library(ggplot2) # graphics library
library(MASS) # contains data sets
library(ISLR) # contains code and data from the textbook
library(knitr) # contains kable() function
library(boot) # contains cross-validation functions
library(gam) # needed for additive models
## Loading required package: splines
## Loading required package: foreach
## Loaded gam 1.14-4
options(scipen = 4) # Suppresses scientific notation
You will need the
Auto
data set from theISLR
library in order to complete this exercise.
Please run all of the code indicated in §5.3.1 of ISLR, even if I don’t explicitly ask you to do so in this document.
View()
command on the Auto
data to see what the data set looks like.qplot
to construct a scatterplot of mpg
vs horsepower
. Use stat_smooth()
to overlay a linear, quadratic, and cubic polynomial fit to the data.# Edit me
set.seed(1)
to set the seed of the random number generator. This will ensure that your answers match those in the text.# Edit me
sample()
command to construct train
, a vector of observation indexes to be used for the purpose of training your model.# Edit me
mpg
on horsepower
, specifying subset = train
. Save this in a variable called lm.fit
.# Edit me
subset = train
?lm.fit
on the test set (i.e., all points that are not in train
)# Edit me
poly()
command to fit a quadratic regression model of mpg
on horsepower
, specifying subset = train
. Save this in a variable called lm.fit2
.# Edit me
# Edit me
lm.fit2
. How does it compare to the linear regression fit?# Edit me
poly()
command to fit a cubic regression model of mpg
on horsepower
, specifying subset = train
. Save this in a variable called lm.fit3
.# Edit me
# Edit me
# Edit me
# Edit me
set.seed(5)
. You do not have to retype all of the code. You can just change the initial set.seed()
command and see what happens.This exercise introduces you to the
cv.glm()
command from theboot
library, which automates K-fold cross-validation for Generalized Linear Models (GLMs). Linear regression is one example of a GLM. Logistic regression is another. GLMs are not the same as Generalized Additive Models (GAMs).
Please run all of the code indicated in §5.3.2 of ISLR, even if I don’t explicitly ask you to do so in this document.
glm
command to fit a linear regression of mpg
on horsepower
. Call the resulting model glm.fit
Confirm that this gives the same coefficient estimates as a linear model fit with the lm
command.Note: You should fit the model to the entire data, not just to the training data.
# Edit me
cv.error
, the vector of LOOCV error estimates for polynomials of degree 1-5.Note: The computations take some time to run, so I’ve set cache = TRUE
in the code chunk header to make sure that the code below doesn’t re-execute at knit time unless this particular chunk has changed.
# Edit me (insert cross-validation code here)
# Edit me (insert plot code here)
# Edit me
Please run all of the code indicated in §5.3.3 of ISLR
# Edit me
# Edit me
# Edit me
Please run all of the code indicated in §7.8.3 of ISLR, up until the loading of the
akima
package. We have not yet studied logistic regression, so you are not asked to do the logistic regression analysis that starts at the bottom of p. 297. The material on ANOVA testing may also be unfamiliar to you. You may skip it.
# Your comment here
gam1 <- lm(wage ~ ns(year, 4) + ns(age, 5) + education, data=Wage)
# Your comment here
gam.m3 <- gam(wage ~ s(year, 4) + s(age, 5) + education, data=Wage)
# Your comment here
par(mfrow=c(1,3))
# Your comment here
plot(gam.m3, se=TRUE,col="blue")
# Your comment here
plot.gam(gam1, se=TRUE, col="red")
# Your comment here
gam.m1 <- gam(wage ~ s(age,5) + education, data=Wage)
# Your comment here
gam.m2 <- gam(wage ~ year + s(age, 5) + education, data=Wage)
# Your comment here
anova(gam.m1, gam.m2, gam.m3, test="F")
## Analysis of Deviance Table
##
## Model 1: wage ~ s(age, 5) + education
## Model 2: wage ~ year + s(age, 5) + education
## Model 3: wage ~ s(year, 4) + s(age, 5) + education
## Resid. Df Resid. Dev Df Deviance F Pr(>F)
## 1 2990 3711731
## 2 2989 3693842 1 17889.2 14.4771 0.0001447 ***
## 3 2986 3689770 3 4071.1 1.0982 0.3485661
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
# Your comment here
summary(gam.m3)
##
## Call: gam(formula = wage ~ s(year, 4) + s(age, 5) + education, data = Wage)
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -119.43 -19.70 -3.33 14.17 213.48
##
## (Dispersion Parameter for gaussian family taken to be 1235.69)
##
## Null Deviance: 5222086 on 2999 degrees of freedom
## Residual Deviance: 3689770 on 2986 degrees of freedom
## AIC: 29887.75
##
## Number of Local Scoring Iterations: 2
##
## Anova for Parametric Effects
## Df Sum Sq Mean Sq F value Pr(>F)
## s(year, 4) 1 27162 27162 21.981 0.000002877 ***
## s(age, 5) 1 195338 195338 158.081 < 2.2e-16 ***
## education 4 1069726 267432 216.423 < 2.2e-16 ***
## Residuals 2986 3689770 1236
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Anova for Nonparametric Effects
## Npar Df Npar F Pr(F)
## (Intercept)
## s(year, 4) 3 1.086 0.3537
## s(age, 5) 4 32.380 <2e-16 ***
## education
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
# Your comment here
preds <- predict(gam.m2, newdata=Wage)
# Your comment here
gam.lo <- gam(wage ~ s(year, df=4) + lo(age, span=0.7) + education, data=Wage)
# Your comment here
plot.gam(gam.lo, se=TRUE, col="green")
# Your comment here
gam.lo.i <- gam(wage ~ lo(year, age, span=0.5) + education, data=Wage)
## Warning in lo.wam(x, z, wz, fit$smooth, which, fit$smooth.frame,
## bf.maxit, : liv too small. (Discovered by lowesd)
## Warning in lo.wam(x, z, wz, fit$smooth, which, fit$smooth.frame,
## bf.maxit, : lv too small. (Discovered by lowesd)
## Warning in lo.wam(x, z, wz, fit$smooth, which, fit$smooth.frame,
## bf.maxit, : liv too small. (Discovered by lowesd)
## Warning in lo.wam(x, z, wz, fit$smooth, which, fit$smooth.frame,
## bf.maxit, : lv too small. (Discovered by lowesd)
wage
and age
in the gam.lo
model. How do you interpret the dependence plot between wage
and education
?# Edit me
# Edit me
The
splines
library has asmooth.spline()
command with built-in cross-validated smoothness selection. We will now give an example of using this command.
agelims <- range(Wage$age)
# Your comment here
with(Wage, plot(age, wage, xlim = agelims, cex=0.5, col = "darkgrey"))
title("Smoothing Spline")
# Your comment here
fit <- with(Wage, smooth.spline(age, wage, df=16))
# Your comment here
fit2 <- with(Wage, smooth.spline(age, wage, cv=TRUE))
## Warning in smooth.spline(age, wage, cv = TRUE): cross-validation with non-
## unique 'x' values seems doubtful
# Your comment here
fit2$df
## [1] 6.794596
# Your comment here
lines(fit, col="red", lwd=2)
lines(fit2, col="blue", lwd=2)
smooth.spline
function, can you figure out what kind of cross-validation is done when cv = TRUE
? i.e., It’s \(K\)-fold CV for what choice of \(K\)?