Data Mining Homework 3
John Thomas
February 20, 2019
Data Mining Homework 3
Problem 10
This question should be answered using the “Weekly” data set, which is part of the “ISLR” package. This data is similar in nature to the “Smarket” data from this chapter’s lab, except that it contains 1089 weekly returns for 21 years, from the beginning of 1990 to the end of 2010.
Part 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
## Min. :-18.1950 Min. :-18.1950 Min. :0.08747
## 1st Qu.: -1.1580 1st Qu.: -1.1660 1st Qu.:0.33202
## Median : 0.2380 Median : 0.2340 Median :1.00268
## Mean : 0.1458 Mean : 0.1399 Mean :1.57462
## 3rd Qu.: 1.4090 3rd Qu.: 1.4050 3rd Qu.:2.05373
## Max. : 12.0260 Max. : 12.0260 Max. :9.32821
## Today Direction
## Min. :-18.1950 Down:484
## 1st Qu.: -1.1540 Up :605
## Median : 0.2410
## Mean : 0.1499
## 3rd Qu.: 1.4050
## Max. : 12.0260pairs(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 1.000000000With data from the “Weekly” set we can observer that Year and Volume have some sort of relationship whereas the lags don’t appear to have one with anything, similarly Today does not have any substainstial relationship.
Part 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)
glm.fit = glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Weekly, family = binomial)
summary(glm.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: 4With a significance level of .05, the only predictor that appears to be statistically significant is Lag2 with it’s value being less than .05.
Part 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.
probs <- predict(glm.fit, type = "response")
pred.glm <- rep("Down", length(probs))
pred.glm[probs > 0.5] <- "Up"
table(pred.glm, Direction)## Direction
## pred.glm Down Up
## Down 54 48
## Up 430 557From our training data we can see that the percentage of correct predictions is (54+557)/(54+557+48+430) = 56.11%. This means we have training error rate of 43.89%. We could also see that when the market goes up the prediction rate becomes 557/(48+557) = 92.07% of the time. When the market goes down however, our model is right only 54/(54+430) = 11.16% of the time.
Part 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.20092010 <- Weekly[!train, ]
Direction.20092010 <- Direction[!train]
fit.glm2 <- glm(Direction ~ Lag2, data = Weekly, family = binomial, subset = train)
summary(fit.glm2)##
## Call:
## glm(formula = Direction ~ Lag2, family = binomial, data = Weekly,
## subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.536 -1.264 1.021 1.091 1.368
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.20326 0.06428 3.162 0.00157 **
## Lag2 0.05810 0.02870 2.024 0.04298 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 1354.7 on 984 degrees of freedom
## Residual deviance: 1350.5 on 983 degrees of freedom
## AIC: 1354.5
##
## Number of Fisher Scoring iterations: 4probs2 <- predict(fit.glm2, Weekly.20092010, type = "response")
pred.glm2 <- rep("Down", length(probs2))
pred.glm2[probs2 > 0.5] <- "Up"
table(pred.glm2, Direction.20092010)## Direction.20092010
## pred.glm2 Down Up
## Down 9 5
## Up 34 56From our matrix we can calculate that the overall percentage of correct prediction from test data to be (9+54)/104 or 62.5%.
Part E
Repeat (d) using LDA.
library(MASS)
fit.lda <- lda(Direction ~ Lag2, data = Weekly, subset = train)
fit.lda## Call:
## lda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581
##
## Coefficients of linear discriminants:
## LD1
## Lag2 0.4414162pred.lda <- predict(fit.lda, Weekly.20092010)
table(pred.lda$class, Direction.20092010)## Direction.20092010
## Down Up
## Down 9 5
## Up 34 56From our matrix we can calculate that the overall percentage of correct prediction from test data to be 62.5%.
Part F
Repeat (d) using QDA.
fit.qda <- qda(Direction ~ Lag2, data = Weekly, subset = train)
fit.qda## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581pred.qda <- predict(fit.qda, Weekly.20092010)
table(pred.qda$class, Direction.20092010)## Direction.20092010
## Down Up
## Down 0 0
## Up 43 61From our matrix we can calculate that the overall percentage of correct prediction from test data to be 58.65%.
