I constructed a binary logistic model. The response variable is binary. There are 4 regressors - 2 binary and 2 integers. I want to find the outliers and delete them. For this i have create some plots:
par(mfrow = c(2,2))
plot(hat.ep,rstudent.ep,col="#E69F00", main="hat-values versus studentized residuals",
xlab="Hat value", ylab="Studentized residual")
dffits.ep <- dffits(model_logit)
plot(id,dffits.ep,type="l", col="#E69F00", main="Index Plot",
xlab="Identification", ylab="Diffits")
cov.ep <- covratio(model_logit)
plot(id,cov.ep,type="l",col="#E69F00", main="Covariance Ratio",
xlab="Identification", ylab="Covariance Ratio")
cook.ep <- cooks.distance(model_logit)
plot(id,cook.ep,type="l",col="#E69F00", main="Cook's Distance",
xlab="Identification", ylab="Cook's Distance")
According to the plots there is an outlier. How can I identify which observation is the outlier?
I have tried :
> outlierTest(model_logit)
No Studentized residuals with Bonferonni p < 0.05
Largest |rstudent|:
rstudent unadjusted p-value Bonferonni p
1061 1.931043 0.053478 NA
Are there some other functions for outlier detection?
Well this answer comes quite late. I'm unsure if you have found the answer or not. Continuing further, in the absence of a minimum reproducible example, I'll attempt to answer the question using some dummy data and two custom functions. For a given continuous variable, outliers are those observations that lie outside of 1.5*IQR, where IQR, the ‘Inter Quartile Range’ is the difference between the 75th and 25th quartiles. I also recommend you to see this post containing far better solutions than my crude answer.
> df <- data.frame(X = c(NA, rnorm(1000), runif(20, -20, 20)), Y = c(runif(1000),rnorm(20, 2), NA), Z = c(rnorm(1000, 1), NA, runif(20)))
> head(df)
X Y Z
1 NA 0.8651 0.2784
2 -0.06838 0.4700 2.0483
3 -0.18734 0.9887 1.8353
4 -0.05015 0.7731 2.4464
5 0.25010 0.9941 1.3979
6 -0.26664 0.6778 1.1277
> boxplot(df$Y) # notice the outliers above the top whisker
Now, I'll create a custom function to detect the outliers and the other function will replace the outlier values with NA.
# this function will return the indices of the outlier values
> findOutlier <- function(data, cutoff = 3) {
## Calculate the sd
sds <- apply(data, 2, sd, na.rm = TRUE)
## Identify the cells with value greater than cutoff * sd (column wise)
result <- mapply(function(d, s) {
which(d > cutoff * s)
}, data, sds)
result
}
# check for outliers
> outliers <- findOutlier(df)
# custom function to remove outliers
> removeOutlier <- function(data, outliers) {
result <- mapply(function(d, o) {
res <- d
res[o] <- NA
return(res)
}, data, outliers)
return(as.data.frame(result))
}
> filterData<- removeOutlier(df, outliers)
> boxplot(filterData$Y)
Related
This function needs to take a data frame with three variables and four observations (a, b, c and d) and calculate (a/c) / (b/d).
for example:
df <- data.frame(female = c("White", "White", "non-White", "non-White"),
male = c("White", "non-White", "White", "non-White"),
n = c(85, 5, 5, 10))
xtabs(n ~ female + male, df)
the function would have to calculate (85 * 10) / (5 * 5) and return a result of 34.
I have previously tried this:
oddsRatio <- function(x){
x %>%
summarise(oddsratio = (n[1] * n[4]) / (n[2] * n[3]))
}
oddsRatio(df)
but this produced the answer in a table and also didn't work universally on other data frames with 4 observations and 3 variables in the way that I wanted it to.
A dplyr way of doing it
df %>%
summarize(oddsRatio = prod(n[female == male])/prod(n[female != male]))
As a function that returns a number
oddsRatio <- function(x) {
x %>%
summarize(oddsRatio = prod(n[female == male])/prod(n[female != male])) %>%
pull(oddsRatio)
}
oddsRatio(df)
# 34
Here are some ways. No packages are needed.
