ISLR Assignment 3
Chapter 4 (page 168) Problems 10, 11, 13
Misty Stultz
Due 3/5/2022
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 1,089 weekly returns for 21 years, from the beginning of 1990 to the end of 2010.
(a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
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 1.000000000plot(Volume)
plot(log(Volume))
There appears to be a positive correlation between Volume and Year. It looks like the Volume variable could benefit from a transformation so that it may become a better predictor. Taking the log of Volume gives us a more uniform variable to work with.
(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?
logist.full <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Weekly, family = binomial)
summary(logist.full)##
## 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: 4Lag2 appears to be moderately significant.
(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.
full.probs <- predict(logist.full, type = 'response')
full.pred <- rep('Down', 1089)
full.pred[full.probs > 0.5] <- 'Up'
table(full.pred, Direction)## Direction
## full.pred Down Up
## Down 54 48
## Up 430 557mean(full.pred == Direction)## [1] 0.5610652The percentage of correct predictions is approximately 56.1%.
The majority of the mistakes made involve predicting that the market will be up when it actually goes down.
(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).
training <- (Year <= 2008)
Weekly.Held <- Weekly[!training, ]
Direction.Held <- Direction[!training]
log.train.fit <- glm(Direction ~ Lag2, data = Weekly, subset = training, family = 'binomial')
log.train.prob <- predict(log.train.fit, Weekly.Held, type = 'response')
log.train.pred <- rep('Down', 104)
log.train.pred[log.train.prob > 0.5] <- 'Up'
table(log.train.pred, Direction.Held)## Direction.Held
## log.train.pred Down Up
## Down 9 5
## Up 34 56mean(log.train.pred == Direction.Held)## [1] 0.625Adjusting the model so that we use the last 2 years of data as test data to compare to the training data and only including the significant variable results in a model that is accurate 62.5% of the time.
This model results in more inaccuracies related to predicting the market will be up, but in reality it goes down.
(e) Repeat (d) using LDA.
lda.train.fit <- lda(Direction ~ Lag2, data = Weekly, subset = training)
plot(lda.train.fit)
lda.train.pred <- predict(lda.train.fit, Weekly.Held)
names(lda.train.pred)## [1] "class" "posterior" "x"lda.train.class <- lda.train.pred$class
table(lda.train.class, Direction.Held)## Direction.Held
## lda.train.class Down Up
## Down 9 5
## Up 34 56mean(lda.train.class == Direction.Held)## [1] 0.625Using LDA we get the exact same results that we got using Logistic Regression. The model is accurate 62.5% of the time.
(f) Repeat (d) using QDA.
qda.train.fit <- qda(Direction ~ Lag2, data = Weekly, subset = training)
qda.train.fit## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = training)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581qda.train.class <- predict(qda.train.fit, Weekly.Held)$class
table(qda.train.class, Direction.Held)## Direction.Held
## qda.train.class Down Up
## Down 0 0
## Up 43 61mean(qda.train.class == Direction.Held)## [1] 0.5865385Using quadratic discriminant analysis our accuraccy has decreased to 58.65%.
This model predicts that the market will be up 100% of the time, so it appears that it’s not really predicting much of anything in relation to this data set.
(g) Repeat (d) using KNN with K = 1.
training.Weekly <- as.matrix(Lag2[training])
test.Weekly <- as.matrix(Lag2[!training])
training.Direction <- Direction[training]
set.seed(1)
knn.pred <- knn(training.Weekly, test.Weekly, training.Direction, k = 1)
table(knn.pred, Direction.Held)## Direction.Held
## knn.pred Down Up
## Down 21 30
## Up 22 31mean(knn.pred == Direction.Held)## [1] 0.5Using kNN we get a model that is only accurate 50% of the time.
While the model is less accurate than other models we had, there is a little more balance on which direction the predictions will
(h) Which of these methods appears to provide the best results on this data?
Both Logistic Regression and Linear Discriminant Analysis give us the highest result of 62.5% accuracy.
