similar to: Random Forest confusion matrix

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

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,

Hi, I need help with the randomForest prediction. i run the folowing code: > iris.rf <- randomForest(Species ~ ., data=iris, > importance=TRUE,keep.forest=TRUE, proximity=TRUE) > pr<-predict(iris.rf,iris,predict.all=T) > iris.rf$votes[53,] setosa versicolor virginica 0.0000000 0.8074866 0.1925134 > table(pr$individual[53,])/500 versicolor virginica 0.928

Hi, Since I've had no replies on my previous post about my problem I am posting it again in the hope someone notice it. The problem is that the randomForest function doesn't take datasets which has instances only containing a subset of all the classes. So the dataset with instances that either belong to class "a" or "b" from the levels "a", "b" and

Thanks Duncan. Awesome ideas! I think we're getting closer! I tried what you suggested and got a possibly better error... . . . rConnection.assign("boxMVariable", myDf); String resultBV = "str(boxMVariable)"; // your suggestion. RESULTING ERROR: Error in format.default(nam.ob, width = max(ncn), justify = "left") : invalid 'width' argument (No idea

Hey Duncan, Hard to debug? That's an understatement. Eyes bleeding.... In any case, I tried all your suggestions. To get "integer" for the final column, I had to change the code to get integers instead of strings. double[] d1 = ((REXPVector) ((RList) tableRead).get(0)).asDoubles(); double[] d2 = ((REXPVector) ((RList) tableRead).get(1)).asDoubles(); double[] d3 = ((REXPVector)

Thanks Duncan. I can't tell you how helpful all your terrific replies have been. I think the biggest surprise is that nobody appears to be using Java and R together like I"m trying to do. I suppose it should be a surprise since there are no books on the subject and almost no technical documentation other than a few sites here and there. ----- I originally had the "int" as the

Hi there: I have a question about generating mean value of a data.frame. Take iris data for example, if I have a data.frame looking like the following: --------------------- Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 5.1 3.5 1.4 0.2 setosa 2 4.9 3.0 1.4 0.2

If do: > library("e1071") > example(svm) I get: svm> data(iris) svm> attach(iris) svm> ## classification mode svm> # default with factor response: svm> model <- svm(Species ~ ., data = iris) svm> # alternatively the traditional interface: svm> x <- subset(iris, select = -Species) svm> y <- Species svm> model <- svm(x, y) svm>

I'm confused, hope someone can point out what is not obvious to me. I thought I was creating a new data frame by 'deleting' rows from an existing dataframe - I've tried 2 methods. But this new data frame seems to remember values from its parent - even though there are no occurences. Where does it get the values versicolor and virginica from and give then a count of 0? What

I'm not sure what you mean. Could you please be more specific? If I print the string, I get: boxM(boxMVariable[, -5], boxMVariable[, 5]) From this code: . . . // assign the data to a variable.rConnection.assign("boxMVariable", myDf); // create a string command with that variable name.String boxVariable = "boxM(boxMVariable[, -5], boxMVariable[, 5])";

On Fri, Mar 23, 2018 at 6:43 PM, Rui Barradas <ruipbarradas at sapo.pt> wrote: > Hello, > > Not exactly an answer but here it goes. > If you use the formula interface the names will be retained. Also if you pass named arguments: aggregate(iris["Sepal.Length"], by = iris["Species"], FUN = foo) # Species Sepal.Length # 1 setosa 5.006 # 2

Hi, I was wondering if it is possible to get the rowindices without using the function "which" because I don't have a restriction criteria. Here's an example of what I mean: # take 10 randomly selected instances iris[sample(1:nrow(iris), 10),] # output Sepal.Length Sepal.Width Petal.Length Petal.Width Species 76 6.6 3.0 4.4 1.4

In the examples below, the first loses the name attached by foo(), the second retains names attached by bar(). Is this an intentional difference? I?d prefer that the names be retained in both cases. foo <- function(x) { c(mean = base::mean(x)) } bar <- function(x) { c(mean = base::mean(x), sd = stats::sd(x))} aggregate(iris$Sepal.Length, by = list(iris$Species), FUN = foo) #>

Hi, I tried to use naivebayes in package 'e1071'. when I use following parameter, only one predictor, there is an error. > m<- naiveBayes(iris[,1], iris[,5]) > table(predict(m, iris[,1]), iris[,5]) Error in log(sapply(attribs, function(v) { : Non-numeric argument to mathematical function However, when I use two predictors, there is not error any more. > m<-

Hi, I tried to use naivebayes in package 'e1071'. when I use following parameter, only one predictor, there is an error. > m <- naiveBayes(iris[,1], iris[,5]) > table(predict(m, iris[,1]), iris[,5]) Error in log(sapply(attribs, function(v) { : Non-numeric argument to mathematical function However, when I use two predictors, there is not error any more. > m <-

Greetings R experts, I'm having some difficulty recovering lda objects that I've saved within sublists using the save.image() function. I am running a script that exports a variety of different information as a list, included within that list is an lda object. I then take that list and create a list of that with all the different replications I've run. Unfortunately I've been

Hello all, Ive been playing with nnet (package 'nnet') and Ive come across this problem. nnet doesnt seems to like to have more than 1000 weights. If I do: > data(iris) > names(iris)[5] <- "species" > net <- nnet(species ~ ., data=iris, size=124, maxit=10) # weights: 995 initial value 309.342009 iter 10 value 21.668435 final value 21.668435 stopped after 10

Hi, I have the following problem (library Hmisc loaded, iris data loaded, R Version 1.7.0 (2003-04-16), packages updated, running on a linux Debian i386): > summary(Species~Sepal.Length,method="reverse")->a > a Descriptive Statistics by Species +------------+-----------------+-----------------+-----------------+ | |setosa |versicolor |virginica

Hopefully I will not be flamed for this on the list, but I am starting out with R and having some trouble with combining plots. I am playing with the famous iris dataset (checking out example dataset in R while reading through Introduction to datamining) What I would like to do is create three graphs (combined boxplots) besides each other for each of the three species (Setosa, Versicolour and