Ch6-varselect-lab
Lab: Linear Models and Regularization Methods
Subset Selection Methods
Best Subset Selection
Here we apply the best subset selection approach to the Hitters data. We wish to predict a baseball player’s Salary on the basis of various statistics associated with performance in the previous year.
First of all, we note that the Salary variable is missing for some of the players. The is.na() function can be used to identify the missing observations. It returns a vector of the same length as the input vector, with a TRUE for any elements that are missing, and a FALSE for non-missing elements. The sum() function can then be used to count all of the missing elements.
library(ISLR2)
names(Hitters)## [1] "AtBat" "Hits" "HmRun" "Runs" "RBI" "Walks"
## [7] "Years" "CAtBat" "CHits" "CHmRun" "CRuns" "CRBI"
## [13] "CWalks" "League" "Division" "PutOuts" "Assists" "Errors"
## [19] "Salary" "NewLeague"dim(Hitters)## [1] 263 20sum(is.na(Hitters$Salary))## [1] 0Hence we see that Salary is missing for \(59\) players. The na.omit() function removes all of the rows that have missing values in any variable.
Hitters <- na.omit(Hitters)
dim(Hitters)## [1] 263 20sum(is.na(Hitters))## [1] 0The regsubsets() function (part of the leaps library) performs best subset selection by identifying the best model that contains a given number of predictors, where best is quantified using RSS. The syntax is the same as for lm(). The summary() command outputs the best set of variables for each model size.
library(leaps)
regfit.full <- regsubsets(Salary ~ ., Hitters)
summary(regfit.full)## Subset selection object
## Call: regsubsets.formula(Salary ~ ., Hitters)
## 19 Variables (and intercept)
## Forced in Forced out
## AtBat FALSE FALSE
## Hits FALSE FALSE
## HmRun FALSE FALSE
## Runs FALSE FALSE
## RBI FALSE FALSE
## Walks FALSE FALSE
## Years FALSE FALSE
## CAtBat FALSE FALSE
## CHits FALSE FALSE
## CHmRun FALSE FALSE
## CRuns FALSE FALSE
## CRBI FALSE FALSE
## CWalks FALSE FALSE
## LeagueN FALSE FALSE
## DivisionW FALSE FALSE
## PutOuts FALSE FALSE
## Assists FALSE FALSE
## Errors FALSE FALSE
## NewLeagueN FALSE FALSE
## 1 subsets of each size up to 8
## Selection Algorithm: exhaustive
## AtBat Hits HmRun Runs RBI Walks Years CAtBat CHits CHmRun CRuns CRBI
## 1 ( 1 ) " " " " " " " " " " " " " " " " " " " " " " "*"
## 2 ( 1 ) " " "*" " " " " " " " " " " " " " " " " " " "*"
## 3 ( 1 ) " " "*" " " " " " " " " " " " " " " " " " " "*"
## 4 ( 1 ) " " "*" " " " " " " " " " " " " " " " " " " "*"
## 5 ( 1 ) "*" "*" " " " " " " " " " " " " " " " " " " "*"
## 6 ( 1 ) "*" "*" " " " " " " "*" " " " " " " " " " " "*"
## 7 ( 1 ) " " "*" " " " " " " "*" " " "*" "*" "*" " " " "
## 8 ( 1 ) "*" "*" " " " " " " "*" " " " " " " "*" "*" " "
## CWalks LeagueN DivisionW PutOuts Assists Errors NewLeagueN
## 1 ( 1 ) " " " " " " " " " " " " " "
## 2 ( 1 ) " " " " " " " " " " " " " "
## 3 ( 1 ) " " " " " " "*" " " " " " "
## 4 ( 1 ) " " " " "*" "*" " " " " " "
## 5 ( 1 ) " " " " "*" "*" " " " " " "
## 6 ( 1 ) " " " " "*" "*" " " " " " "
## 7 ( 1 ) " " " " "*" "*" " " " " " "
## 8 ( 1 ) "*" " " "*" "*" " " " " " "An asterisk indicates that a given variable is included in the corresponding model. For instance, this output indicates that the best two-variable model contains only Hits and CRBI. By default, regsubsets() only reports results up to the best eight-variable model. But the nvmax option can be used in order to return as many variables as are desired. Here we fit up to a 19-variable model.
regfit.full <- regsubsets(Salary ~ ., data = Hitters,
nvmax = 19)
reg.summary <- summary(regfit.full)The summary() function also returns \(R^2\), RSS, adjusted \(R^2\), \(C_p\), and BIC. We can examine these to try to select the best overall model.
names(reg.summary)## [1] "which" "rsq" "rss" "adjr2" "cp" "bic" "outmat" "obj"For instance, we see that the \(R^2\) statistic increases from \(32\,\%\), when only one variable is included in the model, to almost \(55\,\%\), when all variables are included. As expected, the \(R^2\) statistic increases monotonically as more variables are included.
reg.summary$rsq## [1] 0.3214501 0.4252237 0.4514294 0.4754067 0.4908036 0.5087146 0.5141227
## [8] 0.5285569 0.5346124 0.5404950 0.5426153 0.5436302 0.5444570 0.5452164
## [15] 0.5454692 0.5457656 0.5459518 0.5460945 0.5461159Plotting RSS, adjusted \(R^2\), \(C_p\), and BIC for all of the models at once will help us decide which model to select. Note the type = "l" option tells R to connect the plotted points with lines.
par(mfrow = c(1, 2))
plot(reg.summary$rss, xlab = "Number of Variables",
ylab = "RSS", type = "l")
plot(reg.summary$adjr2, xlab = "Number of Variables",
ylab = "Adjusted RSq", type = "l")
The points() command works like the plot() command, except that it puts points on a plot that has already been created, instead of creating a new plot. The which.max() function can be used to identify the location of the maximum point of a vector. We will now plot a red dot to indicate the model with the largest adjusted \(R^2\) statistic.