1) Create a logical that picks out the diagonals and then subset xt with that and with its negative.
xt <- xtabs(n ~ female + male, df)
is.diag <- row(xt) == col(xt)
prod(xt[is.diag]) / prod(xt[!is.diag])
## [1] 34
2) or pick out the diagonal and antidiagonal using indexes:
prod(xt[c(1, 4)]) / prod(xt[2:3])
## [1] 34
3) If the values of xt are known to be strictly positive then we could take the log, multiply that by c(1, -1, -1, 1), sum and take exp to get back:
exp(sum(log(xt) * c(1, -1, -1, 1)))
## [1] 34
4) If you are performing this calculation to test independence of the factors you could just directly use fisher.test . Fisher's exact test calculates the maximum likelihood estimate of the odds ratio given the table's marginals using the hypergeometric distribution. The null hypothesis is that the two factors are independent, i.e. the odds ratio equals 1, and in the example below it is rejected, i.e. the factors are not independent. Note that the confidence interval does not contain 1.
fisher.test(xt)
giving:
Fisher's Exact Test for Count Data
data: xt
p-value = 2.435e-07
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
6.951076 174.962113
sample estimates:
odds ratio
31.48572
with(df, prod(ifelse(female==male, n, 1/n)))
[1] 34
This is similar to #cnicol's solution but aviods any need for tidyverse.
I have an array of outputs from hundreds of segmented linear models (made using the segmented package in R). I want to be able to use these outputs on new data, using the predict function. To be clear, I do not have the segmented linear model objects in my workspace; I just saved and reimported the relevant outputs (e.g. the coefficients and breakpoints). For this reason I can't simply use the predict.segmented function from the segmented package.
Below is a toy example based on this link that seems promising, but does not match the output of the predict.segmented function.
library(segmented)
set.seed(12)
xx <- 1:100
zz <- runif(100)
yy <- 2 + 1.5*pmax(xx-35,0) - 1.5*pmax(xx-70,0) +
15*pmax(zz-0.5,0) + rnorm(100,0,2)
dati <- data.frame(x=xx,y=yy,z=zz)
out.lm<-lm(y~x,data=dati)
o<-## S3 method for class 'lm':
segmented(out.lm,seg.Z=~x,psi=list(x=c(30,60)),
control=seg.control(display=FALSE))
# Note that coefficients with U in the name are differences in slopes, not slopes.
# Compare:
slope(o)
coef(o)[2] + coef(o)[3]
coef(o)[2] + coef(o)[3] + coef(o)[4]
# prediction
pred <- data.frame(x = 1:100)
pred$dummy1 <- pmax(pred$x - o$psi[1,2], 0)
pred$dummy2 <- pmax(pred$x - o$psi[2,2], 0)
pred$dummy3 <- I(pred$x > o$psi[1,2]) * (coef(o)[2] + coef(o)[3])
pred$dummy4 <- I(pred$x > o$psi[2,2]) * (coef(o)[2] + coef(o)[3] + coef(o)[4])
names(pred)[-1]<- names(model.frame(o))[-c(1,2)]
# compute the prediction, using standard predict function
# computing confidence intervals further
# suppose that the breakpoints are fixed
pred <- data.frame(pred, predict(o, newdata= pred,
interval="confidence"))
# Try prediction using the predict.segment version to compare
test <- predict.segmented(o)
plot(pred$fit, test, ylim = c(0, 100))
abline(0,1, col = "red")
# At least one segment not being predicted correctly?
Can I use the base r predict() function (not the segmented.predict() function) with the coefficients and break points saved from segmented linear models?
UPDATE
I figured out that the code above has issues (don't use it). Through some reverse engineering of the segmented.predict() function, I produced the design matrix and use that to predict values instead of directly using the predict() function. I do not consider this a full answer of the original question yet because predict() can also produce confidence intervals for the prediction, and I have not yet implemented that--question still open for someone to add confidence intervals.
library(segmented)
## Define function for making matrix of dummy variables (this is based on code from predict.segmented())
dummy.matrix <- function(x.values, x_names, psi.est = TRUE, nameU, nameV, diffSlope, est.psi) {
