R help - Feb 2011 - nls

hi,
I would like to find the x value (independent variable) for a certain dependent
value using the fitted model with nls.
with (predict) I can find y that corresponds to a list of x. I need the other
way around. can it be done?
thanks,
afadda

On Fri, Feb 25, 2011 at 6:09 AM, Abeer Fadda
<a.fadda at zmbh.uni-heidelberg.de> wrote:
> hi,
> I would like to find the x value (independent variable) for a certain
dependent value using the fitted model with nls.
> with (predict) I can find y that corresponds to a list of x. I need the
other way around. can it be done?
> thanks,
> afadda
Yes.

-- Bert




... Oh, if you mean HOW can it be done? -- lots of ways. Analytically,
if y=f(x) is your fitted model, just backsolve x = g(y).

If your algebra isn't up to the task, then just predict y on a
suitably fine x grid, and (assuming monotonicity) find the x
corresponding to the predicted y closest to the y you wish to back
calibrate. This is just a matter of indexing. See ?which.



>
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>


-- 
Bert Gunter
Genentech Nonclinical Biostatistics

I am running the following nls equation. I tried it with data that excel was
fitting and got the error: 
singular gradient matrix at initial parameter estimates
I thought it was due to a low number of points (6), but when I create a
dataset, I get the same problem. If I remove the parameter "a," then
it can
find a solution. Does anyone know what I can do to fit this model?

vardata<- rnorm(73,mean=0,sd=5)
x<-0:72
a<-1
b<-50
k<-0.05
l<-5
startt<-c(a,b,k,l)
yf<-a+b*(1-exp(-k*(x-l)))
y<-yf+vardata

nmodel<-nls(y~a+b*(1-exp(-k*(x-l))),
start=list(a=1,b=50,k=0.05,l=5))

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