similar to: Strange question/result about SVM

Hi This is a question regarding classification performance using different methods. So far I've tried NaiveBayes (klaR package), svm (e1071) package and randomForest (randomForest). What has puzzled me is that randomForest seems to perform far better (32% classification error) than svm and NaiveBayes, which have similar classification errors (45%, 48% respectively). A similar difference in

I have a data set with about 30,000 training cases and 103 variable. I've trained an SVM (using the e1071 package) for a binary classifier {0,1}. The accuracy isn't great. I used a grid search over the C and G parameters with an RBF kernel to find the best settings. I remember that for least squares, R has a nice stepwise function that will try combining subsets of variables to find

Dear Community! I'm using the svm method of package e1071 for classifying my data. This really works fine, but however I have to work after creating the support vectors and the parameters with unscaled data. So the problem is when I try to train the classifier with the option "scale=F" the result is quite poor, so training with scaled data is essential. The rescaling of the support

Hello, I'm using the e1071 library for SVM functions. I can quickly train an SVM with: svm(formula = label ~ ., data = testdata) That works well. I want to tune the parameters, so I tried: tune.svm(label ~ ., data=testdata[1:2000, ], gamma=10^(-6:3), cost=10^(1:2)) THIS FAILS WITH AN ERROR: 'names' attribute [199] must be the same length as the vector [184] I don't

Hello! We are new in using R. We use the SVM module from the library ''e1071'' for training. Problem formulation: a classification has been performed using SVM module (linear kernel). Later, a new data set (test set) comparable to the training data shall be scaled in the same way as the training set (using the same scaling parameter set, but without using the SVM again

Hi, The answers to my previous question about nominal variables has lead me to a more important question. What is the "best practice" way to feed nominal variable to an SVM. For example: color = ("red, "blue", "green") I could translate that into an index so I wind up with color= (1,2,3) But my concern is that the SVM will now think that the values are

Hello. I'm new to R and I'm trying to solve a classification problem. I have a training dataset of about 40,000 rows and 50 columns. When I try to train support vector machine, it gives me this error after a few seconds: Error in predict.svm(ret, xhold) : Model is empty! This is the code I use: ne_span_data <- as.matrix(read.table('ne_span.data.R.txt', header=TRUE,

I am using the "predict" function on a support vector machine (svm) object, and I don't understand why I can't predict on a dataset with more observations than the training dataset. I think this problem is a generic "predict" problem, but I'm not sure. The original svm was fit on 50 observations.

Hi Abbas, Before I try to give you answers, I just want to mention that you should send R related reqests to the R-help list, and not me personally because (i) there's a greater likelihood that it will get answered in a timely manner, and (ii) people who might have a similar problem down the road might benefit from any answer via searching the list archives ... anyway: On Sep 5, 2009, at

Hello, I'm using the e1071 package for training an SVM. It seems to be working well. This question has two parts: 1) Once I've trained an SVM model, I want to USE it within R at a later date to predict various new data. I see the write.svm command, but don't know how to LOAD the model back in so that I can use it tomorrow. How can I do this? 2) I would like to add the

I am using SVM to classify categorical data and I would like the probabilities instead of the classification. ?predict.svm says that its only enabled when you train the model with it enabled, so I did that, but it didn't work. I can't even get it to work with iris. The help file shows that probability = TRUE when training the model, but doesn't show an example. Then I try to

I want to train svm models on increasingly large training data subsets of some zrr as follows: > m <- sapply(1:5,function(i) svm(person_oid~.,data=zrr[1:100*i,])) # (*) However, when I inspect m[1], it literally shows > m[1] [[1]] svm(formula = person_oid ~ ., data = zrr[1:N, ]) -- as opposed to > m1 <- svm(person_oid~.,data=zrr[1:100,]) > m1 > m1 Call:

Hi all, I've encountered an issue using svm (e1071) in the specific case of supplying new data which may not have the full range of levels that were present in the training data. I've constructed this really primitive example to illustrate the point: > library(e1071) > training.data <- data.frame(x = c("yellow","red","yellow","red"), a =

Hi, I am studying using SVM functions of e1071 package to do prediction, and I found during the training data are "factor" type, then svm.predict() can predict data directly by categories; but if response variables are "numerical", the predicted value from svm will be continuous quantitative numbers, then how can I connect these quantitative numbers to categories? (for

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 - 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) =

Dear R:ers, I'm using the svm from the e1071 package to train a model with the option "probabilities = TRUE". I then use "predict" with "probabilities = TRUE" and get the probabilities for the data point belonging to either class. So far all is well. My question is why I get different results each time I train the model, although I use exactly the same data.

Hi I am Aadhithya I am trying to write a code to classify microarray data (AML and ALL) using SVM in R my code goes like this : 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))); model<- svm(train,cl); pred <-

Dear all, I am exploring the possibilities for automated classification of my data. I have successfully used KNN, but was thinking about looking at SVM (which I did nto use before). I have a pairwise distance matrix of training observations which are classified in set classes, and a distance matrix of new observations to the training ones. Is it possible to use distance matrices for SVM, and

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