MATLAB: How to use the regress function for more than 2 predictors

mlrregressx

I need to make a multiple linear regression with 4 predictor, like x1, x2, x3 and x4. So, i discovered the function regress.

In the doc, there is the example bellow:

    load carsmall
  x1 = Weight;
  x2 = Horsepower;    % Contains NaN data
  y = MPG;
  X = [ones(size(x1)) x1 x2 x1.*x2];
  b = regress(y,X)    % Removes NaN data

But, in my case, i have x1, x2, x3 and x4. I don´t know how to use correct for this case, and i don't know how create the array X (showed in the doc of Matlab).

Best Answer

Does this example using 3 variables help?

% Load the data
load carsmall
% Redefine names, to look more like your problem
% Explanatory variables
x1 = Weight;
x2 = Horsepower;
x3 = Displacement;
% Response variable
y = MPG;
% Put all the explanatory variables (including a constant term) into one matrix
X = [ones(size(x1)) x1 x2 x3];
% Estimate the parameters
b = regress(y,X);

Related Solutions

MATLAB: Regression error data

Let

y = a(:,2);
x1 = a(:,1);

I'm assuming the 2nd column is the dependent variable and the first column is your predictor, independent variable.

Then,

   X = [ones(size(x1)) x1];
   [b,bint,r,rint,stats] = regress(y,X);
   rsquared = stats(1) % this is the R-squared
   % plot the data and the fit
   yhat = b(1)+b(2)*x1;
   plot(x1,y,'b*');
   plot(x1,yhat,'r-.');

MATLAB: Multivariate fit with two unknown coefficients and known intercepts

Using the modeling tools you know, this is correct:

% some random data
x1 = randn(100,1);
x2 = randn(100,1);
Y = x1 + 2*x2 + 1 +randn(size(x1))/100;
Y1 = Y-1;
T = table(Y,x1,x2);
lm = fitlm(T,'Y ~ x1 + x2 - 1')
lm = 
Linear regression model:
    Y ~ x1 + x2
Estimated Coefficients:
          Estimate       SE       tStat       pValue  
          ________    ________    ______    __________
    x1     0.799       0.10355    7.7158    1.0195e-11
    x2    1.8409      0.095852    19.206    5.4359e-35
Number of observations: 100, Error degrees of freedom: 98
Root Mean Squared Error: 0.977

Effectively, subtract 1 from Y, then fit a model with no intercept at all. To use the model, just add 1 back into your prediction at the end. So you create a model with no constant term at all, and then put it back in when you use the model.

If you have other tools available, for example, LSQLIN from the optimization toolbox, there are other ways you can do it. Or you can download my own LSE from the file exchange . It can constrain a parameter to a specific given value.

coefs = lse([ones(size(x1)),x1,x2],Y,[1 0 0],1)
coefs =
          1
     1.0009
      1.998

Using lsqlin, I would do this:

lsqlin([ones(size(x1)),x1,x2],Y,[],[],[1 0 0],1)
Minimum found that satisfies the constraints.
Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the default value of the optimality tolerance,
and constraints are satisfied to within the default value of the constraint tolerance.
<stopping criteria details>
ans =
            1
       1.0009
        1.998

In either of the latter two cases, you get the coefficients as just a vector, with less additional information that the stats toolbox gives you.

Best Answer

Related Solutions

MATLAB: Regression error data

MATLAB: Multivariate fit with two unknown coefficients and known intercepts

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