# This function creates a model matrix with dummy variables for a segmented lm with two breakpoints.
# Inputs:
# x.values: the x values of the segmented lm
# x_names: the name of the column of x values
# psi.est: this is legacy from the predict.segmented function, leave it set to 'TRUE'
# obj: the segmented lm object
# nameU: names (class character) of 3rd and 4th coef, which are "U1.x" "U2.x" for lm with two breaks. Example: names(c(obj$coef[3], obj$coef[4]))
# nameV: names (class character) of 5th and 6th coef, which are "psi1.x" "psi2.x" for lm with two breaks. Example: names(c(obj$coef[5], obj$coef[6]))
# diffSlope: the coefficients (class numeric) with the slope differences; called U1.x and U2.x for lm with two breaks. Example: c(o$coef[3], o$coef[4])
# est.psi: the estimated break points (class numeric); these are the estimated breakpoints from segmented.lm. Example: c(obj$psi[1,2], obj$psi[2,2])
#
n <- length(x.values)
k <- length(est.psi)
PSI <- matrix(rep(est.psi, rep(n, k)), ncol = k)
newZ <- matrix(x.values, nrow = n, ncol = k, byrow = FALSE)
dummy1 <- pmax(newZ - PSI, 0)
if (psi.est) {
V <- ifelse(newZ > PSI, -1, 0)
dummy2 <- if (k == 1)
V * diffSlope
else V %*% diag(diffSlope)
newd <- cbind(x.values, dummy1, dummy2)
colnames(newd) <- c(x_names, nameU, nameV)
} else {
newd <- cbind(x.values, dummy1)
colnames(newd) <- c(x_names, nameU)
}
# if (!x_names %in% names(coef(obj.seg)))
# newd <- newd[, -1, drop = FALSE]
return(newd)
}
## Test dummy matrix function----------------------------------------------
set.seed(12)
xx<-1:100
zz<-runif(100)
yy<-2+1.5*pmax(xx-35,0)-1.5*pmax(xx-70,0)+15*pmax(zz-.5,0)+rnorm(100,0,2)
dati<-data.frame(x=xx,y=yy,z=zz)
out.lm<-lm(y~x,data=dati)
#1 segmented variable, 2 breakpoints: you have to specify starting values (vector) for psi:
o<-segmented(out.lm,seg.Z=~x,psi=c(30,60),
control=seg.control(display=FALSE))
slope(o)
plot.segmented(o)
summary(o)
# Test dummy matrix fn with the same dataset
newdata <- dati
nameU1 <- c("U1.x", "U2.x")
nameV1 <- c("psi1.x", "psi2.x")
diffSlope1 <- c(o$coef[3], o$coef[4])
est.psi1 <- c(o$psi[1,2], o$psi[2,2])
test <- dummy.matrix(x.values = newdata$x, x_names = "x", psi.est = TRUE,
nameU = nameU1, nameV = nameV1, diffSlope = diffSlope1, est.psi = est.psi1)
# Predict response variable using matrix multiplication
col1 <- matrix(1, nrow = dim(test)[1])
test <- cbind(col1, test) # Now test is the same as model.matrix(o)
predY <- coef(o) %*% t(test)
plot(predY[1,])
lines(predict.segmented(o), col = "blue") # good, predict.segmented gives same answer
The question is given like this:
Read the file diabetes.csv. There are two variables called BMI and Outcome. The variable Outcome takes on only two values: 0 and 1. Conduct a non-parametric two sample test for the hypothesis that the standard deviation of BMI is the same for both Outcome values
bmi <- diabetes$BMI
bmi
outcome <- diabetes$Outcome
outcome
n <- length(bmi)
# tstat
tstat <- ???
# Describe the population and draw synthetic samples
f1 <- function()
{
x <- c(bmi, outcome)
x <- sample(x)
m1 <- sd(x[1:n])
m2 <- sd(x[(n+1):length(x)])
return(m1 - m2)
}
# Create sampling distribution
sdist <- replicate(10000, f1())
plot(density(sdist))
# Gap
gap <- abs(mean(sdist) - tstat)
abline(v = mean(sdist) + c(-1,1) * gap, col = "dark orange")
s1 <- sdist[sdist <(mean(sdist - gap)) | sdist >(mean(sdist + gap))]
pvalue <- length(s1) / length(sdist)
pvalue
The data is in some dataset called "diabetes". My question is how to represent the "t-statistic" since the outcome is binary?
