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 myself clear enough so let me try again:
the known variables to us in what I am trying to do are the data frame "z' :
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)
and the vector "res":
res<-c(10.000000, 9.296382, 8.642955, 8.036076 ,7.472374, 6.948723,
6.462233, 6.010223 ,5.590211
,5.199896 ,4.837147, 4.499989 ,4.186589, 3.895250 ,3.624397, 3.372570,
3.138415, 2.920675
, 2.718185 ,2.529864 ,2.354708, 2.191786, 2.040233, 1.899247, 1.768084)
and I need to find the probability (probs) , the unknown value, which would
result in creating "res", ie: the probs=c(0.3), from:
res<-apply(z, 1, quantile, probs=c(0.3))...
a more simplified example assuming :
k<-c(1:100)
f<-30
ecdf(k)(f)
would give us the value of 0.3... so same idea as this, but instead of "k" we
have data frame "z", and instead of "f" we have "res", and need to find the
value of 0.3... Does that make sense?
much appreciate the help...
Andras Farkas,
On Thursday, June 15, 2017 6:46 PM, David Winsemius <dwinsem...@comcast.net>
wrote:
> On Jun 15, 2017, at 12:37 PM, Andras Farkas via R-help <r-help@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 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 "res" available to me... For a single vector this would be done using
> ecdf something like this:
>
> x <- rnorm(100)
> #then I know this value:
> quantile(x,0.33)
> #so do this step
> ecdf(x)(quantile(x,0.33))
> #to get 0.33 back...
>
> any suggestions on how I could to that for a data frame?
Can't you just used ecdf and quantile ecdf?
# See ?ecdf page for both functions
> lapply( lapply(z, ecdf), quantile, 0.33)
$a
33%
4.475758
$b
33%
3.245151
$c
33%
2.003595
$d
33%
6.173204
--
David Winsemius
Alameda, CA, USA
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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.
