Generalized Models
2023-09-23
library(ISLR2) # Adding Packages## Warning: package 'ISLR2' was built under R version 4.3.1library(tidyr)## Warning: package 'tidyr' was built under R version 4.3.1library(moments)
library(ggplot2)## Warning: package 'ggplot2' was built under R version 4.3.1library(caret)## Warning: package 'caret' was built under R version 4.3.1## Loading required package: latticelibrary(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:ISLR2':
##
## Bostonlibrary(e1071)## Warning: package 'e1071' was built under R version 4.3.1##
## Attaching package: 'e1071'## The following objects are masked from 'package:moments':
##
## kurtosis, moment, skewnesslibrary(class)
library("corrplot")## Warning: package 'corrplot' was built under R version 4.3.1## corrplot 0.92 loadedEDA
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.
- Produce some numerical and graphical summaries of the Weekly data. Do there appear to be any patterns?
data("Weekly") #adding data
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
##
##
##
## subset_data <- Weekly[, 2:8] #Subset Data
data_long <- gather(subset_data)
# Create side-by-side histograms of lag variables
ggplot(data_long, aes(x = value, fill = key)) +
geom_histogram(position = "dodge", bins = 20) +
facet_wrap(~ key, scales = "free_x") +
labs(title = "Lag 1-5", x = "Value") +
theme_minimal()
kurtosis_lag<-kurtosis(Weekly$Lag1) #Sampling kurtosis Values
print(kurtosis_lag)## [1] 5.668866kurtosis_lag2<-kurtosis(Weekly$Lag2)
print(kurtosis_lag2)## [1] 5.665817week_sub<-subset(Weekly, select = -c(Direction))
week_sd<-sapply(week_sub, sd)
print(week_sd)## Year Lag1 Lag2 Lag3 Lag4 Lag5 Volume Today
## 6.033182 2.357013 2.357254 2.360502 2.360279 2.361285 1.686636 2.356927- Starting out, it seems like the distribution of the today and lag variables generally follows a leptokurtic distribution. Since we will be using these variables for analyses such as QDA, and LDA, this could impact the performance of the models, which make assumptions about the variables being normally distributed. The lag variables themselves share similar Sd values, though not exactly the same. Other variables have different standard deviations that may impact the outcome of LDA models. The mean values of the lag and today variables show a slight increase as time comes to present, with a small difference in the value of the means. Their standard deviations are approximately the same, with a difference of + or - 0.01.
plot(Weekly[-9])
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 1.000000000- Looking at the correlation plot, it doesnt seem like many variables have pronounced correlations with the exception of year and volume.
plot(Weekly$Volume, ylab = "Shares")
#plotting response variable
barplot(table(Weekly$Direction), main = "Direction: Up and Down", xlab = "Direction", col = "blue", border = "white")
# Log Regression (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?
# Log Regression
week.fit <- glm(Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume, data = Weekly, family = binomial)
summary(week.fit)##
## Call:
## glm(formula = Direction ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 +
## Volume, family = binomial, data = Weekly)
##
## 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- 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.
#Creating Confusion Matrix
probs = predict(week.fit, type = "response")
preds_prob=rep("Down", 1089)
preds_prob[probs > 0.5] = "Up"
table(preds_prob, Weekly$Direction)##
## preds_prob Down Up
## Down 54 48
## Up 430 557mean(preds_prob == Weekly$Direction)## [1] 0.5610652The correct prediction percentage, given by (TN + TP)/ (TO) = 56% (roughly), which is slightly better than chance.
