search for: ytest

I'm puzzled by as.matrix. It appears to work differently for Surv objects. Here is a session from my computer: tmt% R --vanilla > library(survival) Loading required package: splines > ytest <- Surv(1:3, c(1,0,1)) > is.matrix(ytest) >[1] TRUE > attr(ytest, 'type') [1] "right" > attr(as.matrix(ytest), 'type') [1] "right" > > y2 <- ytest > class(y2) <- "charlie" > as.matrix.charlie <- survival:::as...

Dear R-help, I was bitten by the behavior of all() when given logical(0): It is TRUE! (And any(logical(0)) is FALSE.) Wouldn't it be better to return logical(0) in both cases? The problem surfaced because some un-named individual called randomForest(x, y, xtest, ytest,...), and gave y as a two-level factor, but ytest as just numeric vector. I thought I check for that in my code by testing for if (!all(levels(y) == levels(ytest))) stop(...) but levels() on a non-factor returns NULL, and the comparison ended up being logical(0). Since all(logical(0)) is TRUE,...

...y help! - Jim CT <- read.table("CT.txt",header=TRUE,sep="\t") CV <- read.table("CV.txt",header=TRUE,sep="\t") # Both CT & CV have the syntaxis X1, X2,...,X97,Y where all variables are numeric x <- CT[,-98] y <- CT[,98] xtest <- CV[,-98] ytest <- CV[,98] library(randomForest) model <- randomForest(x ,y , xtest, ytest,ntree=500,mtry=32,nodesize=5,nPerm=2) model #Call: # randomForest(x = x, y = y, xtest = xtest, ytest = ytest, ntree = 500, mtry = 32, nodesize = 5, # nPerm = 2) # Type of random forest: regression...

Hi all, Can some one tell me the difference between the following two formulas? 1. epiG.rf <-randomForest(gamma~.,data=data, na.action = na.fail,ntree = 300,xtest = NULL, ytest = NULL,replace = T, proximity =F) 2.epiG.rf <-randomForest(gamma~.,data=data, na.action = na.fail,ntree = 300,xtest = NULL, ytest = NULL,replace = T, proximity =F) [[alternative HTML version deleted]]

The standard errors produced by predict.glm with type = "response" seem wrong. Here is an example using R 1.2 windows version along with the same problem in Splus. The standard errors for type = "link" are the same in both systems. R1.2> set.seed(10) R1.2> ytest <- 100*.95^(0:9) + rnorm(10,sd = 5) R1.2> ytest [1] 103.96964 97.60590 88.43220 85.90504 79.18262 76.05762 68.34566 [8] 74.24119 66.80257 62.95880 R1.2> foo <- glm(ytest~I(0:9),family=quasi(link=log)) R1.2> predict(foo,type="link") 1 2 3...

...45,0.1,0.45) would result in fewer cases classified as class 2. Did I understand something wrong? Christian x1rf <- randomForest(x=as.data.frame(mfilters[cvtrain,]), y=as.factor(traingroups), xtest=as.data.frame(mfilters[cvtest,]), ytest=as.factor(testgroups)) > x1rf$test$confusion 1 2 3 class.error 1 9954 30 19 0.00489853 2 139 1854 0 0.06974410 3 420 0 84 0.83333333 x1rf <- randomForest(x=as.data.frame(mfilters[cvtrain,]), y=as.factor(traingroups), xtest=as.data....

Sir, This query is related to randomForest regression using R. I have a dataset called qsar.arff which I use as my training set and then I run the following function - rf=randomForest(x=train,y=trainy,xtest=train,ytest=trainy,ntree=500) where train is a matrix of predictors without the column to be predicted(the target column), trainy is the target column.I feed the same data for xtest and ytest too as shown. On verifying I found, rf$mse[500] and rf$test$mse[500] are different(the r-squares are also dif...

..."datasets.RData") # import traindat and testdat > nlevels(traindat$predictor1) [1] 20 > nlevels(testdat$predictor1) [1] 19 > nrow(traindat) [1] 9838 > nrow(testdat) [1] 3841 > set.seed(10) > rf_orig <- randomForest(x=traindat[,-1], y=traindat[,1], xtest=testdat[,-1], ytest=testdat[,1],ntree=100) > data.frame(rf_orig$test$err.rate)[100,1] # Error on test-dataset [1] 0.3082531 # assign the levels of the training dataset th the test dataset for predictor 1 > levels(testdat$predictor1) <- levels(traindat$predictor1) > nlevels(traindat$predictor1) [1]...

...a similar dataset to test against > length(xcoords) [1] 26257 > length(ycoords) [1] 26257 > mean(xcoords) [1] 670462.4 > mean(ycoords) [1] 5005382 > sd(xcoords) [1] 149.3114 > sd(ycoords) [1] 181.5950 #generate the test data > xtest<-rnorm(26257,670462.4,149.3) > ytest<-rnorm(26257,5005382,181.60) # try ashape routine with success > alpha.shape<-ashape(xtest,ytest,15) > class(alpha.shape) [1] "ashape" Thanks for any insight into this! Murray ps I am able to compute the alpha shapes for this same dataset without problem using CGAL but...

