Homework 5
Ashley Mata
April 5, 2019
2. For parts (a) through (c), indicate which of i. through iv. is correct. Justify your answer.
(a) The lasso, relative to least squares, is:
- Less flexible and hence will give improved prediction accuracy when its increase in bias is less than its decrease in variance.
(b) Repeat (a) for ridge regression relative to least squares.
- Less flexible and hence will give improved prediction accuracy when its increase in bias is less than its decrease in variance.
(c) Repeat (a) for non-linear methods relative to least squares.
- More flexible and hence will give improved prediction accuracy when its increase in variance is less than its decrease in bias.
9. In this exercise, we will predict the number of applications received using the other variables in the College data set.
(a) Split the data set into a training set and a test set.
library(ISLR)## Warning: package 'ISLR' was built under R version 3.5.2set.seed(11)
sum(is.na(College))## [1] 0train.size = dim(College)[1] / 2
train = sample(1:dim(College)[1], train.size)
test = -train
College.train = College[train, ]
College.test = College[test, ](b) Fit a linear model using least squares on the training set, and report the test error obtained.
lm.fit = lm(Apps~., data=College.train)
lm.pred = predict(lm.fit, College.test)
mean((College.test[, "Apps"] - lm.pred)^2)## [1] 1538442(c) Fit a ridge regression model on the training set, with ?? chosen by cross-validation. Report the test error obtained.
library(glmnet)## Warning: package 'glmnet' was built under R version 3.5.3## Loading required package: Matrix## Loading required package: foreach## Warning: package 'foreach' was built under R version 3.5.3## Loaded glmnet 2.0-16train.mat = model.matrix(Apps~., data=College.train)
test.mat = model.matrix(Apps~., data=College.test)
grid = 10 ^ seq(4, -2, length=100)
mod.ridge = cv.glmnet(train.mat, College.train[, "Apps"], alpha=0, lambda=grid, thresh=1e-12)
lambda.best = mod.ridge$lambda.min
lambda.best## [1] 18.73817ridge.pred = predict(mod.ridge, newx=test.mat, s=lambda.best)
mean((College.test[, "Apps"] - ridge.pred)^2)## [1] 1608859(d) Fit a lasso model on the training set, with ?? chosen by crossvalidation. Report the test error obtained, along with the number of non-zero coefficient estimates.
mod.lasso = cv.glmnet(train.mat, College.train[, "Apps"], alpha=1, lambda=grid, thresh=1e-12)
lambda.best = mod.lasso$lambda.min
lambda.best## [1] 21.54435lasso.pred = predict(mod.lasso, newx=test.mat, s=lambda.best)
mean((College.test[, "Apps"] - lasso.pred)^2)## [1] 1635280The Coefficients:
mod.lasso = glmnet(model.matrix(Apps~., data=College), College[, "Apps"], alpha=1)
predict(mod.lasso, s=lambda.best, type="coefficients")## 19 x 1 sparse Matrix of class "dgCMatrix"
## 1
## (Intercept) -6.038452e+02
## (Intercept) .
## PrivateYes -4.235413e+02
## Accept 1.455236e+00
## Enroll -2.003696e-01
## Top10perc 3.367640e+01
## Top25perc -2.403036e+00
## F.Undergrad .
## P.Undergrad 2.086035e-02
## Outstate -5.781855e-02
## Room.Board 1.246462e-01
## Books .
## Personal 1.832910e-05
## PhD -5.601313e+00
## Terminal -3.313824e+00
## S.F.Ratio 4.478684e+00
## perc.alumni -9.796600e-01
## Expend 6.967693e-02
## Grad.Rate 5.159652e+00(e) Fit a PCR model on the training set, with M chosen by crossvalidation. Report the test error obtained, along with the value of M selected by cross-validation.
library(pls)## Warning: package 'pls' was built under R version 3.5.3##
## Attaching package: 'pls'## The following object is masked from 'package:stats':
##
## loadingspcr.fit = pcr(Apps~., data=College.train, scale=T, validation="CV")
validationplot(pcr.fit, val.type="MSEP")
pcr.pred = predict(pcr.fit, College.test, ncomp=10)
mean((College.test[, "Apps"] - data.frame(pcr.pred))^2)## Warning in mean.default((College.test[, "Apps"] - data.frame(pcr.pred))^2):
## argument is not numeric or logical: returning NA## [1] NA(f) Fit a PLS model on the training set, with M chosen by crossvalidation. Report the test error obtained, along with the value of M selected by cross-validation.
pls.fit = plsr(Apps~., data=College.train, scale=T, validation="CV")
validationplot(pls.fit, val.type="MSEP")
pls.pred = predict(pls.fit, College.test, ncomp=10)
mean((College.test[, "Apps"] - data.frame(pls.pred))^2)## Warning in mean.default((College.test[, "Apps"] - data.frame(pls.pred))^2):
## argument is not numeric or logical: returning NA## [1] NA(g) Comment on the results obtained. How accurately can we predict the number of college applications received? Is there much difference among the test errors resulting from these five approaches?
