similar to: e1071 SVM one-classification tune problem

Dear R-list! I am using the e1071 package in R to solve a binary classification problem in a dataset of round 180 predictor variables (blood metabolites) of two groups of subjects (patients and healthy controls). I am confused regarding the correct way to assess the classification accuracy of the trained svm. (A) The svm command allows to specificy via the 'cross=k' parameter to specify a

Hello, When calling several times the svm() function, I get different results. Do I miss something, or is there some random generation in the C library? In this second hypothesis, is it possible to fix an eventual seed? Thank you Pierre ### Example library('e1071') x = rnorm(100) # train set y = rnorm(100) c = runif(100)>0.5 x2 = rnorm(100)# test set y2 = rnorm(100) # learning a

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, 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

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), :

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

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

1. This is _not_ a bug in R itself. Please don't use R's bug reporting system for contributed packages. 2. This is _not_ a bug in svm() in `e1071'. I believe you forgot to take sqrt. 3. You really should use the `tot.MSE' component rather than the mean of the `MSE' component, but this is only a very small difference. So, instead of spread[i] <- mean(mysvm$MSE), you

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

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

Full_Name: Noel O'Boyle Version: 2.1.0 OS: Debian GNU/Linux Sarge Submission from: (NULL) (131.111.8.96) (1) Description of error The 10-fold CV option for the svm function in e1071 appears to give incorrect results for the rmse. The example code in (3) uses the example regression data in the svm documentation. The rmse for internal prediction is 0.24. It is expected the 10-fold CV rmse

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,

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,

I am accustomed to the LIBSVM package, which provides cross-validation on training with the -v option % svm-train -v 5 ... This does 5 fold cross validation while building the model and avoids over-fitting. But I don't see how to accomplish that in the e1071 package. (I learned that svm(... cross=5 ...) only _tests_ using cross-validation -- it doesn't affect the training.) Can

Dear all, I got a code for e1071 package in R for SVM regression. I have used *m$coefs* for extracting the coefficients but I am getting only 72 . How can I extract coefficients of the predictors set? Does it mean that I will get only 72 as *Number of Support Vectors: 72. * ** Thanks in advance Code: -------------- library(e1071) > # create data > x <- seq(0.1, 5,

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,

Dear all, I executed svm calculation using e1071 library with a microarray data (http://www.iu.a.u-tokyo.ac.jp/~kadota/R/data_Singh_RMA_3274.txt). Then, I shuffled the data samples and executed svm calculation again. The results of 2 calculation were different (in SV, coefs and weights). I attached the script below. Could please tell me why this happens? If possible please tell me how to make

This could be a bug in the e1071 svm code, but maybe not -- I guess I'll send it here anyway. It's reproducible. > x <- seq (0.1,5,by=0.05) > y <- log(x) + rnorm (x, sd=0.2) > library(e1071) > m <- svm (x,y) Process R segmentation fault at Tue Nov 20 23:34:19 2001 > version _ platform i686-pc-linux-gnu arch i686 os

Hi All, I'm attempting to use eps-regression or nu-regression SVM to compute probabilities but the predict function applied to an svm model object returns values outside [0, 1]: Variable Data looks like: Present X02 X03 X05 X06 X07 X13 X14 X15 X18 1 0 1634 48 2245.469 -1122.0750 3367.544 11105.013 2017.306 40 23227 2 0 1402 40 2611.519 -811.2500 3422.769 10499.425 1800.475 40 13822 3 0 1379

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