similar to: nls and R scoping rules

hi list, used versions: 2.12.1 and 2.14.0 under ubuntu and macosx. I recently stumbled over a problem with `nls', which occurs if the model is not specified explicitly but via an evaluation of a 'call' object. simple example: 8<-------------------------------------------------------------------------------------- nlsProblem <- function (len = 5) {

A colleague of mine is trying to use nls() to effect an optimization, and is encountering a scoping problem. I should know how to solve it for him but .... well, I just don't. I also had a quick scrounge of the archives --- I know I've seen this topic addressed before --- but I couldn't track it down. So here's a toy example that demonstrates the problem: hhh <-

Dear all I have tried to estimate the confidence intervals for predicted values of a nonlinear model fitted with nls. The function predict gives the predicted values and the lower and upper limits of the prediction, when the class of the object is lm or glm. When the object is derived from nls, the function predict (or predict.nls) gives only the predicted values. The se.fit and interval aguments

Hello I was trying to estimate the weibull model using nls after putting OLS values as the initial inputs to NLS. I tried multiple times but still i m getting the same error of Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial parameter estimates. The Program is as below > vel <- c(1,2,3,4,5,6,7,8,9,10,11,12,13,14) > df <- data.frame(conc, vel) >

I could use some help understanding how nls parses the formula argument to a model.frame and estimates the model. I am trying to utilize the functionality of the nls formula argument to modify garchFit() to handle other variables in the mean equation besides just an arma(u,v) specification. My nonlinear model is y<-nls(t~a*sin(w*2*pi/365*id+p)+b*id+int,data=t1,

hi everybody, is there a canonical way to get hold of the "trace=TRUE" output from nls, i.e. to copy it to a R variable (or at least to an external log file)? I have only found the possibility to "fix(nlsModel)" (and than the correct copy of that: namespace function ...) within the R-session by modifying the trace() definition within nlsModel. not really good for everyday

Hi, I am a non-expert user of R. I am essaying the fit of two different functions to my data, but I receive two different error messages. I suppose I have two different problems here... But, of which nature? In the first instance I did try with some different starting values for the parameters, but without success. If anyone could suggest a sensible way to proceed to solve these I would be

Dear R-users, I am trying to create a model using the NLS function, such that: Y = f(X) + q + e Where f is a nonlinear (Weibull: a*(1-exp(-b*X^c)) function of X and q is a covariate (continous variable) and e is an error term. I know that you can create multiple nonlinear regressions where x is polynomial for example, but is it possible to do this kind of thing when x is a function with unknown

Hello, I have data like the following: datum <- structure(list(Y = c(415.5, 3847.83333325, 1942.833333325, 1215.22222233333, 950.142857325, 2399.5833335, 804.75, 579.5, 841.708333325, 494.053571425 ), X = c(1.081818182, 0.492727273, 0.756363636, 0.896363636, 1.518181818, 0.499166667, 1.354545455, 1.61, 1.706363636, 1.063636364 )), .Names = c("Y", "X"), row.names = c(NA,

I am using nls to fit a non linear function to some data. The non linear function is: y= 1- exp(-(k0+k1*p1+ .... + kn*pn)) I have chosen algorithm "port", with lower boundary is 0 for all of the ki parameters, and I have tried many start values for the parameters ki (including generating them at random). If I fit the non linear function to the same data using an external

Dear all, I noticed the following under R 1.8.1 (when nls was still a separate package) but the same problem occurs under R 1.9.0 (where most (all?) of nls is now in the stats package): > data(Puromycin) > fm <- nls(rate~SSmicmen(conc,b0,b1), Puromycin, subset = state=="treated") > coef(summary(fm)) NULL The problem seems to be that summary.nls uses the name

Greetings, I am struggling a bit with a non-linear regression. The problem is described below with the known values r and D inidcated. I tried to alter the start values but get always following error message: Error in nlsModel(formula, mf, start, wts): singular gradient matrix at initial parameter estimates Calls: nls -> switch -> nlsModel I might be missing something with regard to the

Hello everyone, I'm trying to test the accurracy of R on the Eckerle4 dataset from NIST and I don't understand how the control option of the nls function works. I tought nls(...) was equivalent to nls(...control=nls.control()) i.e nls.control() was the default value of control, but here is the error I get : > n2=nls(V1~(b1/b2) *

Colleagues, I am using "nls" successfully (2.11.1, OS X) but I am having difficulties retrieving part of the output - residual sum of squares. I have assigned the output to FIT: > > FIT > Nonlinear regression model > model: NEWY ~ PMESOR + PAMPLITUDE * cos(2 * pi * (NEWX - POFFSET)/PERIOD) > data: parent.frame() > PMESOR PAMPLITUDE POFFSET >

Hi,all: I met a problem of nls. My data: x y 60 0.8 80 6.5 100 20.5 120 45.9 I want to fit exp curve of data. My code: > nls(y ~ exp(a + b*x)+d,start=list(a=0,b=0,d=1)) Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial parameter estimates I can't find out the reason for the error. Any suggesions are welcome. Many thanks. [[alternative HTML

Hello, I am trying to model a type II functional response of number of prey eaten (Ne) against number supplied (No) with a non-linear least squares regression (nls). I am using a modification of Holling's (1959) disc equation to account for non-replacement of prey; Ne=No{1-exp[a(bNe-T)]} where a is the attack rate, b is the handling time, and T is the experimental period. My script is as

Hi all, I've got a problem with the nls function. I have an adjustment which works when I fix one of the argument of my function (Xo=150) : *Xo*=150 f<- function (tt*,Xo*,a,b) ifelse(tt<*Xo*,a*exp(-b**Xo*),a*exp(-b*tt)) ajust<-nls(RER~f(tt,*Xo*,a,b),data=data.frame(tt=Ph2[,2*k],RER=Ph2[,2*k+1]),start=list(a=0.5,b=0.014)) But, when I use it as a "normal" parameter (and

Do I have the right impression that it's currently not possible to produce confidence intervals for the nls predictions using R? I had a course were we used SAS PROC nlin and there you could get intervals for the parameters and the prediction but I do not have access to SAS. Would it be difficult to implement, I tried to dig into the help pages of nls, vcov and nlsModel but I could not

Hi, I'm trying to make a regression of the form : formula <- y ~ Asym_inf + Asym_sup * ( (1 / (1 + (n1 * (exp( (tmid1-x) / scal1) )^(1/n1) ) ) ) - (1 / (1 + (n2 * (exp( (tmid2-x) / scal2) )^(1/n2) ) ) ) ) which is a sum of the generalized logistic model proposed by richards. with data such as these: x <- c(88,113,128,143,157,172,184,198,210,226,240,249,263,284,302,340) y <-

Hello there, I am trying to fit an exponential fit using Least squares to some data. #data x <- c(1 ,10, 20, 30, 40, 50, 60, 70, 80, 90, 100) y <- c(0.033823, 0.014779, 0.004698, 0.001584, -0.002017, -0.003436, -0.000006, -0.004626, -0.004626, -0.004626, -0.004626) sub <- data.frame(x,y) #If model is y = a*exp(-x) + b then fit <- nls(y ~ a*exp(-x) + b, data = sub, start