All of the models produce slightly different results in the type of inaccuracy though. Decisions on which model is “best” may need to depend on the preference of direction of the error. If predicting innacurately that the market will go up, when it actually goes down, there could be a considerable amount of money lost using that prediction. Depeninding on who these predictions were for, it could be prefferable that we use a model that is slightly less accurate if the innacuracies lean more in a direction where money is not lost.
In truth, the low levels of accuracy of all of these models suggest that using previous weeks as predictors is not reliable for predicting whether or not the stock market will be up or down.
(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.
kNN
#k=2
set.seed(1)
knn.pred <- knn(training.Weekly, test.Weekly, training.Direction, k = 2)
table(knn.pred, Direction.Held)## Direction.Held
## knn.pred Down Up
## Down 19 27
## Up 24 34mean(knn.pred == Direction.Held)## [1] 0.5096154#k=3
set.seed(1)
knn.pred <- knn(training.Weekly, test.Weekly, training.Direction, k = 3)
table(knn.pred, Direction.Held)## Direction.Held
## knn.pred Down Up
## Down 16 20
## Up 27 41mean(knn.pred == Direction.Held)## [1] 0.5480769#k=4
set.seed(1)
knn.pred <- knn(training.Weekly, test.Weekly, training.Direction, k = 4)
table(knn.pred, Direction.Held)## Direction.Held
## knn.pred Down Up
## Down 20 17
## Up 23 44mean(knn.pred == Direction.Held)## [1] 0.6153846#k=5
set.seed(1)
knn.pred <- knn(training.Weekly, test.Weekly, training.Direction, k = 5)
table(knn.pred, Direction.Held)## Direction.Held
## knn.pred Down Up
## Down 16 21
## Up 27 40mean(knn.pred == Direction.Held)## [1] 0.5384615If we start increasing k in the kNN model, we find that k=4 gives us the most accuracy possible, with predictions being correct 61.54% of the time. When k increases to 5 the accuracy starts to decrease again.
Sinc k=4 gives us almost the same level of accuracy as our initial LDA and Logistic Regression models, while still having a better balance of the direction of inaccuracy, it may be that using kNN where k=4 is the best method to model this specific data.
Logistic
logist.exp <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + log(Volume), data = Weekly, family = binomial)
summary(logist.exp)##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 +
## log(Volume), family = binomial, data = Weekly)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.6922 -1.2600 0.9928 1.0847 1.4665
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.22562 0.06224 3.625 0.000289 ***
## Lag1 -0.04127 0.02637 -1.565 0.117578
## Lag2 0.05834 0.02679 2.178 0.029433 *
## Lag3 -0.01607 0.02663 -0.603 0.546213
## Lag4 -0.02790 0.02643 -1.055 0.291218
## Lag5 -0.01457 0.02636 -0.553 0.580433
## log(Volume) -0.05133 0.05607 -0.915 0.359988
## ---
## 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: 1485.9 on 1082 degrees of freedom
## AIC: 1499.9
##
## Number of Fisher Scoring iterations: 4exp.probs <- predict(logist.exp, type = 'response')
exp.pred <- rep('Down', 1089)
exp.pred[full.probs > 0.5] <- 'Up'
table(exp.pred, Direction)## Direction
## exp.pred Down Up
## Down 54 48
## Up 430 557mean(exp.pred == Direction)## [1] 0.5610652As noted in part (a), it looks like the variable Volume would benefit from a Log transformation. Testing this transformation on the overall logistic model (all training data), it does appear to reduce the p-value for Volume, but not enough to make it a significant variable. Including it in the model does not change any of the overall predictions.
LDA
lda.exp.fit <- lda(Direction ~ Lag2 + Lag3, data = Weekly, subset = training)
lda.exp.pred <- predict(lda.exp.fit, Weekly.Held)
names(lda.exp.pred)## [1] "class" "posterior" "x"lda.exp.class <- lda.exp.pred$class
table(lda.exp.class, Direction.Held)## Direction.Held
## lda.exp.class Down Up
## Down 8 4
## Up 35 57mean(lda.exp.class == Direction.Held)## [1] 0.625While additional variables do not seem to make any positive addition to the model, I found it interesting that the addition of the Lag3 variable makes such an insignificant difference that there is no change to the prediction accuracy.
