[R-sig-Geo] pretty in spplot()

[Tord Snäll]

Dear Thierry,
Thanks very much. This was helpful. The code work with some changes.

Tord

ONKELINX, Thierry wrote:
> Dear Tord,
>
> You need to supply the datavalues to the pretty function. An untested example as your example was not reproducible with a (dummy) dataset.
>
> PrettyBreaks <- pretty(as.vector(resdf[,c("TempEst.t","TempEst.tM1")]))
> spplot(SpatialPolygonsDataFrame(karta, as.data.frame(resdf[,c("TempEst.t","TempEst.tM1")])),at=PrettyBreaks)
>
> HTH,
>
> Thierry
>
> ----------------------------------------------------------------------------
> ir. Thierry Onkelinx
> Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest
> Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance
> Gaverstraat 4
> 9500 Geraardsbergen
> Belgium 
> tel. + 32 54/436 185
> Thierry.Onkelinx at inbo.be 
> www.inbo.be 
>
> To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of.
> ~ Sir Ronald Aylmer Fisher
>
> The plural of anecdote is not data.
> ~ Roger Brinner
>
> The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data.
> ~ John Tukey
>
> -----Oorspronkelijk bericht-----
> Van: r-sig-geo-bounces at stat.math.ethz.ch [mailto:r-sig-geo-bounces at stat.math.ethz.ch] Namens Tord Snäll
> Verzonden: vrijdag 14 november 2008 12:14
> Aan: r-sig-geo at stat.math.ethz.ch
> Onderwerp: [R-sig-Geo] pretty in spplot()
>
> Dear all,
> I would like legend cut-offs that are prettier that just executing
> spplot(SpatialPolygonsDataFrame(karta, 
> as.data.frame(resdf[,c("TempEst.t","TempEst.tM1")])))
>
> I try
> spplot(SpatialPolygonsDataFrame(karta, 
> as.data.frame(resdf[,c("TempEst.t","TempEst.tM1")])),at=pretty())
> , though without sucess.
>
> Could someone give a clue on how to do it?
>
> Thanks!
>
> /Tord
>
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