Re: [R] nls and factor
Hi Manuel,
an alternative to the approach pointed out by Prof. Ripley is to use
the package 'drc' which allows one or more parameters in a non-linear
regression model to depend on a factor.
You will need the latest version available at www.bioassay.dk (an older
version is available on CRAN).
Bconc - runif(30,0.1,10)
Aconc - runif(30,0.1,10)
At - runif(30,1,30)
Bt - runif(30,1,30)
BKm - 1
AKm - 0
EBoth - -0.41
yB - 100*exp(EBoth*Bt)*Bconc/(BKm+Bconc)+rnorm(30,0,1)
yA - 75*exp(EBoth*At)*Aconc/(AKm+Aconc)+rnorm(30,0,1)
## Constructing the dataset
tvalues - c(At, Bt)
conc - c(Aconc, Bconc)
yfactor - factor(rep(c(A, B), c(30, 30)))
y - c(yA, yB)
## Defining the non-linear function
twodimMM - function()
{
fct - function(x, parm)
{
parm[, 1]*exp(parm[, 2]*x[, 1])*x[, 2]/(parm[, 3] + x[, 2])
}
ssfct - function(dataset) # a very simple self starter function
{
return(c(90, -0.5, 0.8))
}
names - c(lev, Ev, km) # parameter names
return(list(fct=fct, ssfct=ssfct, names=names))
}
library(drc)
model1 - multdrc(y~cbind(tvalues,conc), yfactor, fct=twodimMM())
summary(model1)
## Collapsing the Ev parameters into a single parameter
model2 - multdrc(y~cbind(tvalues,conc), yfactor,
collapse=data.frame(yfactor,1,yfactor), fct=twodimMM())
anova(model2, model1) # model reduction is insignificant
summary(model2)
Christian
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] nls and factor
Thanks, it was actually p.249, at least in my MASS3. but that solved my doubt. I've have another doubt, can this factor interact with one of the parameters in the model? My problem is basically a Michaelis Menten term, where this factor determines a different Km. The rest of the parameters in the model are the same. But I don't know how to write the nls formula, or if it is possible. This is a toy example, B and A are the two factors, conc is the concentration and t is the temperature: ## Generate independent variables Bconc-runif(30,0.1,10) Aconc-runif(30,0.1,10) At-runif(30,1,30) Bt-runif(30,1,30) ## These are the parameters I want to calculate from ## my real data BKm-1 AKm-0 EBoth--0.41 # These are my simulated dependent variables yB-100*exp(EBoth*Bt)*Bconc/(BKm+Bconc)+rnorm(30,0,1) yA-75*exp(EBoth*At)*Aconc/(AKm+Aconc)+rnorm(30,0,1) #The separate models BModel-nls(Response~lev*exp(Ev*t)*conc/(Km+conc),data=list(Response=yB,t=Bt,conc=Bconc),start=list(lev=90,Ev=-0.5,Km=0.8),trace=TRUE) AModel-nls(Response~lev*exp(Ev*t)*conc/(Km+conc),data=list(Response=yA,t=At,conc=Aconc),start=list(lev=90,Ev=-0.5,Km=0.8),trace=TRUE) ## I want to obtain a combined model of the form: ## Y=Intercept[1:2]*exp(Eboth*t)*conc/(Km[1:2]+conc) ## where I have a common E but two intercepts and two ## Kms (one of them should in fact be zero) yBoth-c(yB,yA) concBoth-c(Bconc,Aconc) tBoth-c(At,Bt) AorB-as.factor(c(rep(0,length(yA)),rep(1,length(yB ## Amongst other things I've tried FullModel-nls(Response~lev[AorB]*exp(Ev*t)*conc/(Km[AorB]+conc),data=list(Response=yBoth,t=tBoth,conc=concBoth),start=list(lev=c(90,70),Ev=-0.5,Km=c(0.8,0)),trace=TRUE) ## but i get to a singular gradient Any other pointers, thanks Manuel --- Prof Brian Ripley [EMAIL PROTECTED] escribió: On Thu, 20 Apr 2006, Manuel Gutierrez wrote: Is it possible to include a factor in an nls formula? Yes. What do you intend by it? If you mean what it would mean for a lm formula, you need A[a] and starting values for A. There's an example on p.219 of MASS4. I've searched the help pages without any luck so I guess it is not feasible. I've given it a few attempts without luck getting the message: + not meaningful for factors in: Ops.factor(independ^EE, a) This is a toy example, my realworld case is much more complicated (and can not be solved linearizing an using lm) a-as.factor(c(rep(1,50),rep(0,50))) independ-rnorm(100) respo-rep(NA,100) respo[a==1]-(independ[a==1]^2.3)+2 respo[a==0]-(independ[a==0]^2.1)+3 nls(respo~independ^EE+a,start=list(EE=1.8),trace=TRUE) Any pointers welcomed Many Thanks, Manu __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] nls and factor