Use this code:
# Sort the table diabetes on accending order of Outcome to separate the BMI
# values with outcome = 0 and BMI values with outcome = 1
diabetes = diabetes[order(diabetes$Outcome),]
View(diabetes)
# Find the number of values with outcome = 0
n = length(which(diabetes$Outcome == 0))
# Find total number of rows
l = length(diabetes$BMI)
# Find BMI values to create the sample later on
g = diabetes$BMI
# Create function to take the values of BMI and shuffle it every time and
# to find the difference between the standard deviations
f1 = function()
{
x = sample(g)
z = abs(sd(x[1:n]) - sd(x[(n+1):l]))
return(z)
}
# Replicate the function several times
dist = replicate(100000,f1())
# Plot density of distribution
plot(density(dist))
polygon(density(dist),col="green")
diabetes0 = diabetes[diabetes$Outcome == 0,]
diabetes1 = diabetes[diabetes$Outcome == 1,]
View(diabetes0)
View(diabetes1)
# Find the difference between standard deviation of BMI when outcome = 0 and
# when outcome = 1
tstat = abs(sd(diabetes0$BMI) - sd(diabetes1$BMI))
tstat
abline(v=tstat)
rside = dist[dist>tstat]
pvalue = length(rside)/length(dist)
pvalue
I am trying to build a rolling regression function based on the example here, but in addition to returning the predicted values, I would like to return the some rolling model diagnostics (i.e. coefficients, t-values, and mabye R^2). I would like the results to be returned in discrete objects based on the type of results. The example provided in the link above sucessfully creates thr rolling predictions, but I need some assistance packaging and writing out the rolling model diagnostics:
In the end, I would like the function to return three (3) objects:
Predictions
Coefficients
T values
R^2
Below is the code:
require(zoo)
require(dynlm)
## Create Some Dummy Data
set.seed(12345)
x <- rnorm(mean=3,sd=2,100)
y <- rep(NA,100)
y[1] <- x[1]
for(i in 2:100) y[i]=1+x[i-1]+0.5*y[i-1]+rnorm(1,0,0.5)
int <- 1:100
dummydata <- data.frame(int=int,x=x,y=y)
zoodata <- as.zoo(dummydata)
rolling.regression <- function(series) {
mod <- dynlm(formula = y ~ L(y) + L(x), data = as.zoo(series)) # get model
nextOb <- max(series[,'int'])+1 # To get the first row that follows the window
if (nextOb<=nrow(zoodata)) { # You won't predict the last one
# 1) Make Predictions
predicted <- predict(mod,newdata=data.frame(x=zoodata[nextOb,'x'],y=zoodata[nextOb,'y']))
attributes(predicted) <- NULL
c(predicted=predicted,square.res <-(predicted-zoodata[nextOb,'y'])^2)
# 2) Extract coefficients
#coefficients <- coef(mod)
# 3) Extract rolling coefficient t values
#tvalues <- ????(mod)
# 4) Extract rolling R^2
#rsq <-
}
}
rolling.window <- 20
results.z <- rollapply(zoodata, width=rolling.window, FUN=rolling.regression, by.column=F, align='right')
So after figuring out how to extract t values from model (i.e. mod) , what do I need to do to make the function return three (3) seperate objects (i.e. Predictions, Coefficients, and T-values)?
I am fairly new to R, really new to functions, and extreemly new to zoo, and I'm stuck.
Any assistance would be greatly appreciated.
I hope I got you correctly, but here is a small edit of your function:
rolling.regression <- function(series) {
mod <- dynlm(formula = y ~ L(y) + L(x), data = as.zoo(series)) # get model
nextOb <- max(series[,'int'])+1 # To get the first row that follows the window
if (nextOb<=nrow(zoodata)) { # You won't predict the last one
# 1) Make Predictions
predicted=predict(mod,newdata=data.frame(x=zoodata[nextOb,'x'],y=zoodata[nextOb,'y']))
attributes(predicted)<-NULL
#Solution 1; Quicker to write
# c(predicted=predicted,
# square.res=(predicted-zoodata[nextOb,'y'])^2,
# summary(mod)$coef[, 1],
# summary(mod)$coef[, 3],
# AdjR = summary(mod)$adj.r.squared)
#Solution 2; Get column names right
