Your best option is to read the relevant help files.
A simple (untested) example to find R when P, T and scal.fn=Z is given,
is to do this:
my.fun <- function(P, R, T, Z) scal.fn(P, R, T) - Z
uniroot( fn, R=rr, T=tt, Z=zz, lower=-1000000, upper=1000000 )$root
You have to make an intelligent guess on the upper and lower ranges for
the parameter R. I have used +/- 1 million as a silly example.
HOWEVER, I do not think this works when P,R,T,Z are scalars. Try it to
be sure. If not, then you may have to write a for or apply loop.
Regards, Adai
On 16/08/2010 13:19, Petar Milin wrote:
Thanks for the answer!
However, if I would have scal.fn() like below, how would I apply
uniroot() or optimize() or the like?
Best,
PM
On 16/08/10 13:24, Adaikalavan Ramasamy wrote:
You probably need to look up on how to write functions.
Try
scal.fn<- function(P, R, T){
out<- ( 1/R - T ) / ( P - T )
return(out)
}
Here is a fake example:
df<- cbind.data.frame( P=rnorm(10), R=rnorm(10), T=rnorm(10) )
scal.fn( df$P, df$R, df$T )
Or are you trying to solve other parameters given scal values? If so,
try having a look at functions like uniroot().
Regards, Adai
On 16/08/2010 11:48, Petar Milin wrote:
Hello!
I need to find a simple scalar value:
Scal = ((1/R) - T) / (P - T),
where R, T, and P are vectors in a data.frame.
Please, can anyone tell me how to solve that in R?
Best,
PM
______________________________________________
R-help@r-project.org mailing list
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.
______________________________________________
R-help@r-project.org mailing list
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.
