Assignment 3
Sarah K Lovaas
6-6-2021
Chapter 04 (page 168): 10, 11, 13
Chapter 4 ex 10:
library(ISLR)## Warning: package 'ISLR' was built under R version 4.0.5fix(Weekly)
summary(Weekly)## Year Lag1 Lag2 Lag3
## Min. :1990 Min. :-18.1950 Min. :-18.1950 Min. :-18.1950
## 1st Qu.:1995 1st Qu.: -1.1540 1st Qu.: -1.1540 1st Qu.: -1.1580
## Median :2000 Median : 0.2410 Median : 0.2410 Median : 0.2410
## Mean :2000 Mean : 0.1506 Mean : 0.1511 Mean : 0.1472
## 3rd Qu.:2005 3rd Qu.: 1.4050 3rd Qu.: 1.4090 3rd Qu.: 1.4090
## Max. :2010 Max. : 12.0260 Max. : 12.0260 Max. : 12.0260
## Lag4 Lag5 Volume Today
## Min. :-18.1950 Min. :-18.1950 Min. :0.08747 Min. :-18.1950
## 1st Qu.: -1.1580 1st Qu.: -1.1660 1st Qu.:0.33202 1st Qu.: -1.1540
## Median : 0.2380 Median : 0.2340 Median :1.00268 Median : 0.2410
## Mean : 0.1458 Mean : 0.1399 Mean :1.57462 Mean : 0.1499
## 3rd Qu.: 1.4090 3rd Qu.: 1.4050 3rd Qu.:2.05373 3rd Qu.: 1.4050
## Max. : 12.0260 Max. : 12.0260 Max. :9.32821 Max. : 12.0260
## Direction
## Down:484
## Up :605
##
##
##
## pairs(Weekly)
par(mfrow=c(2,2))
plot(Weekly$Lag1, Weekly$Year)
plot(Weekly$Lag2, Weekly$Year)
plot(Weekly$Lag3, Weekly$Year)
plot(Weekly$Lag4, Weekly$Year)
plot(Weekly$Lag1, Weekly$Lag2)
plot(Weekly$Lag2, Weekly$Lag3)
plot(Weekly$Lag3, Weekly$Lag4)
plot(Weekly$Lag4, Weekly$Lag5)
par(mfrow=c(1,1))
plot(Weekly$Today, Weekly$Volume)
Lovaas answer ex 10a: there aren’t terribly strong linear patterns, and I’m not sure what we’re going to do with this data. Volume tends to increase a bit when returns are high, and vise versa.
glm.fits=glm(Direction~Lag1+Lag2+Lag3+Lag4+Lag5+Volume,data=Weekly,family=binomial)
summary(glm.fits)##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 +
## Volume, family = binomial, data = Weekly)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.6949 -1.2565 0.9913 1.0849 1.4579
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.26686 0.08593 3.106 0.0019 **
## Lag1 -0.04127 0.02641 -1.563 0.1181
## Lag2 0.05844 0.02686 2.175 0.0296 *
## Lag3 -0.01606 0.02666 -0.602 0.5469
## Lag4 -0.02779 0.02646 -1.050 0.2937
## Lag5 -0.01447 0.02638 -0.549 0.5833
## Volume -0.02274 0.03690 -0.616 0.5377
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1496.2 on 1088 degrees of freedom
## Residual deviance: 1486.4 on 1082 degrees of freedom
## AIC: 1500.4
##
## Number of Fisher Scoring iterations: 4Lovaas answer ex 10b Only lag2 appears to be significant, which is a bummer.
glm.probs = predict(glm.fits, type = "response")
glm.preds = rep("Down", length(glm.probs))
glm.preds[glm.probs > 0.5] = "Up"
attach(Weekly)
table(glm.preds, Direction)#confusion matrix## Direction
## glm.preds Down Up
## Down 54 48
## Up 430 557(54+557)/(54+48+430+557) #56% accuracy.## [1] 0.5610652The confusion matrix is telling us which observations were classified by the model as down or up vs which were actually down and up. The market was down (54 + 430) times, but the model only predicted it being down 54 times. The model was better at predicting market up - the market was actually up (48 + 557) times and the model predicted that 557 times.
