Assignment 3
David Rodriguez
6/24/2020
QUESTION 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?
Using the cor() function we can see that there is almost no correlation among the variables. This is not a surprise given that there is no way to determine if the stock market will go up or down based on past returns. The Volume has increased (for the most part) over time but this is also not out of the ordinary.
library(ISLR)
attach(Weekly)
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(Today)
(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?
Lag2 appears to be the only predictor that is statistically significant.
glm.fit1=glm(Direction~Lag1+Lag2+Lag3+Lag4+Lag5+Volume, data=Weekly, family=binomial)
summary(glm.fit1)##
## 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(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.
The confusion matrix is showing the correct and incorrect predictions for each Direction. So the model guessed correctly 54+557=661 times and incorrectly 48+430=478 tmes.
glm.probs=predict(glm.fit1,type = 'response')
glm.pred=rep("Down", 1089)
glm.pred[glm.probs >0.5]="Up"
table(glm.pred, Direction)## Direction
## glm.pred Down Up
## Down 54 48
## Up 430 557mean(glm.pred==Direction)## [1] 0.5610652(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]
glm.fit2=glm(Direction~Lag2, data=Weekly, family=binomial, subset=train)
glm.probs2=predict(glm.fit2,Weekly.2009, type="response")
glm.pred=rep("Down", length(glm.probs2))
glm.pred[glm.probs2>.5]="Up"
table(glm.pred, Direction.2009)## Direction.2009
## glm.pred Down Up
## Down 9 5
## Up 34 56mean(glm.pred==Direction.2009)## [1] 0.625(e) Repeat (d) using LDA.
library(MASS)
lda.fit=lda(Direction~Lag2, data=Weekly, subset=train)
lda.pred=predict(lda.fit,Weekly.2009)
lda.class=lda.pred$class
table(lda.class, Direction.2009)## Direction.2009
## lda.class Down Up
## Down 9 5
## Up 34 56mean(lda.class==Direction.2009)## [1] 0.625(f) Repeat (d) using QDA.
qda.fit=qda(Direction~Lag2, data=Weekly, subset=train)
qda.fit## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581qda.class=predict(qda.fit, Weekly.2009)$class
table(qda.class, Direction.2009)## Direction.2009
## qda.class Down Up
## Down 0 0
## Up 43 61mean(qda.class==Direction.2009)## [1] 0.5865385(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)
knn.pred=knn(train.X, test.X, train.Direction, k=1)
table(knn.pred, Direction.2009)## Direction.2009
## knn.pred Down Up
## Down 21 30
## Up 22 31mean(knn.pred==Direction.2009)## [1] 0.5(h) Which of these methods appears to provide the best results on this data? *LDA and logistic regression came out to the same accuracy predicion of 62.5%. QDA and KNN were lower at 58.7% and 50% respectively.
(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. I reran all the models using the Lag2:Lag1 interaction. The original logistic regression and LDA models still have the highest accuracy rates. While the KNN model still did not get close to the best accuracy of 62.5%, increasing the value of K to 100 did increase the method from 50% to 57.7% for K=100
glm.fit2=glm(Direction~Lag2:Lag1, data=Weekly, family=binomial, subset=train)
glm.probs2=predict(glm.fit2,Weekly.2009, type="response")
glm.pred=rep("Down", length(glm.probs2))
glm.pred[glm.probs2>.5]="Up"
table(glm.pred, Direction.2009)## Direction.2009
## glm.pred Down Up
## Down 1 1
## Up 42 60mean(glm.pred==Direction.2009)## [1] 0.5865385library(MASS)
lda.fit=lda(Direction~Lag2:Lag1, data=Weekly, subset=train)
lda.pred=predict(lda.fit,Weekly.2009)
lda.class=lda.pred$class
table(lda.class, Direction.2009)## Direction.2009
## lda.class Down Up
## Down 0 1
## Up 43 60mean(lda.class==Direction.2009)## [1] 0.5769231qda.fit=qda(Direction~Lag2:Lag1, data=Weekly, subset=train)
qda.fit## Call:
## qda(Direction ~ Lag2:Lag1, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2:Lag1
## Down -0.8014495
## Up -0.1393632qda.class=predict(qda.fit, Weekly.2009)$class
table(qda.class, Direction.2009)## Direction.2009
