Assignment 3: Classification
Andrew Mazza
Question 10
(a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
library(ISLR)
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)
cor(Weekly[,-9])## Year Lag1 Lag2 Lag3 Lag4
## Year 1.00000000 -0.032289274 -0.03339001 -0.03000649 -0.031127923
## Lag1 -0.03228927 1.000000000 -0.07485305 0.05863568 -0.071273876
## Lag2 -0.03339001 -0.074853051 1.00000000 -0.07572091 0.058381535
## Lag3 -0.03000649 0.058635682 -0.07572091 1.00000000 -0.075395865
## Lag4 -0.03112792 -0.071273876 0.05838153 -0.07539587 1.000000000
## Lag5 -0.03051910 -0.008183096 -0.07249948 0.06065717 -0.075675027
## Volume 0.84194162 -0.064951313 -0.08551314 -0.06928771 -0.061074617
## Today -0.03245989 -0.075031842 0.05916672 -0.07124364 -0.007825873
## Lag5 Volume Today
## Year -0.030519101 0.84194162 -0.032459894
## Lag1 -0.008183096 -0.06495131 -0.075031842
## Lag2 -0.072499482 -0.08551314 0.059166717
## Lag3 0.060657175 -0.06928771 -0.071243639
## Lag4 -0.075675027 -0.06107462 -0.007825873
## Lag5 1.000000000 -0.05851741 0.011012698
## Volume -0.058517414 1.00000000 -0.033077783
## Today 0.011012698 -0.03307778 1.000000000Looking at the plots produced from the pairs function, we can see a notable pattern between the year and volume variables. If we also look at the correlation matrix, we can see that the year and volume variables have a very high correlation of 0.8419.
(b) Use the full data set to perform a logistic regression with Direction as the response and the five lag variables plus Volume as predictors. Use the summary function to print the results. Do any of the predictors appear to be statistically significant? If so, which ones?
attach(Weekly)
Weekly.fit=glm(Direction ~ Lag1+Lag2+Lag3+Lag4+Lag5+Volume, data=Weekly, family=binomial)
summary(Weekly.fit)##
## 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: 4After this model is produced, we can see that the only predictor that is significant at the 0.05 level is the Lag2 variable. For all other variables we fail to reject the null hypothesis.
(c) Compute the confusion matrix and overall fraction of correct predictions. Explain what the confusion matrix is telling you about the types of mistakes made by logistic regression.
Weekly.prob=predict(Weekly.fit, type='response')
Weekly.pred=rep("Down", length(Weekly.prob))
Weekly.pred[Weekly.prob > 0.5] = "Up"
table(Weekly.pred, Direction)## Direction
## Weekly.pred Down Up
## Down 54 48
## Up 430 557(54+557)/1089## [1] 0.5610652Here we have illustrated that this model will correctly predict the market trend roughly 56% of the time. The mistake made by logistic regression is that there is a large difference in the percentage of predicting up trends and down trends. In fact this model will predict up trends correctly 92% of the time, however the model will only predict about 11% of the down trends correctly.
(d) Now fit the logistic regression model using a training data period from 1990 to 2008, with Lag2 as the only predictor. Compute the confusion matrix and the overall fraction of correct predictions for the held out data (that is, the data from 2009 and 2010)
train = (Year<2009)
Weekly.2009 = Weekly[!train,]
Weekly.fit1 = glm(Direction ~ Lag2, data=Weekly, family=binomial, subset=train)
Weekly.prob = predict(Weekly.fit1, Weekly.2009, type='response')
Weekly.pred = rep("Down", length(Weekly.prob))
Weekly.pred[Weekly.prob > 0.5]= "Up"
Direction.2009 = Direction[!train]
table(Weekly.pred, Direction.2009)## Direction.2009
## Weekly.pred Down Up
## Down 9 5
## Up 34 56mean(Weekly.pred == Direction.2009)## [1] 0.625We can see that this model is improved from our prior model, our prediction accuracy is up to 62.5% and it was also improved in minimizing the difference between predicting the up trend vs. the down trend. The up trend is still around 92% while the prediction of the down trend is up to 21%.
