Assignment 3
2023-02-25
Question 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.
(a) 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])## 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 1.000000000ggscatmat(Weekly, color = "Direction")## Warning in ggscatmat(Weekly, color = "Direction"): Factor variables are omitted
## in plot## Warning: The dot-dot notation (`..scaled..`) was deprecated in ggplot2 3.4.0.
## ℹ Please use `after_stat(scaled)` instead.
## ℹ The deprecated feature was likely used in the GGally package.
## Please report the issue at <]8;;https://github.com/ggobi/ggally/issueshttps://github.com/ggobi/ggally/issues]8;;>.
ggscatmat(Weekly, columns = 2:9, color = "Direction")## Warning in ggscatmat(Weekly, columns = 2:9, 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)## `geom_smooth()` using method = 'gam' and formula = 'y ~ s(x, bs = "cs")'
Correlations are all basically zero, except for year and volume which shows an increase in volume over time.
(b) Use the full data set to perform a logistic regression with Direction as the response and the five lag variables plus Volumeas predictors. Use the summary function to print the results. Doany of the predictors appear to be statistically significant? If so, which ones?
glm_fit_wk <- glm(Direction ~
Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,
data = Weekly,
family = binomial)
summary(glm_fit_wk)##
## 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: 4Only Lag2 shows any real significance.
(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.
glm_probs_wk = predict(glm_fit_wk, type = "response")
glm_pred_wk = rep("Down", length(glm_probs_wk))
glm_pred_wk[glm_probs_wk > 0.5] <- "Up"
table(glm_pred_wk, Weekly$Direction)##
## glm_pred_wk Down Up
## Down 54 48
## Up 430 557mean(glm_pred_wk == Weekly$Direction)## [1] 0.5610652It appears as though 56.1% of the responses are predicted correctly.Training error rate = 43.89%, overly optimistic.There are only 54 out of 484 down days accurately predicted but 557 out of 605 of up days are predicted. So its right only ~ 11% of time when market is down but ~ 92% when up.
(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 <- (Weekly$Year < 2009)
Weekly_train <- Weekly[train,]
Weekly_test <- Weekly[!train,]
Direction_train <- Weekly_train$Direction
Direction_test <- Weekly_test$Direction
logistic_wkly <- glm(Direction ~ Lag2,
data = Weekly_train,
family = binomial)(e) Repeat (d) using LDA.
lda_wkly <- lda(Direction ~ Lag2, data = Weekly, subset = train)
lda_wkly## Call:
## lda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581
##
## Coefficients of linear discriminants:
## LD1
## Lag2 0.4414162plot(lda_wkly)
lda_probs <- predict(lda_wkly, Weekly_test)
table(lda_probs$class, Direction_test)## Direction_test
## Down Up
## Down 9 5
## Up 34 56mean(lda_probs$class == Direction_test)## [1] 0.625(f) Repeat (d) using QDA.
qda_wkly <- qda(Direction ~ Lag2, data = Weekly, subset = train)
qda_wkly## Call:
## qda(Direction ~ Lag2, data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2
## Down -0.03568254
## Up 0.26036581qda_pred <- predict(qda_wkly, Weekly_test)
table(qda_pred$class, Direction_test)## Direction_test
## Down Up
## Down 0 0
## Up 43 61mean(qda_pred$class == Direction_test)## [1] 0.5865385(g) Repeat (d) using KNN with K = 1.
train_X <- as.matrix(Weekly$Lag2[train])
test_X <- as.matrix(Weekly$Lag2[!train])
set.seed(1)
knn_pred <- knn(train_X, test_X, Direction_train, k = 1)
table(knn_pred, Direction_test)## Direction_test
## knn_pred Down Up
## Down 21 30
## Up 22 31mean(knn_pred == Direction_test)## [1] 0.5(h)Which of these methods appears to provide the best results on this data?
Logistic and LDA are pretty close. QDA is not very good nor is 1NN,which is the worst.