## qda.class Down Up
## Down 16 32
## Up 27 29mean(qda.class==Direction.2009)## [1] 0.4326923KNN=10
library(class)
train.X=as.matrix(Lag2[train])
test.X=as.matrix(Lag2[!train])
train.Direction =Direction[train]
set.seed(1)
knn.pred=knn(train.X, test.X, train.Direction, k=10)
table(knn.pred, Direction.2009)## Direction.2009
## knn.pred Down Up
## Down 17 21
## Up 26 40mean(knn.pred==Direction.2009)## [1] 0.5480769KNN=50
library(class)
train.X=as.matrix(Lag2[train])
test.X=as.matrix(Lag2[!train])
train.Direction =Direction[train]
set.seed(1)
knn.pred=knn(train.X, test.X, train.Direction, k=50)
table(knn.pred, Direction.2009)## Direction.2009
## knn.pred Down Up
## Down 20 23
## Up 23 38mean(knn.pred==Direction.2009)## [1] 0.5576923KNN=100
library(class)
train.X=as.matrix(Lag2[train])
test.X=as.matrix(Lag2[!train])
train.Direction =Direction[train]
set.seed(1)
knn.pred=knn(train.X, test.X, train.Direction, k=100)
table(knn.pred, Direction.2009)## Direction.2009
## knn.pred Down Up
## Down 10 11
## Up 33 50mean(knn.pred==Direction.2009)## [1] 0.5769231QUESTION 11
(a) Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
library(ISLR)
attach(Auto)
mpg01=rep(0, length(mpg))
mpg01[mpg>median(mpg)] = 1
Auto=data.frame(Auto, mpg01)
summary(Auto)## mpg cylinders displacement horsepower weight
## Min. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0 Min. :1613
## 1st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0 1st Qu.:2225
## Median :22.75 Median :4.000 Median :151.0 Median : 93.5 Median :2804
## Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5 Mean :2978
## 3rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0 3rd Qu.:3615
## Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0 Max. :5140
##
## acceleration year origin name
## Min. : 8.00 Min. :70.00 Min. :1.000 amc matador : 5
## 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000 ford pinto : 5
## Median :15.50 Median :76.00 Median :1.000 toyota corolla : 5
## Mean :15.54 Mean :75.98 Mean :1.577 amc gremlin : 4
## 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000 amc hornet : 4
## Max. :24.80 Max. :82.00 Max. :3.000 chevrolet chevette: 4
## (Other) :365
## mpg01
## Min. :0.0
## 1st Qu.:0.0
## Median :0.5
## Mean :0.5
## 3rd Qu.:1.0
## Max. :1.0
## (b) Explore the data graphically in order to investigate the association between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings. When looking at the predictors graphically, I noticed that box plots were the best way to see the correlation between mpg01 and the other variables. The box plots show that there is a disctinct threshhold where a variables’ values cross the median mpg’s.
cor(Auto[,-9])#eliminate the "name" column## mpg cylinders displacement horsepower weight
## mpg 1.0000000 -0.7776175 -0.8051269 -0.7784268 -0.8322442
## cylinders -0.7776175 1.0000000 0.9508233 0.8429834 0.8975273
## displacement -0.8051269 0.9508233 1.0000000 0.8972570 0.9329944
## horsepower -0.7784268 0.8429834 0.8972570 1.0000000 0.8645377
## weight -0.8322442 0.8975273 0.9329944 0.8645377 1.0000000
## acceleration 0.4233285 -0.5046834 -0.5438005 -0.6891955 -0.4168392
## year 0.5805410 -0.3456474 -0.3698552 -0.4163615 -0.3091199
## origin 0.5652088 -0.5689316 -0.6145351 -0.4551715 -0.5850054
## mpg01 0.8369392 -0.7591939 -0.7534766 -0.6670526 -0.7577566
## acceleration year origin mpg01
## mpg 0.4233285 0.5805410 0.5652088 0.8369392
## cylinders -0.5046834 -0.3456474 -0.5689316 -0.7591939
## displacement -0.5438005 -0.3698552 -0.6145351 -0.7534766
## horsepower -0.6891955 -0.4163615 -0.4551715 -0.6670526
## weight -0.4168392 -0.3091199 -0.5850054 -0.7577566
## acceleration 1.0000000 0.2903161 0.2127458 0.3468215
## year 0.2903161 1.0000000 0.1815277 0.4299042
## origin 0.2127458 0.1815277 1.0000000 0.5136984
## mpg01 0.3468215 0.4299042 0.5136984 1.0000000pairs(Auto)
boxplot(displacement~mpg01)
boxplot(weight~mpg01)
boxplot(horsepower~mpg01)
boxplot(cylinders~mpg01)
(c) Split the data into a training set and a test set.