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.
(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.
summary(mpg)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 9.00 17.00 22.75 23.45 29.00 46.60# median = 22.75
mpg01 <- rep(0, length(mpg))
mpg01[mpg > median(mpg)] <- 1
Auto2 <- data.frame(Auto,mpg01)(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.
par(mfrow = c(2,3))
boxplot(cylinders ~ mpg01, main = 'Cyl/mpg01', col = c('slateblue2','darkslategray3'))
boxplot(displacement ~ mpg01, main = 'Disp/mpg01', col = c('hotpink3','darkslategray3'))
boxplot(horsepower ~ mpg01, main = 'HP/mpg01', col = c('gold1','darkslategray3'))
boxplot(weight ~ mpg01, main = 'Wt/mpg01', col = c('palegreen1','darkslategray3'))
boxplot(acceleration ~ mpg01, main = 'Acc/mpg01', col = c('thistle4','darkslategray3'))
boxplot(year ~ mpg01, main = 'Year/mpg01', col = c('mistyrose','darkslategray3'))
All variables except mpg appear to be useful in predicting high or low gas mileage.
(c) Split the data into a training set and a test set.
set.seed(21)
set.training <- sample(nrow(Auto2), nrow(Auto2) * 0.7)
training.auto <- Auto2[set.training, ]
test.auto <- Auto2[-set.training, ]
dim(training.auto)## [1] 274 10dim(test.auto)## [1] 118 10(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.Auto.fit <- lda(mpg01 ~ cylinders + displacement + horsepower + weight + year, data = training.auto)
lda.Auto.pred <- predict(lda.Auto.fit, test.auto)
table(lda.Auto.pred$class, test.auto$mpg01, dnn = c('Predicted MPG Cat', "Actual MPG Cat"))## Actual MPG Cat
## Predicted MPG Cat 0 1
## 0 52 1
## 1 14 51mean(lda.Auto.pred$class != test.auto$mpg01)## [1] 0.1271186The test error rate for the LDA model is 12.7%
(e) Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qda.Auto.fit <- qda(mpg01 ~ cylinders + displacement + horsepower + weight + year, data = training.auto)
qda.Auto.pred <- predict(qda.Auto.fit, test.auto)
table(qda.Auto.pred$class, test.auto$mpg01, dnn = c('Predicted MPG Cat', "Actual MPG Cat"))## Actual MPG Cat
## Predicted MPG Cat 0 1
## 0 55 3
## 1 11 49mean(qda.Auto.pred$class != test.auto$mpg01)## [1] 0.1186441The error rate of the model for QDA is 11.86%.
(f) Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
logi.Auto.fit <- glm(mpg01 ~ cylinders + displacement + horsepower + weight + year, data = training.auto, family = 'binomial')
#summary(logi.Auto.fit)
logi.Auto.pred <- predict(logi.Auto.fit, test.auto, type = 'response')
logi.Auto.class <- rep(0, dim(test.auto)[1])
logi.Auto.class[logi.Auto.pred > 0.5] <- 1
table(logi.Auto.class, test.auto$mpg01,dnn = c('Predicted MPG Cat', "Actual MPG Cat"))## Actual MPG Cat
## Predicted MPG Cat 0 1
## 0 56 6
## 1 10 46mean(logi.Auto.class != test.auto$mpg01)## [1] 0.1355932The model error rate for logistic regression is 13.56%
(g) Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
training.auto.knn <- cbind(training.auto[,c(2:7,10)])
test.auto.knn <- cbind(test.auto[,c(2:7,10)])
training.mpg.knn <- cbind(training.auto$mpg01)k=1
set.seed(1)
knn.auto.pred <- knn(training.auto.knn, test.auto.knn, training.mpg.knn, k=1)
