Classification
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library(tidyverse)## Warning: package 'tidyverse' was built under R version 4.1.3## -- Attaching packages --------------------------------------- tidyverse 1.3.2 --
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## The following object is masked from 'package:dplyr':
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## selectlibrary(class)
library(ISLR2)## Warning: package 'ISLR2' was built under R version 4.1.3##
## Attaching package: 'ISLR2'
##
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## BostonQ13
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
##(a) Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
head(Weekly)## Year Lag1 Lag2 Lag3 Lag4 Lag5 Volume Today Direction
## 1 1990 0.816 1.572 -3.936 -0.229 -3.484 0.1549760 -0.270 Down
## 2 1990 -0.270 0.816 1.572 -3.936 -0.229 0.1485740 -2.576 Down
## 3 1990 -2.576 -0.270 0.816 1.572 -3.936 0.1598375 3.514 Up
## 4 1990 3.514 -2.576 -0.270 0.816 1.572 0.1616300 0.712 Up
## 5 1990 0.712 3.514 -2.576 -0.270 0.816 0.1537280 1.178 Up
## 6 1990 1.178 0.712 3.514 -2.576 -0.270 0.1544440 -1.372 Downcor(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 1.000000000#The only high/moderate correlations we see are cor(Year, volume)= 0.84, the rest of the correlations seem to be very low
attach(`Weekly`)
plot(Volume)
# we also see an exponential trend with volume throughout the 21 years.##(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?
logm <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Weekly, family = binomial)
summary(logm)##
## 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# Only one predictior variable seems to be significant which is Lag2##(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.
logm_probs <- predict(logm, type = "response")
contrasts(Direction)## Up
## Down 0
## Up 1logm_pred <- rep("Down", 1089)
logm_pred[logm_probs > .5]= "Up"
table(logm_pred, Direction)## Direction
## logm_pred Down Up
## Down 54 48
## Up 430 557(54+557)/1089## [1] 0.5610652# We see that our model makes correct predictions about 56% of the time.
# We also see that our model predicts up 557 times when the data was actually up and down 54 times whent he data was actually 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).
train=(Year<2009)
Weekly.2009 <- Weekly[!train ,]
Direction.2009 <- Direction[!train]
dim(Weekly.2009)## [1] 104 9Direction.2009 <- Direction[!train]
logm2 <- glm(Direction ~ Lag2, data = Weekly, family = binomial, subset = train)
summary(logm2)##
## 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: 4logm2_probs <- predict(logm2, Weekly.2009, type = "response")
logm2_pred <- rep("Down", 104)
logm2_pred[logm2_probs > .5]= "Up"
table(logm2_pred, Direction.2009)## Direction.2009
## logm2_pred Down Up
## Down 9 5
## Up 34 56(9+56)/104## [1] 0.625#We see that our model predicts 62.5% of the data correctly##(e) Repeat (d) using LDA.
lda.fit=lda(Direction~Lag2 ,data=`Weekly` ,subset=train)
lda.fit## 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 0.4414162lda.pred=predict (lda.fit , Weekly.2009)
names(lda.pred)## [1] "class" "posterior" "x"lda.class=lda.pred$class
table(lda.class ,Direction.2009)## Direction.2009
## lda.class Down Up
## Down 9 5
## Up 34 56(9+56)/104## [1] 0.625# The model predicts 62.5% of the data correctly##(f) Repeat (d) using QDA.
qda.fit=qda(Direction~Lag2 ,data=Weekly ,subset=train)
qda.fit## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581qda.class=predict(qda.fit ,Weekly.2009)$class
table(qda.class ,Direction.2009)## Direction.2009
## qda.class Down Up
## Down 0 0
## Up 43 6161/104## [1] 0.5865385# Our model predicts 58.7% of our data correctly##(g) Repeat (d) using KNN with K = 1.
train.X=as.matrix(Lag2[train])
test.X=as.matrix(Lag2[!train])
train.Direction =Direction[train]
set.seed(1)
Weekknn.pred=knn(train.X, test.X, train.Direction, k=1)
table(Weekknn.pred, Direction.2009)## Direction.2009
## Weekknn.pred Down Up
## Down 21 30
## Up 22 31(21+31)/104## [1] 0.5# Our model predicts 50% of our data correctly###(i) Which of these methods appears to provide the best results on this data?