- 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<-subset(Weekly, Year < 2009)
test<- subset(Weekly, Year > 2008)
lag2_reg = glm(Direction ~ Lag2, data= train, family = "binomial")
print(lag2_reg)##
## Call: glm(formula = Direction ~ Lag2, family = "binomial", data = train)
##
## Coefficients:
## (Intercept) Lag2
## 0.2033 0.0581
##
## Degrees of Freedom: 984 Total (i.e. Null); 983 Residual
## Null Deviance: 1355
## Residual Deviance: 1351 AIC: 1355lag2_pred = predict(lag2_reg, type="response", newdata = test)
lag2_dir = Weekly$Direction[Weekly$Year>2008]
length(lag2_dir)## [1] 104lag2_prob = rep("Down", 104)
lag2_prob[lag2_pred > 0.5] = "Up"
table(lag2_prob, test$Direction)##
## lag2_prob Down Up
## Down 9 5
## Up 34 56mean(lag2_prob == test$Direction)## [1] 0.625The error rate for this model was slightly higher than with the inclusion of other predictors. This indicates that it correctly predicted the movement of the market by 62.2%.It correctly predicted that the direction would go down 9 times, while predicting it would go up 56 times correctly. It falsely predicted that the direction would go up 5 times, when it actually went down, It falsely prediction the direction would go down 34 times, when it actually went up. Thus, it correctly predicted the direction would go down for 21% of the down observations, and up 91% of the time. Considering that some of the up observations were incorrectly categorized, it means that of all the occurrences of up (which were distributed correctly and incorrectly) the model identified 62% of them correctly (the others being distributed into the incorrect down category). Alternatively, it correctly identified down 64% of the time, incorrectly distributing the rest to up.
- Repeat (d) using LDA.
- Repeat (d) using QDA.
- Repeat (d) using KNN with K = 1.
- Repeat (d) using naive Bayes.
- Which of these methods appears to provide the best results on this data?
lag2.fit <- lda(Direction ~ Lag2, data = train)
lag2.fit## Call:
## lda(Direction ~ Lag2, data = 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 0.4414162lag2.pred<-predict(lag2.fit, test)
names(lag2.pred)## [1] "class" "posterior" "x"lda.class <-lag2.pred$class
table(lda.class, test$Direction)##
## lda.class Down Up
## Down 9 5
## Up 34 56mean(lda.class == test$Direction)## [1] 0.625qda.fit<-qda(Direction ~ Lag2 , data = train)
qda.fit## Call:
## qda(Direction ~ Lag2, data = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581qda.class<-predict(qda.fit, test)$class
table(qda.class, test$Direction)##
## qda.class Down Up
## Down 0 0
## Up 43 61mean(qda.class == test$Direction)## [1] 0.5865385nb.fit <-naiveBayes(Direction ~ Lag2 , data = train)
nb.fit##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## Down Up
## 0.4477157 0.5522843
##
## Conditional probabilities:
## Lag2
## Y [,1] [,2]
## Down -0.03568254 2.199504
## Up 0.26036581 2.317485nb.class<-predict(nb.fit, test)
table(nb.class, test$Direction)##
## nb.class Down Up
## Down 0 0
## Up 43 61mean(nb.class == test$Direction)## [1] 0.5865385nb.preds<-predict(nb.fit, Weekly, type = "raw")
nb.preds[1:5, ]## Down Up
## [1,] 0.4342284 0.5657716
## [2,] 0.4492059 0.5507941
## [3,] 0.4657684 0.5342316
## [4,] 0.4810977 0.5189023
## [5,] 0.3835799 0.6164201# Splitting Data
train.X <-cbind(train$Lag2)
test.X <- cbind(test$Lag2)
train.Y = cbind(train$Direction)
test.Y = cbind(test$Direction)
knn.pred = knn(train.X, test.X, train.Y, k=1)
table(knn.pred, test$Direction)##
## knn.pred Down Up
## 1 21 29
## 2 22 32mean(knn.pred == test.Y)## [1] 0.5096154Reviewing the mean errors calculated from each model, in rank, log, lda, qda are the top 3 performing models.
14.In this problem, you will develop a model to predict whether a given car gets high or low gas mileage based on the Auto data set.
- Create a binary variable, mpg01, that contains a 1 if mpg contains a value above its median, and a 0 if mpg contains a value below its median. You can compute the median using the median() function. Note you may find it helpful to use the data.frame() function to create a single data set containing both mpg01 and the other Auto variables.
