similar to: binning results

I've completed an experiment and want to summarize the results. There are two things I like to create. 1) A simple count of things from the data.frame with predictions 1a) Number of predictions with probability greater than x 1b) Number of predictions with probability greater than x that are really true In SQL, this would be, "Select count(predictions) from

Hi, I am wondering if it is possible to get an estimate of standard error of the predicted posterior probability from LDA using lda() from MASS? Logistic regression using glm() would generate a standard error for predicted probability with se.fit=T argument in predict(), so would it make sense to get standard error for posterior probability from lda() and how? Another question about standard

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

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 +

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

If do: > library("e1071") > example(svm) I get: svm> data(iris) svm> attach(iris) svm> ## classification mode svm> # default with factor response: svm> model <- svm(Species ~ ., data = iris) svm> # alternatively the traditional interface: svm> x <- subset(iris, select = -Species) svm> y <- Species svm> model <- svm(x, y) svm>

Hi, I have a strange one for the group. We have a system that predicts probabilities using a fairly standard svm (e1017). We are looking at probabilities of a binary outcome. The input data is generated by a perl script that calculates a bunch of things, fetches data from a database, etc. We train the system on 30,000 examples and then test the system on an unseen set of 5,000 records.

Hi I am doing this sort of thing: POLY: > > obj = best.tune(svm, similarity ~., data = training, kernel = "polynomial") > summary(obj) Call: best.tune(svm, similarity ~ ., data = training, kernel = "polynomial") Parameters: SVM-Type: eps-regression SVM-Kernel: polynomial cost: 1 degree: 3 gamma: 0.04545455 coef.0: 0

I am trying to understand what is the difference between linear and polynomial kernel: linear: u'*v polynomial: (gamma*u'*v + coef0)^degree It would seem that polynomial kernel with gamma = 1; coef0 = 0 and degree = 1 should be identical to linear kernel, however it gives me significantly different results for very simple data set, with linear kernel

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 - I am trying to tune an SVM model by optimizing the cross-validation accuracy. Maximizing this value doesn't necessarily seem to minimize the number of misclassifications. Can anyone tell me how the cross-validation accuracy is defined? In the output below, for example, cross-validation accuracy is 92.2%, while the number of correctly classified samples is (1476+170)/(1476+170+4) =

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)

svm.fit<-svm(as.factor(out) ~ ., data=all_h, method="C-classification", kernel="radial", cost=bestc, gamma=bestg, cross=10) # model fitting svm.pred<-predict(svm.fit, hh, decision.values = TRUE, probability = TRUE) # find the probability, but can not find. attr(svm.pred, "probabilities") > attr(svm.pred, "probabilities") 1 0 1 0 0 2 0

Hi, I'm working with support vector machine for the classification purpose, and I have a problem about the accuracy of prediction. I divided my data set in train (1/3 of enteire data set) and test (2/3 of data set) using the "sample" function. Each time I perform the svm model I obtain different result, according with the result of the "sample" function. I would like

Hi all, This feature was throwing me for a loop for quite some time until I played with the names of the slots. Consider exhibit A: ============== setClass("SVM", representation=representation( x='numeric', y='numeric', C='numeric', eps='numeric'), prototype=prototype( x=numeric(),

Dear R people, I am trying to plot the results from running svm in library(e1071). I use plot.svm. After searching through the help archives and FAQ, I still have several questions: 1. In default, crosses indicate support vectors. But why are there two colors of crosses? What do they represent? 2. I want to draw a white-gray colored plot and modify the different colored crosses or circles by

Hi everyone, Can someone please help me in these questions?: 1)if I use crossvalidation with svm, do I have to use this equation to calculate RMSE?: mymodel <- svm(myformula,data=mydata,cross=10) sqrt(mean(mymodel$MSE)) But if I don’t use crossvalidation, I have to use the following to calculate RMSE: mymodel <- svm(myformula,data=mydata) mytest

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,

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