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61 changes: 46 additions & 15 deletions Assignment7.Rmd
Original file line number Diff line number Diff line change
@@ -1,7 +1,7 @@
---
title: "Assignment 7 - Answers"
author: "Charles Lang"
date: "11/30/2016"
author: "Beibei Cao"
date: "12/09/2019"
output: html_document
---

Expand All @@ -11,24 +11,35 @@ In the following assignment you will be looking at data from an one level of an

#Upload data
```{r}

library(readr)
library(dplyr)
library(tidyr)
library(ggplot2)
f<-file.choose("~/Desktop/hudk4050/Assignment 7/online.data.csv")
D1 <- read.csv(f, header = TRUE)
```

#Visualization
```{r}
#Start by creating histograms of the distributions for all variables (#HINT: look up "facet" in the ggplot documentation)

D1$level.up <- ifelse(D1$level.up == "yes", 1,0)
D2 <- gather(D1, "measure", "score", 2:7)
p <- ggplot(D2, aes(score)) + facet_wrap(~measure, scales = "free")
p + geom_histogram()
#Then visualize the relationships between variables

pairs(D1)
#Try to capture an intution about the data and the relationships

```
#Classification tree
```{r}
#Create a classification tree that predicts whether a student "levels up" in the online course using three variables of your choice (As we did last time, set all controls to their minimums)

library(rpart)
library(rpart.plot)
rp <- rpart(level.up ~ post.test.score + av.assignment.score + forum.posts, method = "class", data = D1, control = rpart.control(minsplit = 1, minbucket = 1, cp = 0.001))
#Plot and generate a CP table for your tree

printcp(rp)
rpart.plot(rp)
#Generate a probability value that represents the probability that a student levels up based your classification tree

D1$pred <- predict(rp, type = "prob")[,2]#Last class we used type = "class" which predicted the classification for us, this time we are using type = "prob" to see the probability that our classififcation is based on.
Expand All @@ -44,24 +55,34 @@ plot(performance(pred.detail, "tpr", "fpr"))
abline(0, 1, lty = 2)

#Calculate the Area Under the Curve
unlist(slot(performance(Pred2,"auc"), "y.values"))#Unlist liberates the AUC value from the "performance" object created by ROCR
unlist(slot(performance(pred.detail,"auc"), "y.values"))#Unlist liberates the AUC value from the "performance" object created by ROCR

#Now repeat this process, but using the variables you did not use for the previous model and compare the plots & results of your two models. Which one do you think was the better model? Why?
rp2 <- rpart(level.up ~ pre.test.score + messages, method = "class", data = D1, control = rpart.control(minsplit = 1, minbucket = 1, cp = 0.001))
printcp(rp2)
rpart.plot(rp2)
D1$pred2 <- predict(rp2, type = "prob")[,2]
pred.detail2 <- prediction(D1$pred2, D1$level.up)
plot(performance(pred.detail2, "tpr", "fpr"))
abline(0, 1, lty = 2)
unlist(slot(performance(pred.detail2,"auc"), "y.values"))
#I think the first one is a better model since the auc is 1, which means that there is 100% percent to distinguish between positive class and negative class.
```
## Part III
#Thresholds
```{r}
#Look at the ROC plot for your first model. Based on this plot choose a probability threshold that balances capturing the most correct predictions against false positives. Then generate a new variable in your data set that classifies each student according to your chosen threshold.

threshold.pred1 <-
D1$threshold.pred1 <- ifelse(D1$pred2 >= 0.707, "yes", "no")

#Now generate three diagnostics:

D1$accuracy.model1 <-

D1$precision.model1 <-

D1$recall.model1 <-
accuracy.model1 <- mean(ifelse(D1$level.up == D1$threshold.pred1,1,0))
D1$truepos.model1 <- ifelse(D1$level.up == "yes"&D1$threshold.pred1 == "yes",1,0)
D1$falsepos.model1 <- ifelse(D1$level.up == "no" & D1$threshold.pred1 == "yes", 1,0)
D1$falseneg.model1 <- ifelse(D1$level.up == "yes" & D1$threshold.pred1 == "no", 1,0)
precision.model1 <- sum(D1$truepos.model1)/(sum(D1$truepos.model1) + sum(D1$falsepos.model1))
recall.model1 <- sum(D1$truepos.model1)/(sum(D1$truepos.model1) + sum(D1$falseneg.model1))

#Finally, calculate Kappa for your model according to:

Expand All @@ -75,7 +96,17 @@ matrix1 <- as.matrix(table1)
kappa(matrix1, exact = TRUE)/kappa(matrix1)

#Now choose a different threshold value and repeat these diagnostics. What conclusions can you draw about your two thresholds?

D1$threshold.pred2 <- ifelse(D1$pred2 >= 0.606, "yes", "no")
accuracy.model2 <- mean(ifelse(D1$level.up == D1$threshold.pred2, 1, 0))
D1$truepos.model2 <- ifelse(D1$level.up == "yes" & D1$threshold.pred2 == "yes", 1, 0)
D1$falsepos.model2 <- ifelse(D1$level.up == "no" & D1$threshold.pred2 == "yes", 1,0)
D1$falseneg.model2 <- ifelse(D1$level.up == "yes" & D1$threshold.pred2 == "no", 1,0)
precision.model2 <- sum(D1$truepos.model2)/(sum(D1$truepos.model2) + sum(D1$falsepos.model2))
recall.model2 <- sum(D1$truepos.model2)/(sum(D1$truepos.model2) + sum(D1$falseneg.model2))
table2 <- table(D1$level.up, D1$threshold.pred2)
matrix2 <- as.matrix(table2)
kappa(matrix2, exact = TRUE)/kappa(matrix2)
# The kappa value are very close with different thresholds.
```

### To Submit Your Assignment
Expand Down
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