summary(Auto)## mpg cylinders displacement horsepower weight
## Min. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0 Min. :1613
## 1st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0 1st Qu.:2225
## Median :22.75 Median :4.000 Median :151.0 Median : 93.5 Median :2804
## Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5 Mean :2978
## 3rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0 3rd Qu.:3615
## Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0 Max. :5140
##
## acceleration year origin name
## Min. : 8.00 Min. :70.00 Min. :1.000 amc matador : 5
## 1st Qu.:13.78 1st Qu.:73.00 1st Qu.:1.000 ford pinto : 5
## Median :15.50 Median :76.00 Median :1.000 toyota corolla : 5
## Mean :15.54 Mean :75.98 Mean :1.577 amc gremlin : 4
## 3rd Qu.:17.02 3rd Qu.:79.00 3rd Qu.:2.000 amc hornet : 4
## Max. :24.80 Max. :82.00 Max. :3.000 chevrolet chevette: 4
## (Other) :365median_mpg<-median(Auto$mpg)
Auto$mpg01<-ifelse(Auto$mpg > median_mpg, 1, 0)auto01<-subset(Auto, select = -mpg)
corrplot(cor(auto01[,-8]), method= "number")
1) It seems that MPG has some negative correlations. The most pronounced
of them include cylinder, displacement, horsepower, and weight. I’ll
move forward with reviewing these variables.
- Explore the data graphically in order to investigate the association between mpg01 and the other features. Which of the other features seem most likely to be useful in predicting mpg01? Scatterplots and boxplots may be useful tools to answer this question. Describe your findings.
par(mfrow=c(1,2))
# cylinder
boxplot(cylinders ~ mpg01, data = auto01, col = c("lavender"), main = "mpg vs cyl", xlab = "mpg", ylab = "cyl")
# weight
boxplot(weight ~ mpg01, data = auto01, col = c("lightgreen"), main = "mpg vs weight", xlab = "mpg", ylab = "weight")
# displacement
boxplot(displacement ~ mpg01, data = auto01, col = c("orange"), main = "mpg vs cyl", xlab = "mpg", ylab = "displacement")
# horsepower
boxplot(horsepower ~ mpg01, data = auto01, col = c("magenta"), main = "mpg vs horsepower", xlab = "mpg", ylab = "horsepower")
- It looks like most occurrences within these variables of MPG pertain to instances that are under the median for MPG. This might suggest that there would be reason to support that these variables result in lower mpg when they are higher in value.
- Split the data into a training set and a test set.
# index
set.seed(42)
ind<-sample(1:nrow(auto01), size = 0.7*nrow(auto01))
# splitting data
train<-auto01[ind, ]
test<-auto01[-ind, ]- 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 ~ cylinders + horsepower + displacement + weight, data = auto01)
auto_lda## Call:
## lda(mpg01 ~ cylinders + horsepower + displacement + weight, data = auto01)
##
## Prior probabilities of groups:
## 0 1
## 0.5 0.5
##
## Group means:
## cylinders horsepower displacement weight
## 0 6.765306 130.11224 273.1582 3620.403
## 1 4.178571 78.82653 115.6658 2334.765
##
## Coefficients of linear discriminants:
## LD1
## cylinders -0.4708126926
## horsepower 0.0044241441
## displacement -0.0014339977
## weight -0.0009846242predict_auto<-predict(auto_lda, test)
table(predict_auto$class, test$mpg01)##
## 0 1
## 0 42 4
## 1 5 67mean(predict_auto$class != test$mpg01)## [1] 0.07627119The output from this model shows that the group means for all predictors tend to be higher for all predictor variables. This may suggest that increased values of these predictors are associated with a decrease in mpg performance. The rate of error is .07%, which is great.
- 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 ~ cylinders + horsepower + displacement + weight, data = auto01)
auto_qda## Call:
## qda(mpg01 ~ cylinders + horsepower + displacement + weight, data = auto01)
##
## Prior probabilities of groups:
## 0 1
## 0.5 0.5
##
## Group means:
## cylinders horsepower displacement weight
## 0 6.765306 130.11224 273.1582 3620.403
## 1 4.178571 78.82653 115.6658 2334.765predict_qda<-predict(auto_qda, test)$class
table(predict_qda, test$mpg01)##
## predict_qda 0 1
## 0 43 6
## 1 4 65mean(predict_qda != test$mpg01)## [1] 0.08474576qda_coef<-coef(auto_qda)The output from this model has a similar trend in group means as the LDA model, but the error rate is slightly higher at .08%
- Perform logistic regression on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
auto_reg = glm(mpg01 ~ cylinders + horsepower + displacement + weight, data = auto01, family = "binomial")
summary(auto_reg)##
## Call:
## glm(formula = mpg01 ~ cylinders + horsepower + displacement +
## weight, family = "binomial", data = auto01)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 11.7966002 1.7091082 6.902 5.12e-12 ***
## cylinders -0.0129303 0.3457347 -0.037 0.97017
## horsepower -0.0421321 0.0139763 -3.015 0.00257 **
## displacement -0.0129820 0.0082203 -1.579 0.11428
## weight -0.0019458 0.0006918 -2.812 0.00492 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 543.43 on 391 degrees of freedom
## Residual deviance: 207.27 on 387 degrees of freedom
## AIC: 217.27
##
## Number of Fisher Scoring iterations: 7auto_pred = predict(auto_reg, type="response", newdata = test)
pred<-rep(0, length(auto_pred))
pred[auto_pred > 0.5] <- 1
table(pred, test$mpg01)##
## pred 0 1
## 0 43 5
## 1 4 66mean(pred != test$mpg01)## [1] 0.07627119As opposed to LDA, the log regression assigns negative coefficients to all of the predictor variables.The test error is .07%.