train = (Year < 2009) #train dataset
Weekly.test = Weekly[!train, ] #just 2009 and 2010
glm.train = glm(Direction ~ Lag2, data = Weekly, family = binomial, subset = train) #note subset
glm.probs = predict(glm.train, Weekly.test, type = "response")#predictions
glm.preds = rep("Down", length(glm.probs))
glm.preds[glm.probs > 0.5] = "Up" #classification
Direction.test = Direction[!train] #Removed training data
table(glm.preds, Direction.test) #Confusion matrix## Direction.test
## glm.preds Down Up
## Down 9 5
## Up 34 56(9+56)/(9+5+34+56) #fraction of correct predictions## [1] 0.625library(MASS)## Warning: package 'MASS' was built under R version 4.0.5lda.train = lda(Direction ~ Lag2, data = Weekly, subset = train) #note subset
lda.pred = predict(lda.train, Weekly.test)
table(lda.pred$class, Direction.test)## Direction.test
## Down Up
## Down 9 5
## Up 34 56(9+56)/(9+5+34+56)#fraction of correct predictions## [1] 0.625Lovaas note ex 10e: very suspicious how the confusion matrix was identical to 10d.
qda.train = qda(Direction ~ Lag2, data = Weekly, subset = train) #note subset
qda.pred = predict(qda.train, Weekly.test)
table(qda.pred$class, Direction.test)## Direction.test
## Down Up
## Down 0 0
## Up 43 61(0+61)/(0+0+43+61)#fraction of correct predictions## [1] 0.5865385Lovaas note ex 10f: Don’t love how the model predicted 0 “down” weeks.
library(class)## Warning: package 'class' was built under R version 4.0.5train.X = as.matrix(Lag2[train])
test.X = as.matrix(Lag2[!train])
train.Direction = Direction[train]
set.seed(21)
knn.pred = knn(train.X, test.X, train.Direction, k = 1)
table(knn.pred, Direction.test)## Direction.test
## knn.pred Down Up
## Down 21 29
## Up 22 32(21+32)/(21+29+22+32)#fraction of correct predictions## [1] 0.509615451% accuracy; that can vary a tiny bit based on the set.seed()
- Which of these methods appears to provide the best results on this data?
Lovaas answer: GLM and LDA are tied for the best accuracy score at 62.5%
- Experiment with different combinations of predictors, including possible transformations and interactions, for each of the methods. Report the variables, method, and associated confusion matrix that appears to provide the best results on the held out data. Note that you should also experiment with values for K in the KNN classifier.
#playing around with diff values of
train.X = as.matrix(Lag2[train])
test.X = as.matrix(Lag2[!train])
train.Direction = Direction[train]
set.seed(21)
knn.pred2 = knn(train.X, test.X, train.Direction, k = 2)
table(knn.pred2, Direction.test)## Direction.test
## knn.pred2 Down Up
## Down 23 33
## Up 20 28(23+28)/(23+33+20+28)#fraction of correct predictions## [1] 0.4903846knn.pred3 = knn(train.X, test.X, train.Direction, k = 3)
table(knn.pred3, Direction.test)## Direction.test
## knn.pred3 Down Up
## Down 16 19
## Up 27 42(16+42)/(16+19+27+42)#fraction of correct predictions## [1] 0.5576923knn.pred4 = knn(train.X, test.X, train.Direction, k = 4)
table(knn.pred4, Direction.test)## Direction.test
## knn.pred4 Down Up
## Down 17 23
## Up 26 38(17+38)/(17+23+26+38)#fraction of correct predictions## [1] 0.5288462#Using volume in place of Lag2
glm.train = glm(Direction ~ Volume, data = Weekly, family = binomial, subset = train) #note subset
glm.probs = predict(glm.train, Weekly.test, type = "response")#predictions
glm.preds = rep("Down", length(glm.probs))
glm.preds[glm.probs > 0.5] = "Up" #classification
Direction.test = Direction[!train] #Removed training data
table(glm.preds, Direction.test)#Confusion matrix## Direction.test
## glm.preds Down Up
## Down 31 47
## Up 12 14(31+14)/(31+47+12+14)#fraction of correct predictions## [1] 0.4326923lda.train = lda(Direction ~ Volume, data = Weekly, subset = train) #note subset
lda.pred = predict(lda.train, Weekly.test)
table(lda.pred$class, Direction.test)## Direction.test
## Down Up
## Down 33 47
## Up 10 14(33+14)/(33+47+10+14)#fraction of correct predictions## [1] 0.4519231Lovaas notes: KNN with k = 3 has the highest success rate of all KNN, but still lower than GLM. Using volume in lieu of Lag2 didn’t improve prediction accuracy.
