similar to: Error using e1071 svm: NA/NaN/Inf in foreign function call

I am trying to understand what is the difference between linear and polynomial kernel: linear: u'*v polynomial: (gamma*u'*v + coef0)^degree It would seem that polynomial kernel with gamma = 1; coef0 = 0 and degree = 1 should be identical to linear kernel, however it gives me significantly different results for very simple data set, with linear kernel

Dear list, In some places (for example, http://en.wikipedia.org/wiki/Support_vector_machine) , the polynomail kernel in SVM is written as (u'*v + 1)^d, while in the document of svm() in e1071 package, the polynomial kernel is written as (gamma*u'*v + coef0)^d. I am a little confused here: When doing parameter optimization (grid search or so) for polynomial kernel, does it need to tune

Hello, I am using SVM under e1071 package for nu-regression with 18 parameters. The variables are ordered factors, factors, date or numeric datatypes. I use the linear kernel. It gives the following error that I cannot solve. I tryed debug, browser and all that stuff, but no way. The error is: Error in get(ctr, mode = "function", envir = parent.frame())(levels(x), :

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Hi, If I try to read the codes of functions in e1071 package, it gives me following error message. >library(e1071) > svm function (x, ...) UseMethod("svm") <environment: namespace:e1071> > predict.svm Error: Object "predict.svm" not found > Can someone help me on this how to read the codes of the functions in the e1071 package? Thanks. Raj

In e1071 package/svm default epsilon value is set to 0.1 and not 0.5 as documentation says. R

Hi, I ran two svm models in R e1071 package: the first without cross-validation and the second with 10-fold cross-validation. I used the following syntax: #Model 1: Without cross-validation: > svm.model <- svm(Response ~ ., data=data.df, type="C-classification", > kernel="linear", cost=1) > predict <- fitted(svm.model) > cm <- table(predict,

Dear David, Dear Friends, After any running svm I receive different results of Error estimation of 'svm' using 10-fold cross validation. What is the reason ? It is caused by the algorithm, libsvm , e1071 or something els? Which value can be optimal one ? How much run can reach to the optimality.And finally, what is difference between Error estimation of svm using 10-fold cross validation

Hi, I am trying to use tune.svm in e1071 package. the command i use is tobj <- tune.svm(labels, data= data, cost = 10^(1:2)) Should the last column of the 'data' contain the labels as well? I want to use the linear kernel. But it gives me the error "Error in model.frame.default(formula, data) : 'data' must be a data.frame, not a matrix or an array" Do you know why

Hi, I have a question about the parameter C (cost) in svm function in e1071. I thought larger C is prone to overfitting than smaller C, and hence leads to more support vectors. However, using the Wisconsin breast cancer example on the link: http://planatscher.net/svmtut/svmtut.html I found that the largest cost have fewest support vectors, which is contrary to what I think. please see the scripts

Dear All , 1- I'm trying to access the values of fitted(model) after model<- tune.svm( ) but seemingly it is not poosible. How can I access to values of fitted ? However ,it is possible only after model<- svm( ) 2- How can I access to the other values such as the number of Support Vectors , gamma, cost , nu , epsilon , after model<- tune.svm( ) ? these are not possible? I

Hi all, I've got the following error message in using e1071 svm routine... Could anybody please help me? Thank you! --------------------------------- model <- svm(y=factor(mytraindata[, 1]), x=mytraindata[, -1], probability=T) Error in if (any(co)) { : missing value where TRUE/FALSE needed In addition: Warning message: In FUN(newX[, i], ...) : NAs introduced by coercion

Hello, I have an SVM model previously calibrated using libsvm R implementation from the e1071 package. I would like to use this SVM to predict values, from a Java program. I first tried to use jlibsvm and the "standard" java implementation of libsvm, without success. Thus, I am now considering writing data in files from my Java code, calling an R program to predict values, then gather

Hi. I'm trying to plot a pair of intertwined spirals and an svm that separates them. I'm having some trouble. Here's what I tried. > library(mlbench) > library(e1071) Loading required package: class > raw <- mlbench.spirals(200,2) > spiral <- data.frame(class=as.factor(raw$classes), x=raw$x[,1], y=raw$x[,2]) > m <- svm(class~., data=spiral) > plot(m,

Full_Name: Julien Gagneur Version: 2.2.1 OS: Linux (Suse 9.3) Submission from: (NULL) (194.94.44.4) Using the SparseM library (SparseM_0.66) and the e1071 library (e1071_1.5-12) I fail using svm method with a sparse matrix. Here is a sample example. I experienced the same problem under Windows. > library(SparseM) [1] "SparseM library loaded" > library("e1071")

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

Hi, Sorry that I have many questions today. I am using svm function on about 180,000 points of training set. It takes very long time to run. However, I would like it to spit out something to make sure that the run is not dead in between. Would you please suggest anyway to do so? And is there anyway to speed up the performance of this svm function? Thank you. - adschai

> 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,]

Dear all, I have a question about using categorical predictors for SVM, using "svm" from library(e1071). If I have multiple categorical predictors, should they just be included as factors? Take a simple artificial data example: x1<-rnorm(500) x2<-rnorm(500) #Categorical Predictor 1, with 5 levels x3<-as.factor(rep(c(1,2,3,4,5),c(50,150,130,70,100))) #Catgegorical Predictor

Hi, I run the following tuning function for svm. It's very strange that every time i run this function, the best.parameters give different values. [A] >svm.tune <- tune(svm, train.x, train.y, validation.x=train.x, validation.y=train.y, ranges = list(gamma = 2^(-1:2), cost = 2^(-3:2))) # where train.x and train.y are matrix