- Perform naive Bayes on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
nb_auto<-naiveBayes(mpg01 ~ cylinders + horsepower + displacement + weight, data = auto01)
nb_auto##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## 0 1
## 0.5 0.5
##
## Conditional probabilities:
## cylinders
## Y [,1] [,2]
## 0 6.765306 1.4200110
## 1 4.178571 0.6746319
##
## horsepower
## Y [,1] [,2]
## 0 130.11224 37.35564
## 1 78.82653 15.91969
##
## displacement
## Y [,1] [,2]
## 0 273.1582 89.52399
## 1 115.6658 38.42951
##
## weight
## Y [,1] [,2]
## 0 3620.403 676.9322
## 1 2334.765 397.1924nba_pred<-predict(nb_auto, test)
table(nba_pred, test$mpg01)##
## nba_pred 0 1
## 0 43 5
## 1 4 66mean(nba_pred != test$mpg01)## [1] 0.07627119This model follows trends from previous models, with an error rate of .07%
- Perform KNN on the training data, with several values of K, in order to predict mpg01. Use only the variables that seemed most associated with mpg01 in (b). What test errors do you obtain? Which value of K seems to perform the best on this data set?
# Splitting Data for KNN
train.auto <-cbind(train$cylinders, train$displacement, train$horsepower, train$weight)
test.auto <- cbind(test$cylinders, test$displacement, test$horsepower, test$weight)
train.Y_auto = cbind(train$mpg01)
test.Y_auto = cbind(test$mpg01)
knn_auto = knn(train.auto, test.auto, train.Y_auto, k=1)
table(knn_auto, test.Y_auto)## test.Y_auto
## knn_auto 0 1
## 0 42 9
## 1 5 62mean(knn_auto != test.Y_auto)## [1] 0.1186441knn_5 = knn(train.auto, test.auto, train.Y_auto, k=5)
table(knn_5, test.Y_auto)## test.Y_auto
## knn_5 0 1
## 0 42 10
## 1 5 61mean(knn_5 != test.Y_auto)## [1] 0.1271186knn_10 = knn(train.auto, test.auto, train.Y_auto, k=10)
table(knn_10, test.Y_auto)## test.Y_auto
## knn_10 0 1
## 0 42 11
## 1 5 60mean(knn_10 != test.Y_auto)## [1] 0.1355932It looks like knn=1 is the best model, with increasing the value of k resulting in increased rate of error.
IN PROGRESS
- This problem involves writing functions.
- Write a function, Power(), that prints out the result of raising 2 to the 3rd power. In other words, your function should compute 23 and print out the results.
power<-(2^3)
print(power)## [1] 8- Create a new function, Power2(), that allows you to pass any two numbers, x and a, and prints out the value of x^a. You can do this by beginning your function with the line Power2 <- function(x, a) { 4.8 Exercises 195 You should be able to call your function by entering, for instance, Power2(3, 8) on the command line. This should output the value of 38, namely, 6, 561.
power2<-function(x, a) {x^a}
power2(3, 8)## [1] 6561- Using the Power2() function that you just wrote, compute 103, 817, and 1313.
power2(10, 3)## [1] 1000power2(81, 7)## [1] 2.287679e+13power2(131, 3)## [1] 2248091- Now create a new function, Power3(), that actually returns the result x^a as an R object, rather than simply printing it to the screen. That is, if you store the value x^a in an object called result within your function, then you can simply return() this return() result, using the following line: return(result) The line above should be the last line in your function, before the } symbol.
