Regularize Logistic Regression
This example shows how to regularize binomial regression. The default (canonical) link function for binomial regression is the logistic function.
Step 1. Prepare the data.
ionosphere data. The response
Y is a cell array of
'b' characters. Convert the cells to logical values, with
'g'. Remove the first two columns of
X because they have some awkward statistical properties, which are beyond the scope of this discussion.
load ionosphere Ybool = strcmp(Y,'g'); X = X(:,3:end);
Step 2. Create a cross-validated fit.
Construct a regularized binomial regression using 25
Lambda values and 10-fold cross validation. This process can take a few minutes.
rng('default') % for reproducibility [B,FitInfo] = lassoglm(X,Ybool,'binomial',... 'NumLambda',25,'CV',10);
Step 3. Examine plots to find appropriate regularization.
lassoPlot can give both a standard trace plot and a cross-validated deviance plot. Examine both plots.
lassoPlot(B,FitInfo,'PlotType','CV'); legend('show','Location','best') % show legend
The plot identifies the minimum-deviance point with a green circle and dashed line as a function of the regularization parameter
Lambda. The blue circled point has minimum deviance plus no more than one standard deviation.
The trace plot shows nonzero model coefficients as a function of the regularization parameter
Lambda. Because there are 32 predictors and a linear model, there are 32 curves. As
Lambda increases to the left,
lassoglm sets various coefficients to zero, removing them from the model.
The trace plot is somewhat compressed. Zoom in to see more detail.
xlim([.01 .1]) ylim([-3 3])
Lambda increases toward the left side of the plot, fewer nonzero coefficients remain.
Find the number of nonzero model coefficients at the
Lambda value with minimum deviance plus one standard deviation point. The regularized model coefficients are in column
FitInfo.Index1SE of the
indx = FitInfo.Index1SE; B0 = B(:,indx); nonzeros = sum(B0 ~= 0)
nonzeros = 14
When you set
lassoglm removes over half of the 32 original predictors.
Step 4. Create a regularized model.
The constant term is in the
FitInfo.Index1SE entry of the
FitInfo.Intercept vector. Call that value
The model is logit(mu) = log(mu/(1 - mu)) =
X*B0 + cnst . Therefore, for predictions, mu =
exp(X*B0 + cnst)/(1+exp(x*B0 + cnst)).
glmval function evaluates model predictions. It assumes that the first model coefficient relates to the constant term. Therefore, create a coefficient vector with the constant term first.
cnst = FitInfo.Intercept(indx); B1 = [cnst;B0];
Step 5. Examine residuals.
Plot the training data against the model predictions for the regularized
preds = glmval(B1,X,'logit'); histogram(Ybool - preds) % plot residuals title('Residuals from lassoglm model')
Step 6. Alternative: Use identified predictors in a least-squares generalized linear model.
Instead of using the biased predictions from the model, you can make an unbiased model using just the identified predictors.
predictors = find(B0); % indices of nonzero predictors mdl = fitglm(X,Ybool,'linear',... 'Distribution','binomial','PredictorVars',predictors)
mdl = Generalized linear regression model: y ~ [Linear formula with 15 terms in 14 predictors] Distribution = Binomial Estimated Coefficients: Estimate SE tStat pValue _________ _______ ________ __________ (Intercept) -2.9367 0.50926 -5.7666 8.0893e-09 x1 2.492 0.60795 4.099 4.1502e-05 x3 2.5501 0.63304 4.0284 5.616e-05 x4 0.48816 0.50336 0.9698 0.33215 x5 0.6158 0.62192 0.99015 0.3221 x6 2.294 0.5421 4.2317 2.3198e-05 x7 0.77842 0.57765 1.3476 0.1778 x12 1.7808 0.54316 3.2786 0.0010432 x16 -0.070993 0.50515 -0.14054 0.88823 x20 -2.7767 0.55131 -5.0365 4.7402e-07 x24 2.0212 0.57639 3.5067 0.00045372 x25 -2.3796 0.58274 -4.0835 4.4363e-05 x27 0.79564 0.55904 1.4232 0.15467 x29 1.2689 0.55468 2.2876 0.022162 x32 -1.5681 0.54336 -2.8859 0.0039035 351 observations, 336 error degrees of freedom Dispersion: 1 Chi^2-statistic vs. constant model: 262, p-value = 1e-47
Plot the residuals of the model.
As expected, residuals from the least-squares model are slightly smaller than those of the regularized model. However, this does not mean that
mdl is a better predictor for new data.