similar to: Extraction of rules from Support Vector Machines

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

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, 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

Hello List, I am having a great trouble using svm function in e1071 package. I have 4gb of data that i want to use to train svm. I am using Amazon cloud, my Amazon Machine Image(AMI) has 34.2 GB of memory. my R process was killed several times when i tried to use 4GB of data for svm. Now I am using a subset of that data and it is only 1.4 GB. i remove all unnecessary objects before calling

Dear R Community- I am a new user of support vector machines for species distribution modeling and am using package e1071 to run svm() and predict.svm(). Briefly, I want to create an svm model for classification of a factor response (species presence or absence) based on climate predictor variables. I have used a training dataset to train the model, and tested it against a validation data set

Hi, I have some difficulties in interpreting the prediction of a svm model using the package e1071. y1 is the variable I want to predict. It is of type factor and has got two levels: "< 50%" and "> 50%". z is the dataset. > model <- svm(y1 ~ ., data = z,type="C-classification", cross=10) > model Call: svm(formula = y1 ~ ., data = z, type =

HI, GUYS, I used the following codes to run SVM and get prediction on new data set hh. dim(all_h) [1] 2034 24 dim(hh) # it contains all the variables besides the variables in all_h data set. [1] 640 415 require(e1071) svm.tune<-tune(svm, as.factor(out) ~ ., data=all_h, ranges=list(gamma=2^(-5:5), cost=2^(-5:5)))# find the best parameters. bestg<-svm.tune$best.parameters[[1]]

I'm running SVM from e1071 package on a data with ~150 columns (variables) and 50000 lines of data (it takes a bit of time) for radial kernel for different gamma and cost values. I get a very large models with at least 30000 vectors and the prediction I get is not the best one. What does it mean and what could I do to ameliorate my model ? Jerzy Orlowski

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,

Dear all, I am currently looking for a book about support vector machines for regression and classification and am a bit lost since they are plenty of books dealing with this subject. I am not totally new to the field and would like to get more information on that subject for later use with the e1071 <http://cran.r-project.org/web/packages/e1071/index.html> package for instance. Does

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,

Hello all, I am trying to use the "svm" method provided by e1071 (Version: 1.5-16) together with a matrix provided by the SparseM package (Version: 0.73) but it fails with this message: > model <- svm(lm, lv, scale = TRUE, type = 'C-classification', kernel = 'linear') Error in t.default(x) : argument is not a matrix although lm was created before with

Hola, Puedes poner un RStudio en Amazon, poner "caret" y a correr.... No sé si tendrás suficiente con lo que te pueda ofrecer Amazon para tu problema... creo que sí... ;-).... O directamente hacerlo aquí, que toda esta instalación ya la tienen hecha: http://www.teraproc.com/front-page-posts/r-on-demand/ Gracias, Carlos. El 10 de diciembre de 2015, 14:43, MªLuz Morales <mlzmrls

Hello, I have an SVM model previously calibrated using libsvm R implementation from the e1071 package. I would like to use this SVM to predict values, from a Java program. I first tried to use jlibsvm and the "standard" java implementation of libsvm, without success. Thus, I am now considering writing data in files from my Java code, calling an R program to predict values, then gather

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

I run the following code, all other is ok, but plot(m.svm,p5.new,As~Cur) is not ok Anyone know why? install.packages("e1071") library(e1071) library(MASS) p5 <- read.csv("http://www.public.iastate.edu/~aiminy/data/p_5_2.csv") p5.new<-subset(p5,select=-Ms) p5.new$Y<-factor(p5.new$Y) levels(p5.new$Y) <- list(Out=c(1), In=c(0)) attach(p5.new)

I run the following code, all other is ok, but plot(m.svm,p5.new,As~Cur) is not ok Anyone know why? install.packages("e1071") library(e1071) library(MASS) p5 <- read.csv("http://www.public.iastate.edu/~aiminy/data/p_5_2.csv") p5.new<-subset(p5,select=-Ms) p5.new$Y<-factor(p5.new$Y) levels(p5.new$Y) <- list(Out=c(1), In=c(0)) attach(p5.new)

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.

Hola Mª Luz, Te cuento un poco mi visión: Lo primero de todo es tener claro qué quiero hacer exactamente en paralelo, se me ocurren 3 escenarios: (1) Aplicar un modelo en este caso SVM sobre unos datos muy grandes y por eso necesito hadoop/spark (2) Realizar muchos modelos SVM sobre datos pequeños (por ejemplo uno por usuario) y por eso necesito hadoop/spark para parelilizar estos procesos

Estimados Un día leí algo en el siguiente hipervínculo, pero nunca lo use. http://blog.revolutionanalytics.com/2015/06/using-hadoop-with-r-it-depends.html Javier Rubén Marcuzzi De: Carlos J. Gil Bellosta Enviado: miércoles, 9 de diciembre de 2015 14:33 Para: MªLuz Morales CC: r-help-es Asunto: Re: [R-es] SVM hadoop No, no correrán en paralelo si usas los SVM de paquetes como e1071. No