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...) ## 50 being the number of observations in the minority variable ## The other option which there seems to be some confusion over is establish some class weights ## to balance the error rate. This approach I've mostly drawn from here: ## http://stat-www.berkeley.edu/users/breiman/RandomForests/cc_home.htm#balance ## This might not be appropriate, however, as of September it looks like Breiman method wasn't used in R df.rf.weights<-randomForest(cls~var1+var2+var3, data=df,classwt=c(1, 600), importance=TRUE) ## Nevertheless, what I am concerned about is the effect of an unbalanced data set...

I am new to R and I'd like to use the randomForest package for my thesis (identifying important variables for more detailed analysis with other software). I have found extremely well written and helpful information on the usage of R. Unfortunately it seems to be very difficult to find similarly detailed tutorials for randomForest, and I just can't get it work with the information on

Does anyone know a reason why, in principle, a call to randomForest cannot accept a data frame with missing predictor values? If each individual tree is built using CART, then it seems like this should be possible. (I understand that one may impute missing values using rfImpute or some other method, but I would like to avoid doing that.) If this functionality were available, then when the trees

...soy un experto en estas técnicas, pero hasta donde yo se, el algoritmo Random Forest no requiere cross validation. Lo dice el mismo Leo Breiman que creo ha sido uno de los investigadores que más ha contribuido al desarrollo de Random Forest (https://www.stat.berkeley.edu/~breiman/RandomForests/cc_home.htm) <https://www.stat.berkeley.edu/~breiman/RandomForests/cc_home.htm> Random forests - classification description www.stat.berkeley.edu <http://www.stat.berkeley.edu> Contents. Introduction Overview Features of random forests Remarks How Random Forests work The oob error estimat...

Dear all, I am trying to train a randomForest using all my control data (12,000 cases, ~ 20 explanatory variables, 2 classes). Because of memory constraints, I have split my data into 7 subsets and trained a randomForest for each, hoping that using combine() afterwards would solve the memory issue. Unfortunately, combine() still runs out of memory. Is there anything else I can do? (I am not using

...s, pero hasta donde yo se, el algoritmo > Random Forest no requiere cross validation. > > > > Lo dice el mismo Leo Breiman que creo ha sido uno de los investigadores > que más ha contribuido al desarrollo de Random Forest ( > https://www.stat.berkeley.edu/~breiman/RandomForests/cc_home.htm) > > Random forests - classification description > <https://www.stat.berkeley.edu/~breiman/RandomForests/cc_home.htm> > > www.stat.berkeley.edu > > Contents. Introduction Overview Features of random forests Remarks How > Random Forests work The oob error estimate V...

...l algoritmo >> Random Forest no requiere cross validation. >> >> >> >> Lo dice el mismo Leo Breiman que creo ha sido uno de los investigadores >> que más ha contribuido al desarrollo de Random Forest ( >> https://www.stat.berkeley.edu/~breiman/RandomForests/cc_home.htm) >> >> Random forests - classification description >> <https://www.stat.berkeley.edu/~breiman/RandomForests/cc_home.htm> >> >> www.stat.berkeley.edu >> >> Contents. Introduction Overview Features of random forests Remarks How >> Random Fores...

...soy un experto en estas técnicas, pero hasta donde yo se, el algoritmo Random Forest no requiere cross validation. Lo dice el mismo Leo Breiman que creo ha sido uno de los investigadores que más ha contribuido al desarrollo de Random Forest (https://www.stat.berkeley.edu/~breiman/RandomForests/cc_home.htm) Manuel El 2 de junio de 2017, 6:35, Isidro Hidalgo Arellano <ihidalgo en jccm.es <mailto:ihidalgo en jccm.es> > escribió: Una vez que tienes la técnica y los parámetros óptimos resultantes de la validación cruzada, ya tienes el modelo que necesitas, NO tienes que hacer nad...

...dom Forest no requiere cross validation. >>> >>> >>> >>> Lo dice el mismo Leo Breiman que creo ha sido uno de los investigadores >>> que más ha contribuido al desarrollo de Random Forest ( >>> https://www.stat.berkeley.edu/~breiman/RandomForests/cc_home.htm) >>> >>> Random forests - classification description >>> <https://www.stat.berkeley.edu/~breiman/RandomForests/cc_home.htm> >>> >>> www.stat.berkeley.edu >>> >>> Contents. Introduction Overview Features of random forests Remar...

Una vez que tienes la técnica y los parámetros óptimos resultantes de la validación cruzada, ya tienes el modelo que necesitas, NO tienes que hacer nada más. Si vuelves a modelar con todos los datos todo el trabajo de validación que has hecho lo envías a hacer gárgaras. Estarías construyendo un modelo con sobreajuste. Para quedarte tranquilo, haz la prueba, coge el modelo resultante de la

I want to take the first row of each unique ID value from a data frame. For instance > ddTable <- data.frame(Id=c(1,1,2,2),name=c("Paul","Joe","Bob","Larry")) I want a dataset that is Id Name 1 Paul 2 Bob > unique(ddTable) Will give me all 4 rows, and > unique(ddTable$Id) Will give me c(1,2), but not accompanied by the name column.