Hi
As I said I have no experience with kernlab but I can read in manual:
"probabilities matrix of class probabilities (one column for each class and one
row for each input)"
from which I understand that pred is matrix, which should have the same number
of rows as df$cons but several columns. And matrix is a vector with dimensions
what means that it is column times longer than df$cons, hence it is longer than
df$cons.
Plotting error suggeststs that ksvc object has to be classification object and
> type = "C-bsvc",
is not the same as "C-svc".
Cheers
Petr
> -----Original Message-----
> From: Luigi Marongiu <marongiu.lu...@gmail.com>
> Sent: Tuesday, January 8, 2019 4:40 PM
> To: PIKAL Petr <petr.pi...@precheza.cz>
> Cc: r-help <r-help@r-project.org>
> Subject: Re: [R] error in plotting model from kernlab
>
> Hi,
> the maintainer hasn't answered yet. The problem with 'acc' is that yes the
> objects are not of the same length but they should be: according to the
> manual,
> ' table(pred, df$cons)' would return a 2x2 matrix of the results. This is not
> the
> case, so there is a problem with the model -- that is why there is no plotting
> either -- even if an object of class ksvm had been created.
>
> On Tue, Jan 8, 2019 at 4:12 PM PIKAL Petr <petr.pi...@precheza.cz> wrote:
> >
> > Hi
> >
> > I cannot help you with kernlab
> >
> > > > pred = predict(mod, df, type = "probabilities") acc =
> > > > table(pred, df$cons)
> > > Error in table(pred, df$cons) : all arguments must have the same
> > > length which again is weird since mod, df and df$cons are made from
> > > the same dataframe.
> >
> > Why not check length of those objects?
> >
> > length(pred)
> > length(df$cons)
> >
> > > > plot(mod, data = df)
> > > > kernlab::plot(mod, data = df)
> > > but I get this error:
> > >
> > > Error in .local(x, ...) :
> > > Only plots of classification ksvm objects supported
> > >
> >
> > seems to me selfexplanatory. What did maintainer said about it?
> >
> > Cheers
> > Petr
> >
> >
> > > -----Original Message-----
> > > From: R-help <r-help-boun...@r-project.org> On Behalf Of Luigi
> > > Marongiu
> > > Sent: Monday, January 7, 2019 1:26 PM
> > > To: r-help <r-help@r-project.org>
> > > Subject: [R] error in plotting model from kernlab
> > >
> > > Dear all,
> > > I have a set of data in this form:
> > > > str <data>
> > > 'data.frame': 1574 obs. of 14 variables:
> > > $ serial: int 12751 14157 7226 15663 11088 10464 1003 10427 11934
> > > 3999 ...
> > > $ plate : int 43 46 22 50 38 37 3 37 41 11 ...
> > > $ well : int 79 333 314 303 336 96 235 59 30 159 ...
> > > $ sample: int 266 295 151 327 231 218 21 218 249 84 ...
> > > $ target: chr "HEV 2-AI5IQWR" "Dientamoeba fragilis-AIHSPMK"
> > > "Astro
> > > 2 Liu-AI20UKB" "C difficile GDH-AIS086J" ...
> > > $ ori.ct: num 0 33.5 0 0 0 ...
> > > $ ct.out: int 0 1 0 0 0 0 0 1 0 0 ...
> > > $ mr : num -0.002 0.109 0.002 0 0.001 0.006 0.015 0.119 0.003 0.004
> > > ...
> > > $ fcn : num 44.54 36.74 6.78 43.09 44.87 ...
> > > $ mr.out: int 0 1 0 0 0 0 0 1 0 0 ...
> > > $ oper.a: int 0 1 0 0 0 0 0 1 0 0 ...
> > > $ oper.b: int 0 1 0 0 0 0 0 1 0 0 ...
> > > $ oper.c: int 0 1 0 0 0 0 0 1 0 0 ...
> > > $ cons : int 0 1 0 0 0 0 0 1 0 0 ...
> > > from which I have selected two numerical variables correspondig to x
> > > and y in a Cartesian plane and one outcome variable (z):
> > > > df = subset(t.data, select = c(mr, fcn, cons)) df$cons =
> > > > factor(c("negative", "positive"))
> > > > head(df)
> > > mr fcn cons
> > > 1 -0.002 44.54 negative
> > > 2 0.109 36.74 positive
> > > 3 0.002 6.78 negative
> > > 4 0.000 43.09 positive
> > > 5 0.001 44.87 negative
> > > 6 0.006 2.82 positive
> > >
> > > I created an SVM the method with the KERNLAB package with:
> > > > mod = ksvm(cons ~ mr+fcn, # i prefer it to the more canonical "."
> > > > but the
> > > outcome is the same
> > > data = df,
> > > type = "C-bsvc",
> > > kernel = "rbfdot",
> > > kpar = "automatic",
> > > C = 10,
> > > prob.model = TRUE)
> > >
> > > > mod
> > > Support Vector Machine object of class "ksvm"
> > >
> > > SV type: C-bsvc (classification)
> > > parameter : cost C = 10
> > >
> > > Gaussian Radial Basis kernel function.
> > > Hyperparameter : sigma = 42.0923201429106
> > >
> > > Number of Support Vectors : 1439
> > >
> > > Objective Function Value : -12873.45 Training error : 0.39263
> > > Probability model included.
> > >
> > > First of all, I am not sure if the model worked because 1439 support
> > > vectors out of 1574 data points means that over 90% of the data is
> > > required to fix the hyperplane. this does not look like a model but
> > > a patch. Secondly, the prediction is rubbish -- but this is another
> > > story -- and when I try to create a confusion table of the processed
> > > data I get:
> > > > pred = predict(mod, df, type = "probabilities") acc =
> > > > table(pred, df$cons)
> > > Error in table(pred, df$cons) : all arguments must have the same
> > > length which again is weird since mod, df and df$cons are made from
> > > the same dataframe.
> > >
> > > Coming to the actual error, I tried to plot the model with:
> > > > plot(mod, data = df)
> > > > kernlab::plot(mod, data = df)
> > > but I get this error:
> > >
> > > Error in .local(x, ...) :
> > > Only plots of classification ksvm objects supported
> > >
> > > Would you know what I am missing?
> > > Thank you
> > > --
> > > Best regards,
> > > Luigi
> > >
> > > ______________________________________________
> > > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > PLEASE do read the posting guide
> > > http://www.R-project.org/posting-guide.html
> > > and provide commented, minimal, self-contained, reproducible code.
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>
> --
> Best regards,
> Luigi
Osobní údaje: Informace o zpracování a ochraně osobních údajů obchodních
partnerů PRECHEZA a.s. jsou zveřejněny na:
https://www.precheza.cz/zasady-ochrany-osobnich-udaju/ | Information about
processing and protection of business partner’s personal data are available on
website: https://www.precheza.cz/en/personal-data-protection-principles/
Důvěrnost: Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a
podléhají tomuto právně závaznému prohláąení o vyloučení odpovědnosti:
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it may be confidential and are subject to the legally binding disclaimer:
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______________________________________________
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