Re: [R] R function for percentrank

From: Marc Schwartz <marc_schwartz_at_comcast.net>
Date: Wed, 05 Dec 2007 12:43:50 -0600

On Wed, 2007-12-05 at 18:42 +0100, Martin Maechler wrote:
> I'm coming late to this, but this *does* need a correction
> just for the archives !
>
> >>>>> "MS" == Marc Schwartz <marc_schwartz_at_comcast.net>

> >>>>> on Sat, 01 Dec 2007 13:33:21 -0600 writes:
>
> MS> On Sat, 2007-12-01 at 18:40 +0000, David Winsemius wrote:
> >> David Winsemius <dwinsemius_at_comcast.net> wrote in
> >> news:Xns99F989B3A3057dNOTwinscomcast_at_80.91.229.13:
> >>
> >> > "tom soyer" <tom.soyer_at_gmail.com> wrote in
> >> > news:65cc7bdf0712010951p451a993i70da89f285d801de_at_mail.gmail.com:
> >> >
> >> >> John,
> >> >>
> >> >> The Excel's percentrank function works like this: if one has a number,
> >> >> x for example, and one wants to know the percentile of this number in
> >> >> a given data set, dataset, one would type =percentrank(dataset,x) in
> >> >> Excel to calculate the percentile. So for example, if the data set is
> >> >> c(1:10), and one wants to know the percentile of 2.5 in the data set,
> >> >> then using the percentrank function one would get 0.166, i.e., 2.5 is
> >> >> in the 16.6th percentile.
> >> >>
> >> >> I am not sure how to program this function in R. I couldn't find it as
> >> >> a built-in function in R either. It seems to be an obvious choice for
> >> >> a built-in function. I am very surprised, but maybe we both missed it.
> >> >
> >> > My nomination for a function with a similar result would be ecdf(), the
> >> > empirical cumulative distribution function. It is of class "function"
> >> so
> >> > efforts to index ecdf(.)[.] failed for me.
>
> I think you did not understand ecdf() !!!
> It *returns* a function,
> that you can then apply to old (or new) data; see below
>
> MS> You can use ls.str() to look into the function environment:
>
> >> ls.str(environment(ecdf(x)))
> MS> f : num 0
> MS> method : int 2
> MS> n : int 25
> MS> x : num [1:25] -2.215 -1.989 -0.836 -0.820 -0.626 ...
> MS> y : num [1:25] 0.04 0.08 0.12 0.16 0.2 0.24 0.28 0.32 0.36 0.4 ...
> MS> yleft : num 0
> MS> yright : num 1
>
>
>
> MS> You can then use get() or mget() within the function environment to
> MS> return the requisite values. Something along the lines of the following
> MS> within the function percentrank():
>
> MS> percentrank <- function(x, val)
> MS> {
> MS> env.x <- environment(ecdf(x))
> MS> res <- mget(c("x", "y"), env.x)
> MS> Ind <- which(sapply(seq(length(res$x)),
> MS> function(i) isTRUE(all.equal(res$x[i], val))))
> MS> res$y[Ind]
> MS> }
>
> sorry Marc, but "Yuck !!"
>
> - this percentrank() only works when you apply it to original x[i] values
> - only works for 'val' of length 1
> - is a complicated hack
>
> and absolutely unneeded (see below)
>
> MS> Thus:
>
> MS> set.seed(1)
> MS> x <- rnorm(25)
>
> >> x
> MS> [1] -0.62645381 0.18364332 -0.83562861 1.59528080 0.32950777
> MS> [6] -0.82046838 0.48742905 0.73832471 0.57578135 -0.30538839
> MS> [11] 1.51178117 0.38984324 -0.62124058 -2.21469989 1.12493092
> MS> [16] -0.04493361 -0.01619026 0.94383621 0.82122120 0.59390132
> MS> [21] 0.91897737 0.78213630 0.07456498 -1.98935170 0.61982575
>
>
> >> percentrank(x, 0.48742905)
> MS> [1] 0.56
>
> [gives 0.52 in my version of R ]
>
> Well, that is *THE SAME* as using ecdf() the way you
> should have used it :
>
> ecdf(x)(0.48742905)
>
> {in two lines, that is
>
> mypercR <- ecdf(x)
> mypercR(0.48742905)
>
> which maybe easier to understand, if you have never used the
> nice concept that underlies all of
>
> approxfun(), splinefun() or ecdf()
> }
>
> You can also use
>
> ecdf(x)(x)
>
> and indeed check that it is identical to the convoluted
> percentrank() function above :
>
> > ecdf(x)(0.48742905)
> [1] 0.52
> > ecdf(x)(x)
> [1] 0.20 0.44 0.12 1.00 0.48 0.16 0.56 0.72 0.60 0.28 0.96 0.52 0.24 0.04 0.92
> [16] 0.32 0.36 0.88 0.80 0.64 0.84 0.76 0.40 0.08 0.68
> > all(ecdf(x)(x) == sapply(x, function(v) percentrank(x,v)))
> [1] TRUE
> >
>
>
> Regards (and apologies for my apparent indignation ;-)
> by the author of ecdf() ,
>
> Martin Maechler, ETH Zurich

Martin,

Thanks for the corrections. In hindsight, now seeing the intended use of ecdf() in the fashion you describe above, it is now clear that my approach in response to David's query was un-needed and "over the top". "Yuck" is quite appropriate... :-)

As I was going through this "exercise", it did seem overly complicated, given R's usual elegant philosophy about such things. I suppose if I had looked at the source for plot.stepfun(), it would have been more evident as to how the y values are acquired.

In reviewing the examples in ?ecdf, I think that an example using something along the lines of the discussion here more explicitly, would be helpful. It is not crystal clear from the examples, that one can use ecdf() in this fashion, though the use of "12 * Fn(tt)" hints at it.

Perhaps:

##-- Simple didactical ecdf example:
x <- rnorm(12)
Fn <- ecdf(x)
Fn
Fn(x) # returns the percentiles for x
...

Thanks again Martin and no offense taken... :-)

Regards,

Marc



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