Part G
Repeat (d) using KNN with K = 1.
library(class)
train.X <- as.matrix(Lag2[train])
test.X <- as.matrix(Lag2[!train])
train.Direction <- Direction[train]
set.seed(1)
pred.knn <- knn(train.X, test.X, train.Direction, k = 1)
table(pred.knn, Direction.20092010)## Direction.20092010
## pred.knn Down Up
## Down 21 30
## Up 22 31From our matrix we can calculate that the overall percentage of correct prediction from test data to be 50%.
Part H
Which of these methods appears to provide the best results on this data?
If we compare the test error rates, we see that logistic regression and LDA have the minimum error rates, followed by QDA and KNN. This is seen by the fact that the LDA and logistic regression have the highest correct prediction rate so their error rates are smaller.
Part I
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.
# Logistic regression with Lag2:Lag1
fit.glm3 <- glm(Direction ~ Lag2:Lag1, data = Weekly, family = binomial, subset = train)
probs3 <- predict(fit.glm3, Weekly.20092010, type = "response")
pred.glm3 <- rep("Down", length(probs3))
pred.glm3[probs3 > 0.5] = "Up"
table(pred.glm3, Direction.20092010)## Direction.20092010
## pred.glm3 Down Up
## Down 1 1
## Up 42 60mean(pred.glm3 == Direction.20092010)## [1] 0.5865385# LDA with Lag2 interaction with Lag1
fit.lda2 <- lda(Direction ~ Lag2:Lag1, data = Weekly, subset = train)
pred.lda2 <- predict(fit.lda2, Weekly.20092010)
mean(pred.lda2$class == Direction.20092010)## [1] 0.5769231# QDA with sqrt(abs(Lag2))
fit.qda2 <- qda(Direction ~ Lag2 + sqrt(abs(Lag2)), data = Weekly, subset = train)
pred.qda2 <- predict(fit.qda2, Weekly.20092010)
table(pred.qda2$class, Direction.20092010)## Direction.20092010
## Down Up
## Down 12 13
## Up 31 48mean(pred.qda2$class == Direction.20092010)## [1] 0.5769231# KNN k = 10, 100, 200
pred.knn2 <- knn(train.X, test.X, train.Direction, k = 10)
table(pred.knn2, Direction.20092010)## Direction.20092010
## pred.knn2 Down Up
## Down 17 18
## Up 26 43mean(pred.knn2 == Direction.20092010)## [1] 0.5769231pred.knn3 <- knn(train.X, test.X, train.Direction, k = 100)
table(pred.knn3, Direction.20092010)## Direction.20092010
## pred.knn3 Down Up
## Down 9 12
## Up 34 49mean(pred.knn3 == Direction.20092010)## [1] 0.5576923pred.knn4 <- knn(train.X, test.X, train.Direction, k = 200)
table(pred.knn4, Direction.20092010)## Direction.20092010
## pred.knn4 Down Up
## Down 3 0
## Up 40 61mean(pred.knn4 == Direction.20092010)## [1] 0.6153846Out of all of these we can see that the KNN with k=200 is the best predictor found.
Problem 11
In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set.
Part 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.
library(ISLR)
summary(Auto)## mpg cylinders displacement horsepower
## Min. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0
## 1st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0
## Median :22.75 Median :4.000 Median :151.0 Median : 93.5
## Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5
## 3rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0
## Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0
##
## weight acceleration year origin
## Min. :1613 Min. : 8.00 Min. :70.00 Min. :1.000
## 1st Qu.:2225 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000
## Median :2804 Median :15.50 Median :76.00 Median :1.000
## Mean :2978 Mean :15.54 Mean :75.98 Mean :1.577
## 3rd Qu.:3615 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000
## Max. :5140 Max. :24.80 Max. :82.00 Max. :3.000
##
## name
## amc matador : 5
## ford pinto : 5
## toyota corolla : 5
## amc gremlin : 4
## amc hornet : 4
## chevrolet chevette: 4
## (Other) :365attach(Auto)
l=length(mpg)
mpg01<-rep(0,l)
mpg01[mpg > median(mpg)]<- 1
Auto<-data.frame(Auto,mpg01)Part 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? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.
pairs(Auto)
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 1.0000000boxplot(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")
Through various boxplots and the pairs along with correlations, we can see that there is a correlation between mpg01 and: cylinders, weight, horsepower and displacement.