which.max(reg.summary$adjr2)## [1] 11plot(reg.summary$adjr2, xlab = "Number of Variables",
ylab = "Adjusted RSq", type = "l")
points(11, reg.summary$adjr2[11], col = "red", cex = 2,
pch = 20)
In a similar fashion we can plot the \(C_p\) and BIC statistics, and indicate the models with the smallest statistic using which.min().
plot(reg.summary$cp, xlab = "Number of Variables",
ylab = "Cp", type = "l")
which.min(reg.summary$cp)## [1] 10points(10, reg.summary$cp[10], col = "red", cex = 2,
pch = 20)
which.min(reg.summary$bic)## [1] 6plot(reg.summary$bic, xlab = "Number of Variables",
ylab = "BIC", type = "l")
points(6, reg.summary$bic[6], col = "red", cex = 2,
pch = 20)
The regsubsets() function has a built-in plot() command which can be used to display the selected variables for the best model with a given number of predictors, ranked according to the BIC, \(C_p\), adjusted \(R^2\), or AIC. To find out more about this function, type ?plot.regsubsets.
plot(regfit.full, scale = "r2")
plot(regfit.full, scale = "adjr2")
plot(regfit.full, scale = "Cp")
plot(regfit.full, scale = "bic")
The top row of each plot contains a black square for each variable selected according to the optimal model associated with that statistic. For instance, we see that several models share a BIC close to \(-150\). However, the model with the lowest BIC is the six-variable model that contains only AtBat, Hits, Walks, CRBI, DivisionW, and PutOuts. We can use the coef() function to see the coefficient estimates associated with this model.
coef(regfit.full, 6)## (Intercept) AtBat Hits Walks CRBI DivisionW
## 91.5117981 -1.8685892 7.6043976 3.6976468 0.6430169 -122.9515338
## PutOuts
## 0.2643076Forward and Backward Stepwise Selection
We can also use the regsubsets() function to perform forward stepwise or backward stepwise selection, using the argument method = "forward" or method = "backward".
regfit.fwd <- regsubsets(Salary ~ ., data = Hitters,
nvmax = 19, method = "forward")
summary(regfit.fwd)## Subset selection object
## Call: regsubsets.formula(Salary ~ ., data = Hitters, nvmax = 19, method = "forward")
## 19 Variables (and intercept)
## Forced in Forced out
## AtBat FALSE FALSE
## Hits FALSE FALSE
## HmRun FALSE FALSE
## Runs FALSE FALSE
## RBI FALSE FALSE
## Walks FALSE FALSE
## Years FALSE FALSE
## CAtBat FALSE FALSE
## CHits FALSE FALSE
## CHmRun FALSE FALSE
## CRuns FALSE FALSE
## CRBI FALSE FALSE
## CWalks FALSE FALSE
## LeagueN FALSE FALSE
## DivisionW FALSE FALSE
## PutOuts FALSE FALSE
## Assists FALSE FALSE
## Errors FALSE FALSE
## NewLeagueN FALSE FALSE
## 1 subsets of each size up to 19
## Selection Algorithm: forward
## AtBat Hits HmRun Runs RBI Walks Years CAtBat CHits CHmRun CRuns CRBI
## 1 ( 1 ) " " " " " " " " " " " " " " " " " " " " " " "*"
## 2 ( 1 ) " " "*" " " " " " " " " " " " " " " " " " " "*"
## 3 ( 1 ) " " "*" " " " " " " " " " " " " " " " " " " "*"
## 4 ( 1 ) " " "*" " " " " " " " " " " " " " " " " " " "*"
## 5 ( 1 ) "*" "*" " " " " " " " " " " " " " " " " " " "*"
## 6 ( 1 ) "*" "*" " " " " " " "*" " " " " " " " " " " "*"
## 7 ( 1 ) "*" "*" " " " " " " "*" " " " " " " " " " " "*"
## 8 ( 1 ) "*" "*" " " " " " " "*" " " " " " " " " "*" "*"
## 9 ( 1 ) "*" "*" " " " " " " "*" " " "*" " " " " "*" "*"
## 10 ( 1 ) "*" "*" " " " " " " "*" " " "*" " " " " "*" "*"
## 11 ( 1 ) "*" "*" " " " " " " "*" " " "*" " " " " "*" "*"
## 12 ( 1 ) "*" "*" " " "*" " " "*" " " "*" " " " " "*" "*"
## 13 ( 1 ) "*" "*" " " "*" " " "*" " " "*" " " " " "*" "*"
## 14 ( 1 ) "*" "*" "*" "*" " " "*" " " "*" " " " " "*" "*"
## 15 ( 1 ) "*" "*" "*" "*" " " "*" " " "*" "*" " " "*" "*"
## 16 ( 1 ) "*" "*" "*" "*" "*" "*" " " "*" "*" " " "*" "*"
## 17 ( 1 ) "*" "*" "*" "*" "*" "*" " " "*" "*" " " "*" "*"
## 18 ( 1 ) "*" "*" "*" "*" "*" "*" "*" "*" "*" " " "*" "*"
## 19 ( 1 ) "*" "*" "*" "*" "*" "*" "*" "*" "*" "*" "*" "*"
## CWalks LeagueN DivisionW PutOuts Assists Errors NewLeagueN
## 1 ( 1 ) " " " " " " " " " " " " " "