train=sample(1:nrow(Auto),0.7*nrow(Auto))
test = !train
train.Auto=Auto[train,]
test.Auto = Auto[-train,]
mpg01.test=mpg01[-train]
dim(Auto)## [1] 392 10dim(train.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? Test error for the model is 11.9%
library(MASS)
lda.auto=lda(mpg01~displacement+weight+cylinders+horsepower, data=Auto, subset=train)
lda.pred=predict(lda.auto, test.Auto)
lda.class=lda.pred$class
mean(lda.class != mpg01.test)## [1] 0.1355932(e) Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained? Test error for the model is 11.9%
qda.auto=qda(mpg01~displacement+weight+cylinders+horsepower, data=Auto, subset=train)
qda.pred=predict(qda.auto, test.Auto)
mean(qda.pred$class != mpg01.test)## [1] 0.1440678(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? The test error for the model is 11.9%
glm.auto=glm(mpg01~displacement+weight+cylinders+horsepower, data=Auto, family=binomial, subset=train)
glm.probauto=predict(glm.auto, test.Auto, type="response")
glm.predauto=rep(0, length(glm.probauto))
glm.predauto[glm.probauto > 0.5] = 1
mean(glm.predauto != mpg01.test)## [1] 0.1271186(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? The test errors I got for different K values are: K=1 - 16.9% K=10 - 13.6% K=50 - 14.4% K=100 - 12.7% The best K value out of these tested is K=1
library(class)
train.X=cbind(displacement, weight, cylinders, horsepower)[train, ]
test.X=cbind(displacement, weight, cylinders, horsepower)[-train, ]
train.mpg01=mpg01[train]
set.seed(1)
knn.pred=knn(train.X, test.X, train.mpg01, k=1)
mean(knn.pred!=mpg01.test)## [1] 0.1694915K=10
knn.pred=knn(train.X, test.X, train.mpg01, k=10)
mean(knn.pred!=mpg01.test)## [1] 0.1525424K=50
knn.pred=knn(train.X, test.X, train.mpg01, k=50)
mean(knn.pred!=mpg01.test)## [1] 0.1440678K=100
knn.pred=knn(train.X, test.X, train.mpg01, k=100)
mean(knn.pred!=mpg01.test)## [1] 0.1440678##QUESTION 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)
names(Boston)## [1] "crim" "zn" "indus" "chas" "nox" "rm" "age"
## [8] "dis" "rad" "tax" "ptratio" "black" "lstat" "medv"attach(Boston)
#create new variable and assign boolean based on median crime rate
crim01=rep(0, length(crim))
crim01[crim>median(crim)]=1
Boston=data.frame(Boston, crim01)
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#Create train and test data sets
train=sample(1:nrow(Boston),0.7*nrow(Boston))
test = !train
train.Boston=Boston[train, ]
test.Boston=Boston[-train, ]
crim01.test=crim01[-train]
#Verify data sets
dim(Boston)## [1] 506 15dim(train.Boston)## [1] 354 15dim(test.Boston)## [1] 152 15#Check for 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 1.00000000Logistic Regression
glm.Boston=glm(crim01~. -crim01 -crim -chas -tax, data=Boston, family=binomial, subset=train)
glm.probBoston=predict(glm.Boston, test.Boston, type="response")
glm.predBoston=rep(0, length(glm.probBoston))
glm.predBoston[glm.probBoston > 0.5] = 1
LR1 = mean(glm.predBoston == crim01.test)Another set of predictors for logistic regression. Due to the high accuracy of the first model, this will just be for exploration purposes
glm.Boston=glm(crim01~nox+rad+medv+dis, data=Boston, family=binomial, subset=train)
glm.probBoston=predict(glm.Boston, test.Boston, type="response")
glm.predBoston=rep(0, length(glm.probBoston))
glm.predBoston[glm.probBoston > 0.5] = 1
LR2 = mean(glm.predBoston == crim01.test)LDA
lda.Boston=lda(crim01~. -crim01 -crim -chas -tax, data=Boston, subset=train)
lda.predBoston=predict(lda.Boston, test.Boston)
#lda.class=lda.predBoston$class