#The methods that had the best results are Logistic regression and linear Discriminant Analysis with both having a 62.5% accuracy on the test data set.###(j) 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 with interactions
logm3 <- glm(Direction ~ Lag2:Lag4+Lag2, data = Weekly, family = binomial, subset = train)
summary(logm2)##
## 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: 4logm3_probs <- predict(logm3, Weekly.2009, type = "response")
logm3_pred <- rep("Down", 104)
logm3_pred[logm3_probs > .5]= "Up"
table(logm3_pred, Direction.2009)## Direction.2009
## logm3_pred Down Up
## Down 3 4
## Up 40 57(3+57)/104## [1] 0.5769231# we see our model is 57% accurate on our test data
#we will try a knn model with K=100
train.X=as.matrix(Lag2[train])
test.X=as.matrix(Lag2[!train])
train.Direction =Direction[train]
set.seed(1)
Weekknn.pred=knn(train.X, test.X, train.Direction, k=100)
table(Weekknn.pred, Direction.2009)## Direction.2009
## Weekknn.pred Down Up
## Down 10 11
## Up 33 5060/104## [1] 0.5769231# we see our model is 57% accurate on our test data
detach(Weekly)#Q14 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.
Auto <- Auto
attach(Auto)## The following object is masked from package:ggplot2:
##
## mpgmpg01 <- rep(0, length(mpg))
mpg01[mpg > median(mpg)] <- 1
Auto = 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.
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 1.0000000plot(mpg01, weight)
plot(mpg01, displacement)
# We see strong correlations with mpg and displacement, weight, and horsepower.##(c) Split the data into a training set and a test set.
train <- (year %% 2 == 0)
train.auto <- Auto[train,]
test.auto <- Auto[-train,]##(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 <- lda(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.auto)
lda.pred <- predict(lda, test.auto)
table(lda.pred$class, test.auto$mpg01)##
## 0 1
## 0 169 7
## 1 26 189mean(lda.pred$class != test.auto$mpg01)## [1] 0.08439898# The test error of our lda model is 8.43%##(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 <- qda(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.auto)
qda.pred <- predict(qda, test.auto)
table(qda.pred$class, test.auto$mpg01)##
## 0 1
## 0 176 20
## 1 19 176mean(qda.pred$class != test.auto$mpg01)## [1] 0.09974425# The test error of our qda model is 9.97%##(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?
log<-glm(mpg01~displacement+horsepower+weight+year+cylinders+origin, data=train.auto,family=binomial)
log.probs = predict(log, test.auto, type = "response")
log.pred = rep(0, length(log.probs))
log.pred[log.probs > 0.5] = 1
table(log.pred, test.auto$mpg01)##
## log.pred 0 1
## 0 174 12
## 1 21 184mean(log.pred != test.auto$mpg01)## [1] 0.08439898# The test error of our qda model is 8.44%##(h) 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?
#with K=1
train.K= cbind(displacement,horsepower,weight,cylinders,year, origin)[train,]
test.K=cbind(displacement,horsepower,weight,cylinders, year, origin)[-train,]
set.seed(1)
autok.pred=knn(train.K,test.K,train.auto$mpg01,k=1)
mean(autok.pred != test.auto$mpg01)## [1] 0.07161125#with K=3
autok.pred=knn(train.K,test.K,train.auto$mpg01,k=3)
mean(autok.pred != test.auto$mpg01)## [1] 0.09462916#withK=5
autok.pred=knn(train.K,test.K,train.auto$mpg01,k=5)
mean(autok.pred != test.auto$mpg01)## [1] 0.112532#We see that the lowest test errors rate we obtain was k=1 with 7.16%
detach(Auto)#Q16 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
head(Boston)## crim zn indus chas nox rm age dis rad tax ptratio lstat medv