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

...an 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 confusion matrices for the two forests I thought would be the same by virtue of the same seed settings, but they differ as do the predicted values as well as the votes. At first I thought it was just the way ties were broken, so I changed the number of trees to an odd number so the...

...orest) set.seed(2863) x<-matrix(runif(1000),ncol=10) colnames(x)<-1:10 beta<-matrix(c(1,2,3,4,5,0,0,0,0,0),ncol=1) y<-drop(x %*% beta + rnorm(100)) newx<-matrix(runif(1000),ncol=10) newy<-drop(newx %*% beta + rnorm(100)) set.seed(2863) rf.fit <- randomForest(x=x,y=y,xtest=newx,ytest=newy,importance=F) print(rf.fit$test$mse[500]) set.seed(2863) rf.fit <- randomForest(x=x,y=y,xtest=newx,ytest=newy,importance=T) print(rf.fit$test$mse[500])

...(1000),ncol=10) > colnames(x)<-1:10 > beta<-matrix(c(1,2,3,4,5,0,0,0,0,0),ncol=1) > y<-drop(x %*% beta + rnorm(100)) > newx<-matrix(runif(1000),ncol=10) > newy<-drop(newx %*% beta + rnorm(100)) > > set.seed(2863) > rf.fit <- randomForest(x=x,y=y,xtest=newx,ytest=newy,importance=F) > print(rf.fit$test$mse[500]) > > set.seed(2863) > rf.fit <- randomForest(x=x,y=y,xtest=newx,ytest=newy,importance=T) > print(rf.fit$test$mse[500]) > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > http...

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) [1] 7361 2 which has the same length as my training data. but if I use mg.rf<-randomForest(x=trn3[,1:208], y=trn3[,209], data=trn3, importance=T) followed by > mg.pred<-predict(mg.r...

Hello, does any one know how to interpret this output in R? > Classification with logitboost > fit <- logitboost(xlearn, ylearn, xtest, presel=50, mfinal=20) > summarize(fit, ytest) Minimal mcr: 0 achieved after 6 boosting step(s) Fixed mcr: 0 achieved after 20 boosting step(s) What is "mcr" mean? Thanks [[alternative HTML version deleted]]

...ed the different values into the matrices to use. Please refer to the code below. I want to use 'xtrain' and 'ytrain' to train the data (the 60% of the observations) and I want to simulate the NN with 'xtest' and then compare the predicted Y values from the NN with the 'ytest' to get a value of MSE. ***************************************************************************** data<-read.table(file="C:/data.dat",sep=",") headings<- c("Class","x1","x2","x3","x4","x5","x6",...

...er, after running superpc.predict.red, I do not find this value in the output. The same is true when I run the example script provided in the documentation, as below: library(superpc) set.seed(332) #generate some data x<-matrix(rnorm(1000*40),ncol=40) y<-10+svd(x[1:60,])$v[,1]+ .1*rnorm(40) ytest<-10+svd(x[1:60,])$v[,1]+ .1*rnorm(40) censoring.status<- sample(c(rep(1,30),rep(0,10))) censoring.status.test<- sample(c(rep(1,30),rep(0,10))) featurenames <- paste("feature",as.character(1:1000),sep="") data<-list(x=x,y=y, censoring.status=censoring.status, feat...

...: \item{err.rate}{final error rate of the prediction on the input data.} ... For regression problem, the following are included: \item{mse}{mean square error: sum of squared residuals divided by \code{n}.} ... If test set is given (through the \code{xtest} or additionally \code{ytest} arguments), then there is also a \code{test} component, which is a list with the following components: \item{predicted}{predicted classes/values for the test set.} ... Note: The \code{forest} structure is slightly different between classification and regression. =============== In...

...iors and size of training data. Because I was expecting to see little difference in the perfomance of these methods I am worried that I may have made a mistake in my randomForest call: my.rf=randomForest(x=train.df[,-response_index], y=train.df[,response_index], xtest=test.df[,-response_index], ytest=test.df[,response_index], importance=TRUE,proximity=FALSE, keep.forest=FALSE) (where train.df and test.df are my train and test data.frames and response_index is the column number specifiying the class) My main question is: could there be a legitimate reason why random forest would outperform the...

...rainInd <- sample(SampInd, nSample*(CVtype-1)/CVtype ) TestInd <- SampInd[ which(!(SampInd %in% TrainInd ))] } } nTests <- nTests + length(TestInd) ## in each level, train a SVM model and record test error xTrain <- x[TrainInd, ] yTrain <- y[TrainInd] xTest <- x[TestInd,] yTest <- y[TestInd] ## index of the genes used in the SelInd <- seq(1, nGene) for( gLevel in 1:length(ladder) ) { ## record the genes selected in this ladder SelFreq[SelInd, gLevel] <- SelFreq[SelInd, gLevel] +1 ## train SVM model and test error svmres <- svm(xTrain[, SelInd], yTrain...