## 2 ( 1 ) " " " " " " " " " " " " " "
## 3 ( 1 ) " " " " " " "*" " " " " " "
## 4 ( 1 ) " " " " "*" "*" " " " " " "
## 5 ( 1 ) " " " " "*" "*" " " " " " "
## 6 ( 1 ) " " " " "*" "*" " " " " " "
## 7 ( 1 ) "*" " " "*" "*" " " " " " "
## 8 ( 1 ) "*" " " "*" "*" " " " " " "
## 9 ( 1 ) "*" " " "*" "*" " " " " " "
## 10 ( 1 ) "*" " " "*" "*" "*" " " " "
## 11 ( 1 ) "*" "*" "*" "*" "*" " " " "
## 12 ( 1 ) "*" "*" "*" "*" "*" " " " "
## 13 ( 1 ) "*" "*" "*" "*" "*" "*" " "
## 14 ( 1 ) "*" "*" "*" "*" "*" "*" " "
## 15 ( 1 ) "*" "*" "*" "*" "*" "*" " "
## 16 ( 1 ) "*" "*" "*" "*" "*" "*" " "
## 17 ( 1 ) "*" "*" "*" "*" "*" "*" "*"
## 18 ( 1 ) "*" "*" "*" "*" "*" "*" "*"
## 19 ( 1 ) "*" "*" "*" "*" "*" "*" "*"regfit.bwd <- regsubsets(Salary ~ ., data = Hitters,
nvmax = 19, method = "backward")
summary(regfit.bwd)## Subset selection object
## Call: regsubsets.formula(Salary ~ ., data = Hitters, nvmax = 19, method = "backward")
## 19 Variables (and intercept)
## Forced in Forced out
## AtBat FALSE FALSE
## Hits FALSE FALSE
## HmRun FALSE FALSE
## Runs FALSE FALSE
## RBI FALSE FALSE
## Walks FALSE FALSE
## Years FALSE FALSE
## CAtBat FALSE FALSE
## CHits FALSE FALSE
## CHmRun FALSE FALSE
## CRuns FALSE FALSE
## CRBI FALSE FALSE
## CWalks FALSE FALSE
## LeagueN FALSE FALSE
## DivisionW FALSE FALSE
## PutOuts FALSE FALSE
## Assists FALSE FALSE
## Errors FALSE FALSE
## NewLeagueN FALSE FALSE
## 1 subsets of each size up to 19
## Selection Algorithm: backward
## AtBat Hits HmRun Runs RBI Walks Years CAtBat CHits CHmRun CRuns CRBI
## 1 ( 1 ) " " " " " " " " " " " " " " " " " " " " "*" " "
## 2 ( 1 ) " " "*" " " " " " " " " " " " " " " " " "*" " "
## 3 ( 1 ) " " "*" " " " " " " " " " " " " " " " " "*" " "
## 4 ( 1 ) "*" "*" " " " " " " " " " " " " " " " " "*" " "
## 5 ( 1 ) "*" "*" " " " " " " "*" " " " " " " " " "*" " "
## 6 ( 1 ) "*" "*" " " " " " " "*" " " " " " " " " "*" " "
## 7 ( 1 ) "*" "*" " " " " " " "*" " " " " " " " " "*" " "
## 8 ( 1 ) "*" "*" " " " " " " "*" " " " " " " " " "*" "*"
## 9 ( 1 ) "*" "*" " " " " " " "*" " " "*" " " " " "*" "*"
## 10 ( 1 ) "*" "*" " " " " " " "*" " " "*" " " " " "*" "*"
## 11 ( 1 ) "*" "*" " " " " " " "*" " " "*" " " " " "*" "*"
## 12 ( 1 ) "*" "*" " " "*" " " "*" " " "*" " " " " "*" "*"
## 13 ( 1 ) "*" "*" " " "*" " " "*" " " "*" " " " " "*" "*"
## 14 ( 1 ) "*" "*" "*" "*" " " "*" " " "*" " " " " "*" "*"
## 15 ( 1 ) "*" "*" "*" "*" " " "*" " " "*" "*" " " "*" "*"
## 16 ( 1 ) "*" "*" "*" "*" "*" "*" " " "*" "*" " " "*" "*"
## 17 ( 1 ) "*" "*" "*" "*" "*" "*" " " "*" "*" " " "*" "*"
## 18 ( 1 ) "*" "*" "*" "*" "*" "*" "*" "*" "*" " " "*" "*"
## 19 ( 1 ) "*" "*" "*" "*" "*" "*" "*" "*" "*" "*" "*" "*"
## CWalks LeagueN DivisionW PutOuts Assists Errors NewLeagueN
## 1 ( 1 ) " " " " " " " " " " " " " "
## 2 ( 1 ) " " " " " " " " " " " " " "
## 3 ( 1 ) " " " " " " "*" " " " " " "
## 4 ( 1 ) " " " " " " "*" " " " " " "
## 5 ( 1 ) " " " " " " "*" " " " " " "
## 6 ( 1 ) " " " " "*" "*" " " " " " "
## 7 ( 1 ) "*" " " "*" "*" " " " " " "
## 8 ( 1 ) "*" " " "*" "*" " " " " " "
## 9 ( 1 ) "*" " " "*" "*" " " " " " "
## 10 ( 1 ) "*" " " "*" "*" "*" " " " "
## 11 ( 1 ) "*" "*" "*" "*" "*" " " " "
## 12 ( 1 ) "*" "*" "*" "*" "*" " " " "
## 13 ( 1 ) "*" "*" "*" "*" "*" "*" " "
## 14 ( 1 ) "*" "*" "*" "*" "*" "*" " "
## 15 ( 1 ) "*" "*" "*" "*" "*" "*" " "
## 16 ( 1 ) "*" "*" "*" "*" "*" "*" " "
## 17 ( 1 ) "*" "*" "*" "*" "*" "*" "*"
## 18 ( 1 ) "*" "*" "*" "*" "*" "*" "*"
## 19 ( 1 ) "*" "*" "*" "*" "*" "*" "*"For instance, we see that using forward stepwise selection, the best one-variable model contains only CRBI, and the best two-variable model additionally includes Hits. For this data, the best one-variable through six-variable models are each identical for best subset and forward selection. However, the best seven-variable models identified by forward stepwise selection, backward stepwise selection, and best subset selection are different.
