**R-english – Freakonometrics**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

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For my ACT6100 weekly quiz, I usually generate some datasets, and then ask students to compare various predictive algorithms. Last week, it was about classification trees and random forests. And students were surprised to have such differences (they had to estimate the probability to have a specific label, for the barycenter of the covariates).

Usually, I use the following to generate some (here 12) covariates that could be correlated

library(FactoMineR) n=279 library(clusterGeneration) library(mnormt) k=12 S=genPositiveDefMat("unifcorrmat",dim=k) X=round(rmnorm(n,varcov=S$Sigma)+8,2) rownames(X)=1:n colnames(X)=LETTERS[1:k]

Then I need to generate some data, based on some covariates (5 out of 12), with various strengths

idx = sample(1:k,size=5) u = sample(c(-(4:1),1:4),5) beta = rep(0,k) beta[idx] = u U = X%*%beta U = U-min(U) U = U/max(U)*6-3 p = exp(( U))/(1+exp((U ))) Y = rbinom(n,size=1,prob=p) df = data.frame(Y=as.factor(Y),X) levels(df$Y)=levels=c("blue","red")

We can run a classification tree

library(rpart) arbre = rpart(Y~., data=df)

and a random forest,

library(randomForest) set.seed(1) arbres = randomForest(Y~., data=df)

Here are the partial plots for 4 of the explanatory variables that actually have an impact

partialPlot(arbres,pred.data = df, x.var = "A")

Predictions for the “average” point of the dataset is here

(parbre = predict(arbre,newdata=data.frame(t(apply(df[,-1],2,mean))),type = "prob")) blue red 1 0.8064516 0.1935484 (parbres = predict(arbres,newdata=data.frame(t(apply(df[,-1],2,mean))),type = "prob")) blue red 1 0.422 0.578 attr(,"class") [1] "matrix" "votes"

and there is a substantial difference, with a probability of 19% with a single tree, 58% with 500 trees (the default value of the function).

To understand why we can have such a difference, we should not only focus on the bagging stratgy, but look at the variability of the predictions, obtained with trees,

B=1e4 parbres = rep(NA,B) m=data.frame(t(apply(df[,-1],2,mean))) for(b in 1:B){ idx = sample(1:nrow(df),size=nrow(df),replace=TRUE) arbre = rpart(Y~., data=df[idx,]) parbres[b] = predict(arbre,newdata=m,type = "prob")[2] } hist(parbres)

Surprisingly, we have here a bimodal function for \(\hat{y}\) which is either very small for some trees, of very large for others. On average, we have a value close to 55%… I think I will use more that generative algorithm for future quiz…

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