Assignment #3, Chapter 4
April Jacob
2022-10-21
Questions 13, 14, 16
13. This question should be answered using the Weekly data set, which
is part of the ISLR2 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.
library(dplyr)## Warning: package 'dplyr' was built under R version 4.1.3##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionlibrary(ggplot2)## Warning: package 'ggplot2' was built under R version 4.1.3library(ISLR2)## Warning: package 'ISLR2' was built under R version 4.1.3library(corrplot) #visual Correlation Plot## corrplot 0.92 loadedlibrary(GGally) #Scatterplot Matrix## Warning: package 'GGally' was built under R version 4.1.3## Registered S3 method overwritten by 'GGally':
## method from
## +.gg ggplot2library(gmodels) #Confusion Matrix calculations## Warning: package 'gmodels' was built under R version 4.1.3library(MASS) #LDA##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## Boston## The following object is masked from 'package:dplyr':
##
## selectlibrary(class) #KNN
library(e1071) #Naive Bayes## Warning: package 'e1071' was built under R version 4.1.3attach(Weekly)
names(Weekly)## [1] "Year" "Lag1" "Lag2" "Lag3" "Lag4" "Lag5"
## [7] "Volume" "Today" "Direction"- Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
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
##
##
##
## cor(Weekly[, -9]) #correlation between Today and Lag1-Lag5 are close to zero and have little correlation between this weeks returns and previous weeks.## 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 1.000000000corrplot(cor(Weekly[,-9]), method = "number", title = "Correlation of the Weekly Data Set") #correlation between Year & Volume
ggscatmat(Weekly, color = "Direction")## Warning in ggscatmat(Weekly, color = "Direction"): Factor variables are omitted
## in plot
Weekly %>% mutate(row = row_number()) %>%
ggplot(aes(x = row, y = Volume)) +
geom_point() +
geom_smooth(se = FALSE) #average number of shares traded in billions; shows an increase in volume over time## `geom_smooth()` using method = 'gam' and formula 'y ~ s(x, bs = "cs")'
- 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?
Weekly.glm = glm(
Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,
data = Weekly,
family = binomial)
summary(Weekly.glm)##
## 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: 4#Lag2 shows the smallest p-value with a positive Lag2 coefficient- 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.glm, type = "response")
Weekly.pred = rep("Down", length(Weekly.prob))
Weekly.pred[Weekly.prob > 0.5] = "Up"
CrossTable(Weekly.pred, Direction)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 1089
##
##
## | Direction
## Weekly.pred | Down | Up | Row Total |
## -------------|-----------|-----------|-----------|
## Down | 54 | 48 | 102 |
## | 1.657 | 1.325 | |
## | 0.529 | 0.471 | 0.094 |
## | 0.112 | 0.079 | |
## | 0.050 | 0.044 | |
## -------------|-----------|-----------|-----------|
## Up | 430 | 557 | 987 |
## | 0.171 | 0.137 | |
## | 0.436 | 0.564 | 0.906 |
## | 0.888 | 0.921 | |
## | 0.395 | 0.511 | |
## -------------|-----------|-----------|-----------|
## Column Total | 484 | 605 | 1089 |
## | 0.444 | 0.556 | |
## -------------|-----------|-----------|-----------|
##
## #The model accurately predicts only 11.2% when the market is down but accurately predicts 92.1% when the market is up.
0.511+0.050## [1] 0.561#The accuracy rate is 56.1%.
0.395+0.044## [1] 0.439#The error rate is 43.9%.- 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.train = Weekly[train,]
Weekly.test = Weekly[!train,]
Direction.train = Weekly.train$Direction
Direction.test = Weekly.test$Direction
Weekly.lrm = glm(Direction ~ Lag2,
data = Weekly,
family = binomial,
subset = train)
lrm.prob = predict(Weekly.lrm, Weekly.test, type = "response")
lrm.pred = rep("Down", length(lrm.prob))
lrm.pred[lrm.prob > 0.5] = "Up"
CrossTable(lrm.pred, Direction.test)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 104