(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.
logistic_wkly3 <- glm(Direction ~ Lag2:Lag1,
data = Weekly_train,
family = binomial)
summary(logistic_wkly3)##
## Call:
## glm(formula = Direction ~ Lag2:Lag1, family = binomial, data = Weekly_train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.368 -1.269 1.077 1.089 1.353
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.21333 0.06421 3.322 0.000893 ***
## Lag2:Lag1 0.00717 0.00697 1.029 0.303649
## ---
## 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: 1353.6 on 983 degrees of freedom
## AIC: 1357.6
##
## Number of Fisher Scoring iterations: 4logistic_probs3 <- predict(logistic_wkly3, Weekly_test, type = "response")
logistic_pred3 = rep("Down", length(Direction_test))
logistic_pred3[logistic_probs3 > 0.5] <- "Up"
table(logistic_pred3, Direction_test)## Direction_test
## logistic_pred3 Down Up
## Down 1 1
## Up 42 60mean(logistic_pred3 == Direction_test)## [1] 0.5865385qda_wkly2 <- qda(Direction ~ Lag2 + sqrt(abs(Lag2)),
data = Weekly,
subset = train)
qda_wkly2## Call:
## qda(Direction ~ Lag2 + sqrt(abs(Lag2)), data = Weekly, subset = train)
##
## Prior probabilities of groups:
## Down Up
## 0.4477157 0.5522843
##
## Group means:
## Lag2 sqrt(abs(Lag2))
## Down -0.03568254 1.140078
## Up 0.26036581 1.169635qda_pred2 <- predict(qda_wkly2, Weekly_test)
table(qda_pred2$class, Direction_test)## Direction_test
## Down Up
## Down 12 13
## Up 31 48mean(qda_pred2$class == Direction_test)## [1] 0.5769231train_X <- as.matrix(Weekly$Lag2[train])
test_X <- as.matrix(Weekly$Lag2[!train])
set.seed(1)
knn_pred3 <- knn(train_X, test_X, Direction_train, k = 3)
table(knn_pred3, Direction_test)## Direction_test
## knn_pred3 Down Up
## Down 16 20
## Up 27 41mean(knn_pred3 == Direction_test)## [1] 0.5480769train_X <- as.matrix(Weekly$Lag2[train])
test_X <- as.matrix(Weekly$Lag2[!train])
set.seed(1)
knn_pred100 <- knn(train_X, test_X, Direction_train, k = 100)
table(knn_pred100, Direction_test)## Direction_test
## knn_pred100 Down Up
## Down 10 11
## Up 33 50mean(knn_pred100 == Direction_test)## [1] 0.5769231Question 13
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.
library(GGally)
Auto <- as_tibble(ISLR::Auto)(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_mpg01 <- Auto %>%
mutate(mpg01 = ifelse(mpg > median(mpg), 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.
summary(Auto_mpg01)## 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
## ggscatmat(as.data.frame(Auto_mpg01))## Warning in ggscatmat(as.data.frame(Auto_mpg01)): Factor variables are omitted in
## plot
Auto_mpg01 %>%
ggplot(aes(cut_number(mpg01, 2), displacement)) +
geom_boxplot()
Auto_mpg01 %>%
ggplot(aes(cut_number(mpg01, 2), horsepower)) +
geom_boxplot()
Auto_mpg01 %>%
ggplot(aes(cut_number(mpg01, 2), weight)) +
geom_boxplot()
Auto_mpg01 %>%
ggplot(aes(cut_number(mpg01, 2), acceleration)) +
geom_boxplot()
ggplot(Auto_mpg01) +
geom_count(aes(cut_number(mpg01, 2), cylinders))
ggplot(Auto_mpg01) +
geom_boxplot(aes(cut_number(mpg01, 2), cylinders))
Auto_mpg01 %>%
ggplot(aes(cut_number(mpg01, 2), mpg)) +
geom_boxplot()
To predict mpg01, can use mpg. there seems to be neg corr with cylnders and displacement, horsepower, and weight but positive correlation with acceleration.
