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David, thanks for the response. In your response the quantile function (if I see correctly) runs on the columns versus I need to run it on the rows, which is an easy fix, but that is not exactly what I had in mind... essentially we can remove t() from my original code to make "res" look like this: res<-apply(z, 1, quantile, probs=c(0.3)) but after all maybe I did not explain

It would depend on which one of the 9 quantile definitions you are using. The discontinuous ones aren't invertible, and the continuous ones won't be either, if there are ties in the data. This said, it should just be a matter of setting up the inverse of a piecewise linear function. To set ideas, try x <- rnorm(5) curve(quantile(x,p), xname="p") The breakpoints for the

> On Jun 15, 2017, at 12:37 PM, Andras Farkas via R-help <r-help at r-project.org> wrote: > > Dear All, > > we have: > > t<-seq(0,24,1) > a<-10*exp(-0.05*t) > b<-10*exp(-0.07*t) > c<-10*exp(-0.1*t) > d<-10*exp(-0.03*t) > z<-data.frame(a,b,c,d) > > res<-t(apply(z, 1, quantile, probs=c(0.3))) > > > > my

Dear All, we have: t<-seq(0,24,1) a<-10*exp(-0.05*t) b<-10*exp(-0.07*t) c<-10*exp(-0.1*t) d<-10*exp(-0.03*t) z<-data.frame(a,b,c,d) res<-t(apply(z, 1, quantile, probs=c(0.3))) my goal is to do a 'reverse" of the function here that produces "res" on a data frame, ie: to get the answer 0.3 back for the percentile location when I have

Peter, thanks, very nice, this will work for me... could you also help with setting up the code to run the on liner "approx(sort(x), seq(0,1,,length(x)), q)$y" on the rows of a data frame using my example above? So if I cbind z and res,? df<-cbind(z,res) the "x" in your one liner would be the first 4 column values of each row and "q" is the last (5fth) column