similar to: ksvm question -- help! cannot get program to run...

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,

Hi, I have a process using svm from the e1071 library. it works. I want to try using the KSVM library instead. The same data used wiht e1071 gives me an error with KSVM. My data is a data.frame. sample code: svm_formula <- formula(y ~ a + B + C) svm_model <- ksvm(formula, data=train_data, type="C-svc", kernel="rbfdot", C=1) I get the following error:

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 all, I have a seemingly simple question which I have searched for an answer for for a few hours without luck. I have a matrix of both values and characters with thousands of rows. I would like to run a search of this matrix for certain values and would like the search to retrieve the entire row. How would I accomplish this? Thanks! Mehdi Khan [[alternative HTML version deleted]]

Hello, I am using kernlab to do some binary classification on aminoacid strings. I am using a custom kernel, so i use the kernel="matrix" option of the ksvm method. My (normalized) kernel matrix is of size 1309*1309, my results vector has the same length. I am using C-svc. My kernlab call is something similiar to this: ksvm(kernel="matrix", kernelMatrix, trainingDataYs,

Hello all, I'm trying to run a gird parameter search for a svm. Therefore I'M using the ksvm function from the kernlab package. ---- svp <- ksvm(Ktrain,ytrain,type="nu-svc",nu=C) ---- The problem is that the optimization algorithm does not return for certain parameters. I tried to use setTimeLimit() but that doesn't seem to help. I suspect that ksvm() calls c code that

Not sure if there is an R way to do this or a regular express way, but here is what I am trying to do. I've got lots of data where the format is HH:MM:SS, but I need to format it like HH:MM:00, i.e. round the second down to zero. What is the best way to do this? Thanks again. Jason

Dear list, I am using the ksvm function from kernlab as follows: (1) learning > svm.pol4 <- ksvm(class.labs ~ ., data = train.data, prob.model = T, scale = T, kernel = "polydot") (2) prediction > svm.pol.prd4 <- predict(svm.pol4, train.data, type = "probabilities")[,2] But unfortunately, when calling the prediction, once in every 10s of times (using the exact

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 <=

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!

Hi, I'm working on a project using the kernlab library. For one phase, I want the "decision values" from the SVM prediction, not the class label. the e1071 library has this function, but I can't find the equivalent in ksvm. In general, when an SVM is used for classification, the label of an unknown test-case is decided by the "sign" of its resulting value as

Hi, I drew a plot with R of my data and now I would like to display the median values of each group above the respective datapoint in the plot. I already search for ages, but maybe I am just blind. I am sure that this has to be an option somewhere... I appreciate any kind of input! Best, Christian

Is there a command as "mat.or.vec(nr,nc)" to create an array that I must calculate with more cicle? [[alternative HTML version deleted]]

Hi, Is there a function in R to do a-b where a and b are two non-numeric sets (or intersection complement of these two sets). Kind Regards, Praveen. [[alternative HTML version deleted]]

Hi I want to simulate random numbers normal distributed with this size (2000,10000). I tried this but my computer exhaust there is a fast way to make it? randz<-matrix(rnorm(2000000),2000,10000) [[alternative HTML version deleted]]

Hi all, could someone send me examples of the EM Clustering algorithm and/or some pdf describing it? Thanks, Douglas Sousa.

Dear R People: I'm running from R to a grid computer and getting some unusual results with the system command: > e7 <- system("globus-job-run xxxxx /bin/sh -c 'cd $OSG_APP/engage;chmod 777 oops'",intern=TRUE) chmod: changing permissions of `oops': Operation not permitted > e7 character(0) > I xx'ed out the site name. Anyhow, I would expect that the e7

Hello, I have a question about lars package, probably basic. The returned values of lars function include R squares along the variable selection path. However, such values are always slightly different from the R squares returned by the regression function lm using the same models. Anyone know the reasons? Very important, and needs quick answers. Thanks a million! -- View this message in

Hi, I was wondering if I could get R to warn me, or give me a rude awakening somehow, if I'm accessing a variable that is out of my function's scope. For example, often times I'm creating a function as I'm testing it in the REPL, copying and pasting between both. As a simple example, I might end up with a function like: f <- function(a, b) { a + b.test } Where

Hi list, Could someone help me to explain why the leave-one-out cross validation results I got from svm using the internal option "cross" are different from those I got manually? It seems using "cross" to do cross validation, the results are always better. Please see the code below. I also include lda as a comparison. I'm using WinXP, R-2.9.0, and e1071_1.5-19. Many