Part C
Split the data into a training set and a test set.
train <- (year %% 2 == 0)
Auto.train <- Auto[train, ]
Auto.test <- Auto[!train, ]
mpg01.test <- mpg01[!train]Part 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?
fit.lda <- lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, subset = train)
fit.lda## Call:
## lda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto,
## subset = train)
##
## Prior probabilities of groups:
## 0 1
## 0.4571429 0.5428571
##
## Group means:
## cylinders weight displacement horsepower
## 0 6.812500 3604.823 271.7396 133.14583
## 1 4.070175 2314.763 111.6623 77.92105
##
## Coefficients of linear discriminants:
## LD1
## cylinders -0.6741402638
## weight -0.0011465750
## displacement 0.0004481325
## horsepower 0.0059035377pred.lda <- predict(fit.lda, Auto.test)
table(pred.lda$class, mpg01.test)## mpg01.test
## 0 1
## 0 86 9
## 1 14 73mean(pred.lda$class != mpg01.test)## [1] 0.1263736From out output we can see that performing LDA on our data will have a test error rate of 12.64%.
Part E
Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
fit.qda <- qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, subset = train)
fit.qda## Call:
## qda(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto,
## subset = train)
##
## Prior probabilities of groups:
## 0 1
## 0.4571429 0.5428571
##
## Group means:
## cylinders weight displacement horsepower
## 0 6.812500 3604.823 271.7396 133.14583
## 1 4.070175 2314.763 111.6623 77.92105pred.qda <- predict(fit.qda, Auto.test)
table(pred.qda$class, mpg01.test)## mpg01.test
## 0 1
## 0 89 13
## 1 11 69mean(pred.qda$class != mpg01.test)## [1] 0.1318681Using QDA with our data shows us that we’d have a test error rate of 13.19%
Part 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?
fit.glm <- glm(mpg01 ~ cylinders + weight + displacement + horsepower, data = Auto, family = binomial, subset = train)
summary(fit.glm)##
## Call:
## glm(formula = mpg01 ~ cylinders + weight + displacement + horsepower,
## family = binomial, data = Auto, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.48027 -0.03413 0.10583 0.29634 2.57584
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 17.658730 3.409012 5.180 2.22e-07 ***
## cylinders -1.028032 0.653607 -1.573 0.1158
## weight -0.002922 0.001137 -2.569 0.0102 *
## displacement 0.002462 0.015030 0.164 0.8699
## horsepower -0.050611 0.025209 -2.008 0.0447 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 289.58 on 209 degrees of freedom
## Residual deviance: 83.24 on 205 degrees of freedom
## AIC: 93.24
##
## Number of Fisher Scoring iterations: 7probs <- predict(fit.glm, Auto.test, 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 89 11
## 1 11 71mean(pred.glm != mpg01.test)## [1] 0.1208791Using a Logistic regression on our data will result in a test error rate of 12.09%, lower than both KDA and QDA.
Part 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?
train.X <- cbind(cylinders, weight, displacement, horsepower)[train, ]
test.X <- cbind(cylinders, weight, displacement, horsepower)[!train, ]
train.mpg01 <- mpg01[train]
set.seed(1)
#Use K= 1, 10, 100, 200
pred.knn1 <- knn(train.X, test.X, train.mpg01, k = 1)
table(pred.knn1, mpg01.test)## mpg01.test
## pred.knn1 0 1
## 0 83 11
## 1 17 71mean(pred.knn1 != mpg01.test)## [1] 0.1538462pred.knn10 <- knn(train.X, test.X, train.mpg01, k = 10)
table(pred.knn10, mpg01.test)## mpg01.test
## pred.knn10 0 1
## 0 77 7
## 1 23 75mean(pred.knn10 != mpg01.test)## [1] 0.1648352pred.knn100 <- knn(train.X, test.X, train.mpg01, k = 100)
table(pred.knn100, mpg01.test)## mpg01.test
## pred.knn100 0 1
## 0 81 7
## 1 19 75mean(pred.knn100 != mpg01.test)## [1] 0.1428571pred.knn200 <- knn(train.X, test.X, train.mpg01, k = 200)
table(pred.knn200, mpg01.test)## mpg01.test
## pred.knn200 0 1
## 0 0 0
## 1 100 82mean(pred.knn200 != mpg01.test)## [1] 0.5494505We can see for various K values we have lower and lower test error rates untill we go to 200. Our Lowest test error rate being for K=100 at 14.29% versus, K=1 with 15.38%, K=10 with 16.48% and significantly better than K=200 with a 54.95% error rate.
