search for: trainingset

...39;m using the Iris data set to create a training set of 30 instances (10 per species) and use the remaining 120 instances as my test set. This is what I do: trainingIndices <- rep(1:10, 3) + rep(0:2, each=10) * 50 testIndices <- c(1:150)[-(trainingIndices)] testSet <- iris[testIndices,] trainingSet <- iris[trainingIndices,] t <- J48(trainingSet ~ ., data=trainingSet) So far, so good. I can even do predict(t, testSet). Now I want to get more detailed statistics about the performance of my classifier. evaluate_Weka_classifier(t, testSet) This is when I get Error in model.frame.default...

Hi, For another trainingset I get this error message, which again is rather cryptic to me: Setting default kernel parameters Error in array(0, c(n, p)) : 'dim' specifies too large an array RMate stopped at line 0 of selection Calls: rvm ... .local -> backsolve -> as.matrix -> chol -> diag -> array...

Hi all, After trial and error by myself for a few hours, I decide to ask for your help. I have a training set which is a matrix of size 200 x 2, where the two columns denote each independent variable. I have 200 observations. ----------------- ss=data.frame(trainingSet); result=lm(trainingClass~ss$X1+ss$X2); ----------------- where trainingClass denotes the true classes of the training data. Now I want to apply the model to predict new data: ----------------- > gg=predict(result, data.frame(X1=1, X2=2)) Warning message: 'newdata' had 1 rows but vari...

Hi, Could you please help me in solving the following error message: Error in `[.data.frame`(mytestdata, fp_temp == 1) : undefined columns selected when I use scan instead on read.table, I reicieve this message: Error in names(ret2) <- rowns : 'names' attribute [172] must be the same length as the vector [152] Many thanks, Nancy

...1048576.0, 2) gives 1023.48 If such an increase of memory can not succeed, perhaps the size of the rpart object could be reduced by not storing unnecessary information. The rpart function call is (the calls of FALSE is to try to reduce the size of the fit object): fit <- rpart(price ~ ., data = trainingset, control=rpart.control(maxdepth=2,cp=0.001),model=FALSE,x=FALSE,y=FALSE) This fit object is later called in 2 predict functions, for example: predict(fit,newdata=sample) Can anybody please help me by letting R use more memory (for example swap) or can anybody help me reducing the size of the fit...

Hi Abbas, Before I try to give you answers, I just want to mention that you should send R related reqests to the R-help list, and not me personally because (i) there's a greater likelihood that it will get answered in a timely manner, and (ii) people who might have a similar problem down the road might benefit from any answer via searching the list archives ... anyway: On Sep 5, 2009, at

Good day, I'm using a VAR model to forecast sales with some extra variables (google trends data). I have divided my dataset into a trainingset (weekly sales + vars in 2006 and 2007) and a holdout set (2008). It is unclear to me how I should predict the out-of-sample data, because using the predict() function in the vars package seems to estimate my google trends vars as well. However, I want to forecast the sales figures, with knowledge o...

Hi everyone, Can anyone please tell whether there is a difference between the code for using svm in regression and code for using svm in classification? This is my code for regression, should I change it to do classification?: train <- read.table("trainingset.txt",sep=";") test <- read.table("testset.txt",sep=";") svmmodelfitness <- function(myformula,mydata,mytestdata) { mymodel <- svm(myformula,data=mydata) mytest <- predict(mymodel, mytestdata) error <- myt...

...RMate stopped at line 0 of selection missing value where TRUE/FALSE needed Calls: rvm ... .local -> backsolve -> as.matrix -> chol -> diag -> array can someone explain this error message? It works for other data-sets with the same feature / example-space. But for one particular trainingset I get this error message... thanks!