Exercise 1
 In the ISwR data set alkfos, do a PCA of the placebo and
Tamoxifen groups separately, then together. Plot the first two
principal components of the whole group with color coding for
the treatment and control subjects. For this and other parts to
this assignment, omit the patients with missing data.
 Conduct a linear discriminant analysis of the two groups using
the 7 variables. How well can you predict the treatment? Is this
the usual kind of analysis you would see?
 Use logistic regression to predict the group based on the
measurements. Compare the in-sample error rates.
 Use cross-validation with repeated training subsamples of 30/35
and test sets of size 5/35. What can you now conclude about the
two methods?
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Exercise 2
 In the ISwR data set alkfos, cluster the data based on
the 7 measurements using hclust(), kmeans(), and
Mclust().
 Compare the 2-group clustering with the
placebo/Tamoxifen classification.
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> alkfos2 <- na.omit(alkfos)
# omits missing values
> pc1 <- prcomp(alkfos2[alkfos2[,1]==1,2:8],scale=T)
> pc2 <- prcomp(alkfos2[alkfos2[,1]==2,2:8],scale=T)
> pc.all <- prcomp(alkfos2[,2:8],scale=T)
Standard deviations:
[1] 2.3731316 0.7123154 0.6122286 0.4955545 0.3208553 0.2954631 0.2240720
Rotation:
PC1
c0 0.3484871
c3 0.3887022
c6 0.3418856
c9 0.4064443
c12 0.3809874
c18 0.3913700
c24 0.3834877
PC2
-0.54632917
0.16108868
0.76808477
-0.02858575
-0.22109261
0.07215108
-0.17525921
PC3
PC4
PC5
PC6
PC7
0.6016554 0.21896409 0.39427330 -0.106360261 0.05816212
0.4384481 0.06722018 -0.75501296 0.184741080 -0.14843167
0.2118390 -0.03486734 0.48709933 0.097438042 -0.01756660
-0.1832107 -0.27713072 -0.12380851 -0.132875647 0.83104381
-0.1586836 -0.74610277 0.08094328 -0.004681993 -0.46641516
-0.3589670 0.40921911 -0.07419710 -0.689796022 -0.25294738
-0.4618380 0.38129961 0.09926056 0.671987832 -0.04601444
> plot(pc.all)
> plot(pc.all$x,col=alkfos2[,1])
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> library(MASS)
> alkfos.lda <- lda(alkfos2[,2:8],grouping=alkfos2[,1])
> alkfos.lda
Call:
lda(alkfos2[, 2:8], grouping = alkfos2[, 1])
Prior probabilities of groups:
1
2
0.6 0.4
Group means:
c0
c3
c6
c9
c12
c18
c24
1 156.7143 161.8571 173.9048 158.4286 163.8571 164.3333 163.2857
2 164.2143 125.1429 123.7143 118.7143 117.2857 130.7857 134.8571
Coefficients of linear discriminants:
LD1
c0
0.0618073455
c3 -0.0329471378
c6
0.0004421163
c9 -0.0232320119
c12 -0.0248954902
c18 0.0113410946
c24 0.0003473940
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> plot(alkfos.lda)
> alkfos.pred <- predict(alkfos.lda)
> table(alkfos2$grp,alkfos.pred$class)
1 2
1 20 1
2 0 14
34 in 35 correct.
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> alkfos.glm <- glm(as.factor(grp) ~ 1,data=alkfos2,family=binomial)
> step(alkfos.glm,scope=formula(~ c0+c3+c6+c9+c12+c18+c24),steps=2)
Start: AIC=49.11
as.factor(grp) ~ 1
Df Deviance
AIC
+ c6
1
38.475 42.475
+ c12
1
39.116 43.116
+ c9
1
39.484 43.484
+ c3
1
41.721 45.721
+ c18
1
43.291 47.291
+ c24
1
44.093 48.093
<none>
47.111 49.111
+ c0
1
46.869 50.869
Step: AIC=42.47
as.factor(grp) ~ c6
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> alkfos.glm <- glm(as.factor(grp) ~ 1,data=alkfos2,family=binomial)
> step(alkfos.glm,scope=formula(~ c0+c3+c6+c9+c12+c18+c24),steps=2)
Step: AIC=42.47
as.factor(grp) ~ c6
Df Deviance
AIC
+ c0
1
24.281 30.281
<none>
38.475 42.475
+ c18
1
37.286 43.286
+ c24
1
37.509 43.509
+ c12
1
37.545 43.545
+ c3
1
38.113 44.113
+ c9
1
38.128 44.128
- c6
1
47.111 49.111
Step: AIC=30.28
as.factor(grp) ~ c6 + c0
We used step limited to two steps to avoid a model with undetermined coefficients.