Chapter 4 ex 11:
mpg01 = rep(0, length(Auto$mpg))
Auto1<- Auto
mpg01[Auto$mpg > median(Auto$mpg)] = 1
Auto1 = data.frame(Auto1, mpg01)par(mfrow=c(2,2))
plot(Auto1$mpg01, Auto1$Year)
plot(Auto1$mpg01, Auto1$Cylinders)
plot(Auto1$mpg01, Auto1$Acceleration)
plot(Auto1$mpg01, Auto1$Horsepower)
Lovaas answer 11b: honestly, I didn’t find too much. We’ll be able to set up a model but I doubt the rsquared values will be above 50%.
attach(Auto1)## The following object is masked _by_ .GlobalEnv:
##
## mpg01test = (year%%5 == 0) # if the year is divisible by 5; this leaves 20% of the data for testing.
train = !test
train.x = Auto1[train, ]
test.x = Auto1[test, ]
mpg01.test = mpg01[test]lda.fit=lda(mpg01~year + cylinders + acceleration + horsepower,data=Auto1,subset=train)
lda.fit## Call:
## lda(mpg01 ~ year + cylinders + acceleration + horsepower, data = Auto1,
## subset = train)
##
## Prior probabilities of groups:
## 0 1
## 0.503268 0.496732
##
## Group means:
## year cylinders acceleration horsepower
## 0 74.91558 6.675325 14.74416 127.4545
## 1 77.67763 4.217105 16.48092 78.3750
##
## Coefficients of linear discriminants:
## LD1
## year 0.11329929
## cylinders -0.57588930
## acceleration -0.10862945
## horsepower -0.01686372lda.pred=predict(lda.fit, test.x)
mean(lda.pred$class != mpg01.test)## [1] 0.01162791Lovaas answer ex 11d: test error using this size training/test data is 0.0116, which is lower than I expected. I guess since we’re just testing whether MPG is below or above the median it’s easier to fit a model.
qda.fit=qda(mpg01~year + cylinders + acceleration + horsepower,data=Auto1,subset=train)
qda.fit## Call:
## qda(mpg01 ~ year + cylinders + acceleration + horsepower, data = Auto1,
## subset = train)
##
## Prior probabilities of groups:
## 0 1
## 0.503268 0.496732
##
## Group means:
## year cylinders acceleration horsepower
## 0 74.91558 6.675325 14.74416 127.4545
## 1 77.67763 4.217105 16.48092 78.3750qda.pred=predict(qda.fit,test.x)
mean(qda.pred$class !=mpg01.test)## [1] 0.06976744Lovaas answer 11e: Error is 0.0698, which is higher than in 11d.
glm.fit=glm(mpg01~year + cylinders + acceleration + horsepower,data=Auto1,subset=train,family=binomial)
summary(glm.fit)##
## Call:
## glm(formula = mpg01 ~ year + cylinders + acceleration + horsepower,
## family = binomial, data = Auto1, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.0399 -0.1944 -0.0004 0.2986 3.2065
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -4.54409 5.11283 -0.889 0.37413
## year 0.32944 0.06788 4.854 1.21e-06 ***
## cylinders -0.68342 0.23622 -2.893 0.00381 **
## acceleration -0.37853 0.11931 -3.173 0.00151 **
## horsepower -0.11651 0.02209 -5.274 1.34e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 424.19 on 305 degrees of freedom
## Residual deviance: 151.74 on 301 degrees of freedom
## AIC: 161.74
##
## Number of Fisher Scoring iterations: 7glm.prob = predict(glm.fit, test.x, type = "response")
glm.pred = rep(0, length(glm.prob))
glm.pred[glm.prob > 0.5] = 1
mean(glm.pred != mpg01.test)## [1] 0.127907glm.pred[glm.prob > 0.6] = 1
mean(glm.pred != mpg01.test)## [1] 0.127907Lovaas answer ex 11f: test error is 12.8%; regardless of where I set the classification (0.5 or 0.6)
library(class)
train.X2 = cbind(cylinders, weight, displacement, horsepower)[train, ]
test.X2 = cbind(cylinders, weight, displacement, horsepower)[test, ]
train.mpg01 = mpg01[train]
set.seed(1)
knn.pred = knn(train.X2, test.X2, train.mpg01, k = 1)
mean(knn.pred != mpg01.test)## [1] 0.1046512knn.pred = knn(train.X2, test.X2, train.mpg01, k = 10)
mean(knn.pred != mpg01.test)## [1] 0.1046512knn.pred = knn(train.X2, test.X2, train.mpg01, k = 50)
mean(knn.pred != mpg01.test)## [1] 0.1046512knn.pred = knn(train.X2, test.X2, train.mpg01, k = 100)
mean(knn.pred != mpg01.test)## [1] 0.1046512knn.pred = knn(train.X2, test.X2, train.mpg01, k = 200)
mean(knn.pred != mpg01.test)## [1] 0.09302326Lovaas answer ex 11g: many test errors, see above. 1 through 100 all got the same result! I got an error where I mistakenly selected 500 nearest neighbors since there are only 306 data items in my training dataset. k=200 seemed to work best but I feel like that’s overfitting.
