similar to: How to use tune.svm or tune(svm...) for regression

Hello, I hope you can help me! I`m trying to tune svm parameters: cost and gamma for a landsat image classification, but I get an error and I can't understand it. I write this: > tune(svm, Class~., data = mdt01bis, ranges = list(gamma = 2^(-15:3), cost > = 2^(-5:15))) and R gives: Error en predict.svm(model, if (!is.null(validation.x)) validation.x else if (useFormula)

Hi R users, I'm trying to run a SVM - regression using e1071 package but the function svm() all the time apply a classification method rather than a regression. svm.m1 <- svm(st ~ ., data = train, cost = 1000, gamma = 1e-03) Parameters: SVM-Type: C-classification SVM-Kernel: radial cost: 1000 gamma: 0.001 Number of Support Vectors: 209

Dear Sir, I got a problem with my program. I would like to classify my data using bagging support vector machine ensemble. I split my data into training data and test data. For a given data sets TR(X), K replicated training data sets are first randomly generated by bootstrapping technique with replacement. Next, Support Vector Mechine (SVM) is applied for each bootstrap data sets. Finally, the

Hello, if I try to tune my svm with the code: Tune <- tune.svm(Data.Train, Class.Train, type="C-classification", kernel="radial", gamma = 2^(-1:1), cost = 2^(2:4)) i get a windows Messagebox with a error in the application "Rgui.exe" and the message: "Die Anweisung in 0x6c48174d verweist auf Speicher 0x00000000. Der Vorgang "read" konnte nicht auf

Hi all - (Re my previous question to the list, I managed to generate an ROC plot for my SVM by ranking the data using the Decision.Values property. Thanks for your responses) I have now started tuning the SVM to get optimal parameters for the RBF kernel and I ran into a problem. Whatever parameter ranges I specify, I always get the same error values for all combinations of parameters it

Hello all, I am using the e1071 SVM with the tune options for classification, which work pretty well, given the examples of using tune.svm function for classification. But I have not found any example to tune the SVM novelty detection (one-classification) parameters (gamma, cost, nu), for example this are some of the options I have tried with no success: obj<-tune(svm, x,y, type

Hi I am doing this sort of thing: POLY: > > obj = best.tune(svm, similarity ~., data = training, kernel = "polynomial") > summary(obj) Call: best.tune(svm, similarity ~ ., data = training, kernel = "polynomial") Parameters: SVM-Type: eps-regression SVM-Kernel: polynomial cost: 1 degree: 3 gamma: 0.04545455 coef.0: 0

Hi all, Sorry to be bothering again with probably an easy error to fix, but I've been trying to solve the problem and haven't been able yet to do it. So I'm doing this: > dados<-read.table("b.txt",sep="",nrows=30000) >

HI, GUYS, I used the following codes to run SVM and get prediction on new data set hh. dim(all_h) [1] 2034 24 dim(hh) # it contains all the variables besides the variables in all_h data set. [1] 640 415 require(e1071) svm.tune<-tune(svm, as.factor(out) ~ ., data=all_h, ranges=list(gamma=2^(-5:5), cost=2^(-5:5)))# find the best parameters. bestg<-svm.tune$best.parameters[[1]]

Dear list I've a generic question about how to tune an SVM I'm trying to classify with caret package some population data from a case-control study . In each column of my matrix there are the SNP genotypes , in each row there are the individuals. I correctly splitted my total dataset in training(132 individuals) and test (50 individuals) (respecting the total observed genotypic

Hi all, I don't know how to choose the formula to use when plotting an svm model, I think I'm using the wrong one and so that is why I'm having trouble. I should be very grateful if someone could help me on this.. > dados<-read.table("b.txt",sep="",nrows=30000) >

Hi all - I am trying to tune an SVM model by optimizing the cross-validation accuracy. Maximizing this value doesn't necessarily seem to minimize the number of misclassifications. Can anyone tell me how the cross-validation accuracy is defined? In the output below, for example, cross-validation accuracy is 92.2%, while the number of correctly classified samples is (1476+170)/(1476+170+4) =

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 expected, that I will get the same prediction, if I multiply the weights for all classes with a constant factor, but I got different results. Please look for the following code. > library(e1071) > data(Glass, package = "mlbench") > index <- 1:nrow(Glass) > testindex <- sample(index, trunc(length(index)/5)) > testset <- Glass[testindex, ] > trainset <-

Hi, I have some difficulties in interpreting the prediction of a svm model using the package e1071. y1 is the variable I want to predict. It is of type factor and has got two levels: "< 50%" and "> 50%". z is the dataset. > model <- svm(y1 ~ ., data = z,type="C-classification", cross=10) > model Call: svm(formula = y1 ~ ., data = z, type =

svm.fit<-svm(as.factor(out) ~ ., data=all_h, method="C-classification", kernel="radial", cost=bestc, gamma=bestg, cross=10) # model fitting svm.pred<-predict(svm.fit, hh, decision.values = TRUE, probability = TRUE) # find the probability, but can not find. attr(svm.pred, "probabilities") > attr(svm.pred, "probabilities") 1 0 1 0 0 2 0

I run the following code, all other is ok, but plot(m.svm,p5.new,As~Cur) is not ok Anyone know why? install.packages("e1071") library(e1071) library(MASS) p5 <- read.csv("http://www.public.iastate.edu/~aiminy/data/p_5_2.csv") p5.new<-subset(p5,select=-Ms) p5.new$Y<-factor(p5.new$Y) levels(p5.new$Y) <- list(Out=c(1), In=c(0)) attach(p5.new)

Hello all, this is the first time I'm using R and e1071 package and SVM multiclass (and I'm not a statistician)! I'm very confused, then. The goal is: I have a sentence with sunny; it will be classified as "yes" sentence; I have a sentence with cloud, it will be classified as "maybe"; I have a sentence with rainy il will be classified as "no". The

> x <- train[,c( 2:18, 20:21, 24, 27:31)] > y <- train$out > > svm.pr <- svm(x, y, probability = TRUE, method="C-classification", kernel="radial", cost=bestc, gamma=bestg, cross=10) > > pred <- predict(svm.pr, valid[,c( 2:18, 20:21, 24, 27:31)], decision.values = TRUE, probability = TRUE) > attr(pred, "decision.values")[1:4,]

I run the following code, all other is ok, but plot(m.svm,p5.new,As~Cur) is not ok Anyone know why? install.packages("e1071") library(e1071) library(MASS) p5 <- read.csv("http://www.public.iastate.edu/~aiminy/data/p_5_2.csv") p5.new<-subset(p5,select=-Ms) p5.new$Y<-factor(p5.new$Y) levels(p5.new$Y) <- list(Out=c(1), In=c(0)) attach(p5.new)