(c) Split the data into a training set and a test set.
train <- (Auto_mpg01$year %% 2 == 0)
train_auto <- Auto_mpg01[train,]
test_auto <- Auto_mpg01[!train,](d) Perform LDA on the training data in order to predict mpg01 sing the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
lda_auto <- lda(mpg01 ~ cylinders + displacement + horsepower + weight,
data = Auto_mpg01,
subset = train)
lda_auto## Call:
## lda(mpg01 ~ cylinders + displacement + horsepower + weight, data = Auto_mpg01,
## subset = train)
##
## Prior probabilities of groups:
## 0 1
## 0.4571429 0.5428571
##
## Group means:
## cylinders displacement horsepower weight
## 0 6.812500 271.7396 133.14583 3604.823
## 1 4.070175 111.6623 77.92105 2314.763
##
## Coefficients of linear discriminants:
## LD1
## cylinders -0.6741402638
## displacement 0.0004481325
## horsepower 0.0059035377
## weight -0.0011465750lda_auto_pred <- predict(lda_auto, test_auto)
table(lda_auto_pred$class, test_auto$mpg01)##
## 0 1
## 0 86 9
## 1 14 73mean(lda_auto_pred$class == test_auto$mpg01)## [1] 0.8736264# test error rate is
1 - mean(lda_auto_pred$class == test_auto$mpg01)## [1] 0.1263736(e) Perform QDA on the training data in order to predict mpg01 using the variables that seemed most associated with mpg01 in (b). What is the test error of the model obtained?
qda_auto <- qda(mpg01 ~ cylinders + displacement + horsepower + weight,
data = Auto_mpg01,
subset = train)
qda_auto## Call:
## qda(mpg01 ~ cylinders + displacement + horsepower + weight, data = Auto_mpg01,
## subset = train)
##
## Prior probabilities of groups:
## 0 1
## 0.4571429 0.5428571
##
## Group means:
## cylinders displacement horsepower weight
## 0 6.812500 271.7396 133.14583 3604.823
## 1 4.070175 111.6623 77.92105 2314.763qda_auto_pred <- predict(qda_auto, test_auto)
table(qda_auto_pred$class, test_auto$mpg01)##
## 0 1
## 0 89 13
## 1 11 69mean(qda_auto_pred$class == test_auto$mpg01)## [1] 0.8681319# test error rate is
1 - mean(qda_auto_pred$class == test_auto$mpg01)## [1] 0.1318681(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?
logistic_auto <- glm(mpg01 ~ cylinders + displacement + horsepower + weight,
data = Auto_mpg01,
subset = train,
family = binomial)
summary(logistic_auto)##
## Call:
## glm(formula = mpg01 ~ cylinders + displacement + horsepower +
## weight, family = binomial, data = Auto_mpg01, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.48027 -0.03413 0.10583 0.29634 2.57584
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 17.658730 3.409012 5.180 2.22e-07 ***
## cylinders -1.028032 0.653607 -1.573 0.1158
## displacement 0.002462 0.015030 0.164 0.8699
## horsepower -0.050611 0.025209 -2.008 0.0447 *
## weight -0.002922 0.001137 -2.569 0.0102 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 289.58 on 209 degrees of freedom
## Residual deviance: 83.24 on 205 degrees of freedom
## AIC: 93.24
##
## Number of Fisher Scoring iterations: 7logistic_probs_auto <- predict(logistic_auto, test_auto, type = "response")
logistic_pred_auto <- rep(0,length(test_auto$mpg01))
logistic_pred_auto[logistic_probs_auto > 0.5] <- 1
table(logistic_pred_auto, test_auto$mpg01)##
## logistic_pred_auto 0 1
## 0 89 11
## 1 11 71mean(logistic_pred_auto == test_auto$mpg01)## [1] 0.87912091 - mean(logistic_pred_auto == test_auto$mpg01)## [1] 0.1208791(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?
train_auto_X <- cbind(
Auto_mpg01$cylinders,
Auto_mpg01$displacement,
Auto_mpg01$horsepower,
Auto_mpg01$weight
)[train,]
test_auto_X <- cbind(
Auto_mpg01$cylinders,
Auto_mpg01$displacement,
Auto_mpg01$horsepower,
Auto_mpg01$weight
)[!train,]
set.seed(1)
knn_auto1 <- knn(train_auto_X, test_auto_X, train_auto$mpg01, k = 1)
table(knn_auto1, test_auto$mpg01)##
## knn_auto1 0 1
## 0 83 11
## 1 17 71mean(knn_auto1 == test_auto$mpg01)## [1] 0.8461538#test error
1 - mean(knn_auto1 == test_auto$mpg01)## [1] 0.1538462knn_auto10 <- knn(train_auto_X, test_auto_X, train_auto$mpg01, k = 10)
table(knn_auto10, test_auto$mpg01)##
## knn_auto10 0 1
## 0 77 7
## 1 23 75mean(knn_auto10 == test_auto$mpg01)## [1] 0.8351648#test error
1 - mean(knn_auto10 == test_auto$mpg01)## [1] 0.1648352##Question 16
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.