Is it possible to include a factor in an nls formula? I've searched the help pages without any luck so I guess it is not feasible. I've given it a few attempts without luck getting the message: + not meaningful for factors in: Ops.factor(independ^EE, a) This is a toy example, my realworld case is much more complicated (and can not be solved linearizing an using lm) a-as.factor(c(rep(1,50),rep(0,50))) independ-rnorm(100) respo-rep(NA,100) respo[a==1]-(independ[a==1]^2.3)+2 respo[a==0]-(independ[a==0]^2.1)+3 nls(respo~independ^EE+a,start=list(EE=1.8),trace=TRUE) Any pointers welcomed Many Thanks, Manu __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] nls and factor
Manuel, I don't think that it works very easily. Instead, try gnls() in the nlme package. Cheers Andrew On Thu, Apr 20, 2006 at 11:18:02AM +0200, Manuel Gutierrez wrote: Is it possible to include a factor in an nls formula? I've searched the help pages without any luck so I guess it is not feasible. I've given it a few attempts without luck getting the message: + not meaningful for factors in: Ops.factor(independ^EE, a) This is a toy example, my realworld case is much more complicated (and can not be solved linearizing an using lm) a-as.factor(c(rep(1,50),rep(0,50))) independ-rnorm(100) respo-rep(NA,100) respo[a==1]-(independ[a==1]^2.3)+2 respo[a==0]-(independ[a==0]^2.1)+3 nls(respo~independ^EE+a,start=list(EE=1.8),trace=TRUE) Any pointers welcomed Many Thanks, Manu __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Andrew Robinson Department of Mathematics and StatisticsTel: +61-3-8344-9763 University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599 Email: [EMAIL PROTECTED] http://www.ms.unimelb.edu.au __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] nls and factor
Thanks Andrew. I am now trying but without much success. I don't now how to give start values for the factor?. Could you give me an example solution with my toy example? a-as.factor(c(rep(1,50),rep(0,50))) independ-1:100 respo-rep(NA,100) respo[a==1]-(independ[a==1]^2.3)+2 respo[a==0]-(independ[a==0]^2.1)+3 library(nlme) gnls(respo~independ^b+a,start=list(b=1.8)) Many thanks. Manu --- Andrew Robinson [EMAIL PROTECTED] escribió: Manuel, I don't think that it works very easily. Instead, try gnls() in the nlme package. Cheers Andrew On Thu, Apr 20, 2006 at 11:18:02AM +0200, Manuel Gutierrez wrote: Is it possible to include a factor in an nls formula? I've searched the help pages without any luck so I guess it is not feasible. I've given it a few attempts without luck getting the message: + not meaningful for factors in: Ops.factor(independ^EE, a) This is a toy example, my realworld case is much more complicated (and can not be solved linearizing an using lm) a-as.factor(c(rep(1,50),rep(0,50))) independ-rnorm(100) respo-rep(NA,100) respo[a==1]-(independ[a==1]^2.3)+2 respo[a==0]-(independ[a==0]^2.1)+3 nls(respo~independ^EE+a,start=list(EE=1.8),trace=TRUE) Any pointers welcomed Many Thanks, Manu __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Andrew Robinson Department of Mathematics and Statistics Tel: +61-3-8344-9763 University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599 Email: [EMAIL PROTECTED] http://www.ms.unimelb.edu.au __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] nls and factor
On Thu, 20 Apr 2006, Manuel Gutierrez wrote: Is it possible to include a factor in an nls formula? Yes. What do you intend by it? If you mean what it would mean for a lm formula, you need A[a] and starting values for A. There's an example on p.219 of MASS4. I've searched the help pages without any luck so I guess it is not feasible. I've given it a few attempts without luck getting the message: + not meaningful for factors in: Ops.factor(independ^EE, a) This is a toy example, my realworld case is much more complicated (and can not be solved linearizing an using lm) a-as.factor(c(rep(1,50),rep(0,50))) independ-rnorm(100) respo-rep(NA,100) respo[a==1]-(independ[a==1]^2.3)+2 respo[a==0]-(independ[a==0]^2.1)+3 nls(respo~independ^EE+a,start=list(EE=1.8),trace=TRUE) Any pointers welcomed Many Thanks, Manu __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