##
##
## | Direction.test
## lrm.pred | Down | Up | Row Total |
## -------------|-----------|-----------|-----------|
## Down | 9 | 5 | 14 |
## | 1.782 | 1.256 | |
## | 0.643 | 0.357 | 0.135 |
## | 0.209 | 0.082 | |
## | 0.087 | 0.048 | |
## -------------|-----------|-----------|-----------|
## Up | 34 | 56 | 90 |
## | 0.277 | 0.195 | |
## | 0.378 | 0.622 | 0.865 |
## | 0.791 | 0.918 | |
## | 0.327 | 0.538 | |
## -------------|-----------|-----------|-----------|
## Column Total | 43 | 61 | 104 |
## | 0.413 | 0.587 | |
## -------------|-----------|-----------|-----------|
##
## mean(lrm.pred == Direction.test)## [1] 0.625#62.5%- Repeat (d) using LDA.
Weekly.lda = lda(Direction ~ Lag2,
data = Weekly,
subset = train)
lda.prob = predict(Weekly.lda, Weekly.test)$class
CrossTable(lda.prob, Direction.test)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 104
##
##
## | Direction.test
## lda.prob | Down | Up | Row Total |
## -------------|-----------|-----------|-----------|
## Down | 9 | 5 | 14 |
## | 1.782 | 1.256 | |
## | 0.643 | 0.357 | 0.135 |
## | 0.209 | 0.082 | |
## | 0.087 | 0.048 | |
## -------------|-----------|-----------|-----------|
## Up | 34 | 56 | 90 |
## | 0.277 | 0.195 | |
## | 0.378 | 0.622 | 0.865 |
## | 0.791 | 0.918 | |
## | 0.327 | 0.538 | |
## -------------|-----------|-----------|-----------|
## Column Total | 43 | 61 | 104 |
## | 0.413 | 0.587 | |
## -------------|-----------|-----------|-----------|
##
## mean(lda.prob == Direction.test)## [1] 0.625#62.5%- Repeat (d) using QDA.
Weekly.qda = qda(Direction ~ Lag2,
data = Weekly,
subset = train)
qda.prob = predict(Weekly.qda, Weekly.test)$class
CrossTable(qda.prob, Direction.test)##
##
## Cell Contents
## |-------------------------|
## | N |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 104
##
##
## | Direction.test
## qda.prob | Down | Up | Row Total |
## -------------|-----------|-----------|-----------|
## Up | 43 | 61 | 104 |
## | 0.413 | 0.587 | |
## -------------|-----------|-----------|-----------|
## Column Total | 43 | 61 | 104 |
## -------------|-----------|-----------|-----------|
##
## mean(qda.prob == Direction.test)## [1] 0.5865385#58.7%- Repeat (d) using KNN with K = 1.
knn.train = as.matrix(Lag2[train])
knn.test = as.matrix(Lag2[!train])
set.seed(1)
knn.pred = knn(knn.train, knn.test, Direction.train, k = 1)
CrossTable(knn.pred, Direction.test)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 104
##
##
## | Direction.test
## knn.pred | Down | Up | Row Total |
## -------------|-----------|-----------|-----------|
## Down | 21 | 30 | 51 |
## | 0.000 | 0.000 | |
## | 0.412 | 0.588 | 0.490 |
## | 0.488 | 0.492 | |
## | 0.202 | 0.288 | |
## -------------|-----------|-----------|-----------|
## Up | 22 | 31 | 53 |
## | 0.000 | 0.000 | |
## | 0.415 | 0.585 | 0.510 |
## | 0.512 | 0.508 | |
## | 0.212 | 0.298 | |
## -------------|-----------|-----------|-----------|
## Column Total | 43 | 61 | 104 |
## | 0.413 | 0.587 | |
## -------------|-----------|-----------|-----------|
##
## mean(knn.pred == Direction.test)## [1] 0.5#50%- Repeat (d) using naive Bayes.
nb.fit = naiveBayes(Direction ~ Lag2,
data = Weekly,
subset = train)
nb.pred = predict(nb.fit, Weekly.test)
mean(nb.pred == Direction.test)## [1] 0.5865385#58.7%- Which of these methods appears to provide the best results on this data?
#The logistic regression model accurately predicted the market with results of 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.
#GLM interaction withLag5
Weekly.lrm2 = glm(
Direction ~ Lag5,
data = Weekly,
family=binomial,
subset = train)
Weekly.prob2 = predict(Weekly.lrm2, Weekly.test, type = "response")
Weekly.pred2 = rep("Down", length(Weekly.prob2))
Weekly.pred2[Weekly.prob2 > 0.5] = "Up"
CrossTable(Weekly.pred2, Direction.test)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 104
##
##
## | Direction.test
## Weekly.pred2 | Down | Up | Row Total |
## -------------|-----------|-----------|-----------|
## Down | 0 | 3 | 3 |
## | 1.240 | 0.874 | |
## | 0.000 | 1.000 | 0.029 |
## | 0.000 | 0.049 | |
## | 0.000 | 0.029 | |
## -------------|-----------|-----------|-----------|
## Up | 43 | 58 | 101 |
## | 0.037 | 0.026 | |
## | 0.426 | 0.574 | 0.971 |
## | 1.000 | 0.951 | |
## | 0.413 | 0.558 | |
## -------------|-----------|-----------|-----------|
## Column Total | 43 | 61 | 104 |
## | 0.413 | 0.587 | |
## -------------|-----------|-----------|-----------|
##
## #LDA with Interaction Lag5
Weekly.lda2 = lda(Direction ~ Lag5,