## 1 0.00632 18 2.31 0 0.538 6.575 65.2 4.0900 1 296 15.3 4.98 24.0
## 2 0.02731 0 7.07 0 0.469 6.421 78.9 4.9671 2 242 17.8 9.14 21.6
## 3 0.02729 0 7.07 0 0.469 7.185 61.1 4.9671 2 242 17.8 4.03 34.7
## 4 0.03237 0 2.18 0 0.458 6.998 45.8 6.0622 3 222 18.7 2.94 33.4
## 5 0.06905 0 2.18 0 0.458 7.147 54.2 6.0622 3 222 18.7 5.33 36.2
## 6 0.02985 0 2.18 0 0.458 6.430 58.7 6.0622 3 222 18.7 5.21 28.7summary(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 lstat
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 1.73
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.: 6.95
## Median : 5.000 Median :330.0 Median :19.05 Median :11.36
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :12.65
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:16.95
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :37.97
## medv
## Min. : 5.00
## 1st Qu.:17.02
## Median :21.20
## Mean :22.53
## 3rd Qu.:25.00
## Max. :50.00attach(Boston)
# lets create our new binary variable for crime
crime01 <- rep(0, length(crim))
crime01[crim > median(crim)] <- 1
Boston= data.frame(Boston,crime01)
# lets create a train and test dataset
train = 1:(dim(Boston)[1]/2)
test = (dim(Boston)[1]/2 + 1):dim(Boston)[1]
Boston.train = Boston[train, ]
Boston.test = Boston[test, ]
crime01.test = crime01[test]
# lets look at all correlations
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
## lstat 0.45562148 -0.41299457 0.60379972 -0.053929298 0.59087892
## medv -0.38830461 0.36044534 -0.48372516 0.175260177 -0.42732077
## crime01 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
## 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
## crime01 -0.15637178 0.61393992 -0.61634164 0.619786249 0.60874128 0.2535684
## lstat medv crime01
## crim 0.4556215 -0.3883046 0.40939545
## zn -0.4129946 0.3604453 -0.43615103
## indus 0.6037997 -0.4837252 0.60326017
## chas -0.0539293 0.1752602 0.07009677
## nox 0.5908789 -0.4273208 0.72323480
## rm -0.6138083 0.6953599 -0.15637178
## age 0.6023385 -0.3769546 0.61393992
## dis -0.4969958 0.2499287 -0.61634164
## rad 0.4886763 -0.3816262 0.61978625
## tax 0.5439934 -0.4685359 0.60874128
## ptratio 0.3740443 -0.5077867 0.25356836
## lstat 1.0000000 -0.7376627 0.45326273
## medv -0.7376627 1.0000000 -0.26301673
## crime01 0.4532627 -0.2630167 1.00000000#lets first try a logistic regression
set.seed(1)
Boston.fit <-glm(crime01~ indus+nox+age+dis+rad+tax, data=Boston.train,family=binomial)
Boston.probs = predict(Boston.fit, Boston.test, type = "response")
Boston.pred = rep(0, length(Boston.probs))
Boston.pred[Boston.probs > 0.5] = 1
table(Boston.pred, crime01.test)## crime01.test
## Boston.pred 0 1
## 0 75 8
## 1 15 155(75+155)/253## [1] 0.9090909mean(Boston.pred != crime01.test)## [1] 0.09090909#lets try linear disriminat analysis
Boston.ldafit <-lda(crime01~ indus+nox+age+dis+rad+tax, data=Boston.train,family=binomial)
Bostonlda.pred = predict(Boston.ldafit, Boston.test)
table(Bostonlda.pred$class, crime01.test)## crime01.test
## 0 1
## 0 81 18
## 1 9 145(81+145)/253## [1] 0.8932806#now k nearest neighbors at k=1 and k=10
#K=1
train.K=cbind(indus,nox,age,dis,rad,tax)[train,]
test.K=cbind(indus,nox,age,dis,rad,tax)[test,]
Bosknn.pred=knn(train.K, test.K, crime01.test, k=1)
table(Bosknn.pred,crime01.test)## crime01.test
## Bosknn.pred 0 1
## 0 31 155
## 1 59 8(31+8)/253## [1] 0.1541502train.K=cbind(indus,nox,age,dis,rad,tax)[train,]
test.K=cbind(indus,nox,age,dis,rad,tax)[test,]
Bosknn.pred=knn(train.K, test.K, crime01.test, k=10)
table(Bosknn.pred,crime01.test)## crime01.test
## Bosknn.pred 0 1
## 0 42 8
## 1 48 155(45+155)/(253)## [1] 0.7905138# we see that our best method was using logistic model because the test error rate was the lowest at 9.09%. For the Logictice model we saw that the only variable that was significant was indus, nox and tax.