LDA1 = mean(lda.predBoston$class == crim01.test)LDA2
lda.Boston=lda(crim01~nox+rad+medv+dis, data=Boston, subset=train)
lda.predBoston=predict(lda.Boston, test.Boston)
lda.class=lda.predBoston$class
LDA2 = mean(lda.class == crim01.test)QDA
qda.Boston=qda(crim01~. -crim01 -crim -chas -tax, data=Boston, subset=train)
qda.pred=predict(qda.Boston, test.Boston)
QDA1 = mean(qda.pred$class == crim01.test)QDA2
qda.Boston=qda(crim01~nox+rad+medv+dis, data=Boston, subset=train)
qda.pred=predict(qda.Boston, test.Boston)
QDA2 = mean(qda.pred$class == crim01.test)KNN K = 1
library(class)
train.X=cbind(zn, indus, nox, rm, age, dis, rad, ptratio, black, lstat, medv)[train, ]
test.X=cbind(zn, indus, nox, rm, age, dis, rad, ptratio, black, lstat, medv)[-train, ]
train.crim01=crim01[train]
set.seed(1)
knn.pred=knn(train.X, test.X, train.crim01, k=1)
KNN1.1 = mean(knn.pred==crim01.test)KNN K = 10
library(class)
train.X=cbind(zn, indus, nox, rm, age, dis, rad, ptratio, black, lstat, medv)[train, ]
test.X=cbind(zn, indus, nox, rm, age, dis, rad, ptratio, black, lstat, medv)[-train, ]
train.crim01=crim01[train]
set.seed(1)
knn.pred=knn(train.X, test.X, train.crim01, k=10)
KNN1.10 = mean(knn.pred==crim01.test)KNN with second set of predictors and K = 1
library(class)
train.X=cbind(nox, rad, medv, dis)[train, ]
test.X=cbind(nox, rad, medv, dis)[-train, ]
train.crim01=crim01[train]
set.seed(1)
knn.pred=knn(train.X, test.X, train.crim01, k=1)
KNN2.1 = mean(knn.pred==crim01.test)KNN with second set of predictors and K = 10
library(class)
train.X=cbind(nox, rad, medv, dis)[train, ]
test.X=cbind(nox, rad, medv, dis)[-train, ]
train.crim01=crim01[train]
set.seed(1)
knn.pred=knn(train.X, test.X, train.crim01, k=10)
KNN2.10 = mean(knn.pred==crim01.test)I decided to run the same 2 sets of predictors with each method. The first set of predictors I chose is: zn, indus, nox, rm, age, dis, rad, ptratio, black, lstat, medv and the second set is: nox, rad, medv, dis. Below are the results from each model Logistic Regression Results The first set of predictors provided a better model in this method.
cat("The logistic regression model results are", LR1, "for the first set of predictors and ", LR2, "for the second set.")## The logistic regression model results are 0.875 for the first set of predictors and 0.8486842 for the second set.LDA Results The second set of predictors provided a better model in this method
cat("The LDA model results are", LDA1, "for the first set of predictors and ", LDA2, "for the second set.")## The LDA model results are 0.8815789 for the first set of predictors and 0.8947368 for the second set.QDA Results Both sets of predictors provided the same results in this method
cat("The QDA model results are", QDA1, "for the first set of predictors and ", QDA2, "for the second set.")## The QDA model results are 0.8947368 for the first set of predictors and 0.8815789 for the second set.KNN Results For KNN, I ran each set of predictors for K=1 and K=10. The best result came back with the first set of predictors and K=1.
cat("The KNN for K=1 model results are", KNN1.1, "for the first set of predictors and ", KNN2.1, "for the second set.")## The KNN for K=1 model results are 0.8421053 for the first set of predictors and 0.9144737 for the second set.cat("The KNN for K=10 model results are", KNN1.10, "for the first set of predictors and ", KNN2.10, "for the second set.")## The KNN for K=10 model results are 0.8947368 for the first set of predictors and 0.8486842 for the second set.