(e) Repeat (d) using LDA.
library(MASS)
Weeklylda.fit = lda(Direction~Lag2, data=Weekly, family=binomial, subset=train)
Weeklylda.pred = predict(Weeklylda.fit, Weekly.2009)
table(Weeklylda.pred$class, Direction.2009)## Direction.2009
## Down Up
## Down 9 5
## Up 34 56mean(Weeklylda.pred$class==Direction.2009)## [1] 0.625From the model using the linear discriminant analysis, we can see that we obtain the same result as the logistic model.
(f) Repeat (d) using QDA.
Weeklyqda.fit = qda(Direction~Lag2, data=Weekly, subset=train)
Weeklyqda.pred = predict(Weeklyqda.fit, Weekly.2009)$class
table(Weeklyqda.pred, Direction.2009)## Direction.2009
## Weeklyqda.pred Down Up
## Down 0 0
## Up 43 61mean(Weeklyqda.pred==Direction.2009)## [1] 0.5865385From the QDA model we can see that we do not get a better prediction percentage than the previous two models. This model only yields a 59% accuracy.
(g) Repeat (d) using KNN with K = 1.
library(class)
attach(Weekly)## The following objects are masked from Weekly (pos = 5):
##
## Direction, Lag1, Lag2, Lag3, Lag4, Lag5, Today, Volume, YearWeek.train=as.matrix(Lag2[train])
Week.test=as.matrix(Lag2[!train])
train.Direction=Direction[train]
set.seed(1)
Weeklyknn.pred=knn(Week.train,Week.test,train.Direction, k=1)
table(Weeklyknn.pred, Direction.2009)## Direction.2009
## Weeklyknn.pred Down Up
## Down 21 30
## Up 22 31mean(Weeklyknn.pred == Direction.2009)## [1] 0.5We can see that the knn test is not as good as any of the three previous tests. We get an accuracy rate of 50%.
(h) Which of these methods appears to provide the best results on this data?
After running all the different methods of analysis, we can conclude that the LDA and the Logistic regression give the same accuracy rate of 62.5%. Therefore either of these models would be sufficient in predicting the trend of the market.
Question 11
(a)Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
data("Auto")
mpg01 <- rep(0, length(Auto$mpg))
mpg01[Auto$mpg > median(Auto$mpg)] <- 1
Auto <- data.frame(Auto, mpg01)
summary(Auto)## mpg cylinders displacement horsepower weight
## Min. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0 Min. :1613
## 1st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0 1st Qu.:2225
## Median :22.75 Median :4.000 Median :151.0 Median : 93.5 Median :2804
## Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5 Mean :2978
## 3rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0 3rd Qu.:3615
## Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0 Max. :5140
##
## acceleration year origin name
## Min. : 8.00 Min. :70.00 Min. :1.000 amc matador : 5
## 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000 ford pinto : 5
## Median :15.50 Median :76.00 Median :1.000 toyota corolla : 5
## Mean :15.54 Mean :75.98 Mean :1.577 amc gremlin : 4
## 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000 amc hornet : 4
## Max. :24.80 Max. :82.00 Max. :3.000 chevrolet chevette: 4
## (Other) :365
## mpg01
## Min. :0.0
## 1st Qu.:0.0
## Median :0.5
## Mean :0.5
## 3rd Qu.:1.0
## Max. :1.0
## (b) Explore the data graphically in order to investigate the association between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01? Scatter plots and boxplots may be useful tools to answer this question. Describe your findings.