c(predicted=predicted,
square.res=(predicted-zoodata[nextOb,'y'])^2,
coef_intercept = summary(mod)$coef[1, 1],
coef_Ly = summary(mod)$coef[2, 1],
coef_Lx = summary(mod)$coef[3, 1],
tValue_intercept = summary(mod)$coef[1, 3],
tValue_Ly = summary(mod)$coef[2, 3],
tValue_Lx = summary(mod)$coef[3, 3],
AdjR = summary(mod)$adj.r.squared)
}
}
rolling.window <- 20
results.z <- rollapply(zoodata, width=rolling.window, FUN=rolling.regression, by.column=F, align='right')
head(results.z)
predicted square.res coef_intercept coef_Ly coef_Lx tValue_intercept tValue_Ly tValue_Lx AdjR
20 10.849344 0.721452 0.26596465 0.5798046 1.049594 0.38309211 7.977627 13.59831 0.9140886
21 12.978791 2.713053 0.26262820 0.5796883 1.039882 0.37741499 7.993014 13.80632 0.9190757
22 9.814676 11.719999 0.08050796 0.5964808 1.073941 0.12523824 8.888657 15.01353 0.9340732
23 5.616781 15.013297 0.05084124 0.5984748 1.077133 0.08964998 9.881614 16.48967 0.9509550
24 3.763645 6.976454 0.26466039 0.5788949 1.068493 0.51810115 11.558724 17.22875 0.9542983
25 9.433157 31.772658 0.38577698 0.5812665 1.034862 0.70969330 10.728395 16.88175 0.9511061
To see how it works, make a small example with a regression:
x <- rnorm(1000); y <- 2*x + rnorm(1000)
reg <- lm(y ~ x)
summary(reg)$coef
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.02694322 0.03035502 0.8876033 0.374968
x 1.97572544 0.03177346 62.1816310 0.000000
As you can see, calling summary first and then getting the coefficients of it (coef(summary(reg)) works as well) gives you a table with estimates, standard errors, and t-values. So estimates are saved in column 1 of that table, t-values in column 3. And that's how I obtain them in the updated rolling.regression function.
EDIT
I updated my solution; now it also contains the adjusted R2. If you just want the normal R2, get rid of the .adj.
EDIT 2
Quick and dirty hack how to name the columns:
rolling.regression <- function(series) {
mod <- dynlm(formula = y ~ L(y) + L(x), data = as.zoo(series)) # get model
nextOb <- max(series[,'int'])+1 # To get the first row that follows the window
if (nextOb<=nrow(zoodata)) { # You won't predict the last one
# 1) Make Predictions
predicted=predict(mod,newdata=data.frame(x=zoodata[nextOb,'x'],y=zoodata[nextOb,'y']))
attributes(predicted)<-NULL
#Get variable names
strVar <- c("Intercept", paste0("L", 1:(nrow(summary(mod)$coef)-1)))
vec <- c(predicted=predicted,
square.res=(predicted-zoodata[nextOb,'y'])^2,
AdjR = summary(mod)$adj.r.squared,
summary(mod)$coef[, 1],
summary(mod)$coef[, 3])
names(vec)[4:length(vec)] <- c(paste0("Coef_", strVar), paste0("tValue_", strVar))
vec
}
}
Given a vector of scores and a vector of actual class labels, how do you calculate a single-number AUC metric for a binary classifier in the R language or in simple English?
Page 9 of "AUC: a Better Measure..." seems to require knowing the class labels, and here is an example in MATLAB where I don't understand
R(Actual == 1))
Because R (not to be confused with the R language) is defined a vector but used as a function?
With the package pROC you can use the function auc() like this example from the help page:
> data(aSAH)
>
> # Syntax (response, predictor):
> auc(aSAH$outcome, aSAH$s100b)
Area under the curve: 0.7314
The ROCR package will calculate the AUC among other statistics:
auc.tmp <- performance(pred,"auc"); auc <- as.numeric(auc.tmp#y.values)
As mentioned by others, you can compute the AUC using the ROCR package. With the ROCR package you can also plot the ROC curve, lift curve and other model selection measures.
You can compute the AUC directly without using any package by using the fact that the AUC is equal to the probability that a true positive is scored greater than a true negative.
