Dear UseRs,
maybe is a silly question: how can I get Empirical CDF
values from an object created with ecdf()?? Using
print I obtain:
Empirical CDF
Call: ecdf(t)
x[1:57] = 4.1, 4.4, 4.5, ..., 491.3,
671.27
Thanks in advance.
Regards,
Vito
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Sorry, I solved by myself. Thanks. Vito From: Vito Ricci <vito_ricci <at> yahoo.com> Subject: [R] ECDF values Date: 2005-11-17 07:20:35 GMT (33 minutes ago) Dear UseRs, maybe is a silly question: how can I get Empirical CDF values from an object created with ecdf()?? Using print I obtain: Empirical CDF Call: ecdf(t) x[1:57] = 4.1, 4.4, 4.5, ..., 491.3, 671.27 Thanks in advance. Regards, Vito
On 11/17/05, Vito Ricci <vito_ricci at yahoo.com> wrote:> Dear UseRs, > > maybe is a silly question: how can I get Empirical CDF > values from an object created with ecdf()??The return value of ecdf is a function. Use it as you would any other function. -Deepayan
>>>>> "Vito" == Vito Ricci <vito_ricci at yahoo.com> >>>>> on Thu, 17 Nov 2005 08:20:35 +0100 (CET) writes:Vito> Dear UseRs, Vito> maybe is a silly question: how can I get Empirical CDF Vito> values from an object created with ecdf()?? Using Vito> print I obtain: Vito> Empirical CDF Vito> Call: ecdf(t) Vito> x[1:57] = 4.1, 4.4, 4.5, ..., 491.3, 671.27 help(ecdf) supposedly has enough explanation; however it seems the author of the help page (guess who) was mistaken about the "simple" in>> Examples: >> >> ##-- Simple didactical ecdf example: >> Fn <- ecdf(rnorm(12)) >> Fn; summary(Fn) >> 12*Fn(knots(Fn)) == 1:12 ## == 1:12 if and only if there are no ties !The clue is that the result of ecdf() { and also of stepfun(), approxfun() or splinefun() ! } is a *function* (with some additional attributes); hence it's as simple as tt <- seq(-3, 3, by = 0.1) Fn(tt) Vito> Thanks in advance. you're welcome. Martin Maechler, ETH Zurich