Linear best fit model

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Aniruddha Das
Aniruddha Das 2022-7-14
评论： William Rose 2022-7-15
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
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Adam Danz
Adam Danz 2022-7-14
Thanks for the follow-up. Looks like @William Rose nailed it.
William Rose
William Rose 2022-7-15
@Adam Danz, thank you. Means a lot, coming from you!

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回答（1 个）

William Rose
William Rose 2022-7-14
Ran in:
@Aniruddha Das
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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