finsih!

Assignment7.Rmd

@@ -1,7 +1,7 @@
 ---
 title: "Assignment 7 - Answers"
-author: "Charles Lang"
-date: "11/30/2016"
+author: "Zifan Cao"
+date: "12/03/2019"
 output: html_document
 ---
 
@@ -11,7 +11,7 @@ In the following assignment you will be looking at data from an one level of an
 
 #Upload data
 ```{r}
-
+A1 <- read.csv("online.data.csv")
 ```
 
 #Visualization 
@@ -21,61 +21,104 @@ In the following assignment you will be looking at data from an one level of an
 #Then visualize the relationships between variables
 
 #Try to capture an intution about the data and the relationships
-
+library(ggplot2)
+library(tidyr)
+library(dplyr)
+A1$level.up <- ifelse(A1$level.up == "yes", 1,0)
+A2 <- gather(A1, "level", "score", 2:7)
+plot1 <- ggplot(A2, aes(score)) + facet_wrap(~level, scales = "free")
+plot1 + geom_histogram(stat = "count")
+pairs(A1)
 ```
 #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.plot)
+c.treea <- rpart(level.up ~ forum.posts + pre.test.score, method = "class", data = A1, control=rpart.control(minsplit=1, minbucket=1, cp=0.001))
 #Plot and generate a CP table for your tree 
+printcp(c.treea)
 
+plot(c.treea)
 #Generate a probability value that represents the probability that a student levels up based your classification tree 
-
-D1$pred <- predict(rp, type = "prob")[,2]#Last class we used type = "class" which predicted the classification for us, this time we are using type = "prob" to see the probability that our classififcation is based on.
+A1$pred <- predict(c.treea, A1, 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.
 ```
 ## Part II
 #Now you can generate the ROC curve for your model. You will need to install the package ROCR to do this.
 ```{r}
+
 library(ROCR)
 
 #Plot the curve
-pred.detail <- prediction(D1$pred, D1$level.up) 
+pred.detail <- prediction(A1$pred, A1$level.up) 
 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"))
+#as this question, you need to run both plot and bline together
 
 #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?
+c.tree2<-rpart(level.up~pre.test.score+post.test.score+forum.posts, method="class",data=A1)
+printcp(c.tree2)
+post(c.tree2, file = "dtree2.ps", title = "tree2")
+rpart.plot(c.tree2, type=3, box.palette = c("red", "green"), fallen.leaves = TRUE)
+A1$pred2 <- predict(c.tree2, A1, type="prob")[,2]
+pred.detail2 <- prediction(A1$pred2, A1$level.up) 
+plot(performance(pred.detail2, "tpr", "fpr"))
+abline(0, 1, lty = 2)
+unlist(slot(performance(pred.detail2, "auc"), "y.values"))
+# from the ROC curve, we think the first model is better because the auc value is higher.  
 ```
 ## 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 <- 
-
-#Now generate three diagnostics:
+A1$threshold.pred1 <- ifelse(A1$pred >= 0.8, "yes", "no")
+A1$threshold.pred2 <- ifelse(A1$pred >= 0.95, "yes", "no")
+A1$threshold.pred3 <- ifelse(A1$pred >= 0.25, "yes", "no")
 
-D1$accuracy.model1 <-
 
-D1$precision.model1 <- 
+#Now generate three diagnostics:
+accuracy.model1 <- mean(ifelse(A1$level.up == A1$threshold.pred1, 1, 0))
+A1$truepos.model1 <- ifelse(A1$level.up == "yes" & A1$threshold.pred1 == "yes", 1, 0)
+A1$falsepos.model1 <- ifelse(A1$level.up == "no" & A1$threshold.pred1 == "yes", 1,0)
+A1$falseneg.model1 <- ifelse(A1$level.up == "yes" & A1$threshold.pred1 == "no", 1,0)
+precision.model1 <- sum(A1$truepos.model1)/(sum(A1$truepos.model1) + sum(A1$falsepos.model1))
+recall.model1 <- sum(A1$truepos.model1)/(sum(A1$truepos.model1) + sum(A1$falseneg.model1))
 
-D1$recall.model1 <- 
 
 #Finally, calculate Kappa for your model according to:
 
 #First generate the table of comparisons
-table1 <- table(D1$level.up, D1$threshold.pred1)
+table1 <- table(A1$level.up, A1$threshold.pred1)
+table1 
 
 #Convert to matrix
 matrix1 <- as.matrix(table1)
 
 #Calculate kappa
 kappa(matrix1, exact = TRUE)/kappa(matrix1)
-
+#1.087797
 #Now choose a different threshold value and repeat these diagnostics. What conclusions can you draw about your two thresholds?
 
+accuracy.model3 <- mean(ifelse(A1$level.up == A1$threshold.pred3, 1, 0))
+A1$truepos.model3 <- ifelse(A1$level.up == "yes" & A1$threshold.pred3 == "yes", 1, 0)
+A1$falsepos.model3 <- ifelse(A1$level.up == "no" & A1$threshold.pred3 == "yes", 1,0)
+A1$falseneg.model3 <- ifelse(A1$level.up == "yes" & A1$threshold.pred3 == "no", 1,0)
+precision.model3 <- sum(A1$truepos.model3)/(sum(A1$truepos.model3) + sum(A1$falsepos.model3))
+recall.model3 <- sum(A1$truepos.model3)/(sum(A1$truepos.model3) + sum(A1$falseneg.model3))
+
+#First generate the table of comparisons
+table3 <- table(A1$level.up, A1$threshold.pred3)
+table3 
+
+#Convert to matrix
+matrix3 <- as.matrix(table3)
+
+#Calculate kappa
+kappa(matrix3, exact = TRUE)/kappa(matrix3)
+#1.030205
+#with higher threshold, the kappa value is higher and we can conclude that with higher threshold, our prediction is better 
 ```
 
 ### To Submit Your Assignment

assignment7.Rproj

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