data = Weekly,
subset = train)
lda.prob2 = predict(Weekly.lda2, Weekly.test)$class
CrossTable(lda.prob2, Direction.test)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 104
##
##
## | Direction.test
## lda.prob2 | Down | Up | Row Total |
## -------------|-----------|-----------|-----------|
## Down | 0 | 3 | 3 |
## | 1.240 | 0.874 | |
## | 0.000 | 1.000 | 0.029 |
## | 0.000 | 0.049 | |
## | 0.000 | 0.029 | |
## -------------|-----------|-----------|-----------|
## Up | 43 | 58 | 101 |
## | 0.037 | 0.026 | |
## | 0.426 | 0.574 | 0.971 |
## | 1.000 | 0.951 | |
## | 0.413 | 0.558 | |
## -------------|-----------|-----------|-----------|
## Column Total | 43 | 61 | 104 |
## | 0.413 | 0.587 | |
## -------------|-----------|-----------|-----------|
##
## mean(lda.prob2 == Direction.test)## [1] 0.5576923#K=10, Lag5
knn.train2 = as.matrix(Lag5[train])
knn.test2 = as.matrix(Lag5[!train])
set.seed(1)
knn.pred2 = knn(knn.train2, knn.test2, Direction.train, k = 10)
CrossTable(knn.pred2, Direction.test)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 104
##
##
## | Direction.test
## knn.pred2 | Down | Up | Row Total |
## -------------|-----------|-----------|-----------|
## Down | 21 | 25 | 46 |
## | 0.206 | 0.145 | |
## | 0.457 | 0.543 | 0.442 |
## | 0.488 | 0.410 | |
## | 0.202 | 0.240 | |
## -------------|-----------|-----------|-----------|
## Up | 22 | 36 | 58 |
## | 0.164 | 0.115 | |
## | 0.379 | 0.621 | 0.558 |
## | 0.512 | 0.590 | |
## | 0.212 | 0.346 | |
## -------------|-----------|-----------|-----------|
## Column Total | 43 | 61 | 104 |
## | 0.413 | 0.587 | |
## -------------|-----------|-----------|-----------|
##
## mean(knn.pred2 == Direction.test)## [1] 0.5480769#K=100, Lag5
knn.train3 = as.matrix(Lag5[train])
knn.test3 = as.matrix(Lag5[!train])
set.seed(1)
knn.pred3 = knn(knn.train3, knn.test3, Direction.train, k = 100)
CrossTable(knn.pred3, Direction.test)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 104
##
##
## | Direction.test
## knn.pred3 | Down | Up | Row Total |
## -------------|-----------|-----------|-----------|
## Down | 10 | 21 | 31 |
## | 0.619 | 0.437 | |
## | 0.323 | 0.677 | 0.298 |
## | 0.233 | 0.344 | |
## | 0.096 | 0.202 | |
## -------------|-----------|-----------|-----------|
## Up | 33 | 40 | 73 |
## | 0.263 | 0.185 | |
## | 0.452 | 0.548 | 0.702 |
## | 0.767 | 0.656 | |
## | 0.317 | 0.385 | |
## -------------|-----------|-----------|-----------|
## Column Total | 43 | 61 | 104 |
## | 0.413 | 0.587 | |
## -------------|-----------|-----------|-----------|
##
## mean(knn.pred3 == Direction.test)## [1] 0.4807692detach(Weekly)14. 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.
- 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(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) :365Auto$country = factor(Auto$origin, labels = c("American", "European", "Japanese")) #converting the origin numbers to the countries
with(Auto, table(country, origin)) #amount of each country/origin## origin
## country 1 2 3
## American 245 0 0
## European 0 68 0
## Japanese 0 0 79auto = Auto %>%
mutate(mpg01 = ifelse(mpg > median(mpg), 1, 0)) #binary variable created
attach(auto)## The following object is masked from package:ggplot2:
##
## mpgsummary(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
## country mpg01
## American:245 Min. :0.0
## European: 68 1st Qu.:0.0
## Japanese: 79 Median :0.5
## Mean :0.5
## 3rd Qu.:1.0
## Max. :1.0
## - 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.
cor(auto[, -c(9, 10)])## 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 1.0000000corrplot(cor(auto[, -c(9, 10)]),
method = 'color',
order = 'hclust', addrect = 2,
tl.col = 'black', addCoef.col = 'black', number.cex = 0.65)
#cylinders, displacement, and weight seem to have a high negative correlation with MPG01; horsepower also shows a negative correlation
ggscatmat(auto, color = "mpg01")## Warning in ggscatmat(auto, color = "mpg01"): Factor variables are omitted in
## plot
auto %>%
ggplot(aes(cut_number(mpg01, 2), horsepower)) +
geom_boxplot()
auto %>%
ggplot(aes(cut_number(mpg01, 2), cylinders)) +
geom_boxplot() 
auto %>%
ggplot(aes(cut_number(mpg01, 2), displacement)) +
geom_boxplot()
auto %>%
ggplot(aes(cut_number(mpg01, 2), weight)) +
geom_boxplot()
- Split the data into a training set and a test set.
train = (year %% 2 == 0)
auto.train = auto[train,]
auto.test = auto[!train,]- 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?
