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)$y" on the
rows of a data frame using my example above? So if I cbind z and res,
df<-cbind(z,res)
the "x" in your one liner would be the fi
Never mind, I think i figured:
z<-df
apply(df,1,function(x) approx(sort(x[1:4]), seq(0,1,,length(x[1:4])), x[5])$y)
thanks again for the help
Andras Farkas,
On Friday, June 16, 2017 5:34 AM, Andras Farkas via R-help
wrote:
Peter,
thanks, very nice, this will work for me... could you
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 t
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 ide
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 t
> On Jun 15, 2017, at 12:37 PM, Andras Farkas via R-help
> 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 goa
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
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