Because you do not pass an (initial) nugget value to function vgm(), you
are fitting an exponential model without a nugget. Without a nugget
model, the nugget is (implicitly) zero. Compare:
> vgm(1, "Exp", 300)
model psill range
1 Exp 1 300
> vgm(1, "Exp", 300, 1)
model psill range
1 Nug 1 0
2 Exp 1 300
You are also wrong if you think that you now fix (as of: not fit)
parameters in the fit.variogram step; for this you have to pass
parameters fit.ranges and/or fit.sills; e.g. to fit the nugget but fix
the partial sill, pass fit.sills = c(TRUE,FALSE), to fix the nugget but
fit the partial sill, pass fit.sill=c(FALSE, TRUE).
On 10/04/2010 11:07 AM, Arantzazu Blanco Bernardeau wrote:
> Hello
> I am really glad of your answer, but for me it does not work as I would like
> So, my data are called cc (is a subset of a bigger spatialdataframe) and the
> variogram would be
> varcc=variogram(pf1_3atm~1,cc, width=500,cutoff=10000)
>
> Then I fix the sill and the range of the variogram
> inisill=var(c...@data$pf1_3atm,na.rm=T)
> iniran=var$dist[varcc$gamma>0.9*inisill][1]
> varcc.fit=fit.variogram(varcc,vgm(psill=inisill, "Exp", iniran)) # I can not
> fix the nugget because the output gives errors ("Singular model")
> plot(varcc,varcc.fit)
>> varcc[1,"gamma"] this would be the nugget (where the model cuts the axis)
> [1] 11.97
> The problem in the indicator you suggest is:
> varcc.fit$psill[1] / sum(varcc.fit$psill)= 1, because varcc.fit$psill[1] =
> sum(varcc.fit$psill)
> as the output of varcc.fit gives:
> varcc.fit
> model psill range
> 1 Exp 41.46455 1297.040
>
> I have tried (don't know if it is correct):
>> varcc[1,"gamma"]/sum(varcc.fit$psill)
> [1] 0.2886803
> (and this would mean that the spatial correlation is weak because we have a
> relative big nugget effect)
> Can this be done?
> Thanks a lot again for your attention.
>
>> Date: Thu, 30 Sep 2010 18:39:22 +0200
>> From: edzer.pebe...@uni-muenster.de
>> To: r-sig-geo@stat.math.ethz.ch
>> Subject: Re: [R-sig-Geo] nugget in variogram
>>
>>
>>
>> On 09/30/2010 04:13 PM, Arantzazu Blanco Bernardeau wrote:
>>> Hello everybody
>>> I am new on geostatistics so maybe my question is trivial or is
>>> already answered (I can not find the solution in the thread list). I
>>> have a variogram where I did not fix the nugget, because doing it
>>> was giving "singular models". I would like to know the nugget ratio
>>> of my variograms but I don't know how to get then the nugget value.
>>> So I did look in the output of variogram and did take the minimum
>>> value of gamma. Is this correct?
>>
>> Well, it's not the usual way, where you would fit a model:
>>
>> library(gstat)
>> data(meuse)
>> coordinates(meuse) = ~x+y
>> v = variogram(log(zinc)~1, meuse)
>> v.m = fit.variogram(v, vgm(1, "Sph", 500, 1))
>> v[1, "gamma"] # your estimator:
>> plot(v, v.m)
>> v.m$psill[1] / sum(v.m$psill) # the usual estimator:
>>
>> Only when you have plenty of measurements at very short distances
>> apart, I can imagine you would use the sample variogram (v) value.
>>
>>> Thanks very much for your attention.
>>>
>>> [[alternative HTML version deleted]]
>>>
>>> _______________________________________________
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>>> R-sig-Geo@stat.math.ethz.ch
>>> https://stat.ethz.ch/mailman/listinfo/r-sig-geo
>>
>> --
>> Edzer Pebesma
>> Institute for Geoinformatics (ifgi), University of Münster
>> Weseler Straße 253, 48151 Münster, Germany. Phone: +49 251
>> 8333081, Fax: +49 251 8339763 http://ifgi.uni-muenster.de
>> http://www.52north.org/geostatistics e.pebe...@wwu.de
>>
>> _______________________________________________
>> R-sig-Geo mailing list
>> R-sig-Geo@stat.math.ethz.ch
>> https://stat.ethz.ch/mailman/listinfo/r-sig-geo
>
--
Edzer Pebesma
Institute for Geoinformatics (ifgi), University of Münster
Weseler Straße 253, 48151 Münster, Germany. Phone: +49 251
8333081, Fax: +49 251 8339763 http://ifgi.uni-muenster.de
http://www.52north.org/geostatistics e.pebe...@wwu.de
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