power3<-function(x, a) return({x^a})
result = power3(5,5)
result## [1] 3125- Now using the Power3() function, create a plot of f(x) = x2. The x-axis should display a range of integers from 1 to 10, and the y-axis should display x2. Label the axes appropriately, and use an appropriate title for the figure. Consider displaying either the x-axis, the y-axis, or both on the log-scale. You can do this by using log = “x”, log = “y”, or log = “xy” as arguments to the plot() function.
x<-1:10
plot(x, power3(x, 2), log = "x", xlab = "Log of x", ylab = "x^2", main = "Log of x versus x^2")
(f) Create a function, PlotPower(), that allows you to create a plot of
x against x^a for a fixed a and for a range of values of x. For
instance, if you call PlotPower(1:10, 3) then a plot should be created
with an x-axis taking on values 1, 2, . . . , 10, and a y-axis taking on
values 13, 23, . . . , 103.
plotpower<-function(x, a) {plot(x, x^2, xlab = "x", ylab = "x^a", main = "x^a")}
plotpower(1:10, 3)
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.
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. :50.00crime_median<-median(Boston$crim)
Boston$crim<-ifelse(Boston$crim > crime_median, 1, 0)barplot(table(Boston$crim), main = "Crime: High vs Low", xlab = "Crime", col = "red", border = "white")
cor(Boston)## crim zn indus chas nox
## crim 1.00000000 -0.43615103 0.60326017 0.070096774 0.72323480
## zn -0.43615103 1.00000000 -0.53382819 -0.042696719 -0.51660371
## indus 0.60326017 -0.53382819 1.00000000 0.062938027 0.76365145
## chas 0.07009677 -0.04269672 0.06293803 1.000000000 0.09120281
## nox 0.72323480 -0.51660371 0.76365145 0.091202807 1.00000000
## rm -0.15637178 0.31199059 -0.39167585 0.091251225 -0.30218819
## age 0.61393992 -0.56953734 0.64477851 0.086517774 0.73147010
## dis -0.61634164 0.66440822 -0.70802699 -0.099175780 -0.76923011
## rad 0.61978625 -0.31194783 0.59512927 -0.007368241 0.61144056
## tax 0.60874128 -0.31456332 0.72076018 -0.035586518 0.66802320
## ptratio 0.25356836 -0.39167855 0.38324756 -0.121515174 0.18893268
## black -0.35121093 0.17552032 -0.35697654 0.048788485 -0.38005064
## lstat 0.45326273 -0.41299457 0.60379972 -0.053929298 0.59087892
## medv -0.26301673 0.36044534 -0.48372516 0.175260177 -0.42732077
## rm age dis rad tax ptratio
## crim -0.15637178 0.61393992 -0.61634164 0.619786249 0.60874128 0.2535684
## 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
## black lstat medv
## crim -0.35121093 0.4532627 -0.2630167
## zn 0.17552032 -0.4129946 0.3604453
## indus -0.35697654 0.6037997 -0.4837252
## chas 0.04878848 -0.0539293 0.1752602
## nox -0.38005064 0.5908789 -0.4273208
## rm 0.12806864 -0.6138083 0.6953599
## age -0.27353398 0.6023385 -0.3769546
## dis 0.29151167 -0.4969958 0.2499287
## rad -0.44441282 0.4886763 -0.3816262
## tax -0.44180801 0.5439934 -0.4685359
## ptratio -0.17738330 0.3740443 -0.5077867
## black 1.00000000 -0.3660869 0.3334608
## lstat -0.36608690 1.0000000 -0.7376627
## medv 0.33346082 -0.7376627 1.0000000correlations<-cor(Boston$crim, Boston[2:14])
print(correlations)## zn indus chas nox rm age dis
## [1,] -0.436151 0.6032602 0.07009677 0.7232348 -0.1563718 0.6139399 -0.6163416
## rad tax ptratio black lstat medv
## [1,] 0.6197862 0.6087413 0.2535684 -0.3512109 0.4532627 -0.2630167Some correlations with outome, but also correlations with other predictors, which may be a conflict for the naive bayes analysis that assumes independence of predictors. The variables that meet criteria for high correlation include nox (Nitric oxides concentration (parts per 10 million), indus (proportion of non-retail business acres per town), age (proportion of owner-occupied units built prior to 1940), dis (weighted distances to five Boston employment centres), rad (index of accessibility to radial highways), tax (full-value property-tax rate per $10,000), There are also some moderately correlated variables including zn (proportion of residential land zoned for lots over 25,000 sq.ft), black (1000(Bk - 0.63)^2 where Bk is the proportion of blacks by town), and lstat (% lower status of the population). We will include both highly correlated and moderately correlated variables for this analysis. Note that some of the variables do show correlation with each other such as lstat and medv (-0.7376627). This analysis will not manipulate (i.e. combine/ change variable dimensions) predictors, but future endeavors could benefit by exploring these areas within the paramaters of classification models.