Problem 13
Using the Boston data set, fit classification models in order to predict whether a given suburb has a crime rate above or below the median. Explore logistic regression, LDA, and KNN models using various subsets of the predictors. Describe your findings.
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]fit.glm1 <- glm(crim01 ~ . - crim01 - crim, data = Boston, family = binomial, subset = train)## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredprobs1 <- predict(fit.glm1, Boston.test, type = "response")
pred.glm1 <- rep(0, length(probs1))
pred.glm1[probs1 > 0.5] <- 1
table(pred.glm1, crim01.test)## crim01.test
## pred.glm1 0 1
## 0 68 24
## 1 22 139mean(pred.glm1 != crim01.test)## [1] 0.1818182For a Logistic Regression with the data in regards to Crime Rates the test error rate is 18.18%.
fit.glm2 <- glm(crim01 ~ . - crim01 - crim - chas - nox, data = Boston, family = binomial, subset = train)## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredprobs2 <- predict(fit.glm2, Boston.test, type = "response")
pred.glm2 <- rep(0, length(probs2))
pred.glm2[probs2 > 0.5] <- 1
table(pred.glm2, crim01.test)## crim01.test
## pred.glm2 0 1
## 0 78 28
## 1 12 135mean(pred.glm2 != crim01.test)## [1] 0.1581028For this Logistic Regression were were able to get the test error rate lower to 15.81%
fit.lda1 <- lda(crim01 ~ . - crim01 - crim, data = Boston, subset = train)
pred.lda1 <- predict(fit.lda1, Boston.test)
table(pred.lda1$class, crim01.test)## crim01.test
## 0 1
## 0 80 24
## 1 10 139mean(pred.lda1$class != crim01.test)## [1] 0.1343874Using LDA we can find a test error rate of 13.44%
fit.lda2 <- lda(crim01 ~ . - crim01 - crim - chas - nox, data = Boston, subset = train)
pred.lda2 <- predict(fit.lda2, Boston.test)
table(pred.lda2$class, crim01.test)## crim01.test
## 0 1
## 0 82 30
## 1 8 133mean(pred.lda2$class != crim01.test)## [1] 0.1501976Using LDA a second time with additions, we get a test error rate of 15.02%
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)
pred.knn1 <- knn(train.X, test.X, train.crim01, k = 1)
table(pred.knn1, crim01.test)## crim01.test
## pred.knn1 0 1
## 0 85 111
## 1 5 52mean(pred.knn1 != crim01.test)## [1] 0.458498pred.knn10 <- knn(train.X, test.X, train.crim01, k = 10)
table(pred.knn10, crim01.test)## crim01.test
## pred.knn10 0 1
## 0 83 23
## 1 7 140mean(pred.knn10 != crim01.test)## [1] 0.1185771pred.knn100 <- knn(train.X, test.X, train.crim01, k = 100)
table(pred.knn100, crim01.test)## crim01.test
## pred.knn100 0 1
## 0 86 120
## 1 4 43mean(pred.knn100 != crim01.test)## [1] 0.4901186pred.knn200 <- knn(train.X, test.X, train.crim01, k = 200)
table(pred.knn200, crim01.test)## crim01.test
## pred.knn200 0 1
## 0 90 163
## 1 0 0mean(pred.knn200 != crim01.test)## [1] 0.6442688Using KNN with various Ks, we get the following errors rates:
K=1 : 45.85%
K=10 : 11.86%
K=100 : 49.01%
K=200 : 64.43%
With these tests we can see that the best testing model we had was KNN with K=10. The rest of the tests were also good test values except for the other KNN models we did. This means we can predict with 88.14% accuracy if a specific suburb has a crime rate above or below the median.