cor(Auto[,-9])## mpg cylinders displacement horsepower weight
## mpg 1.0000000 -0.7776175 -0.8051269 -0.7784268 -0.8322442
## cylinders -0.7776175 1.0000000 0.9508233 0.8429834 0.8975273
## displacement -0.8051269 0.9508233 1.0000000 0.8972570 0.9329944
## horsepower -0.7784268 0.8429834 0.8972570 1.0000000 0.8645377
## weight -0.8322442 0.8975273 0.9329944 0.8645377 1.0000000
## acceleration 0.4233285 -0.5046834 -0.5438005 -0.6891955 -0.4168392
## year 0.5805410 -0.3456474 -0.3698552 -0.4163615 -0.3091199
## origin 0.5652088 -0.5689316 -0.6145351 -0.4551715 -0.5850054
## mpg01 0.8369392 -0.7591939 -0.7534766 -0.6670526 -0.7577566
## acceleration year origin mpg01
## mpg 0.4233285 0.5805410 0.5652088 0.8369392
## cylinders -0.5046834 -0.3456474 -0.5689316 -0.7591939
## displacement -0.5438005 -0.3698552 -0.6145351 -0.7534766
## horsepower -0.6891955 -0.4163615 -0.4551715 -0.6670526
## weight -0.4168392 -0.3091199 -0.5850054 -0.7577566
## acceleration 1.0000000 0.2903161 0.2127458 0.3468215
## year 0.2903161 1.0000000 0.1815277 0.4299042
## origin 0.2127458 0.1815277 1.0000000 0.5136984
## mpg01 0.3468215 0.4299042 0.5136984 1.0000000pairs(Auto[,-9])
par(mfrow=c(2,3))
boxplot(cylinders ~ mpg01, data = Auto, main = "Cylinders vs mpg01")
boxplot(displacement ~ mpg01, data = Auto, main = "Displacement vs mpg01")
boxplot(horsepower ~ mpg01, data = Auto, main = "Horsepower vs mpg01")
boxplot(weight ~ mpg01, data = Auto, main = "Weight vs mpg01")
boxplot(acceleration ~ mpg01, data = Auto, main = "Acceleration vs mpg01")
boxplot(year ~ mpg01, data = Auto, main = "Year vs mpg01")
From the above plots we can see there does appear to be a relationship between mpg01, cylinders, weight, displacement, and horsepower.
(c) Split the data into a training set and a test set.
set.seed(123)
train <- sample(1:dim(Auto)[1], dim(Auto)[1]*.7, rep=FALSE)
test <- -train
training_data<- Auto[train, ]
testing_data= Auto[test, ]
mpg01.test <- mpg01[test](d) Perform LDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
lda.fit = lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = training_data)
lda.pred = predict(lda.fit, testing_data)
pred.lda = predict(lda.fit, testing_data)
table(pred.lda$class, mpg01.test)## mpg01.test
## 0 1
## 0 50 3
## 1 10 55mean(pred.lda$class != mpg01.test)## [1] 0.1101695For our first model we can see we obtain a test error rate of roughly 11%
(e) Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qda.fit = qda(mpg01 ~ cylinders + horsepower + weight + acceleration, data=training_data)
qda.class=predict(qda.fit, testing_data)$class
table(qda.class, testing_data$mpg01)##
## qda.class 0 1
## 0 53 4
## 1 7 54mean(qda.class != testing_data$mpg01)## [1] 0.09322034For the QDA model we obtain an error rate of about 9.3%
(f) Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
glm.fit = glm(mpg01 ~ cylinders + weight + displacement + horsepower, data = training_data, family = binomial)
probs = predict(glm.fit, testing_data, type = "response")
pred.glm = rep(0, length(probs))
pred.glm[probs > 0.5] <- 1
table(pred.glm, mpg01.test)## mpg01.test
## pred.glm 0 1
## 0 53 6
## 1 7 52mean(pred.glm != mpg01.test)## [1] 0.1101695From our logistic model we obtain an error rate of roughly 11%
(g) Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
data = scale(Auto[,-c(9,10)])
