similar to: Question about randomForest

Hi Sacha, On second thought, perhaps this is more the direction that you want ... X2 = cbind(X_train,y_train) colnames(X2)[3] = "y" regr2<-randomForest(y~x1+x2, data=X2,maxnodes=10, ntree=10) regr regr2 #Make prediction predictions= predict(regr, X_test) predictions2= predict(regr2, X_test) HTH, Eric On Tue, May 9, 2023 at 6:40?AM Eric Berger <ericjberger at gmail.com>

Gracias Carlos. No uso caret, pero lo miraré. Quoting Carlos Ortega <cof en qualityexcellence.es>: > Hola, > > Creo que si utilizas "caret" y en la función "trainControl()" defines "oob" > como criterio de randomización, puedes luego recuperar del objeto del > modelo, las predicciones individuales... > > Saludos, > Carlos Ortega >

I have a question on the output generated by randomForest in classification mode, specifically, the confusion matrix. The confusion matrix lists the various classes and how the forest classified each one, plus the classification error. Are these numbers essentially averages over all the trees in the forest? If so, is there a way I can get the standard deviation values out of the randomForest,

Hello, When I plot or look at the error rate vector for a random forest (rf$err.rate) it looks like a descending function except for a few first points of the vector with error rates values lower(sometimes much lower) than the general level of error rates for a forest with such number of trees when the error rates stop descending. Does it mean that there is a tree(s) (that is built the first in

Dear Random Forests gurus, I have a question about R^2 provided by randomForest (for regression). I don't succeed in finding this information. In the help file for randomForest under "Value" it says: rsq: (regression only) - "pseudo R-squared'': 1 - mse / Var(y). Could someone please explain in somewhat more detail how exactly R^2 is calculated? Is "mse"

Dear R-users, While making a prediction using the randomForest function (package randomForest) I'm getting the following error message: "Error in predict.randomForest(model, newdata = CV) : No forest component in the object" Here's my complete code. For reproducing this task, please find my 2 data sets attached ( http://www.nabble.com/file/p17855119/data.rar data.rar ).

Dear All, I have been using the randomForest package for a couple of difficult prediction problems (which also share p >> n). The performance is good, but since all the variables in the data set are used, interpretation of what is going on is not easy, even after looking at variable importance as produced by the randomForest run. I have tried a simple "variable selection"

Hi all, I've been using a custom summary function to optimise regression model methods using the caret package. This has worked smoothly. I've been using the default bootstrapping resampling method. For bagging models (specifically randomForest in this case) caret can, in theory, uses the out-of-bag (oob) error estimate from the model instead of resampling, which (in theory) is largely

I would like to ask a general question about the randomForest predict function and how it handles No Data values. I understand that you can omit No Data values while developing the randomForest object, but how does it handle No Data in the prediction phase? I would like the output to be NA if any (not just all) of the input data have an NA value. It is not clear to me if this is the default or

Just out of curiosity, I took the default "iris" example in the RF helpfile... but seeing the admonition against using the formula interface for large data sets, I wanted to play around a bit to see how the various options affected the output. Found something interesting I couldn't find documentation for... Just like the example... > set.seed(12) # to be sure I have

Hi I am new to R, randomForest and I have read about how to use it in your old mails. I have also run the predictions examples from CRAN. But I still don't understand how to use it right. The thing that I don't understand is how to run the result from the randomForest on one line (post) with newdata to get a good guess. What I mean is if I put in a new observation of iris how do I

I've built two RF objects (RF1 and RF2) and have tried to combine them, but I get the following error: Error in rf$votes + ifelse(is.na(rflist[[i]]$votes), 0, rflist[[i]]$votes) : non-conformable arrays In addition: Warning message: In rf$oob.times + rflist[[i]]$oob.times : longer object length is not a multiple of shorter object length Both RF models use the same variables, although

I have a data.frame with a series of variables tagged to a binary response ('present'/'absent'). I am trying to use randomForest to predict present/absent in a second dataset. After a lot a fiddling (using two data frames, making sure data types are the same, lots of testing with data that works such as data(iris)) I've settled on combining all my data into one data.frame

Dear mailinglist members, I was wondering if there was a way to re-evaluate the instances of a tree (in the forest) again after I have manually changed a splitpoint (or split variable) of a decision node. Here's an illustration: library("randomForest") forest.rf <- randomForest(formula = Species ~ ., data = iris, do.trace = TRUE, ntree = 3, mtry = 2, norm.votes = FALSE) # I am

Hello R Gurus, I am perplexed by the different results I obtained when I ran code like this: set.seed(100) test1<-randomForest(BinaryY~., data=Xvars, trees=51, mtry=5, seed=200) predict(test1, newdata=cbind(NewBinaryY, NewXs), type="response") and this code: set.seed(100) test2<-randomForest(BinaryY~., data=Xvars, trees=51, mtry=5, seed=200, xtest=NewXs, ytest=NewBinarY) The

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

Muy buenas, ¿sabe alguien cómo obtener las predicciones sobre el out of bag que hace randomForest? Manuel . -- Dr Manuel Mendoza Department of Biogeography and Global Change National Museum of Natural History (MNCN) Spanish Scientific Council (CSIC) C/ Serrano 115bis, 28006 MADRID Spain

Hi all, When running a randomForest run using the following command: forestplas=randomForest(Prev~.,data=plas,ntree=200000) print(forestplas) I get the following result: Call: randomForest(formula = Prev ~ ., data = plas, ntree = 2e+05, importance = TRUE) Type of random forest: regression Number of trees: 2e+05 No. of variables tried at each split: 5

Greetings. I'm trying to determine whether to use rpart or randomForest for a classification tree. Has anybody tested efficacy formally? I've run both and the confusion matrix for rf beats rpart. I've looking at the rf help page and am unable to figure out how to extract the tree. But more than that I'm looking for a more comprehensive user's guide for randomForest including

Hello everybody, I'm testing the randomForest package in order to do some simulations and I get some trouble with the prediction of new values. The random forest computation is fine but each time I try to predict values with the newly created object, I get an error message. I thought I was because NA values in the dataframe, but I cleaned them and still got the same error. What am I