Boston_tb <- as_tibble(Boston)
Boston_tb <- Boston_tb %>%
mutate(crim01 = ifelse(crim > median(crim), 1, 0))
cor(Boston_tb)## 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 1.00000000ggscatmat(as.data.frame(Boston_tb))
Need to add factor to seperate ‘high crime’ (crim01 = 1) vs ‘low crime’
(crim01 = 0)
Boston_tb_f <- Boston_tb %>%
mutate(crim_gp =
as.factor(
ifelse(crim < median(crim), "below median", "above median")
)
) %>%
select(crim_gp, everything())
Boston_tb_f <- as.data.frame(Boston_tb_f)
ggscatmat(Boston_tb_f)## Warning in ggscatmat(Boston_tb_f): Factor variables are omitted in plot
ggscatmat(Boston_tb_f, color = "crim_gp")## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero
## Warning in cor(xvalue, yvalue, use = "pairwise.complete.obs", method =
## corMethod): the standard deviation is zero## Warning in ggscatmat(Boston_tb_f, color = "crim_gp"): Factor variables are
## omitted in plot
Will give NA when use color since the crim01 and factors are 1-1 correspondence. Remember it calculates correlation by factor
train <- 1:(length(Boston_tb$crim)/2)
test <- (length(Boston_tb$crim)/2 + 1):length(Boston_tb$crim)
Boston_train <- Boston_tb[train,]
Boston_test <- Boston_tb[test,]
glm_crim <- glm(crim01 ~ . -crim,
data = Boston_tb,
family = binomial,
subset = train)## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredsummary(glm_crim)##
## Call:
## glm(formula = crim01 ~ . - crim, family = binomial, data = Boston_tb,
## subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.83229 -0.06593 0.00000 0.06181 2.61513
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -91.319906 19.490273 -4.685 2.79e-06 ***
## zn -0.815573 0.193373 -4.218 2.47e-05 ***
## indus 0.354172 0.173862 2.037 0.04164 *
## chas 0.167396 0.991922 0.169 0.86599
## nox 93.706326 21.202008 4.420 9.88e-06 ***
## rm -4.719108 1.788765 -2.638 0.00833 **
## age 0.048634 0.024199 2.010 0.04446 *
## dis 4.301493 0.979996 4.389 1.14e-05 ***
## rad 3.039983 0.719592 4.225 2.39e-05 ***
## tax -0.006546 0.007855 -0.833 0.40461
## ptratio 1.430877 0.359572 3.979 6.91e-05 ***
## black -0.017552 0.006734 -2.606 0.00915 **
## lstat 0.190439 0.086722 2.196 0.02809 *
## medv 0.598533 0.185514 3.226 0.00125 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 329.367 on 252 degrees of freedom
## Residual deviance: 69.568 on 239 degrees of freedom
## AIC: 97.568
##
## Number of Fisher Scoring iterations: 10# note that chas and tax have no significance
crim_probs <- predict(glm_crim, Boston_test, type = "response")
crim_pred <- rep(0, length(test))
crim_pred[crim_probs > .5] <- 1
table(crim_pred, Boston_test$crim01)##
## crim_pred 0 1
## 0 68 24
## 1 22 139mean(crim_pred == Boston_test$crim01)## [1] 0.81818181 - mean(crim_pred == Boston_test$crim01) # test error## [1] 0.1818182glm2_crim <- glm(crim01 ~ . -crim - chas - tax,
data = Boston_tb,
family = binomial,
subset = train)## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurredsummary(glm2_crim)##