set.seed(1234)
train <- sample(1:dim(Auto)[1], 392*.7, rep=FALSE)
test <- -train
training_data = data[train,c("cylinders","horsepower","weight","acceleration")]
testing_data = data[test, c("cylinders", "horsepower","weight","acceleration")]
train.mpg01 = Auto$mpg01[train]
test.mpg01= Auto$mpg01[test]
library(class)
set.seed(1234)
knn_pred_y = knn(training_data, testing_data, train.mpg01, k = 1)
table(knn_pred_y, test.mpg01)## test.mpg01
## knn_pred_y 0 1
## 0 57 5
## 1 7 49mean(knn_pred_y != test.mpg01)## [1] 0.1016949When K=1 we get an error rate of roughly 10%
test.mpg01= Auto$mpg01[test]
library(class)
set.seed(1234)
knn_pred_y = knn(training_data, testing_data, train.mpg01, k = 2)
table(knn_pred_y, test.mpg01)## test.mpg01
## knn_pred_y 0 1
## 0 55 5
## 1 9 49mean(knn_pred_y != test.mpg01)## [1] 0.1186441When K=2 we get an error rate of roughly 12%, therefore our accuracy is not as good for k=2
test.mpg01= Auto$mpg01[test]
library(class)
set.seed(1234)
knn_pred_y = knn(training_data, testing_data, train.mpg01, k = 3)
table(knn_pred_y, test.mpg01)## test.mpg01
## knn_pred_y 0 1
## 0 55 3
## 1 9 51mean(knn_pred_y != test.mpg01)## [1] 0.1016949When K=3 our error rate comes back down to roughly 10%. So we can see that the best model for knn is when k=1 or 3
Problem 13
First we will run a logistic regression using all of the predictor variables.
library(MASS)
attach(Boston)
crim01 <- rep(0, length(crim))
crim01[crim > median(crim)] <- 1
Boston <- data.frame(Boston, crim01)
train <- 1:(length(crim) / 2)
test <- (length(crim) / 2 + 1):length(crim)
Boston.train <- Boston[train, ]
Boston.test <- Boston[test, ]
crim01.test <- crim01[test]
glm.fit <- glm(crim01 ~ . - crim01 - crim, data = Boston, family = binomial, subset = train)## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredsummary(glm.fit)##
## Call:
## glm(formula = crim01 ~ . - crim01 - crim, family = binomial,
## data = Boston, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.83229 -0.06593 0.00000 0.06181 2.61513
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -91.319906 19.490273 -4.685 2.79e-06 ***
## zn -0.815573 0.193373 -4.218 2.47e-05 ***
## indus 0.354172 0.173862 2.037 0.04164 *
## chas 0.167396 0.991922 0.169 0.86599
## nox 93.706326 21.202008 4.420 9.88e-06 ***
## rm -4.719108 1.788765 -2.638 0.00833 **
## age 0.048634 0.024199 2.010 0.04446 *
## dis 4.301493 0.979996 4.389 1.14e-05 ***
## rad 3.039983 0.719592 4.225 2.39e-05 ***
## tax -0.006546 0.007855 -0.833 0.40461
## ptratio 1.430877 0.359572 3.979 6.91e-05 ***
## black -0.017552 0.006734 -2.606 0.00915 **
## lstat 0.190439 0.086722 2.196 0.02809 *
## medv 0.598533 0.185514 3.226 0.00125 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 329.367 on 252 degrees of freedom
## Residual deviance: 69.568 on 239 degrees of freedom
## AIC: 97.568
##
## Number of Fisher Scoring iterations: 10glm.probs <- predict(glm.fit, Boston.test, type = "response")
glm.pred <- rep(0, length(glm.probs))
glm.pred[glm.probs > 0.5] <- 1
table(glm.pred, crim01.test)## crim01.test
## glm.pred 0 1
## 0 68 24
## 1 22 139mean(glm.pred != crim01.test)## [1] 0.1818182From our first model we can see we achieve an error rate of just over 18%, next we will try the logistic model with some of the non significant variables removed from the model.