auto.lda = lda(mpg01 ~ horsepower + cylinders + displacement + weight,
data = auto,
subset = train)
auto.pred = predict(auto.lda, auto.test)
CrossTable(auto.pred$class, auto.test$mpg01)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 182
##
##
## | auto.test$mpg01
## auto.pred$class | 0 | 1 | Row Total |
## ----------------|-----------|-----------|-----------|
## 0 | 86 | 9 | 95 |
## | 21.890 | 26.695 | |
## | 0.905 | 0.095 | 0.522 |
## | 0.860 | 0.110 | |
## | 0.473 | 0.049 | |
## ----------------|-----------|-----------|-----------|
## 1 | 14 | 73 | 87 |
## | 23.902 | 29.149 | |
## | 0.161 | 0.839 | 0.478 |
## | 0.140 | 0.890 | |
## | 0.077 | 0.401 | |
## ----------------|-----------|-----------|-----------|
## Column Total | 100 | 82 | 182 |
## | 0.549 | 0.451 | |
## ----------------|-----------|-----------|-----------|
##
## mean(auto.pred$class == auto.test$mpg01)## [1] 0.87362641-mean(auto.pred$class == auto.test$mpg01) #test error rate## [1] 0.1263736- Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
auto.qda = qda(mpg01 ~ horsepower + cylinders + displacement + weight,
data = auto,
subset = train)
qda.pred = predict(auto.qda, auto.test)
CrossTable(qda.pred$class, auto.test$mpg01)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 182
##
##
## | auto.test$mpg01
## qda.pred$class | 0 | 1 | Row Total |
## ---------------|-----------|-----------|-----------|
## 0 | 89 | 13 | 102 |
## | 19.379 | 23.633 | |
## | 0.873 | 0.127 | 0.560 |
## | 0.890 | 0.159 | |
## | 0.489 | 0.071 | |
## ---------------|-----------|-----------|-----------|
## 1 | 11 | 69 | 80 |
## | 24.709 | 30.133 | |
## | 0.138 | 0.863 | 0.440 |
## | 0.110 | 0.841 | |
## | 0.060 | 0.379 | |
## ---------------|-----------|-----------|-----------|
## Column Total | 100 | 82 | 182 |
## | 0.549 | 0.451 | |
## ---------------|-----------|-----------|-----------|
##
## mean(qda.pred$class == auto.test$mpg01)## [1] 0.86813191-mean(qda.pred$class == auto.test$mpg01) #test error rate## [1] 0.1318681- Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
auto.lrm = glm(mpg01 ~ horsepower + cylinders + displacement + weight,
data = auto,
family = binomial,
subset = train)
auto.prob = predict(auto.lrm, auto.test, type = "response")
auto.pred = rep(0, length(auto.prob))
auto.pred[auto.prob > 0.5] = 1
CrossTable(auto.pred, auto.test$mpg01)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 182
##
##
## | auto.test$mpg01
## auto.pred | 0 | 1 | Row Total |
## -------------|-----------|-----------|-----------|
## 0 | 89 | 11 | 100 |
## | 21.107 | 25.741 | |
## | 0.890 | 0.110 | 0.549 |
## | 0.890 | 0.134 | |
## | 0.489 | 0.060 | |
## -------------|-----------|-----------|-----------|
## 1 | 11 | 71 | 82 |
## | 25.741 | 31.391 | |
## | 0.134 | 0.866 | 0.451 |
## | 0.110 | 0.866 | |
## | 0.060 | 0.390 | |
## -------------|-----------|-----------|-----------|
## Column Total | 100 | 82 | 182 |
## | 0.549 | 0.451 | |
## -------------|-----------|-----------|-----------|
##
## mean(auto.pred == auto.test$mpg01)## [1] 0.87912091-mean(auto.pred == auto.test$mpg01) #test error rate## [1] 0.1208791- Perform naive Bayes on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
nb.fit = naiveBayes(mpg01 ~ horsepower + cylinders + displacement + weight,
data = auto,
subset = train)
nb.pred = predict(nb.fit, auto.test)
CrossTable(nb.pred, auto.test$mpg01)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 182
##
##
## | auto.test$mpg01
## nb.pred | 0 | 1 | Row Total |
## -------------|-----------|-----------|-----------|
## 0 | 88 | 11 | 99 |
## | 20.760 | 25.317 | |
## | 0.889 | 0.111 | 0.544 |
## | 0.880 | 0.134 | |
## | 0.484 | 0.060 | |
## -------------|-----------|-----------|-----------|
## 1 | 12 | 71 | 83 |
## | 24.762 | 30.198 | |
## | 0.145 | 0.855 | 0.456 |
## | 0.120 | 0.866 | |
## | 0.066 | 0.390 | |
## -------------|-----------|-----------|-----------|
## Column Total | 100 | 82 | 182 |
## | 0.549 | 0.451 | |
## -------------|-----------|-----------|-----------|
##
## mean(nb.pred == auto.test$mpg01)## [1] 0.87362641 - mean(nb.pred == auto.test$mpg01) #test error rate## [1] 0.1263736- Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
knn.train = cbind(horsepower, cylinders, displacement, weight)[train,]
