On Jan 21, 2013, at 10:30 AM, Katherine Gobin wrote:
> Dear R forum
>
>> From the given data, I have estimated the parameters of Wakeby distribution
>> using lmomco package as
>
> library(lmomco)
>
> (amounts <- read.csv("input_S.csv")$amount)
>
> # ___________________________________________________________
>
> # Wakeby distribution - Parameter estimation
>
> N =
> length(amounts)
> lmr = lmom.ub(amounts)
> parameters_of_Wakeby = parwak(lmr)
It appears you have a) not included the code that produced that output and b)
failed to read the Index page for that package
help(package="lmomco")
help(package="lmomco")
?rlmomco # Random Deviates of a Distribution
So on the assumption that you have an object in your workspace named
"parameters_of_Wakeby" and it is an lmomco produced object like that returned
by lmom2par() I would try:
rlmomco(100, parameters_of_Wakeby)
>
>> parameters_of_Wakeby
>
> $type
> [1]
> "wak"
>
> $para
> xi alpha
> 1.18813927666405e+04 0.00000000000000e+00
> beta gamma
> 0.00000000000000e+00 8.11391042554567e+04
> delta
> 9.57554297149062e-01
>
> This means the scale parameters are 0.
>
> However, assuming, all the five parameters of Wakeby distribution (viz.
> location parameter m (xi), the scale parameters a, b, and shape parameters g
> and d are available.
>
> Then, how do I generate say 100 random no.s using Wakeby distribution w.r.t.
> these
> 5 available parameters.
>
> I couldn't find any information about this in lmomco. Kindly guide if random
> no.s can be generated or not and if yes, how it can be done in r.
You should have been able to find this with:
help.search("random", package="lmomco")
--
David Winsemius
Alameda, CA, USA
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