similar to: model.frame() call from inside a function (PR#3671)

I have found some "issues" (bugs?) with nls confidence intervals ... some with the relatively new "port" algorithm, others more general (but possibly in the "well, don't do that" category). I have corresponded some with Prof. Ripley about them, but I thought I would just report how far I've gotten in case anyone else has thoughts. (I'm finding the code

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,

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 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 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

Hi, I am trying to understand S3 classes. I have read several tutorials about the topics but I am still a bit confused. I guess it is because it is so different from Java OOP. I have pasted my attempt at creating a bank-account class below and my problems are: 1. What should be added some plot.default() calls the account$plot() method ? 2. What should the account$plot() be implemented to

Hello there, I am trying to fit an exponential model using nls to some data. #data t <- c(0,15,30,60,90,120,240,360,480) var <- c(0.36,9.72,15.50,23.50,31.44,40.66,59.81,73.11,81.65) df <- data.frame(t, var) # model # var ~ a+b*(1-exp(-k*t)) # I'm looking for values of a,b and k # formula # mod <- nls(formula = var ~ a+b *(1-exp((-k)*t)), start=list(a=0, b=10,

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'm trying to evaluate a pre-built expression using eval(), e.g. dataset <- data.frame(y = runif(30, 50,100), x = gl(5, 6)) # one like this mf <- expression(model.frame(y~x)) eval(mf, dataset, parent.frame()) # rather than this eval(expression(model.frame(y~x)), dataset, parent.frame()) In the example above there is no problem, the problem comes when I try to do a similar thing

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

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

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 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

Philipp Benner reported a Debian bug report against r-cran-rpart aka rpart. In short, the issue has to do with how rpart evaluates a formula and supporting arguments, in particular 'weights'. A simple contrived example is ----------------------------------------------------------------------------- library(rpart) ## using data from help(rpart), set up simple example myformula <-

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

Dear friends, This is the dataset I am currently working with: >dput(mod14data2_random) structure(list(index = c(14L, 27L, 37L, 33L, 34L, 16L, 7L, 1L, 39L, 36L, 40L, 19L, 28L, 38L, 32L), y = c(0.44, 0.4, 0.4, 0.4, 0.4, 0.43, 0.46, 0.49, 0.41, 0.41, 0.38, 0.42, 0.41, 0.4, 0.4 ), x = c(16, 24, 32, 30, 30, 16, 12, 8, 36, 32, 36, 20, 26, 34, 28)), row.names = c(NA, -15L), class =

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

I'm doing a non linear regression with 8 parameters to be fitted: J.Tl.nls<-nls(Gw~(a1/(1+exp(-a2*Tl+a3))+a4)*(b1/(1+exp(b2*Tl-b3))+b4),data=Enveloppe, start=list(a1=0.88957,a2=0.36298,a3=10.59241,a4=0.26308, b1=0.391268,b2=1.041856,b3=0.391268,b4=0.03439)) First, I fitted my curve on my data by guessing the parameters'

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) *

Dear R-helpers, I am trying to estimate a model that I am proposing, which consists of putting an extra hidden layer in the Markov switching models. In the simplest case the S(t) - Markov states - and w(t) - the extra hidden variables - are independent, and w(t) is constant. Formally the model looks like this: y(t)=c(1,y[t-1])%*%beta0*w+c(1,y[t-1])%*%beta1*(1-w). So I ran some simulations to