Problem with Linear Regression

Hi, everyone. I have a problem with Linear Regression. My code is :

knobSetting = [0 1 2 3 4 5]'; 
flowRate = [0.0 2.3 4.2 7.2 8.4 11.9]' ;
flowfit=fit(knobSetting,flowRate,'poly1');

Matlab gives me the following problem :

'fit' requires one of the following:

- Curve Fitting Toolbox

-Predictive Maintenance Toolbox

Error in Untitled2 (line 3)

flowfit=fit(knobSetting,flowRate,'poly1')

Answer 1

[Stephan]

Hi,

this problem can be solved easily by using mldivide. No additional toolboxes are needed:

knobSetting = [0 1 2 3 4 5]';
flowRate = [0.0 2.3 4.2 7.2 8.4 11.9]';
flowfit = knobSetting\flowRate
flowrate_calc = flowfit .* knobSetting
scatter(knobSetting,flowRate,'*r')
hold on
plot(knobSetting,flowrate_calc)
hold off

The result is a least squares fit of your data, where flowfit represents the slope of the linear function. Since your function goes through (0,0) this is the most simple case.

Best regards

Stephan

Answer 2

[Star Strider]

A first-degree polynomial is a linear fit. You do not need any toolboxes to do a linear fit.

Try this:

knobSetting = [0 1 2 3 4 5]'; 
flowRate = [0.0 2.3 4.2 7.2 8.4 11.9]';

B = [knobSetting, ones(size(knobSetting))] \ flowRate;
regfit = [knobSetting, ones(size(knobSetting))] *B;

figure
plot(knobSetting, flowRate, 'p')
hold on
plot(knobSetting, regfit,' -r')
hold off
grid
text(0.6, 9, sprintf('flowRate = %.3f\\cdotknobSetting%.3f', B))

Answer 3

[Image Analyst]

You can simply use the built-in polyfit(x, y, 1) to get the equation of a line fitting your data, then use polyval() to get the fitted y-values at whatever x-locations you want:

% Create and plot existing data.
knobSetting = [0 1 2 3 4 5]'; 
flowRate = [0.0 2.3 4.2 7.2 8.4 11.9]' ;
coefficients = polyfit(knobSetting, flowRate, 1); % Get coefficients of the line.
plot(knobSetting, flowRate, 'bo');
hold on;
% Now get fitted values at the same Knob settings locations and plot them.
fittedValues = polyval(coefficients, knobSetting);
plot(knobSetting, fittedValues, 'r*-', 'MarkerSize', 10);
grid on;
legend('Actual', 'Fitted', 'Location', 'north');

Answer 4

[Claudia Orlando]

Thank you all for your valuable advice.