similar to: Question in using e1071 svm routine

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

Hi all, Could anybody tell me what happened to my logistic regression in R? mylog=glm(mytraindata$V1 ~ ., data=mytraindata, family=binomial("logit")) It generated the following error message: Error in model.frame.default(Terms, newdata, na.action = na.action, xlev = object$xlevels) : factor 'state1' has new level(s) AP Thank you!

I got the data working, but now I got another problem with KSVM: line search fails -2.793708 -0.5831701 1.870406e-05 -5.728611e-06 -5.059796e-08 -3.761822e-08 -7.308871e-13Error in prob.model(object)[[p]]$A : $ operator is invalid for atomic vectors On Tue, Jul 7, 2009 at 6:45 PM, Steve Lianoglou<mailinglist.honeypot at gmail.com> wrote: > Hi, > > On Jul 7, 2009, at 6:44 PM,

What's wrong? Very sad about this... model <- ksvm(x=mytraindata[, -1], y=factor(mytraindata[, 1]), prob.model=T) Error in .local(x, ...) : x and y don't match.

Here is my code: mygbm<-gbm.fit(y=mytraindata[, 1], x=mytraindata[, -1], interaction.depth=4, shrinkage=0.001, n.trees=20000, bag.fraction=1, distribution="bernoulli") Here is the error: Error in gbm.fit(y = mytraindata[, 1], x = mytraindata[, -1], interaction.depth = 4, : The dataset size is too small or subsampling rate is too large: cRows*train.fraction*bag.fraction <=

Dear users, I'm working on binary classification problem using Support Vector Machines (SVM). My objective is to train a series of SVM models on a grid of hyperparameters and then select those that maximize the AUC based on an independent validation sample. My attempted code is shown below. It runs well on "small" data sets but when I use it on a slightly larger sample (e.g., my

Dear colleagues: Despite my best efforts, I have not been able to understand/overcome an error message I received while running the fclustIndex command in package e1071, which produces validity measures for the fuzzy c-means clustering algorithm. I am relatively new to R-- and still depend heavily on Rcmdr-- and so I apologize in advance if there is an obvious answer to my question:

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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

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,

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

Dear All: I found a problem with the SVM internal cross-validation (CV) accuracy estimation in the e1071 package. File: Rsvm.c Line: 120 Today, it is: int j = rand()%(prob->l-i); Should be: int j = i + rand()%(prob->l-i); The erroneous code doesn't shuffle objects. Instead, it "randomly" moves objects from beginning to the end. In hope for a prompt response from the

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), :

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

Dear all, after having calibrated a svm model through the svm() command of the e1071 package, is there a way to i) represent the modeled relationships between the y and X variables (response variable vs. predictors)? ii) rank the influence of the predictors used in the model? Right now I am more interested in regression models, but I guess this would be useful for classification too. Thank

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,

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

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

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

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