[R] R function for percentrank

[Gabor Grothendieck]

On Dec 6, 2007 6:11 AM, Martin Maechler <maechler at stat.math.ethz.ch> wrote:
> >>>>> "MS" == Marc Schwartz <marc_schwartz at comcast.net>
> >>>>>     on Wed, 05 Dec 2007 12:43:50 -0600 writes:
>
> [............]
>
>    MS> Martin,
>
>    MS> Thanks for the corrections. In hindsight, now seeing the intended use of
>    MS> ecdf() in the fashion you describe above, it is now clear that my
>    MS> approach in response to David's query was un-needed and "over the top".
>    MS> "Yuck" is quite appropriate... :-)
>
>    MS> As I was going through this "exercise", it did seem overly complicated,
>    MS> given R's usual elegant philosophy about such things. I suppose if I had
>    MS> looked at the source for plot.stepfun(), it would have been more evident
>    MS> as to how the y values are acquired.
>
>    MS> In reviewing the examples in ?ecdf, I think that an example using
>    MS> something along the lines of the discussion here more explicitly, would
>    MS> be helpful. It is not crystal clear from the examples, that one can use
>    MS> ecdf() in this fashion, though the use of "12 * Fn(tt)" hints at it.
>
>    MS> Perhaps:
>
>    MS> ##-- Simple didactical  ecdf  example:
>    MS> x <- rnorm(12)
>    MS> Fn <- ecdf(x)
>    MS> Fn
>    MS> Fn(x)  # returns the percentiles for x
>    MS> ...
>
> Thank you, Marc for the above proposal, to make the examples
> more "crystal clear" :-)
> I've now amended the R-devel version of  help(ecdf) accordingly.

Since the above does not actually reproduce percentrank in the case
of ties, the change that would facilitate this particularly would be to add
a ties = "excelpercentrank" to approx.  See my solution earlier in this
thread.