coef(regfit.full, 7)## (Intercept) Hits Walks CAtBat CHits CHmRun
## 79.4509472 1.2833513 3.2274264 -0.3752350 1.4957073 1.4420538
## DivisionW PutOuts
## -129.9866432 0.2366813coef(regfit.fwd, 7)## (Intercept) AtBat Hits Walks CRBI CWalks
## 109.7873062 -1.9588851 7.4498772 4.9131401 0.8537622 -0.3053070
## DivisionW PutOuts
## -127.1223928 0.2533404coef(regfit.bwd, 7)## (Intercept) AtBat Hits Walks CRuns CWalks
## 105.6487488 -1.9762838 6.7574914 6.0558691 1.1293095 -0.7163346
## DivisionW PutOuts
## -116.1692169 0.3028847Choosing Among Models Using the Validation-Set Approach and Cross-Validation
We just saw that it is possible to choose among a set of models of different sizes using \(C_p\), BIC, and adjusted \(R^2\). We will now consider how to do this using the validation set and cross-validation approaches.
In order for these approaches to yield accurate estimates of the test error, we must use only the training observations to perform all aspects of model-fitting—including variable selection. Therefore, the determination of which model of a given size is best must be made using only the training observations. This point is subtle but important. If the full data set is used to perform the best subset selection step, the validation set errors and cross-validation errors that we obtain will not be accurate estimates of the test error.
In order to use the validation set approach, we begin by splitting the observations into a training set and a test set. We do this by creating a random vector, train, of elements equal to TRUE if the corresponding observation is in the training set, and FALSE otherwise. The vector test has a TRUE if the observation is in the test set, and a FALSE otherwise. Note the ! in the command to create test causes TRUEs to be switched to FALSEs and vice versa. We also set a random seed so that the user will obtain the same training set/test set split.
set.seed(1)
train <- sample(c(TRUE, FALSE), nrow(Hitters),
replace = TRUE)
test <- (!train)Now, we apply regsubsets() to the training set in order to perform best subset selection.
regfit.best <- regsubsets(Salary ~ .,
data = Hitters[train, ], nvmax = 19)Notice that we subset the Hitters data frame directly in the call in order to access only the training subset of the data, using the expression Hitters[train, ]. We now compute the validation set error for the best model of each model size. We first make a model matrix from the test data.
test.mat <- model.matrix(Salary ~ ., data = Hitters[test, ])The model.matrix() function is used in many regression packages for building an `X'' matrix from data. Now we run a loop, and for each sizei, we extract the coefficients fromregfit.best` for the best model of that size, multiply them into the appropriate columns of the test model matrix to form the predictions, and compute the test MSE.
val.errors <- rep(NA, 19)
for (i in 1:19) {
coefi <- coef(regfit.best, id = i)
pred <- test.mat[, names(coefi)] %*% coefi
val.errors[i] <- mean((Hitters$Salary[test] - pred)^2)
}We find that the best model is the one that contains seven variables.
val.errors## [1] 164377.3 144405.5 152175.7 145198.4 137902.1 139175.7 126849.0 136191.4
## [9] 132889.6 135434.9 136963.3 140694.9 140690.9 141951.2 141508.2 142164.4
## [17] 141767.4 142339.6 142238.2which.min(val.errors)## [1] 7coef(regfit.best, 7)## (Intercept) AtBat Hits Walks CRuns CWalks
## 67.1085369 -2.1462987 7.0149547 8.0716640 1.2425113 -0.8337844
## DivisionW PutOuts
## -118.4364998 0.2526925This was a little tedious, partly because there is no predict() method for regsubsets(). Since we will be using this function again, we can capture our steps above and write our own predict method.
predict.regsubsets <- function(object, newdata, id, ...) {
form <- as.formula(object$call[[2]])
mat <- model.matrix(form, newdata)
coefi <- coef(object, id = id)
xvars <- names(coefi)
mat[, xvars] %*% coefi
}Our function pretty much mimics what we did above. The only complex part is how we extracted the formula used in the call to regsubsets(). We demonstrate how we use this function below, when we do cross-validation.
Finally, we perform best subset selection on the full data set, and select the best seven-variable model. It is important that we make use of the full data set in order to obtain more accurate coefficient estimates. Note that we perform best subset selection on the full data set and select the best seven-variable model, rather than simply using the variables that were obtained from the training set, because the best seven-variable model on the full data set may differ from the corresponding model on the training set.
regfit.best <- regsubsets(Salary ~ ., data = Hitters,
nvmax = 19)
coef(regfit.best, 7)## (Intercept) Hits Walks CAtBat CHits CHmRun
## 79.4509472 1.2833513 3.2274264 -0.3752350 1.4957073 1.4420538
## DivisionW PutOuts
## -129.9866432 0.2366813In fact, we see that the best seven-variable model on the full data set has a different set of variables than the best seven-variable model on the training set.