glm.fit <- glm(crim01 ~ . - crim01 - crim -chas -nox -tax, data = Boston, family = binomial, subset = train)## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredsummary(glm.fit)##
## Call:
## glm(formula = crim01 ~ . - crim01 - crim - chas - nox - tax,
## family = binomial, data = Boston, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -3.04443 -0.24461 -0.00114 0.38919 2.72999
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -17.291707 6.019497 -2.873 0.004071 **
## zn -0.478891 0.104276 -4.593 4.38e-06 ***
## indus 0.362719 0.082969 4.372 1.23e-05 ***
## rm -2.364642 0.967625 -2.444 0.014535 *
## age 0.063371 0.015457 4.100 4.14e-05 ***
## dis 1.494535 0.397249 3.762 0.000168 ***
## rad 1.756498 0.357330 4.916 8.85e-07 ***
## ptratio 0.575045 0.161917 3.551 0.000383 ***
## black -0.018916 0.006754 -2.801 0.005102 **
## lstat 0.057632 0.053051 1.086 0.277326
## medv 0.237282 0.081326 2.918 0.003527 **
## ---
## 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: 139.59 on 242 degrees of freedom
## AIC: 161.59
##
## Number of Fisher Scoring iterations: 9glm.probs <- predict(glm.fit, Boston.test, type = "response")
glm.pred <- rep(0, length(glm.probs))
glm.pred[glm.probs > 0.5] <- 1
table(glm.pred, crim01.test)## crim01.test
## glm.pred 0 1
## 0 78 28
## 1 12 135mean(glm.pred != crim01.test)## [1] 0.1581028From our second logistic model we can see our error rate has improved to roughly 16%. So we can conclude removing some of the non-significant variables from the model was beneficial.
Now lets try an LDA model.
lda.fit <- lda(crim01 ~ . - crim01 - crim, data = Boston, subset = train)
lda.pred <- predict(lda.fit, Boston.test)
table(lda.pred$class, crim01.test)## crim01.test
## 0 1
## 0 80 24
## 1 10 139mean(lda.pred$class != crim01.test)## [1] 0.1343874From our LDA model we can see we get another improvement in our error rate, it has come down to roughly 13.4%.
Lets try fitting a LDA model removing the same variables.
lda.fit <- lda(crim01 ~ . - crim01 - crim - chas - nox - tax, data = Boston, subset = train)
lda.pred <- predict(lda.fit, Boston.test)
table(lda.pred$class, crim01.test)## crim01.test
## 0 1
## 0 83 28
## 1 7 135mean(lda.pred$class != crim01.test)## [1] 0.1383399Here our model does not improve, after removing some variables our error rate is now 13.8%. Slightly worse than the previous model.
Now we will fit a KNN model. The first model we will have k=1.
library(class)
train.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[train, ]
test.X <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio, black, lstat, medv)[test, ]
train.crim01 <- crim01[train]
set.seed(1)
knn.pred <- knn(train.X, test.X, train.crim01, k = 1)
table(knn.pred, crim01.test)## crim01.test
## knn.pred 0 1
## 0 85 111
## 1 5 521-((85+52)/(85+111+5+52))## [1] 0.458498For our first KNN model when k=1 we obtain an error rate of almost 46% which is far worse than any of the previous models.
Next lets try a model with k=10.
knn.pred <- knn(train.X, test.X, train.crim01, k = 10)
table(knn.pred, crim01.test)## crim01.test
## knn.pred 0 1
## 0 83 23
## 1 7 1401-((83+140)/(83+23+7+140))## [1] 0.1185771When we changed k to be equal to 10, we see a dramatic improvement in our error rate. This model has the lowest error rate of all models at roughly 11.8%.