table(knn.auto.pred, test.auto.knn$mpg01,dnn = c('k=1 Predicted', "Actual MPG Cat"))## Actual MPG Cat
## k=1 Predicted 0 1
## 0 53 3
## 1 13 49mean(knn.auto.pred != test.auto.knn$mpg01)## [1] 0.1355932k=2
set.seed(1)
knn.auto.pred <- knn(training.auto.knn, test.auto.knn, training.mpg.knn, k=2)
table(knn.auto.pred, test.auto.knn$mpg01,dnn = c('k=2 Predicted', "Actual MPG Cat"))## Actual MPG Cat
## k=2 Predicted 0 1
## 0 50 5
## 1 16 47mean(knn.auto.pred != test.auto.knn$mpg01)## [1] 0.1779661k=3
set.seed(1)
knn.auto.pred <- knn(training.auto.knn, test.auto.knn, training.mpg.knn, k=3)
table(knn.auto.pred, test.auto.knn$mpg01,dnn = c('k=3 Predicted', "Actual MPG Cat"))## Actual MPG Cat
## k=3 Predicted 0 1
## 0 50 3
## 1 16 49mean(knn.auto.pred != test.auto.knn$mpg01)## [1] 0.1610169k=4
set.seed(1)
knn.auto.pred <- knn(training.auto.knn, test.auto.knn, training.mpg.knn, k=4)
table(knn.auto.pred, test.auto.knn$mpg01,dnn = c('k=4 Predicted', "Actual MPG Cat"))## Actual MPG Cat
## k=4 Predicted 0 1
## 0 51 1
## 1 15 51mean(knn.auto.pred != test.auto.knn$mpg01)## [1] 0.1355932k=5
set.seed(1)
knn.auto.pred <- knn(training.auto.knn, test.auto.knn, training.mpg.knn, k=5)
table(knn.auto.pred, test.auto.knn$mpg01,dnn = c('k=5 Predicted', "Actual MPG Cat"))## Actual MPG Cat
## k=5 Predicted 0 1
## 0 50 1
## 1 16 51mean(knn.auto.pred != test.auto.knn$mpg01)## [1] 0.1440678k=6
set.seed(1)
knn.auto.pred <- knn(training.auto.knn, test.auto.knn, training.mpg.knn, k=6)
table(knn.auto.pred, test.auto.knn$mpg01,dnn = c('k=6 Predicted', "Actual MPG Cat"))## Actual MPG Cat
## k=6 Predicted 0 1
## 0 52 1
## 1 14 51mean(knn.auto.pred != test.auto.knn$mpg01)## [1] 0.1271186k=7
set.seed(1)
knn.auto.pred <- knn(training.auto.knn, test.auto.knn, training.mpg.knn, k=7)
table(knn.auto.pred, test.auto.knn$mpg01, dnn = c('k=7 Predicted', "Actual MPG Cat"))## Actual MPG Cat
## k=7 Predicted 0 1
## 0 51 2
## 1 15 50mean(knn.auto.pred != test.auto.knn$mpg01)## [1] 0.1440678k=6 seems to have the best prediction error rate, at 12.71%.
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.
Set Training Data
crime01 <- rep(0, length(crim))
crime01[crim > median(crim)] <- 1
Boston2 <- data.frame(Boston, crime01)
set.seed(21)
set.training.Boston <- sample(nrow(Boston2), nrow(Boston2) * 0.7)
training.Boston <- Boston2[set.training.Boston, ]
test.Boston <- Boston2[-set.training.Boston, ]
dim(training.Boston)## [1] 354 15dim(test.Boston)## [1] 152 15Logistic Regression
#Checking for significant variables
logi.Boston.exp <- glm(crime01 ~ . - crim, data = training.Boston, family = 'binomial')
summary(logi.Boston.exp)##
## Call:
## glm(formula = crime01 ~ . - crim, family = "binomial", data = training.Boston)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.1727 -0.1827 0.0000 0.0054 3.3116
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -13.371994 9.765189 -1.369 0.170889