We now try to choose among the models of different sizes using cross-validation. This approach is somewhat involved, as we must perform best subset selection within each of the \(k\) training sets. Despite this, we see that with its clever subsetting syntax, R makes this job quite easy. First, we create a vector that allocates each observation to one of \(k=10\) folds, and we create a matrix in which we will store the results.
k <- 10
n <- nrow(Hitters)
set.seed(1)
folds <- sample(rep(1:k, length = n))
cv.errors <- matrix(NA, k, 19,
dimnames = list(NULL, paste(1:19)))Now we write a for loop that performs cross-validation. In the \(j\)th fold, the elements of folds that equal j are in the test set, and the remainder are in the training set. We make our predictions for each model size (using our new predict() method), compute the test errors on the appropriate subset, and store them in the appropriate slot in the matrix cv.errors. Note that in the following code R will automatically use our predict.regsubsets() function when we call predict() because the best.fit object has class regsubsets.
for (j in 1:k) {
best.fit <- regsubsets(Salary ~ .,
data = Hitters[folds != j, ],
nvmax = 19)
for (i in 1:19) {
pred <- predict(best.fit, Hitters[folds == j, ], id = i)
cv.errors[j, i] <-
mean((Hitters$Salary[folds == j] - pred)^2)
}
}This has given us a \(10 \times 19\) matrix, of which the \((j,i)\)th element corresponds to the test MSE for the \(j\)th cross-validation fold for the best \(i\)-variable model. We use the apply() function to average over the columns of this matrix in order to obtain a vector for which the \(i\)th element is the cross-validation error for the \(i\)-variable model.
mean.cv.errors <- apply(cv.errors, 2, mean)
mean.cv.errors## 1 2 3 4 5 6 7 8
## 143439.8 126817.0 134214.2 131782.9 130765.6 120382.9 121443.1 114363.7
## 9 10 11 12 13 14 15 16
## 115163.1 109366.0 112738.5 113616.5 115557.6 115853.3 115630.6 116050.0
## 17 18 19
## 116117.0 116419.3 116299.1par(mfrow = c(1, 1))
plot(mean.cv.errors, type = "b")
We see that cross-validation selects a 10-variable model. We now perform best subset selection on the full data set in order to obtain the 10-variable model.
reg.best <- regsubsets(Salary ~ ., data = Hitters,
nvmax = 19)
coef(reg.best, 10)## (Intercept) AtBat Hits Walks CAtBat CRuns
## 162.5354420 -2.1686501 6.9180175 5.7732246 -0.1300798 1.4082490
## CRBI CWalks DivisionW PutOuts Assists
## 0.7743122 -0.8308264 -112.3800575 0.2973726 0.2831680Ridge Regression and the Lasso
We will use the glmnet package in order to perform ridge regression and the lasso. The main function in this package is glmnet(), which can be used to fit ridge regression models, lasso models, and more. This function has slightly different syntax from other model-fitting functions that we have encountered thus far in this book. In particular, we must pass in an x matrix as well as a y vector, and we do not use the {} syntax. We will now perform ridge regression and the lasso in order to predict Salary on the Hitters data. Before proceeding ensure that the missing values have been removed from the data, as described in Section 6.5.1.
x <- model.matrix(Salary ~ ., Hitters)[, -1]
y <- Hitters$SalaryThe model.matrix() function is particularly useful for creating x; not only does it produce a matrix corresponding to the \(19\) predictors but it also automatically transforms any qualitative variables into dummy variables. The latter property is important because glmnet() can only take numerical, quantitative inputs.
Ridge Regression
The glmnet() function has an alpha argument that determines what type of model is fit. If alpha=0 then a ridge regression model is fit, and if alpha=1 then a lasso model is fit. We first fit a ridge regression model.
library(glmnet)
grid <- 10^seq(10, -2, length = 100)
ridge.mod <- glmnet(x, y, alpha = 0, lambda = grid)By default the glmnet() function performs ridge regression for an automatically selected range of \(\lambda\) values. However, here we have chosen to implement the function over a grid of values ranging from \(\lambda=10^{10}\) to \(\lambda=10^{-2}\), essentially covering the full range of scenarios from the null model containing only the intercept, to the least squares fit. As we will see, we can also compute model fits for a particular value of \(\lambda\) that is not one of the original grid values. Note that by default, the glmnet() function standardizes the variables so that they are on the same scale. To turn off this default setting, use the argument standardize = FALSE.
Associated with each value of \(\lambda\) is a vector of ridge regression coefficients, stored in a matrix that can be accessed by coef(). In this case, it is a \(20 \times 100\) matrix, with \(20\) rows (one for each predictor, plus an intercept) and \(100\) columns (one for each value of \(\lambda\)).
dim(coef(ridge.mod))## [1] 20 100We expect the coefficient estimates to be much smaller, in terms of \(\ell_2\) norm, when a large value of \(\lambda\) is used, as compared to when a small value of \(\lambda\) is used. These are the coefficients when \(\lambda=11{,}498\), along with their \(\ell_2\) norm:
ridge.mod$lambda[50]## [1] 11497.57coef(ridge.mod)[, 50]## (Intercept) AtBat Hits HmRun Runs
## 407.356050200 0.036957182 0.138180344 0.524629976 0.230701523
## RBI Walks Years CAtBat CHits
## 0.239841459 0.289618741 1.107702929 0.003131815 0.011653637
## CHmRun CRuns CRBI CWalks LeagueN
## 0.087545670 0.023379882 0.024138320 0.025015421 0.085028114
## DivisionW PutOuts Assists Errors NewLeagueN
## -6.215440973 0.016482577 0.002612988 -0.020502690 0.301433531sqrt(sum(coef(ridge.mod)[-1, 50]^2))## [1] 6.360612In contrast, here are the coefficients when \(\lambda=705\), along with their \(\ell_2\) norm. Note the much larger \(\ell_2\) norm of the coefficients associated with this smaller value of \(\lambda\).
