[r] Calculate AUC in R?

You can learn more about AUROC in this blog post by Miron Kursa:

https://mbq.me/blog/augh-roc/

He provides a fast function for AUROC:

# By Miron Kursa https://mbq.me
auroc <- function(score, bool) {
  n1 <- sum(!bool)
  n2 <- sum(bool)
  U  <- sum(rank(score)[!bool]) - n1 * (n1 + 1) / 2
  return(1 - U / n1 / n2)
}

Let's test it:

set.seed(42)
score <- rnorm(1e3)
bool  <- sample(c(TRUE, FALSE), 1e3, replace = TRUE)

pROC::auc(bool, score)
mltools::auc_roc(score, bool)
ROCR::performance(ROCR::prediction(score, bool), "auc")@y.values[[1]]
auroc(score, bool)

0.51371668847094
0.51371668847094
0.51371668847094
0.51371668847094

auroc() is 100 times faster than pROC::auc() and computeAUC().

auroc() is 10 times faster than mltools::auc_roc() and ROCR::performance().

print(microbenchmark(
  pROC::auc(bool, score),
  computeAUC(score[bool], score[!bool]),
  mltools::auc_roc(score, bool),
  ROCR::performance(ROCR::prediction(score, bool), "auc")@y.values,
  auroc(score, bool)
))

Unit: microseconds
                                                             expr       min
                                           pROC::auc(bool, score) 21000.146
                            computeAUC(score[bool], score[!bool]) 11878.605
                                    mltools::auc_roc(score, bool)  5750.651
 ROCR::performance(ROCR::prediction(score, bool), "auc")@y.values  2899.573
                                               auroc(score, bool)   236.531
         lq       mean     median        uq        max neval  cld
 22005.3350 23738.3447 22206.5730 22710.853  32628.347   100    d
 12323.0305 16173.0645 12378.5540 12624.981 233701.511   100   c 
  6186.0245  6495.5158  6325.3955  6573.993  14698.244   100  b  
  3019.6310  3300.1961  3068.0240  3237.534  11995.667   100 ab  
   245.4755   253.1109   251.8505   257.578    300.506   100 a   

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