similar to: Parameter names in nls()

Dear R-help, Could you please examine the following code, and see if I have discovered a bug or not, or am just doing something silly. I am trying to create a package to do fish stock assessment using the nls() function to fit the modelled stock size to the various pieces of information that we have. The main problem with this sort of task is that the number and type of parameters that go into

Dear All, I may look ridiculous, but I am puzzled at the behavior of the nls with a fitting I am currently dealing with. My data are: x N 1 346.4102 145.428256 2 447.2136 169.530634 3 570.0877 144.081627 4 721.1103 106.363316 5 894.4272 130.390552 6 1264.9111 36.727069 7 1788.8544 52.848587 8 2449.4897 25.128742 9 3464.1016 7.531766 10 4472.1360 8.827367 11

Hi, I am trying to fit a 4p logistic to this data, using nls function. The function didn't freely converge; however, it converged if I put a lower and an upper bound (in algorithm port). Also, the b1.A parameter always takes value of the upper bound, which is very strange. Has anyone experienced about non-convergent of nls and how to deal with this kind of problem? Thank you very much.

Dear friends. I use nls() and encounter the following puzzling problem: I have a function f(a,b,c,x), I have a data vector of x and a vectory y of realized value of f. Case1 I tried to estimate c with (a=0.3, b=0.5) fixed: nls(y~f(a,b,c,x), control=list(maxiter = 100000, minFactor=0.5 ^2048),start=list(c=0.5)). The error message is: "number of iterations exceeded maximum of

Dear Friends. I tried to use nls.control() to change the 'minFactor' in nls( ), but it does not seem to work. I used nls( ) function and encountered error message "step factor 0.000488281 reduced below 'minFactor' of 0.000976563". I then tried the following: 1) Put "nls.control(minFactor = 1/(4096*128))" inside the brackets of nls, but the same error message

Hello. I have a script in which I repeatedly fit a nonlinear regression to a series of data sets using nls and the port algorithm from within a loop. The general structure of the loop is: for(i in 1:n){ … extract relevant vectors of dependent and independent variables … … estimate starting values for Amax and Q.LCP…

Dear List, I am not sure how should i optimize a log-likelihood numerically: Here is a Text book example from Statistical Inference by George Casella, 2nd Edition Casella and Berger, Roger L. Berger (2002, pp. 355, ex. 7.4 # 7.2.b): data = x = c(20.0, 23.9, 20.9, 23.8, 25.0, 24.0, 21.7, 23.8, 22.8, 23.1, 23.1, 23.5, 23.0, 23.0) n <- length(x) # likelihood from a 2 parameter Gamma(alpha,

I'm having problems getting nls to agree that convergence has occurred in a toy problem. nls.out never gets defined when there is an error in nls. Reaching the maximum number of iterations is alway an error, so nls.out never gets defined when the maximum number of iterations is reched. >From ?nls.control: tol: A positive numeric value specifying the tolerance level for the

Hello colleagues, I am attempting to determine the nonlinear least-squares estimates of the nonlinear model parameters using nls. I have come across a common problem that R users have reported when I attempt to fit a particular 3-parameter nonlinear function to my dataset: Error in nls(r ~ tlm(a, N.fix, k, theta), data = tlm.data, start = list(a = a.st, : step factor 0.000488281

Hi, I'm trying to fit some data using a logistic function defined as y ~ a * (1+m*exp(-x/tau)) / (1+n*exp(-x/tau) My data is below: x <- 1:100 y <- c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,1,1,1,2,2,2,2,2,3,4,4,4,5, 5,5,5,6,6,6,6,6,8,8,9,9,10,13,14,16,19,21, 24,28,33,40,42,44,50,54,69,70,93,96,110,127,127,141,157,169,

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

All- I am interested in estimating a parameter that is the starting value for an ODE model. That is, in the typical combined fitting procedure using nls and lsoda (alternatively rk4), I first defined the ODE model: minmod <- function(t, y, parms) { G <- y[1] X <- y[2] with(as.list(parms),{ I_t <- approx(time, I.input, t)$y dG <- -1*(p1 + X)*G +p1*G_b dX <-

Dear list, I'm migrating a project from Matlab to R, and I'm facing a relatively complicated problem for nls. My objective function is below: >> objFun <- function(yEx,xEx,tEx,gamma,theta,kappa){ yTh <- pdfDY(xEx,tEx,gamma,theta,kappa) sum(log(yEx/yTh)^2) } The equation is yTh=P(xEx,tEx) + noise. I collect my data in: >> data <-

first, apologies for so many posts yesterday and today. I am wrestling with nls() and nls2(). I have tried to whittle it down to a simple example that still has my problem, yet can be cut-and-pasted into R. here it is: library(nls2) options(digits=12); y= c(0.4334,0.3200,0.5848,0.6214,0.3890,0.5233,0.4753,0.2104,0.3240,0.2827,0.3847,0.5571,0.5432,0.1326,0.3481) x=

Hello, I'm trying to fit experimental data with a model and nls. For some experiments, I have data with x from 0 to 1.2 and the fit is quite good. But it can happen that I have data only the [0,0.8] range (see the example below) and, then, the fit is not correct. I would like to add a constraint, for example : the second derivative must be positive. But I don't know how to add this to

Dear list, I do have a problem with nls. I use the following data: >test time conc dose 0.50 5.40 1 0.75 11.10 1 1.00 8.40 1 1.25 13.80 1 1.50 15.50 1 1.75 18.00 1 2.00 17.00 1 2.50 13.90 1 3.00 11.20 1 3.50 9.90 1 4.00 4.70 1 5.00 5.00 1 6.00 1.90 1 7.00 1.90 1 9.00 1.10 1 12.00 0.95 1 14.00

Dear Colleagues, Our group is also working on implementing the use of R for pharmacokinetic compartmental analysis. Perhaps I have missed something, but > fit <- nls(noisy ~ lsoda(xstart, time, one.compartment.model, c(K1=0.5, k2=0.5)), + data=C1.lsoda, + start=list(K1=0.3, k2=0.7), + trace=T + ) Error in eval(as.name(varName), data) : Object

Dear R-help, Here's a simple example of nonlinear curve fitting where nls seems to get the answer wrong on a very simple exponential fit (my R version 2.7.2). Look at this code below for a very basic curve fit using nls to fit to (a) a logarithmic and (b) an exponential curve. I did the fits using self-start functions and I compared the results with a more simple fit using a straight lm()

Hi I?m trying to fit a nonlinear model to a derivative of the logistic function y = a/(1+exp((b-x)/c)) (this is the parametrization for the SSlogis function with nls) The derivative calculated with D function is: > logis<- expression(a/(1+exp((b-x)/c))) > D(logis, "x") a * (exp((b - x)/c) * (1/c))/(1 + exp((b - x)/c))^2 So I enter this expression in the nls function:

Hi, I have a system of ODE's I can solve with lsoda. Model=function(t,x,parms) { #parameter definitions lambda=parms[1]; beta=parms[2]; d = parms[3]; delta = parms[4]; p=parms[5]; c=parms[6] xdot[1] = lambda - (d*x[1])- (beta*x[3]*x[1]) xdot[2] = (beta*x[3]*x[1]) - (delta*x[2]) xdot[3] = (p*x[2]) - (c*x[3]) return(list(xdot)) } I want