ridge.mod$lambda[60]## [1] 705.4802coef(ridge.mod)[, 60]## (Intercept) AtBat Hits HmRun Runs RBI
## 54.32519950 0.11211115 0.65622409 1.17980910 0.93769713 0.84718546
## Walks Years CAtBat CHits CHmRun CRuns
## 1.31987948 2.59640425 0.01083413 0.04674557 0.33777318 0.09355528
## CRBI CWalks LeagueN DivisionW PutOuts Assists
## 0.09780402 0.07189612 13.68370191 -54.65877750 0.11852289 0.01606037
## Errors NewLeagueN
## -0.70358655 8.61181213sqrt(sum(coef(ridge.mod)[-1, 60]^2))## [1] 57.11001We can use the predict() function for a number of purposes. For instance, we can obtain the ridge regression coefficients for a new value of \(\lambda\), say \(50\):
predict(ridge.mod, s = 50, type = "coefficients")[1:20, ]## (Intercept) AtBat Hits HmRun Runs
## 4.876610e+01 -3.580999e-01 1.969359e+00 -1.278248e+00 1.145892e+00
## RBI Walks Years CAtBat CHits
## 8.038292e-01 2.716186e+00 -6.218319e+00 5.447837e-03 1.064895e-01
## CHmRun CRuns CRBI CWalks LeagueN
## 6.244860e-01 2.214985e-01 2.186914e-01 -1.500245e-01 4.592589e+01
## DivisionW PutOuts Assists Errors NewLeagueN
## -1.182011e+02 2.502322e-01 1.215665e-01 -3.278600e+00 -9.496680e+00We now split the samples into a training set and a test set in order to estimate the test error of ridge regression and the lasso. There are two common ways to randomly split a data set. The first is to produce a random vector of TRUE, FALSE elements and select the observations corresponding to TRUE for the training data. The second is to randomly choose a subset of numbers between \(1\) and \(n\); these can then be used as the indices for the training observations. The two approaches work equally well. We used the former method in Section 6.5.1. Here we demonstrate the latter approach.
We first set a random seed so that the results obtained will be reproducible.
set.seed(1)
train <- sample(1:nrow(x), nrow(x) / 2)
test <- (-train)
y.test <- y[test]Next we fit a ridge regression model on the training set, and evaluate its MSE on the test set, using \(\lambda=4\). Note the use of the predict() function again. This time we get predictions for a test set, by replacing type="coefficients" with the newx argument.
ridge.mod <- glmnet(x[train, ], y[train], alpha = 0,
lambda = grid, thresh = 1e-12)
ridge.pred <- predict(ridge.mod, s = 4, newx = x[test, ])
mean((ridge.pred - y.test)^2)## [1] 142199.2The test MSE is \(142{,}199\). Note that if we had instead simply fit a model with just an intercept, we would have predicted each test observation using the mean of the training observations. In that case, we could compute the test set MSE like this:
mean((mean(y[train]) - y.test)^2)## [1] 224669.9We could also get the same result by fitting a ridge regression model with a very large value of \(\lambda\). Note that 1e10 means \(10^{10}\).
ridge.pred <- predict(ridge.mod, s = 1e10, newx = x[test, ])
mean((ridge.pred - y.test)^2)## [1] 224669.8So fitting a ridge regression model with \(\lambda=4\) leads to a much lower test MSE than fitting a model with just an intercept. We now check whether there is any benefit to performing ridge regression with \(\lambda=4\) instead of just performing least squares regression. Recall that least squares is simply ridge regression with \(\lambda=0\).
ridge.pred <- predict(ridge.mod, s = 0, newx = x[test, ],
exact = T, x = x[train, ], y = y[train])
mean((ridge.pred - y.test)^2)## [1] 168588.6lm(y ~ x, subset = train)##
## Call:
## lm(formula = y ~ x, subset = train)
##
## Coefficients:
## (Intercept) xAtBat xHits xHmRun xRuns xRBI
## 274.0145 -0.3521 -1.6377 5.8145 1.5424 1.1243
## xWalks xYears xCAtBat xCHits xCHmRun xCRuns
## 3.7287 -16.3773 -0.6412 3.1632 3.4008 -0.9739
## xCRBI xCWalks xLeagueN xDivisionW xPutOuts xAssists
## -0.6005 0.3379 119.1486 -144.0831 0.1976 0.6804
## xErrors xNewLeagueN
## -4.7128 -71.0951predict(ridge.mod, s = 0, exact = T, type = "coefficients",
x = x[train, ], y = y[train])[1:20, ]## (Intercept) AtBat Hits HmRun Runs RBI
## 274.0200994 -0.3521900 -1.6371383 5.8146692 1.5423361 1.1241837
## Walks Years CAtBat CHits CHmRun CRuns
## 3.7288406 -16.3795195 -0.6411235 3.1629444 3.4005281 -0.9739405
## CRBI CWalks LeagueN DivisionW PutOuts Assists
## -0.6003976 0.3378422 119.1434637 -144.0853061 0.1976300 0.6804200
## Errors NewLeagueN
## -4.7127879 -71.0898914In general, if we want to fit a (unpenalized) least squares model, then we should use the lm() function, since that function provides more useful outputs, such as standard errors and p-values for the coefficients.
In general, instead of arbitrarily choosing \(\lambda=4\), it would be better to use cross-validation to choose the tuning parameter \(\lambda\). We can do this using the built-in cross-validation function, cv.glmnet(). By default, the function performs ten-fold cross-validation, though this can be changed using the argument nfolds. Note that we set a random seed first so our results will be reproducible, since the choice of the cross-validation folds is random.
set.seed(1)
cv.out <- cv.glmnet(x[train, ], y[train], alpha = 0)
plot(cv.out)
bestlam <- cv.out$lambda.min
bestlam## [1] 326.0828Therefore, we see that the value of \(\lambda\) that results in the smallest cross-validation error is \(326\). What is the test MSE associated with this value of \(\lambda\)?
ridge.pred <- predict(ridge.mod, s = bestlam,
newx = x[test, ])
mean((ridge.pred - y.test)^2)## [1] 139856.6This represents a further improvement over the test MSE that we got using \(\lambda=4\). Finally, we refit our ridge regression model on the full data set, using the value of \(\lambda\) chosen by cross-validation, and examine the coefficient estimates.