Once the predictions are perfect (with three or more variables in this case), nothing
can be distinguished.
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alkfos.lda.cv <- function(ncv,ntrials)
{
require(MASS)
data(alkfos)
alkfos2 <- na.omit(alkfos)
n1 <- dim(alkfos2)[1]
nwrong <- 0
npred <- 0
for (i in 1:ntrials)
{
test <- sample(n1,ncv)
test.set <- data.frame(alkfos2[test,2:8])
train.set <- data.frame(alkfos2[-test,2:8])
lda.ap <- lda(train.set,alkfos2[-test,1])
lda.pred <- predict(lda.ap,test.set)
nwrong <- nwrong + sum(lda.pred$class != alkfos2[test,1])
npred <- npred + ncv
}
print(paste("total number classified = ",npred,sep=""))
print(paste("total number wrong = ",nwrong,sep=""))
print(paste("percent wrong = ",100*nwrong/npred,"%",sep=""))
}
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alkfos.glm.cv <- function(ncv,ntrials)
{
require(MASS)
data(alkfos)
alkfos2 <- na.omit(alkfos)
alkfos2$grp <- as.factor(alkfos2$grp)
n1 <- dim(alkfos2)[1]
nwrong <- 0
npred <- 0
for (i in 1:ntrials)
{
test <- sample(n1,ncv)
test.set <- alkfos2[test,]
train.set <- alkfos2[-test,]
glm.ap <- glm(grp ~ 1,data=train.set,family=binomial)
glmstep.ap <- step(glm.ap,scope=formula(~ c0+c3+c6+c9+c12+c18+c24),steps=2,trace=0)
glm.pred <- predict(glmstep.ap,newdata=test.set,type="response")
grp.pred <- (glm.pred > 0.5)+1
nwrong <- nwrong + sum(grp.pred != test.set$grp)
npred <- npred + ncv
}
print(paste("total number classified = ",npred,sep=""))
print(paste("total number wrong = ",nwrong,sep=""))
print(paste("percent wrong = ",100*nwrong/npred,"%",sep=""))
}
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Results of Cross Validation
 LDA has 1 error in 35 in sample (2.9%)
 Cross-Validated seven-fold this is 720/10000 = 7.2%
 Stepwise logistic regression with two variables has 3
errors in 35 in sample (8.6%)
 Cross-Validated seven-fold this is 1830/10000 = 18.3%
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> ap.hc <- hclust(dist(alkfos2[,2:8]))
> plot(ap.hc)
> cutree(ap.hc, 2)
1 2 3 4 5 7 8 9 10 11 12 13 14 15 16 17 19 20 21 22 23 24 25 27 28 29
1 1 2 2 1 2 2 1 1 2 2 1 2 2 2 1 1 1 1 1 1 1 1 1 1 1
30 31 33 34 35 37 38 40 42
1 1 1 1 1 1 1 2 1
> table(cutree(ap.hc, 2),alkfos2$grp)
1 2
1 12 13
2 9 1
> table(kmeans(alkfos2[,2:8],2)$cluster,alkfos2$grp)
1 2
1 11 11
2 10 3
> library(mclust)
> Mclust(alkfos2[,2:8])
'Mclust' model object:
best model: ellipsoidal, equal shape (VEV) with 6 components
> table(Mclust(alkfos2[,2:8])$class,alkfos2$grp)
1 2
1 4 3
2 6 0
3 4 4
4 6 0
5 1 4
6 0 3
> table(Mclust(alkfos2[,2:8],G=2)$class,alkfos2$grp)
1 2
1 15 14
2 6 0
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