Incorrect y intercept using fitlm

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Sarvesh
Sarvesh 2025-2-28
评论： Sarvesh 2025-2-28
Hi and thanks in advance.
I'm currently working on creating basic plots and obtaining the equation of the best-fit line using 'fitlm' in MATLAB. However, I've observed that the y-intercept of the equation doesn't seem to match the value displayed on the graph. The best-fit line visually crosses around approximately 1 on the y-axis, but the displayed value is approximately -0.21 from the equation. Is there a way to correct this discrepancy to match the displayed value, or is there something else happening that I am unaware of
thanks
% Example data (replace with your actual data)
xData = [1, 2, 3, 4, 5, 6, 7];
yData = [1.1, 1.5, 4, 3.8, 6, 6.2, 8];
% Fit a linear model to the data
mdl = fitlm(xData, yData);
% Plotting the data and the best fit line
figure;
scatter(xData, yData, 'filled', 'o'); % Scatter plot of data points
hold on;
plot(mdl); % Plotting the best fit line
hold off;
xlabel('X Data');
ylabel('Y Data');
title('Best Fit Line');
grid on;
% Display the equation of the line
disp(['Equation of the best fit line: y = ' num2str(mdl.Coefficients.Estimate(2)) 'x + ' num2str(mdl.Coefficients.Estimate(1))]);
  2 个评论
Stephen23
Stephen23 2025-2-28
编辑：Stephen23 2025-2-28
" The best-fit line visually crosses around approximately 1 on the y-axis"
Yes, but at x=1, so it has nothing to do with the y-intercept (which is defined as being at x=0).
"...but the displayed value is approximately -0.21 from the equation."
Which seems about correct.
"Is there a way to correct this discrepancy to match the displayed value"
The only discrepancy that I can see is that you are confusing x=1 for x=0.
"or is there something else happening that I am unaware of"
Plot the line of best fit at x=0 and see what value it has.

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采纳的回答

Sam Chak
Sam Chak 2025-2-28
Ran in:
Extend the line, and you will see the y-intercept at .
% Example data (replace with your actual data)
xData = [1, 2, 3, 4, 5, 6, 7];
yData = [1.1, 1.5, 4, 3.8, 6, 6.2, 8];
% Fit a linear model to the data
mdl = fitlm(xData, yData)
mdl =
Linear regression model: y ~ 1 + x1 Estimated Coefficients: Estimate SE tStat pValue ________ _______ ________ __________ (Intercept) -0.21429 0.50341 -0.42567 0.68805 x1 1.1464 0.11257 10.185 0.00015658 Number of observations: 7, Error degrees of freedom: 5 Root Mean Squared Error: 0.596 R-squared: 0.954, Adjusted R-Squared: 0.945 F-statistic vs. constant model: 104, p-value = 0.000157
% Plotting the data and the best fit line
figure;
scatter(xData, yData, 'filled', 'o'); % Scatter plot of data points
hold on;
plot(mdl); % Plotting the best fit line
xlabel('X Data');
ylabel('Y Data');
title('Best Fit Line');
grid on;
%% ---- plot the extension ----
m = mdl.Coefficients.Estimate(2);
c = mdl.Coefficients.Estimate(1);
x = linspace(0, 1, 101);
y = m*x + c;
plot(x, y, '--')
xlim([0 7])
hold off;
% Display the equation of the line
disp(['Equation of the best fit line: y = ' num2str(mdl.Coefficients.Estimate(2)) 'x + ' num2str(mdl.Coefficients.Estimate(1))]);
Equation of the best fit line: y = 1.1464x + -0.21429

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