similar to: what is Non-numeric argument to mathematical function in prediction ?

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 All, OK, in the "back to the drawing board" department, I found what looks like a much better solution to using R in Java. Renjin. Looking at the docs and then trying a quick example, didn't quite work. Of course I'm missing something. Although I'm telling the engine to require ("biotools") just like I would in R itself, when I get to the line of code that

It can't be this hard, right? I really need a shove in the right direction here. Been spinning wheels for three days. Cannot get past the errors. I'm doing something wrong, obviously, since I can easily compute the Box's M right there in RStudio But I don't see what is wrong below with the coding equivalent. The entire code snippet is below. The code fails below on the call to

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])";

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

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 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

Dear R users, I have a question on the confusion matrix generated by function randomForest. I used the entire data set to generate the forest, for example: > print(iris.rf) Call: randomForest(formula = Species ~ ., data = iris, importance = TRUE, keep.forest = TRUE) confusion setosa versicolor virginica class.error setosa 50 0 0 0.00

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

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

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 have written a code in R for classifying microarray data using naive bayes, the code is given below: library(e1071) train<-read.table("Z:/Documents/train.txt",header=T); test<-read.table("Z:/Documents/test.txt",header=T); cl <- c(c(rep("ALL",10), rep("AML",10))); cl <- factor(cl) model <- NaiveBayes(train,cl);

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) #>

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

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

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, 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