out <- glmnet(x, y, alpha = 0)
predict(out, type = "coefficients", s = bestlam)[1:20, ]## (Intercept) AtBat Hits HmRun Runs RBI
## 15.44383135 0.07715547 0.85911581 0.60103107 1.06369007 0.87936105
## Walks Years CAtBat CHits CHmRun CRuns
## 1.62444616 1.35254780 0.01134999 0.05746654 0.40680157 0.11456224
## CRBI CWalks LeagueN DivisionW PutOuts Assists
## 0.12116504 0.05299202 22.09143189 -79.04032637 0.16619903 0.02941950
## Errors NewLeagueN
## -1.36092945 9.12487767As expected, none of the coefficients are zero—ridge regression does not perform variable selection!
The Lasso
We saw that ridge regression with a wise choice of \(\lambda\) can outperform least squares as well as the null model on the Hitters data set. We now ask whether the lasso can yield either a more accurate or a more interpretable model than ridge regression. In order to fit a lasso model, we once again use the glmnet() function; however, this time we use the argument alpha=1. Other than that change, we proceed just as we did in fitting a ridge model.
lasso.mod <- glmnet(x[train, ], y[train], alpha = 1,
lambda = grid)
plot(lasso.mod)## Warning in regularize.values(x, y, ties, missing(ties), na.rm = na.rm):
## collapsing to unique 'x' values
We can see from the coefficient plot that depending on the choice of tuning parameter, some of the coefficients will be exactly equal to zero. We now perform cross-validation and compute the associated test error.
set.seed(1)
cv.out <- cv.glmnet(x[train, ], y[train], alpha = 1)
plot(cv.out)
bestlam <- cv.out$lambda.min
lasso.pred <- predict(lasso.mod, s = bestlam,
newx = x[test, ])
mean((lasso.pred - y.test)^2)## [1] 143673.6This is substantially lower than the test set MSE of the null model and of least squares, and very similar to the test MSE of ridge regression with \(\lambda\) chosen by cross-validation.
However, the lasso has a substantial advantage over ridge regression in that the resulting coefficient estimates are sparse. Here we see that 8 of the 19 coefficient estimates are exactly zero. So the lasso model with \(\lambda\) chosen by cross-validation contains only eleven variables.
out <- glmnet(x, y, alpha = 1, lambda = grid)
lasso.coef <- predict(out, type = "coefficients",
s = bestlam)[1:20, ]
lasso.coef## (Intercept) AtBat Hits HmRun Runs
## 1.27479059 -0.05497143 2.18034583 0.00000000 0.00000000
## RBI Walks Years CAtBat CHits
## 0.00000000 2.29192406 -0.33806109 0.00000000 0.00000000
## CHmRun CRuns CRBI CWalks LeagueN
## 0.02825013 0.21628385 0.41712537 0.00000000 20.28615023
## DivisionW PutOuts Assists Errors NewLeagueN
## -116.16755870 0.23752385 0.00000000 -0.85629148 0.00000000lasso.coef[lasso.coef != 0]## (Intercept) AtBat Hits Walks Years
## 1.27479059 -0.05497143 2.18034583 2.29192406 -0.33806109
## CHmRun CRuns CRBI LeagueN DivisionW
## 0.02825013 0.21628385 0.41712537 20.28615023 -116.16755870
## PutOuts Errors
## 0.23752385 -0.85629148PCR and PLS Regression
Principal Components Regression
Principal components regression (PCR) can be performed using the pcr() function, which is part of the pls library. We now apply PCR to the Hitters data, in order to predict Salary. Again, we ensure that the missing values have been removed from the data, as described in Section 6.5.1.
library(pls)
set.seed(2)
pcr.fit <- pcr(Salary ~ ., data = Hitters, scale = TRUE,
validation = "CV")The syntax for the pcr() function is similar to that for lm(), with a few additional options. Setting scale = TRUE has the effect of standardizing each predictor, using ( 6.6), prior to generating the principal components, so that the scale on which each variable is measured will not have an effect. Setting validation = "CV" causes pcr() to compute the ten-fold cross-validation error for each possible value of \(M\), the number of principal components used. The resulting fit can be examined using summary().
summary(pcr.fit)## Data: X dimension: 263 19
## Y dimension: 263 1
## Fit method: svdpc
## Number of components considered: 19
##
## VALIDATION: RMSEP
## Cross-validated using 10 random segments.
## (Intercept) 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps
## CV 452 351.9 353.2 355.0 352.8 348.4 343.6
## adjCV 452 351.6 352.7 354.4 352.1 347.6 342.7
## 7 comps 8 comps 9 comps 10 comps 11 comps 12 comps 13 comps
## CV 345.5 347.7 349.6 351.4 352.1 353.5 358.2
## adjCV 344.7 346.7 348.5 350.1 350.7 352.0 356.5
## 14 comps 15 comps 16 comps 17 comps 18 comps 19 comps
## CV 349.7 349.4 339.9 341.6 339.2 339.6
## adjCV 348.0 347.7 338.2 339.7 337.2 337.6
##
## TRAINING: % variance explained
## 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps 7 comps 8 comps
## X 38.31 60.16 70.84 79.03 84.29 88.63 92.26 94.96
## Salary 40.63 41.58 42.17 43.22 44.90 46.48 46.69 46.75
## 9 comps 10 comps 11 comps 12 comps 13 comps 14 comps 15 comps
## X 96.28 97.26 97.98 98.65 99.15 99.47 99.75
## Salary 46.86 47.76 47.82 47.85 48.10 50.40 50.55
## 16 comps 17 comps 18 comps 19 comps
## X 99.89 99.97 99.99 100.00
## Salary 53.01 53.85 54.61 54.61The CV score is provided for each possible number of components, ranging from \(M=0\) onwards. (We have printed the CV output only up to \(M=4\).) Note that pcr() reports the root mean squared error; in order to obtain the usual MSE, we must square this quantity. For instance, a root mean squared error of \(352.8\) corresponds to an MSE of \(352.8^2=124{,}468\).
