kernlab | ksvm error

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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 same data), it returns the following error message:

Error in prob.model(object)[[p]]$A :
   $ operator is invalid for atomic vectors
Calls: predict -> predict -> .local -> .SigmoidPredict
Execution halted

Does anyone know what the problem is? Any help will be much appreciated.
We need a reproducible example.

Best,
Uwe Ligges
Cheers

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Dear Uwe,
I appreciate that if you let me know why, when using the attached file, the
following script (two lines) doesn't work once in 10s of times.
Best, Reza

svm.pol4 <- ksvm(class.labs ~ ., data = train.data, prob.model = T, scale =
T, kernel = "polydot")

svm.pol.prd4 <- predict(svm.pol4, train.data, type = "probabilities")[,2]
Dear Uwe,
I appreciate that if you let me know why, when using the attached file,
the following script (two lines) doesn't work once in 10s of times.
Best, Reza

svm.pol4<- ksvm(class.labs~ ., data= train.data, prob.model= T, scale=
T, kernel= "polydot")

svm.pol.prd4<- predict(svm.pol4, train.data, type= "probabilities")[,2]

Some randomness is involved, and when you get an unfortunate subsample 
(e.g. if in the internal cross-validation one class is not selected at 
all) it won't work anymore.

Best,
Uwe Ligges
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Thanks Uwe,
Am I right that in ksvm's internal cross-validation, there is no
guarantee for having *at least one* of each classes in each subset?
That is my guess, but I haven't read the code. Please read it yourself 
in case you want more details - or ask the author of tht function.

Best,
Uwe Ligges
    Some randomness is involved, and when you get an unfortunate
    subsample (e.g. if in the internal cross-validation one class is not
    selected at all) it won't work anymore.