subset_bos <- Boston[c(2, 3, 5, 7, 8, 9, 10, 12, 13)] #Subset Data
data_bo <- gather(subset_bos)
ggplot(data_bo, aes(x = value, fill = key)) +
geom_histogram(position = "dodge", bins = 20) +
facet_wrap(~ key, scales = "free_x") +
labs(title = "Lag 1-5", x = "Value") +
theme_minimal()
Looking at the distribution of the predictor variblaes, we see that the
predictor distributions are not gaussian, which violates an assumption
of the LDA model and naive bayes.
#Box Plots of Predictors on Crime
predictors<-c("zn", "tax", "black", "rad", "nox", "lstat", "indus", "dis", "age")
outcome<-"crim"
colors<-c("blue", "green", "lavender", "purple", "pink", "orange", "red", "yellow")
layout(matrix(1:9, nrow = 3))
for(i in seq_along(predictors)) {boxplot(Boston[[outcome]], Boston[[predictors[i]]], col=colors[i], xlab=outcome, ylab=predictors[i], main=paste("Boxplot of", outcome, "and", predictors[i]))}
par(mfrow=c(1, length(predictors)), mar=c(5, 4, 4, 2) + 0.1)
layout(1)All of the predictor variables show to be related to higher crime rates excepting nox (Nitric oxides concentration (parts per 10 million), which has the reverse relationship.
# splitting Boston Data
set.seed(42)
ind<-sample(1:nrow(Boston), size = 0.7*nrow(Boston))
# splitting data
train_b<-Boston[ind, ]
test_b<-Boston[-ind, ]#Log Reg
boston_reg = glm(crim ~ zn + tax + black + rad + nox + lstat + indus + dis + age, data = Boston, family = "binomial")
summary(boston_reg)##
## Call:
## glm(formula = crim ~ zn + tax + black + rad + nox + lstat + indus +
## dis + age, family = "binomial", data = Boston)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -20.059653 4.489141 -4.468 7.88e-06 ***
## zn -0.081306 0.030119 -2.699 0.00694 **
## tax -0.007723 0.002511 -3.076 0.00210 **
## black -0.011199 0.006034 -1.856 0.06344 .
## rad 0.665127 0.133407 4.986 6.17e-07 ***
## nox 40.193351 6.570294 6.117 9.51e-10 ***
## lstat -0.017439 0.034637 -0.503 0.61464
## indus -0.040343 0.041719 -0.967 0.33354
## dis 0.432404 0.184260 2.347 0.01894 *
## age 0.015372 0.009573 1.606 0.10833
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 701.46 on 505 degrees of freedom
## Residual deviance: 229.30 on 496 degrees of freedom
## AIC: 249.3
##
## Number of Fisher Scoring iterations: 9boston_pred = predict(boston_reg, type="response", newdata = test_b)
pred<-rep(0, length(boston_pred))
pred[boston_pred > 0.5] <- 1
table(pred, test_b$crim)##
## pred 0 1
## 0 76 6
## 1 2 68mean(pred != test_b$crim)## [1] 0.05263158This model shows that it had an error rate of .05%, very good! The model indicates that variables black, lstat, indus, and age are not significant predictors. The model seems to fit pretty good, given factors such as AIC, and resiudal deviance on degrees of freedom.