## Call:
## glm(formula = crim01 ~ . - crim - chas - tax, family = binomial,
## data = Boston_tb, subset = train)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -3.07309 -0.06280 0.00000 0.04518 2.52250
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -95.004862 20.048738 -4.739 2.15e-06 ***
## zn -0.850179 0.184521 -4.607 4.08e-06 ***
## indus 0.413284 0.157156 2.630 0.008544 **
## nox 89.142048 19.552114 4.559 5.13e-06 ***
## rm -4.631311 1.678561 -2.759 0.005796 **
## age 0.050660 0.023105 2.193 0.028337 *
## dis 4.513311 0.954809 4.727 2.28e-06 ***
## rad 2.968052 0.677653 4.380 1.19e-05 ***
## ptratio 1.483118 0.369531 4.014 5.98e-05 ***
## black -0.016615 0.006504 -2.554 0.010636 *
## lstat 0.209340 0.086666 2.415 0.015714 *
## medv 0.631766 0.183235 3.448 0.000565 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 329.367 on 252 degrees of freedom
## Residual deviance: 70.529 on 241 degrees of freedom
## AIC: 94.529
##
## Number of Fisher Scoring iterations: 10crim2_probs <- predict(glm2_crim, Boston_test, type = "response")
crim2_pred <- rep(0, length(test))
crim2_pred[crim2_probs > .5] <- 1
table(crim2_pred, Boston_test$crim01)##
## crim2_pred 0 1
## 0 67 24
## 1 23 139mean(crim2_pred == Boston_test$crim01)## [1] 0.81422921 - mean(crim2_pred == Boston_test$crim01) # test error## [1] 0.1857708This wouldbe for the Logistic regression
lda_crim <- lda(crim01 ~ . -crim, data = Boston_tb, subset = train)
lda_crim_preds <- predict(lda_crim, Boston_test)
table(lda_crim_preds$class, Boston_test$crim01)##
## 0 1
## 0 80 24
## 1 10 139mean(lda_crim_preds$class == Boston_test$crim01)## [1] 0.86561261 - mean(lda_crim_preds$class == Boston_test$crim01) # test error## [1] 0.1343874lda2_crim <- lda(crim01 ~ . -crim - chas - tax,
data = Boston_tb,
subset = train)
lda2_crim_preds <- predict(lda2_crim, Boston_test)
table(lda2_crim_preds$class, Boston_test$crim01)##
## 0 1
## 0 80 21
## 1 10 142mean(lda2_crim_preds$class == Boston_test$crim01)## [1] 0.87747041 - mean(lda2_crim_preds$class == Boston_test$crim01) # test error## [1] 0.1225296attach(Boston_tb)
train_xb <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio,
black, lstat, medv)[train,]
test_xb <- cbind(zn, indus, chas, nox, rm, age, dis, rad, tax, ptratio,
black, lstat, medv)[test,]
train_crim01 <- crim01[train]
set.seed(1)
knn_pred_bos <- knn(train_xb, test_xb, train_crim01, k = 1)
table(knn_pred_bos, Boston_test$crim01)##
## knn_pred_bos 0 1
## 0 85 111
## 1 5 521 - mean(knn_pred_bos == Boston_test$crim01) # test error k =1## [1] 0.458498knn10_pred_bos <- knn(train_xb, test_xb, train_crim01, k = 10)
table(knn10_pred_bos, Boston_test$crim01)##
## knn10_pred_bos 0 1
## 0 83 23
## 1 7 1401 - mean(knn10_pred_bos == Boston_test$crim01) ## [1] 0.1185771knn_pred_bos <- knn(train_xb, test_xb, train_crim01, k = 1)
table(knn_pred_bos, Boston_test$crim01)##
## knn_pred_bos 0 1
## 0 85 111
## 1 5 521 - mean(knn_pred_bos == Boston_test$crim01) # test error k =1## [1] 0.458498knn100_pred_bos <- knn(train_xb, test_xb, train_crim01, k = 100)
table(knn100_pred_bos, Boston_test$crim01)##
## knn100_pred_bos 0 1
## 0 86 120
## 1 4 431 - mean(knn100_pred_bos == Boston_test$crim01) ## [1] 0.4901186This is for LDA