similar to: questions regarding the svmpath package (functions svmpath and predict)

Hello, I want to use nnet package in R, to train and simulate a NN and get the value of MSE. I am reading in a file which has 19 input variables and one output variable and has a total of 2000 observations. The first column in the file is a column just for giving the serial numbers of the observations. I have already read in the file and also extracted the different values into the matrices to

I am trying to implement a simple r-svm example using the iris data (only two of the classes are taken and data is within the code). I am running into some errors. I am not an expert on svm's. If any one has used it, I would appreciate their help. I am appending the code below. Thanks../Murli ####################################################### ### R-code for R-SVM ### use leave-one-out

Dear members, My name is Seungho Huh. I am a statistician who tries to use the Kernel PLS method in a classification problem. I am sending this email to ask you something about the "kernelpls" function in R (pls.pcr package). I would like to obtain the predicted Y values for test data, using the Kernel PLS method. Let's take the example in the R help: > data(NIR) >

You need to do some extra work if you want to do classification with a regression method. One simple way to do classification with PLS is to code the classes as 0s and 1s (assuming there are only two classes) or -1s and 1s, fit the model, then threshold the prediction; e.g., those with predicted values < 0.5 (in the 0/1 coding) get labeled as 0s. There's a predict() method for mvr

Dear list I am using the mvr function of the package pls.pcr to compute PLS resgression using a X matrix of gene expression variables and a Y matrix of medical varaibles. I would like to obtain the R2 (sum of squares captured by the model) and Q2 (proportion of total sum of squares captured in leave-one-out cross validation) of the model. I am not sure if there are specific slots in the

Dear R-users, While making a prediction using the randomForest function (package randomForest) I'm getting the following error message: "Error in predict.randomForest(model, newdata = CV) : No forest component in the object" Here's my complete code. For reproducing this task, please find my 2 data sets attached ( http://www.nabble.com/file/p17855119/data.rar data.rar ).

Hi, here are three results of random forest (version 4.0-1). The results seem to be more or less the same which is strange because I changed the classwt. I hoped that for example classwt=c(0.45,0.1,0.45) would result in fewer cases classified as class 2. Did I understand something wrong? Christian x1rf <- randomForest(x=as.data.frame(mfilters[cvtrain,]),

Hey! How do I get the confusion matrix after performing 10-fold cross validation from SVM in R? When I try to print it, I get the confusion matrix without cross validation. I need to compute PPV. Should I report PPV without CV and total accuracy with CV? I am confused. > svmtrain <- svm(xtrain,ytrain,kernel="sigmoid",cross=10) > pred <- predict(svmtrain, xtrain) >

Hi all, Can some one tell me the difference between the following two formulas? 1. epiG.rf <-randomForest(gamma~.,data=data, na.action = na.fail,ntree = 300,xtest = NULL, ytest = NULL,replace = T, proximity =F) 2.epiG.rf <-randomForest(gamma~.,data=data, na.action = na.fail,ntree = 300,xtest = NULL, ytest = NULL,replace = T, proximity =F) [[alternative HTML version deleted]]

Sir, This query is related to randomForest regression using R. I have a dataset called qsar.arff which I use as my training set and then I run the following function - rf=randomForest(x=train,y=trainy,xtest=train,ytest=trainy,ntree=500) where train is a matrix of predictors without the column to be predicted(the target column), trainy is the target column.I feed the same data

I use the svm package kernlab .I have two question. In R library(kernlab) m=ksvm(xtrain,ytrain,type="C-svc",kernel=custom function, C=10) alpha(m) alphaindex(m) I can get alpha value and alpha index about package. 1. Assumption that number of sample are 20. number of support vectors are 15. then rest 5`s alphas are 0? 2. I want use kernelMatrix xtrain=as.matrix(xtrain)

Hi, I am writing to seek some guidance regarding using Lasso regression with the R package LARS. I have introductory statistics background but I am trying to learn more. Right now I am trying to duplicate the results in a paper for shRNA prediction "An accurate and interpretable model for siRNA efficacy prediction, Jean-Philippe Vert et. al, Bioinformatics" for a Bioinformatics project

Hello R community, I have cross-posted with r-sig-geo as this issue could fall under either interest group I believe. I just came accross the alphahull package and am very pleased I may not need to use CGAL anymore for this purpose. However, I am having a problem computing alpha shapes with my point data, and it seems to have to do with the spatial configuration of my points (which form

How can I get the list of non-zero features from svmpath at any given lambda? All I get is following information and information about what features were selected. In Iris example, we have 4 features and 60 cases. In my own example which is 200cases by 300 features, I can't figure out how to print the list of non-zero (beta) or features from svmpath. Any help? Thanks. ---------------- >

Hello, I came across a problem when building a randomForest model. Maybe someone can help me. I have a training- and a testdataset with a discrete response and ten predictors (numeric and factor variables). The two datasets are similar in terms of number of predictor, name of variables and datatype of variables (factor, numeric) except that only one predictor has got 20 levels in the training

Hi - When using the randomForest function for regression, I get different results for mean-squared error of the predictions depending on whether or not I specify to calculate variable importance. There is an example below. I looked briefly at the source code, but couldn't find anything that would indicate why calculating variable importance would (or should) change predictions. I'm

Are the results dramatically different? The result would be expected to be somewhat different, as setting importance=TRUE would make many calls to the random number generator (for permuting OOB data in each variable), making all but the first tree in the forest different than if importance=FALSE. Cheers, Andy > From: Scott Gilpin > > Hi - > > When using the randomForest

Dear R-help, I was bitten by the behavior of all() when given logical(0): It is TRUE! (And any(logical(0)) is FALSE.) Wouldn't it be better to return logical(0) in both cases? The problem surfaced because some un-named individual called randomForest(x, y, xtest, ytest,...), and gave y as a two-level factor, but ytest as just numeric vector. I thought I check for that in my code by testing

Hi everyone, I cannot install package "svmpath" on R 2.8.1 on Redhat Linux .In fact I cannot install package externally.I am new to Linux so it would be of great help if you all can help me out. Regards, Subhajit. [[alternative HTML version deleted]]

Hi, there: I spent some time on this but I think I really cannot figure it out, maybe I missed something here: my data looks like this: > dim(trn3) [1] 7361 209 > dim(val3) [1] 7427 209 > mg.rf2<-randomForest(x=trn3[,1:208], y=trn3[,209], data=trn3, xtest=val3[, 1:208], ytest=val3[,209], importance=T) my test data has 7427 observations but after prediction, > dim(mg.rf2$votes)