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vioravis

2010-Oct-22 05:19 UTC

[R] Random Forest AUC

Guys,

I used Random Forest with a couple of data sets I had to predict for binary
response. In all the cases, the AUC of the training set is coming to be 1.
Is this always the case with random forests? Can someone please clarify
this? 

I have given a simple example, first using logistic regression and then
using random forests to explain the problem. AUC of the random forest is
coming out to be 1.

data(iris)
iris <- iris[(iris$Species != "setosa"),]
iris$Species <- factor(iris$Species)
fit <- glm(Species~.,iris,family=binomial)
train.predict <- predict(fit,newdata = iris,type="response")
library(ROCR)
plot(performance(prediction(train.predict,iris$Species),"tpr","fpr"),col
"red")
auc1 <-
performance(prediction(train.predict,iris$Species),"auc")@y.values[[1]]
legend("bottomright",legend=c(paste("Logistic Regression
(AUC=",formatC(auc1,digits=4,format="f"),")",sep="")),
		col=c("red"), lty=1)


library(randomForest)
fit <- randomForest(Species ~ ., data=iris, ntree=50)
train.predict <- predict(fit,iris,type="prob")[,2]          
plot(performance(prediction(train.predict,iris$Species),"tpr","fpr"),col
"red")
auc1 <-
performance(prediction(train.predict,iris$Species),"auc")@y.values[[1]]
legend("bottomright",legend=c(paste("Random Forests
(AUC=",formatC(auc1,digits=4,format="f"),")",sep="")),
		col=c("red"), lty=1)

Thank you.

Regards,
Ravishankar R
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Max Kuhn

2010-Oct-22 14:38 UTC

head link

[R] Random Forest AUC

Ravishankar,
> I used Random Forest with a couple of data sets I had to predict for binary
> response. In all the cases, the AUC of the training set is coming to be 1.
> Is this always the case with random forests? Can someone please clarify
> this?
This is pretty typical for this model.
> I have given a simple example, first using logistic regression and then
> using random forests to explain the problem. AUC of the random forest is
> coming out to be 1.
Logistic regression isn't as flexible as RF and some other methods, so
the ROC curve is likely to be less than one, but much higher than it
really is (since you are re-predicting the same data)

For you example:
>
performance(prediction(train.predict,iris$Species),"auc")@y.values[[1]][1] 0.9972

but using simple 10-fold CV:
> library(caret)
> ctrl <- trainControl(method = "cv",+                      classProbs = TRUE,
+                      summaryFunction =
twoClassSummary)>
> set.seed(1)
> cvEstimate <- train(Species ~ ., data = iris,+                     method = "glm",
+                     metric = "ROC",
+                     trControl = ctrl)
Fitting: parameter=none
Aggregating results
Fitting model on full training set
Warning messages:
1: glm.fit: fitted probabilities numerically 0 or 1 occurred
2: glm.fit: algorithm did not converge
3: glm.fit: fitted probabilities numerically 0 or 1 occurred
4: glm.fit: algorithm did not converge
5: glm.fit: fitted probabilities numerically 0 or 1
occurred> cvEstimate
Call:
train.formula(form = Species ~ ., data = iris, method = "glm",
    metric = "ROC", trControl = ctrl)

100 samples
  4 predictors

Pre-processing:
Resampling: Cross-Validation (10 fold)

Summary of sample sizes: 90, 90, 90, 90, 90, 90, ...

Resampling results

  Sens  Spec  ROC   Sens SD  Spec SD  ROC SD
  0.96  0.98  0.86  0.0843   0.0632   0.126

and for random forest:
> set.seed(1)
> rfEstimate <- train(Species ~ .,+                     data = iris,
+                     method = "rf",
+                     metric = "ROC",
+                     tuneGrid = data.frame(.mtry = 2),
+                     trControl = ctrl)
Fitting: mtry=2
Aggregating results
Selecting tuning parameters
Fitting model on full training set> rfEstimate
Call:
train.formula(form = Species ~ ., data = iris, method = "rf",
    metric = "ROC", tuneGrid = data.frame(.mtry = 2), trControl =
ctrl)

100 samples
  4 predictors

Pre-processing:
Resampling: Cross-Validation (10 fold)

Summary of sample sizes: 90, 90, 90, 90, 90, 90, ...

Resampling results

  Sens  Spec  ROC    Sens SD  Spec SD  ROC SD
  0.94  0.92  0.898  0.0966   0.14     0.00632

Tuning parameter 'mtry' was held constant at a value of 2

-- 

Max

Liaw, Andy

2010-Oct-23 01:18 UTC

head link

[R] Random Forest AUC

Let me expand on what Max showed.

For the most part, performance on training set is meaningless.  (That's
the case for most algorithms, but especially so for RF.)  In the default
(and recommended) setting, the trees are grown to the maximum size,
which means that quite likely there's only one data point in most
terminal nodes, and the prediction at the terminal nodes are determined
by the majority class in the node, or the lone data point.  Suppose that
is the case all the time; i.e., in all trees all terminal nodes have
only one data point.  A particular data point would be "in-bag" in
about
64% of the trees in the forest, and every one of those trees has the
correct prediction for that data point.  Even if all the trees where
that data points are out-of-bag gave the wrong prediction, by majority
vote of all trees, you still get the right answer in the end.  Thus
basically the perfect prediction on train set for RF is "by design".

Generally, good training prediction is just self-fulfilling prophecy.

Andy
> -----Original Message-----
> From: r-help-bounces at r-project.org 
> [mailto:r-help-bounces at r-project.org] On Behalf Of vioravis
> Sent: Friday, October 22, 2010 1:20 AM
> To: r-help at r-project.org
> Subject: [R] Random Forest AUC
> 
> 
> Guys,
> 
> I used Random Forest with a couple of data sets I had to 
> predict for binary
> response. In all the cases, the AUC of the training set is 
> coming to be 1.
> Is this always the case with random forests? Can someone 
> please clarify
> this? 
> 
> I have given a simple example, first using logistic 
> regression and then
> using random forests to explain the problem. AUC of the 
> random forest is
> coming out to be 1.
> 
> data(iris)
> iris <- iris[(iris$Species != "setosa"),]
> iris$Species <- factor(iris$Species)
> fit <- glm(Species~.,iris,family=binomial)
> train.predict <- predict(fit,newdata = iris,type="response")
> library(ROCR)
> plot(performance(prediction(train.predict,iris$Species),"tpr",
> "fpr"),col > "red")
> auc1 <-
> performance(prediction(train.predict,iris$Species),"auc")@y.va
> lues[[1]]
> legend("bottomright",legend=c(paste("Logistic Regression
>
(AUC=",formatC(auc1,digits=4,format="f"),")",sep="")),
> 		col=c("red"), lty=1)
> 
> 
> library(randomForest)
> fit <- randomForest(Species ~ ., data=iris, ntree=50)
> train.predict <- predict(fit,iris,type="prob")[,2]          
> plot(performance(prediction(train.predict,iris$Species),"tpr",
> "fpr"),col > "red")
> auc1 <-
> performance(prediction(train.predict,iris$Species),"auc")@y.va
> lues[[1]]
> legend("bottomright",legend=c(paste("Random Forests
>
(AUC=",formatC(auc1,digits=4,format="f"),")",sep="")),
> 		col=c("red"), lty=1)
> 
> Thank you.
> 
> Regards,
> Ravishankar R
> -- 
> View this message in context: 
> http://r.789695.n4.nabble.com/Random-Forest-AUC-tp3006649p3006649.html
> Sent from the R help mailing list archive at Nabble.com.
> 
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