#LDA
boston_lda<-lda(crim ~ zn + tax + black + rad + nox + lstat + indus + dis + age, data = Boston)
boston_lda## Call:
## lda(crim ~ zn + tax + black + rad + nox + lstat + indus + dis +
## age, data = Boston)
##
## Prior probabilities of groups:
## 0 1
## 0.5 0.5
##
## Group means:
## zn tax black rad nox lstat indus dis
## 0 21.525692 305.7431 388.7061 4.158103 0.4709711 9.419486 7.002292 5.091596
## 1 1.201581 510.7312 324.6420 14.940711 0.6384190 15.886640 15.271265 2.498489
## age
## 0 51.31028
## 1 85.83953
##
## Coefficients of linear discriminants:
## LD1
## zn -0.0026883085
## tax -0.0014982050
## black -0.0007710848
## rad 0.0878244826
## nox 7.2508340136
## lstat -0.0212585580
## indus 0.0107470871
## dis -0.0264048093
## age 0.0128439952predict_bos<-predict(boston_lda, test_b)
table(predict_bos$class, test_b$crim)##
## 0 1
## 0 76 19
## 1 2 55mean(predict_bos$class != test_b$crim)## [1] 0.1381579The output from this model indicates a higher error rate than with the log model. It also seems fairly split in terms of what variables it suggests are indicative of high or low crime rates.
nb_bos<-naiveBayes(crim ~ zn + tax + black + rad + nox + lstat + indus + dis + age, data = Boston)
nb_bos##
## Naive Bayes Classifier for Discrete Predictors
##
## Call:
## naiveBayes.default(x = X, y = Y, laplace = laplace)
##
## A-priori probabilities:
## Y
## 0 1
## 0.5 0.5
##
## Conditional probabilities:
## zn
## Y [,1] [,2]
## 0 21.525692 29.319808
## 1 1.201581 4.798611
##
## tax
## Y [,1] [,2]
## 0 305.7431 87.4837
## 1 510.7312 167.8553
##
## black
## Y [,1] [,2]
## 0 388.7061 22.83774
## 1 324.6420 118.83084
##
## rad
## Y [,1] [,2]
## 0 4.158103 1.659121
## 1 14.940711 9.529843
##
## nox
## Y [,1] [,2]
## 0 0.4709711 0.05559789
## 1 0.6384190 0.09870365
##
## lstat
## Y [,1] [,2]
## 0 9.419486 4.923497
## 1 15.886640 7.546922
##
## indus
## Y [,1] [,2]
## 0 7.002292 5.514454
## 1 15.271265 5.439010
##
## dis
## Y [,1] [,2]
## 0 5.091596 2.081304
## 1 2.498489 1.085521
##
## age
## Y [,1] [,2]
## 0 51.31028 25.88190
## 1 85.83953 17.87423nbb_pred<-predict(nb_bos, test_b)
table(nbb_pred, test_b$crim)##
## nbb_pred 0 1
## 0 74 19
## 1 4 55mean(nbb_pred != test_b$crim)## [1] 0.1513158The error rate shows to be getting higher with each model. We know that with the earlier EDA, some assumptions of this model are violated by elements of the data (i.e. correlations among predictors).
# Splitting Data for KNN
train.bos <-cbind(train_b$zn, train_b$tax, train_b$black, train_b$rad, train_b$nox, train_b$lstat, train_b$indus, train_b$dis, train_b$age)
test.bos <- cbind(test_b$zn, test_b$tax, test_b$black, test_b$rad, test_b$nox, test_b$lstat, test_b$indus, test_b$dis, test_b$age)
trainb.Y = cbind(train_b$crim)
testb.Y = cbind(test_b$crim)
knn_bos = knn(train.bos, test.bos, trainb.Y, k=1)
table(knn_bos, testb.Y)## testb.Y
## knn_bos 0 1
## 0 67 4
## 1 11 70mean(knn_bos != testb.Y)## [1] 0.09868421#knn5
knn_bos5 = knn(train.bos, test.bos, trainb.Y, k=5)
table(knn_bos5, testb.Y)## testb.Y
## knn_bos5 0 1
## 0 70 4
## 1 8 70mean(knn_bos5 != testb.Y)## [1] 0.07894737knn_bos10 = knn(train.bos, test.bos, trainb.Y, k=10)
table(knn_bos10, testb.Y)## testb.Y
## knn_bos10 0 1
## 0 69 6
## 1 9 68mean(knn_bos10 != testb.Y)## [1] 0.09868421Of all the models, it seems like the log regression fits the models the best, with knn 3 being the next best fit. This makes sense, as some assumptions of the other models are violated in the data.