similar to: how to evaluate the significance of attributes in tree growing

FWIW, I wrote a little function to extract variable importance as defined in the CART book a while ago. It's rather limited: Only works for regression problem, and you need to set maxsurrogate=0 and maxcompete=0. It may (or may not) help you: varimp.rpart <- function(x) { dev <- x$frame[, c("var", "dev")] dev <- dev[dev$var != "<leaf>",

Hi, there: I spent some time on this but I think I really cannot figure it out, maybe I missed something here: my data looks like this: > dim(trn3) [1] 7361 209 > dim(val3) [1] 7427 209 > mg.rf2<-randomForest(x=trn3[,1:208], y=trn3[,209], data=trn3, xtest=val3[, 1:208], ytest=val3[,209], importance=T) my test data has 7427 observations but after prediction, > dim(mg.rf2$votes)

Hi, I am trying to print out my confusion matrix after having created my random forest. I have put in this command: fit<-randomForest(MMS_ENABLED_HANDSET~.,data=dat,ntree=500,mtry=14, na.action=na.omit,confusion=TRUE) but I can't get it to give me the confusion matrix, anyone know how this works? Thansk! Ruben [[alternative HTML version deleted]]

You need to give us more details, like how you call randomForest, versions of the package and R itself, etc. Also, see if this helps you: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/32918.html Andy From: Eleni Rapsomaniki > > Dear all, > > I am trying to train a randomForest using all my control data > (12,000 cases, ~ 20 explanatory variables, 2 classes). > Because

Dear Listers: I have a question on handling large dataset. I searched R-Search and I hope I can get more information as to my specific case. First, my dataset has 1.7 billion observations and 350 variables, among which, 300 are float and 50 are integers. My system has 8 G memory, 64bit CPU, linux box. (currently, we don't plan to buy more memory). > R.version _ platform

Hi there, How to solve the conflicts as to the same object between two packages, for example, like margins in both randomForest and supclust? When both libraries are installed, supclust will complain "margins" defined in randomForest. I can only solve it by re-starting R, which is very inconvenient, any clever way? Thanks, Weiwei -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc.

You could break your 3 class problem into several (2 or 3) 2 class problems, and then use Andy's suggestion (see the CART book). There are several ways to break the problem into 2 class problems, and several ways to combine the resulting classifiers. Tom Dietterich, Jerry Friedman, Trevor Hastie and Rob Tibshirani, among others, have articles on the question, in places like Annals of

Hello, I'm trying to create a series of randomForest objects, basically in a loop like this: forests <- list(); for (level in 1:10) { # do some other things here # create a random forest forest <- randomForest( x = x.level, y = z.level, ntree = trees ); forests <- c(forests, forest); } But instead of creating a list of 10 forests, this creates a list

hi, is there some good summary on clustering methods in R? It seems there are many packages involving it. And I have two questions on clustering here: 1. Is there a way of evaluate the effecitives (or seperation) of clustering (rather than by visualization)? 2. Is there a search method (like genetic search) which can help find the best subset of attributes which gives best seperation? Thanks,

> From: Weiwei Shi > > it works. > thanks, > > but: (just curious) > why i tried previously and i got > > > is.vector(sample.size) > [1] TRUE Because a list is also a vector: > a <- c(list(1), list(2)) > a [[1]] [1] 1 [[2]] [1] 2 > is.vector(a) [1] TRUE > is.numeric(a) [1] FALSE Actually, the way I initialize a list of known length is by

Hi, I searched the archives and did not find a good solution to that. assume I have 10 sets and I want to have the common character elements of them. how could i do that? -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. "Did you always know?" "No, I did not. But I believed..." ---Matrix III

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

Hi, there: I am wondering if I can find some detailed explanation on gbm or explanation on examples of gbm. thanks, Ed

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

Stock returns and other financial data have often found to be heavy-tailed. Even Cauchy distributions (without even a first absolute moment) have been entertained as models. Your qq function subtracts numbers on the scale of a normal (0,1) distribution from the input data. When the input data are scaled so that they are insignificant compared to 1, say, then you get essentially the

Hi, there: I am wondering if there is a quick way to "reverse" a list like this: t0 <- list(a=1, b=1, c=2, d=1) reverst t0 to t1 > t1 $`1` [1] "a" "b" "d" $`2` [1] "c" thanks. -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. "Did you always know?" "No, I did not. But I believed..." ---Matrix III

Hello, I'm trying to find out the optimal number of splits (mtry parameter) for a randomForest classification. The classification is binary and there are 32 explanatory variables (mostly factors with each up to 4 levels but also some numeric variables) and 575 cases. I've seen that although there are only 32 explanatory variables the best classification performance is reached when

hi, i have another question on tapply: i have a dataset z like this: 5540 389100307391 2600 5541 389100307391 2600 5542 389100307391 2600 5543 389100307391 2600 5544 389100307391 2600 5546 381300302513 NA 5547 387000307470 NA 5548 387000307470 NA 5549 387000307470 NA 5550 387000307470 NA 5551 387000307470 NA 5552 387000307470

Hi I have a data frame like this: V1 V2 V3 V4 Min. :0.01146 Min. :0.0006714 Min. :0.004912 Min. : 0 1st Qu.:0.03938 1st Qu.:0.0072805 1st Qu.:0.052719 1st Qu.:1150 Median :0.04224 Median :0.0077581 Median :0.056388 Median :1150 Mean :0.04010 Mean :0.0074669 Mean :0.052602 Mean :1173 3rd

Hi, there: After I upgraded my R to 2.4.1, it is my first time of trying to use MASS and found the following error message: > install.packages("MASS") --- Please select a CRAN mirror for use in this session --- trying URL 'http://cran.cnr.Berkeley.edu/bin/macosx/universal/contrib/2.4/VR_7.2-33.tgz' Content type 'application/x-gzip' length 995260 bytes opened URL