For example, if pos.scores is a vector containing a score of the positive examples, and neg.scores is a vector containing the negative examples then the AUC is approximated by:
> mean(sample(pos.scores,1000,replace=T) > sample(neg.scores,1000,replace=T))
[1] 0.7261
will give an approximation of the AUC. You can also estimate the variance of the AUC by bootstrapping:
> aucs = replicate(1000,mean(sample(pos.scores,1000,replace=T) > sample(neg.scores,1000,replace=T)))
Without any additional packages:
true_Y = c(1,1,1,1,2,1,2,1,2,2)
probs = c(1,0.999,0.999,0.973,0.568,0.421,0.382,0.377,0.146,0.11)
getROC_AUC = function(probs, true_Y){
probsSort = sort(probs, decreasing = TRUE, index.return = TRUE)
val = unlist(probsSort$x)
idx = unlist(probsSort$ix)
roc_y = true_Y[idx];
stack_x = cumsum(roc_y == 2)/sum(roc_y == 2)
stack_y = cumsum(roc_y == 1)/sum(roc_y == 1)
auc = sum((stack_x[2:length(roc_y)]-stack_x[1:length(roc_y)-1])*stack_y[2:length(roc_y)])
return(list(stack_x=stack_x, stack_y=stack_y, auc=auc))
}
aList = getROC_AUC(probs, true_Y)
stack_x = unlist(aList$stack_x)
stack_y = unlist(aList$stack_y)
auc = unlist(aList$auc)
plot(stack_x, stack_y, type = "l", col = "blue", xlab = "False Positive Rate", ylab = "True Positive Rate", main = "ROC")
axis(1, seq(0.0,1.0,0.1))
axis(2, seq(0.0,1.0,0.1))
abline(h=seq(0.0,1.0,0.1), v=seq(0.0,1.0,0.1), col="gray", lty=3)
legend(0.7, 0.3, sprintf("%3.3f",auc), lty=c(1,1), lwd=c(2.5,2.5), col="blue", title = "AUC")
I found some of the solutions here to be slow and/or confusing (and some of them don't handle ties correctly) so I wrote my own data.table based function auc_roc() in my R package mltools.
library(data.table)
library(mltools)
preds <- c(.1, .3, .3, .9)
actuals <- c(0, 0, 1, 1)
auc_roc(preds, actuals) # 0.875
auc_roc(preds, actuals, returnDT=TRUE)
Pred CountFalse CountTrue CumulativeFPR CumulativeTPR AdditionalArea CumulativeArea
1: 0.9 0 1 0.0 0.5 0.000 0.000
2: 0.3 1 1 0.5 1.0 0.375 0.375
3: 0.1 1 0 1.0 1.0 0.500 0.875
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
Combining code from ISL 9.6.3 ROC Curves, along with #J. Won.'s answer to this question and a few more places, the following plots the ROC curve and prints the AUC in the bottom right on the plot.
Below probs is a numeric vector of predicted probabilities for binary classification and test$label contains the true labels of the test data.
require(ROCR)
require(pROC)
rocplot <- function(pred, truth, ...) {
predob = prediction(pred, truth)
perf = performance(predob, "tpr", "fpr")
plot(perf, ...)
area <- auc(truth, pred)
area <- format(round(area, 4), nsmall = 4)
text(x=0.8, y=0.1, labels = paste("AUC =", area))
# the reference x=y line
segments(x0=0, y0=0, x1=1, y1=1, col="gray", lty=2)
}
rocplot(probs, test$label, col="blue")
This gives a plot like this:
I usually use the function ROC from the DiagnosisMed package. I like the graph it produces. AUC is returned along with it's confidence interval and it is also mentioned on the graph.
ROC(classLabels,scores,Full=TRUE)
Along the lines of erik's response, you should also be able to calculate the ROC directly by comparing all possible pairs of values from pos.scores and neg.scores:
score.pairs <- merge(pos.scores, neg.scores)
names(score.pairs) <- c("pos.score", "neg.score")
sum(score.pairs$pos.score > score.pairs$neg.score) / nrow(score.pairs)
Certainly less efficient than the sample approach or the pROC::auc, but more stable than the former and requiring less installation than the latter.
Related: when I tried this it gave similar results to pROC's value, but not exactly the same (off by 0.02 or so); the result was closer to the sample approach with very high N. If anyone has ideas why that might be I'd be interested.
Currently top voted answer is incorrect, because it disregards ties. When positive and negative scores are equal, then AUC should be 0.5. Below is corrected example.
computeAUC <- function(pos.scores, neg.scores, n_sample=100000) {
# Args:
# pos.scores: scores of positive observations
# neg.scores: scores of negative observations
# n_samples : number of samples to approximate AUC
pos.sample <- sample(pos.scores, n_sample, replace=T)
neg.sample <- sample(neg.scores, n_sample, replace=T)
mean(1.0*(pos.sample > neg.sample) + 0.5*(pos.sample==neg.sample))
}
Calculating AUC with Metrics package is very easy and straightforward:
library(Metrics)
actual <- c(0, 0, 1, 1)
predicted <- c(.1, .3, .3, .9)
auc(actual, predicted)
0.875