## zn -0.081448 0.038285 -2.127 0.033384 *
## indus -0.112759 0.056635 -1.991 0.046483 *
## chas 1.117820 0.840720 1.330 0.183650
## nox 46.735010 8.910699 5.245 1.56e-07 ***
## rm -0.166443 0.937410 -0.178 0.859071
## age 0.016902 0.013801 1.225 0.220698
## dis 0.532352 0.257082 2.071 0.038382 *
## rad 0.610121 0.172841 3.530 0.000416 ***
## tax -0.005554 0.003552 -1.564 0.117917
## ptratio 0.399709 0.157735 2.534 0.011275 *
## black -0.063328 0.021067 -3.006 0.002646 **
## lstat 0.033245 0.056040 0.593 0.553025
## medv 0.127053 0.084904 1.496 0.134539
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 490.02 on 353 degrees of freedom
## Residual deviance: 147.07 on 340 degrees of freedom
## AIC: 175.07
##
## Number of Fisher Scoring iterations: 9logi.Boston.fit <- glm(crime01 ~ nox + rad, data = training.Boston, family = 'binomial')
logi.Boston.pred <- predict(logi.Boston.fit, test.Boston, type = 'response')
logi.Boston.class <- rep(0, dim(test.Boston)[1])
logi.Boston.class[logi.Boston.pred > 0.5] <- 1
table(logi.Boston.class, test.Boston$crime01,dnn = c('Predicted Crime Cat', "Actual Crime Cat"))## Actual Crime Cat
## Predicted Crime Cat 0 1
## 0 74 10
## 1 10 58mean(logi.Boston.class == test.Boston$crime01)## [1] 0.8684211The logistic regression model has an accuracy rate of 86.84%.
LDA
lda.Boston.fit <- lda(crime01 ~ nox + rad, data = training.Boston)
lda.Boston.pred <- predict(lda.Boston.fit, test.Boston)
table(lda.Boston.pred$class, test.Boston$crime01, dnn = c('Predicted Crime Cat', "Actual Crime Cat"))## Actual Crime Cat
## Predicted Crime Cat 0 1
## 0 81 13
## 1 3 55mean(lda.Boston.pred$class == test.Boston$crime01)## [1] 0.8947368The LDA model has an accuracy rate of 89.47%.
kNN
Setup kNN
training.Boston.knn <- cbind(training.Boston[,c(5,9,15)])
test.Boston.knn <- cbind(test.Boston[,c(5,9,15)])
training.crime.knn <- cbind(training.Boston$crime01)K=1
set.seed(1)
knn.Boston.pred <- knn(training.Boston.knn, test.Boston.knn, training.crime.knn, k=1)
table(knn.Boston.pred, test.Boston.knn$crime01, dnn = c('k=1 Predicted', "Actual Crime Cat"))## Actual Crime Cat
## k=1 Predicted 0 1
## 0 84 0
## 1 0 68mean(knn.Boston.pred == test.Boston.knn$crime01)## [1] 1K=2
set.seed(1)
knn.Boston.pred <- knn(training.Boston.knn, test.Boston.knn, training.crime.knn, k=2)
table(knn.Boston.pred, test.Boston.knn$crime01, dnn = c('k=2 Predicted', "Actual Crime Cat"))## Actual Crime Cat
## k=2 Predicted 0 1
## 0 84 1
## 1 0 67mean(knn.Boston.pred == test.Boston.knn$crime01)## [1] 0.9934211k=3
set.seed(1)
knn.Boston.pred <- knn(training.Boston.knn, test.Boston.knn, training.crime.knn, k=3)
table(knn.Boston.pred, test.Boston.knn$crime01, dnn = c('k=3 Predicted', "Actual Crime Cat"))## Actual Crime Cat
## k=3 Predicted 0 1
## 0 84 1
## 1 0 67mean(knn.Boston.pred == test.Boston.knn$crime01)## [1] 0.9934211k=4
set.seed(1)
knn.Boston.pred <- knn(training.Boston.knn, test.Boston.knn, training.crime.knn, k=4)
table(knn.Boston.pred, test.Boston.knn$crime01, dnn = c('k=4 Predicted', "Actual Crime Cat"))## Actual Crime Cat
## k=4 Predicted 0 1
## 0 84 1
## 1 0 67mean(knn.Boston.pred == test.Boston.knn$crime01)## [1] 0.9934211The best overall accuracy for the Boston Data is kNN with k=1. This gives us as close to perfect predictions as possible.