knn.test = cbind(horsepower, cylinders, displacement, weight)[!train,]
set.seed(1)
#KNN = 1
knn.pred1 = knn(knn.train, knn.test, auto.train$mpg01, k = 1)
CrossTable(knn.pred1, auto.test$mpg01)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 182
##
##
## | auto.test$mpg01
## knn.pred1 | 0 | 1 | Row Total |
## -------------|-----------|-----------|-----------|
## 0 | 83 | 11 | 94 |
## | 19.031 | 23.209 | |
## | 0.883 | 0.117 | 0.516 |
## | 0.830 | 0.134 | |
## | 0.456 | 0.060 | |
## -------------|-----------|-----------|-----------|
## 1 | 17 | 71 | 88 |
## | 20.329 | 24.791 | |
## | 0.193 | 0.807 | 0.484 |
## | 0.170 | 0.866 | |
## | 0.093 | 0.390 | |
## -------------|-----------|-----------|-----------|
## Column Total | 100 | 82 | 182 |
## | 0.549 | 0.451 | |
## -------------|-----------|-----------|-----------|
##
## mean(knn.pred1 == auto.test$mpg01)## [1] 0.84615381 - mean(knn.pred1 == auto.test$mpg01) #test error rate## [1] 0.1538462#KNN = 3
knn.pred3 = knn(knn.train, knn.test, auto.train$mpg01, k = 3)
CrossTable(knn.pred3, auto.test$mpg01)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 182
##
##
## | auto.test$mpg01
## knn.pred3 | 0 | 1 | Row Total |
## -------------|-----------|-----------|-----------|
## 0 | 84 | 9 | 93 |
## | 21.184 | 25.834 | |
## | 0.903 | 0.097 | 0.511 |
## | 0.840 | 0.110 | |
## | 0.462 | 0.049 | |
## -------------|-----------|-----------|-----------|
## 1 | 16 | 73 | 89 |
## | 22.136 | 26.995 | |
## | 0.180 | 0.820 | 0.489 |
## | 0.160 | 0.890 | |
## | 0.088 | 0.401 | |
## -------------|-----------|-----------|-----------|
## Column Total | 100 | 82 | 182 |
## | 0.549 | 0.451 | |
## -------------|-----------|-----------|-----------|
##
## mean(knn.pred3 == auto.test$mpg01)## [1] 0.86263741 - mean(knn.pred3 == auto.test$mpg01) #test error rate## [1] 0.1373626#KNN = 05
knn.pred5 = knn(knn.train, knn.test, auto.train$mpg01, k = 5)
CrossTable(knn.pred5, auto.test$mpg01)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 182
##
##
## | auto.test$mpg01
## knn.pred5 | 0 | 1 | Row Total |
## -------------|-----------|-----------|-----------|
## 0 | 82 | 9 | 91 |
## | 20.480 | 24.976 | |
## | 0.901 | 0.099 | 0.500 |
## | 0.820 | 0.110 | |
## | 0.451 | 0.049 | |
## -------------|-----------|-----------|-----------|
## 1 | 18 | 73 | 91 |
## | 20.480 | 24.976 | |
## | 0.198 | 0.802 | 0.500 |
## | 0.180 | 0.890 | |
## | 0.099 | 0.401 | |
## -------------|-----------|-----------|-----------|
## Column Total | 100 | 82 | 182 |
## | 0.549 | 0.451 | |
## -------------|-----------|-----------|-----------|
##
## mean(knn.pred5 == auto.test$mpg01)## [1] 0.85164841 - mean(knn.pred5 == auto.test$mpg01) #test error rate## [1] 0.1483516#KNN = 10
knn.pred10 = knn(knn.train, knn.test, auto.train$mpg01, k = 10)
CrossTable(knn.pred10, auto.test$mpg01)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 182
##
##
## | auto.test$mpg01
## knn.pred10 | 0 | 1 | Row Total |
## -------------|-----------|-----------|-----------|
## 0 | 79 | 7 | 86 |
## | 21.330 | 26.012 | |
## | 0.919 | 0.081 | 0.473 |
## | 0.790 | 0.085 | |
## | 0.434 | 0.038 | |
## -------------|-----------|-----------|-----------|
## 1 | 21 | 75 | 96 |
## | 19.108 | 23.302 | |
## | 0.219 | 0.781 | 0.527 |
## | 0.210 | 0.915 | |
## | 0.115 | 0.412 | |
## -------------|-----------|-----------|-----------|
## Column Total | 100 | 82 | 182 |
## | 0.549 | 0.451 | |
## -------------|-----------|-----------|-----------|
##
## mean(knn.pred10 == auto.test$mpg01)## [1] 0.84615381 - mean(knn.pred10 == auto.test$mpg01) #test error rate## [1] 0.1538462#KNN = 50
knn.pred50 = knn(knn.train, knn.test, auto.train$mpg01, k = 50)
CrossTable(knn.pred50, auto.test$mpg01)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 182
