similar to: Data format for KSVM

Hello useRs! So, I have a random question. I'm trying to build a character string, then evaluate it. I think an example would be the easiest way to explain: kern.vec = c("rbfdot","polydot") for( j in 1:length( kern.vec ) ) { formula = paste("ksvm( ind ~ . , data=d.temp[,c(ind_col,dep_cols)], kernel =",kern.vec[j],", prob.model=T

Hi!! I am trying to figure out how to use the string kernel "stringdot" in kernlab. k <- function(x,y) { (sum(x*y) +1)*exp(-0.001*sum((x-y)^2)) } class(k) <- "kernel" data(promotergene) ## train svm using custom kernel gene.k <- ksvm(Class~.,data=promotergene,kernel=k,C=10,cross=5) # works fine in this case gene.rbf <-

Hi useR's, I am resending this request since I got no response for my last post and I am new to the list so pardon me if I am violating the protocol. I am trying to use the "Kernlab" package for training and prediction using SVM's. I am getting the following error when I am trying to use the predict function: > predictSvm = predict(modelforSVM, testSeq); Error in

I have a dataset that has many columns which are NA or constant, and so I remove them like so: same <- sapply(dataset, function(.col){ all(is.na(.col)) || all(.col[1L] == .col) }) dataset <- dataset[!same] This works GREAT (thanks to the r-users list archive I found this) however, then when I do my data sampling like so: testSize <- floor(nrow(x) * 10/100) test <-

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 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

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,

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

also from this site: https://plotly.com/ggplot2/geom_ribbon/ I get the answer is geom_ribbon but I am still missing something ``` #! plot p = ggplot(data = trainset, aes(x=x, y=y, color=z)) + geom_point() + scale_color_manual(values = c("red", "blue")) # show support vectors df_sv = trainset[svm_model$index, ] p = p + geom_point(data = df_sv, aes(x=x, y=y),

Thank you, but this split the area into two and distorts the shape of the plot. (compared to ``` p + geom_abline(slope = slope_1, intercept = intercept_1 - 1/w[2], linetype = "dashed", col = "royalblue") + geom_abline(slope = slope_1, intercept = intercept_1 + 1/w[2], linetype = "dashed", col = "royalblue") ``` Why there

Hi What about something like p+geom_ribbon(aes(ymin = slope_1*x + intercept_1 - 1/w[2], ymax = slope_1*x + intercept_1 + 1/w[2], fill = "grey70", alpha=0.1)) Cheers Petr > -----Original Message----- > From: Luigi Marongiu <marongiu.luigi at gmail.com> > Sent: Friday, October 23, 2020 11:11 AM > To: PIKAL Petr <petr.pikal at precheza.cz> > Cc: r-help

Hi Did you try google? I got several answers using your question e.g. https://stackoverflow.com/questions/54687321/fill-area-between-lines-using-g gplot-in-r Cheers Petr > -----Original Message----- > From: R-help <r-help-bounces at r-project.org> On Behalf Of Luigi Marongiu > Sent: Friday, October 23, 2020 9:59 AM > To: r-help <r-help at r-project.org> > Subject:

Hello, I am running SVM and showing the results with ggplot2. The results include the decision boundaries, which are two dashed lines parallel to a solid line. I would like to remove the dashed lines and use a shaded area instead. How can I do that? Here is the code I wrote.. ``` library(e1071) library(ggplot2) set.seed(100) x1 = rnorm(100, mean = 0.2, sd = 0.1) y1 = rnorm(100, mean = 0.7, sd =

Hi Put fill outside aes p+geom_ribbon(aes(ymin = slope_1*x + intercept_1 - 1/w[2], ymax = slope_1*x + intercept_1 + 1/w[2]), fill = "blue", alpha=0.1) The "hole" is because you have two levels of data (red and blue). To get rid of this you should put new data in ribbon call. Something like newdat <- trainset newdat$z <- factor(0) p+geom_ribbon(data=newdat, aes(ymin =

Hello, I would like to know, whether for the support vector classification function ksvm() the response values stored in object at ymatrix are cross validated outputs/predictions: Example code from package kernlab, function ksvm: library(kernlab) ## train a support vector machine filter <- ksvm(type~.,data=spam,kernel="rbfdot",kpar=list(sigma=0.05),C=5,cross=3) filter filter at

Hi, I am reading in a dataframe from a CSV file. It has 70 columns. I do not have any kind of unique "row id". rawdata <- read.table("r_work/train_data.csv", header=T, sep=",", na.strings=0) When training an svm, I keep getting an error So, as an experiment, I wrote the data back out to a new file so that I could see what the svm function sees.

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.

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

I'm using caret to assess classifier performance (and it's great!). However, I've found that my results differ between R2.* and R3.* - reported accuracies are reduced dramatically. I suspect that a code change to kernlab ksvm may be responsible (see version 5.16-24 here: http://cran.r-project.org/web/packages/caret/news.html). I get very different results between caret_5.15-61 +