similar to: SVM probability output variation

Hi, I am using SVM from e1071 package. I can get decision values very easily. But whenever, I try to get the probability measure, it returns NULL. I use the following codes to generate decision.values and probability. Is there anything wrong in it? predictor<-svm(train[,c(x1, x2, x3)], train[,x4], probability=TRUE) pred<-predict(predictor, test[,c(x1, x2, x3)], probability=TRUE,

Hello, Connoisseurs! Please explain to novices, why svm model gives different results in the loop with the same data? As a result, I can not find the best gamma and cost parameters. Also tune.svm yields results that can not be repeated. How can I avoid this? My sessionInfo: R version 2.11.1 (2010-05-31) x86_64-pc-linux-gnu locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C

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

I am using SVM to classify categorical data and I would like the probabilities instead of the classification. ?predict.svm says that its only enabled when you train the model with it enabled, so I did that, but it didn't work. I can't even get it to work with iris. The help file shows that probability = TRUE when training the model, but doesn't show an example. Then I try to

Hi All, I'm attempting to use eps-regression or nu-regression SVM to compute probabilities but the predict function applied to an svm model object returns values outside [0, 1]: Variable Data looks like: Present X02 X03 X05 X06 X07 X13 X14 X15 X18 1 0 1634 48 2245.469 -1122.0750 3367.544 11105.013 2017.306 40 23227 2 0 1402 40 2611.519 -811.2500 3422.769 10499.425 1800.475 40 13822 3 0 1379

> x <- train[,c( 2:18, 20:21, 24, 27:31)] > y <- train$out > > svm.pr <- svm(x, y, probability = TRUE, method="C-classification", kernel="radial", cost=bestc, gamma=bestg, cross=10) > > pred <- predict(svm.pr, valid[,c( 2:18, 20:21, 24, 27:31)], decision.values = TRUE, probability = TRUE) > attr(pred, "decision.values")[1:4,]

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'm seeing some weirdness with svm and tune.svm that I can't figure out- was wondering if anyone else has seen this? Perhaps I'm failing to make something the expected class? Below is my repro case, though it *sometimes* doesn't repro. I'm using R2.3.1 on WindowsXP. I was also seeing it happen with R2.1.1 and have seen it on 2 different machines. data(iris) attach(iris)

Hi- I'm seeing some weirdness with svm and tune.svm that I can't figure out- was wondering if anyone else has seen this? Perhaps I'm failing to make something the expected class? Below is my repro case, though it *sometimes* doesn't repro. I'm using R2.3.1 on WindowsXP. I was also seeing it happen with R2.1.1 and have seen it on 2 different machines. data(iris) attach(iris)

> library(ROCR) > library(e1071) svmres.prob <- svm(traindx, traindy, probability=TRUE) svmpred.prob <- predict(svmres.prob, testdx, probability = TRUE, decision.values = TRUE, type="prob") > print(length(attr(svmpred.prob, "probabilities"))) [1] 0 > print(attr(svmpred.prob, "probabilities")) NULL > print(attributes(svmpred.prob)$decision.values)

Hi Andrew, > Representation: 32-bit unsigned int stored in IR meta-data. why store it in the IR, and not just have it be an analysis that you can query for branch probabilities? If you store it in the IR then it may well get out of date (eg: when an optimizer realizes that some branches is dead and deletes it, resulting in probabilities that don't add up to 1 on the other branches). This

Dear R-helpers, I would like to generate a variable that takes 0 or 1, and each subject has different probabilities of taking the draw. So, which of the following code I should use ? suppose there are 5 subjects, and their probabilities of this Bernoulli variable is p=c(0.2, 0.9, 0.15, 0.8, 0.75) n<-5 Ber.var <- rbimon(n,1,p) ## I doubt if this will take the first probability, which is

Dear R users, This is a KDE beginner's question. I have this distribution: > length(cap) [1] 200 > summary(cap) Min. 1st Qu. Median Mean 3rd Qu. Max. 459.9 802.3 991.6 1066.0 1242.0 2382.0 I need to compute the sum of the values times their probability of occurence. The graph is fine, den <- density(cap, from=min(cap), to=max(cap), give.Rkern=F)

Dear all, I have a questions regarding using the c function rmultinom.c. I got the following error message "rbinom: probability sum should be 1, but is 0.999264" Which is thrown by: if(fabs((double)(p_tot - 1.)) > 1e-7) MATHLIB_ERROR(_("rbinom: probability sum should be 1, but is %g"), (double) p_tot); I understand my probabilities do not sum to one close enough. I

Dear all, I am new to R and would like to run a multinomial logistic regression on my dataset (3 predictors for 1 dependent variables) I have used the vglm function from the VGAM package and got some results. Using the predict() function, I obtained the probability table I was looking for. However, I would like to fully understand how the predict() function generates the probabilities or in

Dear r users, I have 4 variables x1,x2,x3,x4 and each one has two levels, for example Y and N. For x1: prob(Y)=0.6, prob(N)=0.4; For x2: prob(Y)=0.5, prob(N)=0.5; For x3: prob(Y)=0.8, prob(N)=0.2; For x4: prob(Y)=0.9, prob(N)=0.1; Therefore, the sample space for (x1, x2, x3, x4)={YYYY, YYYN, YYNY,......} (16 possible combination) and the corresponding probabilities are {(0.6)(0.5)(0.8)(0.9),

Hello, I have a loop with probability computed from a logistic model like this: for (i in 1:300){ p[i]<-exp(-0.834+0.002*x[i]+0.023*z[i])/(1+exp(-0.834+0.002*x[i]+0.023 +z[i])) x and z generated from normal distribution. I get 300 different probabilities And I want to generate variables from bernulli distribution with P for every observation: T[i]<-rbinom(1,1,p[i]) But i get missing

Hi, I run R 2.10.1 under ubuntu 10.04 LTS (Lucid Lynx) and klaR version 0.6-4. I compute a model over a 2 classes dataset (composed of 700 examples). To that aim, I use the function NaiveBayes provided in the package klaR. When I then use the prediction function : predict(my_model, new_data). I get the following warning : "In FUN(1:747[[747L]], ...) : Numerical 0 probability with

Dear friends, If I have a table like this, first row A B C D ... are different levels of the variable, first column 0 1 2 4 ... are the levels of the "numbers", the numbers inside the table are the probabilities of the "number" occuring. A B C D ... 0 0.2 0.3 0.1 0.05 1 0.1 0.1 0.2 0.2 2 0.02 0.2 0 0.1 4 0.3 0.01 0.01

I have never been able to get class probabilities to work and I am relatively new to using these tools, and I am looking for some insight as to what may be wrong. I am using caret with kernlab/ksvm. I will simplify my problem to a basic data set which produces the same problem. I have read the caret vignettes as well as documentation for ?train. I appreciate any direction you can give. I