##
##
## | auto.test$mpg01
## knn.pred50 | 0 | 1 | Row Total |
## -------------|-----------|-----------|-----------|
## 0 | 80 | 7 | 87 |
## | 21.687 | 26.448 | |
## | 0.920 | 0.080 | 0.478 |
## | 0.800 | 0.085 | |
## | 0.440 | 0.038 | |
## -------------|-----------|-----------|-----------|
## 1 | 20 | 75 | 95 |
## | 19.861 | 24.221 | |
## | 0.211 | 0.789 | 0.522 |
## | 0.200 | 0.915 | |
## | 0.110 | 0.412 | |
## -------------|-----------|-----------|-----------|
## Column Total | 100 | 82 | 182 |
## | 0.549 | 0.451 | |
## -------------|-----------|-----------|-----------|
##
## mean(knn.pred50 == auto.test$mpg01)## [1] 0.85164841 - mean(knn.pred50 == auto.test$mpg01) #test error rate## [1] 0.1483516#KNN = 100
knn.pred100 = knn(knn.train, knn.test, auto.train$mpg01, k = 100)
CrossTable(knn.pred100, auto.test$mpg01)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 182
##
##
## | auto.test$mpg01
## knn.pred100 | 0 | 1 | Row Total |
## -------------|-----------|-----------|-----------|
## 0 | 81 | 7 | 88 |
## | 22.045 | 26.884 | |
## | 0.920 | 0.080 | 0.484 |
## | 0.810 | 0.085 | |
## | 0.445 | 0.038 | |
## -------------|-----------|-----------|-----------|
## 1 | 19 | 75 | 94 |
## | 20.638 | 25.168 | |
## | 0.202 | 0.798 | 0.516 |
## | 0.190 | 0.915 | |
## | 0.104 | 0.412 | |
## -------------|-----------|-----------|-----------|
## Column Total | 100 | 82 | 182 |
## | 0.549 | 0.451 | |
## -------------|-----------|-----------|-----------|
##
## mean(knn.pred100 == auto.test$mpg01)## [1] 0.85714291 - mean(knn.pred100 == auto.test$mpg01) #test error rate## [1] 0.1428571detach(auto)16. Using the Boston data set, fit classification models in order to predict whether a given census tract has a crime rate above or below the median. Explore logistic regression, LDA, naive Bayes, and KNN models using various subsets of the predictors. Describe your findings. Hint: You will have to create the response variable yourself, using the variables that are contained in the Boston data set.
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00names(Boston)## [1] "crim" "zn" "indus" "chas" "nox" "rm" "age"
## [8] "dis" "rad" "tax" "ptratio" "black" "lstat" "medv"boston = Boston %>%
mutate(crim01 = ifelse(crim > median(crim), 1, 0))
names(boston)## [1] "crim" "zn" "indus" "chas" "nox" "rm" "age"
## [8] "dis" "rad" "tax" "ptratio" "black" "lstat" "medv"
## [15] "crim01"attach(boston)
summary(boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv crim01
## Min. : 1.73 Min. : 5.00 Min. :0.0
## 1st Qu.: 6.95 1st Qu.:17.02 1st Qu.:0.0
## Median :11.36 Median :21.20 Median :0.5
## Mean :12.65 Mean :22.53 Mean :0.5
## 3rd Qu.:16.95 3rd Qu.:25.00 3rd Qu.:1.0
## Max. :37.97 Max. :50.00 Max. :1.0#CORRELATION
cor(boston)## crim zn indus chas nox
## crim 1.00000000 -0.20046922 0.40658341 -0.055891582 0.42097171
## zn -0.20046922 1.00000000 -0.53382819 -0.042696719 -0.51660371
## indus 0.40658341 -0.53382819 1.00000000 0.062938027 0.76365145
## chas -0.05589158 -0.04269672 0.06293803 1.000000000 0.09120281
## nox 0.42097171 -0.51660371 0.76365145 0.091202807 1.00000000
## rm -0.21924670 0.31199059 -0.39167585 0.091251225 -0.30218819
## age 0.35273425 -0.56953734 0.64477851 0.086517774 0.73147010
## dis -0.37967009 0.66440822 -0.70802699 -0.099175780 -0.76923011
## rad 0.62550515 -0.31194783 0.59512927 -0.007368241 0.61144056
## tax 0.58276431 -0.31456332 0.72076018 -0.035586518 0.66802320
## ptratio 0.28994558 -0.39167855 0.38324756 -0.121515174 0.18893268
## black -0.38506394 0.17552032 -0.35697654 0.048788485 -0.38005064
## lstat 0.45562148 -0.41299457 0.60379972 -0.053929298 0.59087892
## medv -0.38830461 0.36044534 -0.48372516 0.175260177 -0.42732077
## crim01 0.40939545 -0.43615103 0.60326017 0.070096774 0.72323480
## rm age dis rad tax ptratio
## crim -0.21924670 0.35273425 -0.37967009 0.625505145 0.58276431 0.2899456
## zn 0.31199059 -0.56953734 0.66440822 -0.311947826 -0.31456332 -0.3916785
## indus -0.39167585 0.64477851 -0.70802699 0.595129275 0.72076018 0.3832476
## chas 0.09125123 0.08651777 -0.09917578 -0.007368241 -0.03558652 -0.1215152
## nox -0.30218819 0.73147010 -0.76923011 0.611440563 0.66802320 