ROC Curve Ranger - r

Question

I am trying to calculate ROC Curve and AUC using ranger for a binomial classification problem (0 and 1), where the response variable is defined as BiClass.
Suppose I cast a data frame to Train_Set and Test_Set (75% and 25 % respectively) and compute binary class probabilities using:
library(ranger)
library(ROCR)
library(mlr)
library(pROC)
library(tidyverse)
Biclass.ranger <- ranger(BiClass ~ ., ,data=Train_Set, num.trees = 500, importance="impurity", save.memory = TRUE, probability=TRUE)
pred <- predict(BiClass.ranger, data = Test_Set, num.trees = 500, type='response', verbose = TRUE)
My intention is now to compute ROC curve (and AUC). I tried the following code, through which I get ROC curve (using ROCR and mlr packages):
pred_object <- prediction(pred$predictions[,2], Test_Set$BiClass)
per_measure <- performance(pred_object, "tnr", "fnr")
plot(per_measure, col="red", lwd=1)
abline(a=0,b=1,lwd=1,lty=1,col="gray")
Or, aletrnatively using pROC package:
probabilities <- as.data.frame(predict(Biclass.ranger, data = Test_Set, num.trees = 500, type='response', verbose = TRUE)$predictions)
probabilities$predic <- colnames(probabilities)[max.col(probabilities,ties.method="first")] # For each row, return the column name of the largest value from 0 and 1 columns (prediction column). This will be a character type
probabilities$prednum <- as.numeric(as.character(probabilities$predic)) # create prednum as a numeric data type in probabilities
probabilities <- dplyr::mutate_if(probabilities, is.character, as.factor) # convert character to factor
probabilities <- cbind(probabilities,BiClass=Test_Set$BiClass) # append BiClass. This data frame contains the response variable from the Test_Data, along with prediction (prednum) and probability classes (0 and 1)
ROC_ranger <- pROC::roc(Table$BiClass, pred$predictions[,2])
plot(ROC_ranger, col = "blue", main = "ROC - Ranger")
paste("Accuracy % of ranger: ", mean(Test_Set$BiClass == round(pred$predictions[,2], digits = 0))) # print the performance of each model
The ROC curve obtained is given below:
I have the following questions:
1) How can I set a threshold value and plot confusion matrix for the set threshold?
I compute the confusion matrix presently using:
probabilities <- as.data.frame(predict(Biclass.ranger, data = Test_Set, num.trees = 500, type='response', verbose = TRUE)$predictions)
max.col(probabilities) - 1
confusionMatrix(table(Test_Set$BiClass, max.col(probabilities)-1))
2) How do I calculate the optimal thershold value (global value at which I have more true positives or true negatives) through optimization?
Again, referring to the pROC and the guidelines proposed by its author using:
myroc <- pROC::roc(probabilities$BiClass, probabilities$`1`)
mycoords <- pROC::coords(myroc, "all", transpose = FALSE)
plot(mycoords$threshold, mycoords$specificity, type="l", col="red", xlab="Cutoff", ylab="Performance")
lines(mycoords$threshold, mycoords$sensitivity, type="l", col="blue")
legend(0.23,0.2, c("Specificity", "Sensitivity"), col=c("red", "blue"), lty=1)
best.coords <- coords(myroc, "best", best.method="youden", transpose = FALSE)
abline(v=best.coords$threshold, lty=2, col="grey")
abline(h=best.coords$specificity, lty=2, col="red")
abline(h=best.coords$sensitivity, lty=2, col="blue")
I was able to draw this curve using youden index:
]2
Does it mean there isn't a lot of freedom to vary threshold to play with specificity and sensitivity, since the dashed blue and red lines are not far away from each other?
3) How to evaulate AUC?
I calculated AUC using pROC again following the guidelines of its author. See below:
ROC_ranger <- pROC::roc(probabilities$BiClass, probabilities$`1`)
ROC_ranger_auc <- pROC::auc(ROC_ranger)
paste("Area under curve of random forest: ", ROC_ranger_auc) # AUC of the model
The goal finally is to increase the True Neagtives, which are presently defined by 1 in BiClass and of course True Positives (defined by 0 in BiClass) in the confusion matrix. At present, the Accuracy of my classification algorithm is 0.74 and the AUC is 0.81 respectively.