Linear best fit model

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Aniruddha Das
Aniruddha Das el 14 de Jul. de 2022
Comentada: William Rose el 15 de Jul. de 2022
Hello Community,
I have a query about best fit model.
From a large data set, I could generate a linear best fit line, which has a specific slope.
Now, now I want to compare how deviated this best fit is from another line having a different slope.
How can I force a specific slope value to draw another fit to this dataset?
Thanks in advance,
Ani
  5 comentarios
Adam Danz
Adam Danz el 14 de Jul. de 2022
Thanks for the follow-up. Looks like @William Rose nailed it.
William Rose
William Rose el 15 de Jul. de 2022
@Adam Danz, thank you. Means a lot, coming from you!

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Respuestas (1)

William Rose
William Rose el 14 de Jul. de 2022
Here is an example in which the fixed slope is 1.3 and the best fit slope is approximately 1.0.
x=0:1:100;
N=length(x); %number of data points
y=x+5*randn(1,N); %noisy data
% Find best-fit straight line
P = polyfit(x,y,1);
y1=P(1)*x+P(2);
rmse1=sqrt(sum((y-y1).^2)/N);
% Fit straight line with fixed slope
% The straight line should pass through the mean of the data
fixedslope=1.3;
xm=mean(x); ym=mean(y);
intercept=ym-fixedslope*xm;
y2=fixedslope*x+intercept;
rmse2=sqrt(sum((y-y2).^2)/N);
%% Display results on console
fprintf('Fit 1 (best fit) : y1=%.3fx%+.3f. RMS Error=%.2f\n',P(1),P(2),rmse1)
Fit 1 (best fit) : y1=1.014x-0.274. RMS Error=4.48
fprintf('Fit 2 (fixed slope): y2=%.3fx%+.3f. RMS Error=%.2f\n',fixedslope,intercept,rmse2)
Fit 2 (fixed slope): y2=1.300x-14.588. RMS Error=9.47
%% Plot results
figure; plot(x,y,'rx',x,y1,'-g',x,y2,'-b')
xlabel('X'); ylabel('Y'); grid on; legend('data','fit 1','fit 2')

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