Chapter 4 ex 13:
# Set up
library(MASS)
#fix(Boston)
attach(Boston)
crime1 = rep(0, length(crim))
crime1[crim > median(crim)] = 1 #above or below median.
Boston1 = data.frame(Boston, crime1)
train = 1:(dim(Boston1)[1]/2) #train data
test = (dim(Boston1)[1]/2 + 1):dim(Boston1)[1] #test data
Boston1.train = Boston1[train, ]
Boston1.test = Boston1[test, ]
crime1.test = crime1[test]# Logistic Regression
glm.fit=glm(crime1~medv + tax + ptratio + black,data=Boston1,subset=train,family=binomial)
summary(glm.fit)##
## Call:
## glm(formula = crime1 ~ medv + tax + ptratio + black, family = binomial,
## data = Boston1, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -4.3005 -0.6995 -0.5220 0.9033 1.9041
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 2.772593 3.374992 0.822 0.411355
## medv 0.019422 0.020704 0.938 0.348204
## tax 0.011496 0.002359 4.873 1.1e-06 ***
## ptratio 0.177576 0.079955 2.221 0.026354 *
## black -0.028428 0.008171 -3.479 0.000503 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 329.37 on 252 degrees of freedom
## Residual deviance: 260.52 on 248 degrees of freedom
## AIC: 270.52
##
## Number of Fisher Scoring iterations: 6glm.prob = predict(glm.fit, test.x, type = "response")## Warning: 'newdata' had 86 rows but variables found have 506 rowsglm.pred = rep(0, length(glm.prob))
glm.pred[glm.prob > 0.5] = 1
mean(glm.pred != crime1.test)## [1] 0.326087I used median home value, property tax, pupil to teacher ratio, and race (proportion of african american citizens). All were statistically significant (assuming p = 0.05) except for median home value. The error rate was still 33% though, which is pretty high.
# LDA
lda.fit=lda(crime1~medv + rad + ptratio + black,data=Boston1,subset=train)
lda.pred=predict(lda.fit, test.x)## Warning: 'newdata' had 86 rows but variables found have 506 rowsmean(lda.pred$class != crime1.test)## [1] 0.355731235% misclassification; not great. I used median home value, access to radial highways, pupil to teacher ratio, and race (proportion of african american citizens).
# KNN
train.X2 = cbind(age,dis,ptratio,tax)[train, ]
test.X2 = cbind(age,dis,ptratio,tax)[test, ]
train.crim1 = crime1[train]
set.seed(1)
knn.pred = knn(train.X2, test.X2, train.crim1, k = 1)
mean(knn.pred != crime1.test)## [1] 0.6363636knn.pred = knn(train.X2, test.X2, train.crim1, k = 10)
mean(knn.pred != crime1.test)## [1] 0.1225296knn.pred = knn(train.X2, test.X2, train.crim1, k = 50)
mean(knn.pred != crime1.test)## [1] 0.1304348knn.pred = knn(train.X2, test.X2, train.crim1, k = 20)
mean(knn.pred != crime1.test)## [1] 0.1185771knn.pred = knn(train.X2, test.X2, train.crim1, k = 15)
mean(knn.pred != crime1.test)## [1] 0.1185771I used home age,distance to employment centers, pupil to teach ratio, and property tax. The optimal model is around k = 15 to 20 (nearest neighbors)That makes sense; anything higher seems like an extreme number of neighbors.