test.avg = mean(College.test[, "Apps"])
lm.test.r2 = 1 - mean((College.test[, "Apps"] - lm.pred)^2) /mean((College.test[, "Apps"] - test.avg)^2)
ridge.test.r2 = 1 - mean((College.test[, "Apps"] - ridge.pred)^2) /mean((College.test[, "Apps"] - test.avg)^2)
lasso.test.r2 = 1 - mean((College.test[, "Apps"] - lasso.pred)^2) /mean((College.test[, "Apps"] - test.avg)^2)
pcr.test.r2 = 1 - mean((College.test[, "Apps"] - data.frame(pcr.pred))^2) /mean((College.test[, "Apps"] - test.avg)^2)## Warning in mean.default((College.test[, "Apps"] - data.frame(pcr.pred))^2):
## argument is not numeric or logical: returning NApls.test.r2 = 1 - mean((College.test[, "Apps"] - data.frame(pls.pred))^2) /mean((College.test[, "Apps"] - test.avg)^2)## Warning in mean.default((College.test[, "Apps"] - data.frame(pls.pred))^2):
## argument is not numeric or logical: returning NAbarplot(c(lm.test.r2, ridge.test.r2, lasso.test.r2, pcr.test.r2, pls.test.r2), col="red", names.arg=c("OLS", "Ridge", "Lasso", "PCR", "PLS"), main="Test R-squared")
From the chart we can notice that all of the models except for PCR predict college applications with high accuracy. All of the modles excpet for PCR are around .9.
11. We will now try to predict per capita crime rate in the Boston data set.
(a) Try out some of the regression methods explored in this chapter, such as best subset selection, the lasso, ridge regression, and PCR. Present and discuss results for the approaches that you consider.
set.seed(1)
library(MASS)
library(leaps)## Warning: package 'leaps' was built under R version 3.5.3library(glmnet)Subset Selection
predict.regsubsets = function(object, newdata, id, ...) {
form = as.formula(object$call[[2]])
mat = model.matrix(form, newdata)
coefi = coef(object, id = id)
mat[, names(coefi)] %*% coefi
}
k = 10
p = ncol(Boston) - 1
folds = sample(rep(1:k, length = nrow(Boston)))
cv.errors = matrix(NA, k, p)
for (i in 1:k) {
best.fit = regsubsets(crim ~ ., data = Boston[folds != i, ], nvmax = p)
for (j in 1:p) {
pred = predict(best.fit, Boston[folds == i, ], id = j)
cv.errors[i, j] = mean((Boston$crim[folds == i] - pred)^2)
}
}
rmse.cv = sqrt(apply(cv.errors, 2, mean))
plot(rmse.cv, pch = 19, type = "b")
Lasso
x = model.matrix(crim ~ . - 1, data = Boston)
y = Boston$crim
cv.lasso = cv.glmnet(x, y, type.measure = "mse")
plot(cv.lasso)
coef(cv.lasso)## 14 x 1 sparse Matrix of class "dgCMatrix"
## 1
## (Intercept) 1.0894283
## zn .
## indus .
## chas .
## nox .
## rm .
## age .
## dis .
## rad 0.2643196
## tax .
## ptratio .
## black .
## lstat .
## medv .sqrt(cv.lasso$cvm[cv.lasso$lambda == cv.lasso$lambda.1se])## [1] 7.405176Ridge regression
x = model.matrix(crim ~ . - 1, data = Boston)
y = Boston$crim
cv.ridge = cv.glmnet(x, y, type.measure = "mse", alpha = 0)
plot(cv.ridge)
coef(cv.ridge)## 14 x 1 sparse Matrix of class "dgCMatrix"
## 1
## (Intercept) 1.017516870
## zn -0.002805664
## indus 0.034405928
## chas -0.225250601
## nox 2.249887495
## rm -0.162546004
## age 0.007343331
## dis -0.114928730
## rad 0.059813843
## tax 0.002659110
## ptratio 0.086423005
## black -0.003342067
## lstat 0.044495212
## medv -0.029124577sqrt(cv.ridge$cvm[cv.ridge$lambda == cv.ridge$lambda.1se])## [1] 7.456946PCR
library(pls)pcr.fit = pcr(crim ~ ., data = Boston, scale = TRUE, validation = "CV")
summary(pcr.fit)## Data: X dimension: 506 13
## Y dimension: 506 1
## Fit method: svdpc
## Number of components considered: 13
##
## VALIDATION: RMSEP
## Cross-validated using 10 random segments.
## (Intercept) 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps
## CV 8.61 7.170 7.163 6.733 6.728 6.740 6.765
## adjCV 8.61 7.169 7.162 6.730 6.723 6.737 6.760
## 7 comps 8 comps 9 comps 10 comps 11 comps 12 comps 13 comps
## CV 6.760 6.634 6.652 6.642 6.652 6.607 6.546
## adjCV 6.754 6.628 6.644 6.635 6.643 6.598 6.536
##
## TRAINING: % variance explained
## 1 comps 2 comps 3 comps 4 comps 5 comps 6 comps 7 comps
## X 47.70 60.36 69.67 76.45 82.99 88.00 91.14
## crim 30.69 30.87 39.27 39.61 39.61 39.86 40.14
## 8 comps 9 comps 10 comps 11 comps 12 comps 13 comps
## X 93.45 95.40 97.04 98.46 99.52 100.0
## crim 42.47 42.55 42.78 43.04 44.13 45.4(b) Propose a model (or set of models) that seem to perform well on this data set, and justify your answer. Make sure that you are evaluating model performance using validation set error, crossvalidation, or some other reasonable alternative, as opposed to using training error.
Done above
(c) Does your chosen model involve all of the features in the data set? Why or why not? The 9 parameter best subset model because it had the best cross-validated RMSE, next to PCR, but it was simpler model than the 13 component PCR model.