From: Martin Maechler <maechler_at_stat.math.ethz.ch>

Date: Wed, 5 Dec 2007 18:42:40 +0100

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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Wed 05 Dec 2007 - 17:46:44 GMT

Date: Wed, 5 Dec 2007 18:42:40 +0100

I'm coming late to this, but this *does* need a correction just for the archives !

MS> On Sat, 2007-12-01 at 18:40 +0000, David Winsemius wrote:

* >> David Winsemius <dwinsemius_at_comcast.net> wrote in
*

>> news:Xns99F989B3A3057dNOTwinscomcast@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 toMS> 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

MS> One other approach, which returns the values and their respective rank MS> percentiles is:

>> cumsum(prop.table(table(x)))

[...... snip ........]

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