0.1889327
## rm 1.00000000 -0.24026493 0.20524621 -0.209846668 -0.29204783 -0.3555015
## age -0.24026493 1.00000000 -0.74788054 0.456022452 0.50645559 0.2615150
## dis 0.20524621 -0.74788054 1.00000000 -0.494587930 -0.53443158 -0.2324705
## rad -0.20984667 0.45602245 -0.49458793 1.000000000 0.91022819 0.4647412
## tax -0.29204783 0.50645559 -0.53443158 0.910228189 1.00000000 0.4608530
## ptratio -0.35550149 0.26151501 -0.23247054 0.464741179 0.46085304 1.0000000
## black 0.12806864 -0.27353398 0.29151167 -0.444412816 -0.44180801 -0.1773833
## lstat -0.61380827 0.60233853 -0.49699583 0.488676335 0.54399341 0.3740443
## medv 0.69535995 -0.37695457 0.24992873 -0.381626231 -0.46853593 -0.5077867
## crim01 -0.15637178 0.61393992 -0.61634164 0.619786249 0.60874128 0.2535684
## black lstat medv crim01
## crim -0.38506394 0.4556215 -0.3883046 0.40939545
## zn 0.17552032 -0.4129946 0.3604453 -0.43615103
## indus -0.35697654 0.6037997 -0.4837252 0.60326017
## chas 0.04878848 -0.0539293 0.1752602 0.07009677
## nox -0.38005064 0.5908789 -0.4273208 0.72323480
## rm 0.12806864 -0.6138083 0.6953599 -0.15637178
## age -0.27353398 0.6023385 -0.3769546 0.61393992
## dis 0.29151167 -0.4969958 0.2499287 -0.61634164
## rad -0.44441282 0.4886763 -0.3816262 0.61978625
## tax -0.44180801 0.5439934 -0.4685359 0.60874128
## ptratio -0.17738330 0.3740443 -0.5077867 0.25356836
## black 1.00000000 -0.3660869 0.3334608 -0.35121093
## lstat -0.36608690 1.0000000 -0.7376627 0.45326273
## medv 0.33346082 -0.7376627 1.0000000 -0.26301673
## crim01 -0.35121093 0.4532627 -0.2630167 1.00000000corrplot(cor(boston),
method = "number")
corrplot(cor(boston),
method = "square")
corrplot(cor(boston),
method = 'color',
order = 'hclust', addrect = 2,
tl.col = 'black', addCoef.col = 'black', number.cex = 0.65)
ggscatmat(boston, color = "crim01")## Warning in ggscatmat(boston, color = "crim01"): Factor variables are omitted in
## plot
#SPLITTING THE DATASET
train = 1:(length(boston$crim)/2)
test = (length(boston$crim)/2 + 1):length(boston$crim)
boston.train = boston[train, ]
boston.test = boston[test, ]
set.seed(1)boston.glm = glm(crim01 ~ rad + tax + age + indus + nox + dis,
data = boston,
family = binomial,
subset = train)
boston.probs = predict(boston.glm, boston.test, type = "response")
boston.pred = rep(0, length(boston.probs))
boston.pred[boston.probs > 0.5] = 1
CrossTable(boston.pred, boston.test$crim01)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 253
##
##
## | boston.test$crim01
## boston.pred | 0 | 1 | Row Total |
## -------------|-----------|-----------|-----------|
## 0 | 75 | 8 | 83 |
## | 70.038 | 38.671 | |
## | 0.904 | 0.096 | 0.328 |
## | 0.833 | 0.049 | |
## | 0.296 | 0.032 | |
## -------------|-----------|-----------|-----------|
## 1 | 15 | 155 | 170 |
## | 34.195 | 18.881 | |
## | 0.088 | 0.912 | 0.672 |
## | 0.167 | 0.951 | |
## | 0.059 | 0.613 | |
## -------------|-----------|-----------|-----------|
## Column Total | 90 | 163 | 253 |
## | 0.356 | 0.644 | |
## -------------|-----------|-----------|-----------|
##
## mean(boston.pred == boston.test$crim01)## [1] 0.90909091 - mean(boston.pred == boston.test$crim01) #test error## [1] 0.09090909boston.lda = lda(crim01 ~ rad + tax + age + indus + nox + dis,
data = boston,
subset = train)
boston.pred = predict(boston.lda, boston.test)
CrossTable(boston.pred$class, boston.test$crim01)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 253
##
##
## | boston.test$crim01
## boston.pred$class | 0 | 1 | Row Total |
## ------------------|-----------|-----------|-----------|
## 0 | 81 | 18 | 99 |
## | 59.517 | 32.862 | |
## | 0.818 | 0.182 | 0.391 |
## | 0.900 | 0.110 | |
## | 0.320 | 0.071 | |
## ------------------|-----------|-----------|-----------|
## 1 | 9 | 145 | 154 |
## | 38.261 | 21.126 | |
## | 0.058 | 0.942 | 0.609 |
## | 0.100 | 0.890 | |
## | 0.036 | 0.573 | |
## ------------------|-----------|-----------|-----------|
## Column Total | 90 | 163 | 253 |
## | 0.356 | 0.644 | |
## ------------------|-----------|-----------|-----------|
##
## mean(boston.pred$class == boston.test$crim01)## [1] 0.89328061 - mean(boston.pred$class == boston.test$crim01) #test error rate## [1] 0.1067194nb.fit = naiveBayes(crim01 ~ rad + tax + age + indus + nox + dis,
data = boston,