One can also plot the cross-validation scores using the validationplot() function. Using val.type = "MSEP" will cause the cross-validation MSE to be plotted.
validationplot(pcr.fit, val.type = "MSEP")
We see that the smallest cross-validation error occurs when \(M=18\) components are used. This is barely fewer than \(M=19\), which amounts to simply performing least squares, because when all of the components are used in PCR no dimension reduction occurs. However, from the plot we also see that the cross-validation error is roughly the same when only one component is included in the model. This suggests that a model that uses just a small number of components might suffice.
The summary() function also provides the percentage of variance explained in the predictors and in the response using different numbers of components. This concept is discussed in greater detail in Chapter 12. Briefly, we can think of this as the amount of information about the predictors or the response that is captured using \(M\) principal components. For example, setting \(M=1\) only captures \(38.31\,\%\) of all the variance, or information, in the predictors. In contrast, using \(M=5\) increases the value to \(84.29\,\%\). If we were to use all \(M=p=19\) components, this would increase to \(100\,\%\).
We now perform PCR on the training data and evaluate its test set performance.
set.seed(1)
pcr.fit <- pcr(Salary ~ ., data = Hitters, subset = train,
scale = TRUE, validation = "CV")
validationplot(pcr.fit, val.type = "MSEP")
Now we find that the lowest cross-validation error occurs when \(M=5\) components are used. We compute the test MSE as follows.
pcr.pred <- predict(pcr.fit, x[test, ], ncomp = 5)
mean((pcr.pred - y.test)^2)## [1] 142811.8This test set MSE is competitive with the results obtained using ridge regression and the lasso. However, as a result of the way PCR is implemented, the final model is more difficult to interpret because it does not perform any kind of variable selection or even directly produce coefficient estimates.
Finally, we fit PCR on the full data set, using \(M=5\), the number of components identified by cross-validation.
pcr.fit <- pcr(y ~ x, scale = TRUE, ncomp = 5)
summary(pcr.fit)## Data: X dimension: 263 19
## Y dimension: 263 1
## Fit method: svdpc
## Number of components considered: 5
## TRAINING: % variance explained
## 1 comps 2 comps 3 comps 4 comps 5 comps
## X 38.31 60.16 70.84 79.03 84.29
## y 40.63 41.58 42.17 43.22 44.90Partial Least Squares
We implement partial least squares (PLS) using the plsr() function, also in the pls library. The syntax is just like that of the pcr() function.
set.seed(1)
pls.fit <- plsr(Salary ~ ., data = Hitters, subset = train, scale = TRUE, validation = "CV")
summary(pls.fit)## Data: X dimension: 131 19
## Y dimension: 131 1
## Fit method: kernelpls
## Number of components considered: 19
##
## VALIDATION: RMSEP
## Cross-validated using 10 random segments.
## (Intercept) 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps
## CV 428.3 325.5 329.9 328.8 339.0 338.9 340.1
## adjCV 428.3 325.0 328.2 327.2 336.6 336.1 336.6
## 7 comps 8 comps 9 comps 10 comps 11 comps 12 comps 13 comps
## CV 339.0 347.1 346.4 343.4 341.5 345.4 356.4
## adjCV 336.2 343.4 342.8 340.2 338.3 341.8 351.1
## 14 comps 15 comps 16 comps 17 comps 18 comps 19 comps
## CV 348.4 349.1 350.0 344.2 344.5 345.0
## adjCV 344.2 345.0 345.9 340.4 340.6 341.1
##
## TRAINING: % variance explained
## 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps 7 comps 8 comps
## X 39.13 48.80 60.09 75.07 78.58 81.12 88.21 90.71
## Salary 46.36 50.72 52.23 53.03 54.07 54.77 55.05 55.66
## 9 comps 10 comps 11 comps 12 comps 13 comps 14 comps 15 comps
## X 93.17 96.05 97.08 97.61 97.97 98.70 99.12
## Salary 55.95 56.12 56.47 56.68 57.37 57.76 58.08
## 16 comps 17 comps 18 comps 19 comps
## X 99.61 99.70 99.95 100.00
## Salary 58.17 58.49 58.56 58.62validationplot(pls.fit, val.type = "MSEP")
The lowest cross-validation error occurs when only \(M=1\) partial least squares directions are used. We now evaluate the corresponding test set MSE.
pls.pred <- predict(pls.fit, x[test, ], ncomp = 1)
mean((pls.pred - y.test)^2)## [1] 151995.3The test MSE is comparable to, but slightly higher than, the test MSE obtained using ridge regression, the lasso, and PCR.
Finally, we perform PLS using the full data set, using \(M=1\), the number of components identified by cross-validation.
pls.fit <- plsr(Salary ~ ., data = Hitters, scale = TRUE,
ncomp = 1)
summary(pls.fit)## Data: X dimension: 263 19
## Y dimension: 263 1
## Fit method: kernelpls
## Number of components considered: 1
## TRAINING: % variance explained
## 1 comps
## X 38.08
## Salary 43.05Notice that the percentage of variance in Salary that the one-component PLS fit explains, \(43.05\,\%\), is almost as much as that explained using the final five-component model PCR fit, \(44.90\,\%\). This is because PCR only attempts to maximize the amount of variance explained in the predictors, while PLS searches for directions that explain variance in both the predictors and the response.