[R] ecdf

[ONKELINX, Thierry]

R is case sensitive. ecdf() is in the stats package, Ecdf() is in Hmisc.
So you want Ecdf(x,what='1-F')

Thierry
------------------------------------------------------------------------
----
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie & Kwaliteitszorg
Gaverstraat 4
9500 Geraardsbergen
Belgium

Research Institute for Nature and Forest
team Biometrics & Quality Assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium

tel. + 32 54/436 185
Thierry.Onkelinx op 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-help-bounces op r-project.org 
> [mailto:r-help-bounces op r-project.org] Namens Downey, Patrick
> Verzonden: maandag 19 april 2010 15:04
> Aan: R help
> Onderwerp: [R] ecdf
> 
> Hello,
> 
> I'd like to plot an empirical cumulative distribution
> function, except instead of the fraction of values < x, I'd 
> like the fraction of values > x.
> 
> 
> I think this can be done using the ecdf function in {Hmisc}. 
> I installed the package and loaded it. However, when
> following the example given in the documentation, I get an error:
> 
> x <- rnorm(100)
> ecdf(x,what='1-F')
> Error in ecdf(x, what = "1-F") : unused argument(s) (what = "1-F")
> 
> I believe that this is because R is attempting to access the 
> ecdf function in base R, which does not have the what option. 
> Am I correct, and if so, how can I change that?
> 
> Note: I also tried to "do it myself" without the {Hmisc} ecdf 
> function, and couldn't figure out a way. 
> 
> x2 <- 1-ecdf(x)
> 
> doesn't work, and neither does
> 
> x2 <- rep(0,times=100)
> for(i in 1:100){
>   x2[i] <- 1-ecdf(x)[i]
> }
> 
> Both result in errors.
> 
> Thanks in advance for any suggestions you can offer.
> 
> -Mitch
> 
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> 

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