subset = train)
nb.pred = predict(nb.fit, boston.test)
CrossTable(nb.pred, boston.test$crim01)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 253
##
##
## | boston.test$crim01
## nb.pred | 0 | 1 | Row Total |
## -------------|-----------|-----------|-----------|
## 0 | 77 | 13 | 90 |
## | 63.206 | 34.899 | |
## | 0.856 | 0.144 | 0.356 |
## | 0.856 | 0.080 | |
## | 0.304 | 0.051 | |
## -------------|-----------|-----------|-----------|
## 1 | 13 | 150 | 163 |
## | 34.899 | 19.269 | |
## | 0.080 | 0.920 | 0.644 |
## | 0.144 | 0.920 | |
## | 0.051 | 0.593 | |
## -------------|-----------|-----------|-----------|
## Column Total | 90 | 163 | 253 |
## | 0.356 | 0.644 | |
## -------------|-----------|-----------|-----------|
##
## mean(nb.pred == boston.test$crim01)## [1] 0.89723321 - mean(nb.pred == boston.test$crim01) #test error rate## [1] 0.1027668knn.train = cbind(rad, tax, age, indus, nox, dis)[train,]
knn.test = cbind(rad, tax, age, indus, nox, dis)[test,]
set.seed(1)
train.crim01 = crim01[train]
#KNN = 1
knn.pred1 = knn(knn.train, knn.test, train.crim01, k = 1)
CrossTable(knn.pred1, boston.test$crim01)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 253
##
##
## | boston.test$crim01
## knn.pred1 | 0 | 1 | Row Total |
## -------------|-----------|-----------|-----------|
## 0 | 82 | 151 | 233 |
## | 0.009 | 0.005 | |
## | 0.352 | 0.648 | 0.921 |
## | 0.911 | 0.926 | |
## | 0.324 | 0.597 | |
## -------------|-----------|-----------|-----------|
## 1 | 8 | 12 | 20 |
## | 0.110 | 0.061 | |
## | 0.400 | 0.600 | 0.079 |
## | 0.089 | 0.074 | |
## | 0.032 | 0.047 | |
## -------------|-----------|-----------|-----------|
## Column Total | 90 | 163 | 253 |
## | 0.356 | 0.644 | |
## -------------|-----------|-----------|-----------|
##
## mean(knn.pred1 == boston.test$crim01)## [1] 0.37154151 - mean(knn.pred1 == boston.test$crim01) #test error rate## [1] 0.6284585#KNN = 3
knn.pred1 = knn(knn.train, knn.test, train.crim01, k = 3)
CrossTable(knn.pred1, boston.test$crim01)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 253
##
##
## | boston.test$crim01
## knn.pred1 | 0 | 1 | Row Total |
## -------------|-----------|-----------|-----------|
## 0 | 80 | 23 | 103 |
## | 51.311 | 28.331 | |
## | 0.777 | 0.223 | 0.407 |
## | 0.889 | 0.141 | |
## | 0.316 | 0.091 | |
## -------------|-----------|-----------|-----------|
## 1 | 10 | 140 | 150 |
## | 35.234 | 19.454 | |
## | 0.067 | 0.933 | 0.593 |
## | 0.111 | 0.859 | |
## | 0.040 | 0.553 | |
## -------------|-----------|-----------|-----------|
## Column Total | 90 | 163 | 253 |
## | 0.356 | 0.644 | |
## -------------|-----------|-----------|-----------|
##
## mean(knn.pred1 == boston.test$crim01)## [1] 0.86956521 - mean(knn.pred1 == boston.test$crim01) #test error rate## [1] 0.1304348#KNN = 10
knn.pred1 = knn(knn.train, knn.test, train.crim01, k = 10)
CrossTable(knn.pred1, boston.test$crim01)##
##
## Cell Contents
## |-------------------------|
## | N |
## | Chi-square contribution |
## | N / Row Total |
## | N / Col Total |
## | N / Table Total |
## |-------------------------|
##
##
## Total Observations in Table: 253
##
##
## | boston.test$crim01
## knn.pred1 | 0 | 1 | Row Total |
## -------------|-----------|-----------|-----------|
## 0 | 83 | 23 | 106 |
## | 54.403 | 30.039 | |
## | 0.783 | 0.217 | 0.419 |
## | 0.922 | 0.141 | |
## | 0.328 | 0.091 | |
## -------------|-----------|-----------|-----------|
## 1 | 7 | 140 | 147 |
## | 39.230 | 21.660 | |
## | 0.048 | 0.952 | 0.581 |
## | 0.078 | 0.859 | |
## | 0.028 | 0.553 | |
## -------------|-----------|-----------|-----------|
## Column Total | 90 | 163 | 253 |
## | 0.356 | 0.644 | |
## -------------|-----------|-----------|-----------|
##
## mean(knn.pred1 == boston.test$crim01)## [1] 0.88142291 - mean(knn.pred1 == boston.test$crim01) #test error rate## [1] 0.1185771