some sort of header at the top of the code. Like what is the difference b/t

function [m b] = regressionPractice(x, y)

and

function sumsquares = residualsPractice(T, P, C)

Also why werent the variables I had saved in the workspace not showing up when I entered the loop?

I am defiitely missing some base knowledge that resricts me from fully utilizing MatLAB. Any good resources to get practice with that kind of stuff.

This is the professors example output.

ABC =

1.0e+03 *

0.0236

-4.1789

-0.0281

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Answer 2

Torsten 2023-2-24

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1918540-problems-using-linear-regression-and-syntax#answer_1179490

编辑：Torsten 2023-2-25

在 MATLAB Online 中打开

For the Antoine equation, you usually work with log10:

T = [273.15; 300.00; 310.00; 320.00; 330.00; 340.00; 350.00; 360.00; 370.00; 380.00; 390.00; 400.00];

P = [728.56; 3911.92; 6820.02; 11205.84; 18049.22; 28158.07; 42655.42; 63062.41; 91129.51; 129027.61; 179180.44; 244541.42];

Ca = -10;

Cb = 10;

Cbeta = Cb - 0.8*(Cb-Ca);

Calpha = Ca + 0.8 * (Cb-Ca);

fCalpha = residuals(T, P, Calpha);

fCbeta = residuals(T, P, Cbeta);

while abs(fCalpha-fCbeta) >sqrt(eps('double'))

Cbeta = Cb - 0.8*(Cb-Ca);

Calpha = Ca + 0.8 * (Cb-Ca);

fCalpha = residuals(T, P, Calpha);

fCbeta = residuals(T, P, Cbeta);

if fCbeta >= fCalpha

Ca = Cbeta;

C = min(Ca, Cb);

else

Cb = Calpha;

C = min(Ca, Cb);

end

Y = log10(P);

X = [ones(size(T)), 1./(T+C)];

AB = X\Y; % complete the code

A = AB(1);

B = AB(2);

% Do nonlinear regression

fun = @(x,xdata) 10.^(x(1)+x(2)./(xdata+x(3)));

sol = lsqcurvefit(fun,[A B C],T,P)

Local minimum possible. lsqcurvefit 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 = 1×3

1.0e+03 * 0.0103 -1.8404 -0.0258

A = sol(1)

A = 10.3063

B = sol(2)

B = -1.8404e+03

C = sol(3)

C = -25.7850

hold on

plot(T,P,'o')

plot(T,10.^(A+B./(T+C)))

hold off

grid on

%*************************************

function sumSquares = residuals(T,P,C)

Y = log10(P);

X = [ones(size(T)), 1./(T+C)];

AB = X\Y; % complete the code

A = AB(1);

B = AB(2);

sumSquares = sum((log10(P) - A - (B./(T+C))).^2);

end

%********************************************

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Torsten 2023-2-25

在 MATLAB Online 中打开

The usual way to solve the fitting problem:

syms A B C
T = sym('T',[12 1]);
P = sym('P',[12,1]);
f = sum((log(P)-(A+B./(T+C))).^2);
dfdA = diff(f,A);
dfdB = diff(f,B);
dfdC = diff(f,C);
Tnum = [273.15; 300.00;   310.00;   320.00;   330.00;  340.00;  350.00;    360.00;    370.00;    380.00;    390.00;   400.00];
Pnum = [728.56; 3911.92; 6820.02; 11205.84; 18049.22; 28158.07; 42655.42; 63062.41; 91129.51; 129027.61; 179180.44; 244541.42];
%sol = vpasolve([subs(dfdA,[T P],[Tnum,Pnum])==0,subs(dfdB,[T P],[Tnum,Pnum])==0,subs(dfdC,[T P],[Tnum,Pnum])==0],[A B C])
dfdA_num = subs(dfdA,[T P],[Tnum,Pnum]);
dfdB_num = subs(dfdB,[T P],[Tnum,Pnum]);
dfdC_num = subs(dfdC,[T P],[Tnum,Pnum]);
fun = matlabFunction([dfdA_num,dfdB_num,dfdC_num]);
format long
sol = fsolve(@(x)fun(x(1),x(2),x(3)),[23.6 -4180 -28.1])
Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
sol = 1×3
1.0e+03 *

   0.023642583992292  -4.179437047451260  -0.028039064786185

Askic V 2023-2-25

在 MATLAB Online 中打开

@Torsten Can you please why it is important to include line:

format long

before calling the solve function?

Torsten 2023-2-25

在 MATLAB Online 中打开

Can you please why it is important to include line:

format long

before calling the solve function?

Not important - I just wanted to see the solution with more digits to compare to the solution of the OP's professor.

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Problems using Linear Regression and syntax

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