similar to: nls and rbinom function: step factor 0.000488281 reduced below 'minFactor' of 0.000976562

Hello, I've seen various threads on people reporting: step factor 0.000488281 reduced below `minFactor' of 0.000976563 While I know how to set the minFactor, what I'd like to have happen is for nls to return to me, the last or closest fitted parameters before it errors out. In other words, so I don't get convergence, I'd still like to acquire the values of the parameters

Hello, I am trying to estimate 4 parameters of a non-linear model using nls. My model function is a Fourier integral and is very expensive to calculate. I get the following error: > theta0 <- c(0.045, 1.02*10^(-4), 0.00169, 5.67*10^(-4)) > res <- nls(log(y) ~ log(model(theta,r,t)), data=dataModel, + start=list(theta=theta0), trace=TRUE, + control=nls.control(tol=1e-2))

I am trying to fit the 4-parameter logistic model with SSfpl. The estimation is terminated with a message "step factor 0.000488281 reduced below 'minFactor' of 0.000976562". I tried to change the minFactor with nls.control(minFactor=yy) with no apparent success. A call of nls.control() without arguments always shows the value of 0.000976562. Also the estimation terminates as

Dear all, I am trying to fit a non linear regression model to time series data. If I do this: reg.logis = nls(myVar~SSlogis(myTime,Asym,xmid,scal)) I get this error message (translated to English from French): Erreur in nls(y ~ 1/(1 + exp((xmid - x)/scal)), data = xy, start = list(xmid = aux[1], : le pas 0.000488281 became inferior to 'minFactor' of 0.000976562 I then tried to set

All - I have data: TL age 388 4 418 4 438 4 428 5 539 10 432 4 444 7 421 4 438 4 419 4 463 6 423 4 ... [truncated] and I'm trying to fit a simple Von Bertalanffy growth curve with program: #Creates a Von Bertalanffy growth model VonB=nls(TL~Linf*(1-exp(-k*(age-t0))), data=box5.4, start=list(Linf=1000, k=0.1, t0=0.1), trace=TRUE) #Scatterplot of the data plot(TL~age, data=box5.4,

Hi I want to change a control parameter for an nls () as I am getting an error message "step factor 0.000488281 reduced below 'minFactor' of 0.000976562". Despite all tries, it seems that the control parameter of the nls, does not seem to get handed down to the function itself, or the error message is using a different one. Below system info and an example highlighting the

Hi, I have some data I want to fit with a non-linear function using nls, but it won't solve. > regresjon<-nls(lcfu~lN0+log10(1-(1-10^(k*t))^m), data=cfu_data, > start=(list(lN0 = 7.6, k = -0.08, m = 2))) Error in nls(lcfu ~ lN0 + log10(1 - (1 - 10^(k * t))^m), data = cfu_data, : step factor 0.000488281 reduced below 'minFactor' of 0.000976562 Tried to increase minFactor

Hi to All, I am fitting some models to a data using non linear least square, and whenever i run the command, parameters value have good convergence but I get the error in red as shown below. Kindly how can I fix this problem. Convergence of parameter values 0.2390121 : 0.1952981 0.9999975 1.0000000 0.03716107 : 0.1553976 0.9999910 1.0000000 0.009478433 : 0.2011017 0.9999798 1.0000000

SSweibull() :  problems with step factor and singular gradient Hello I am working with growth data of ~4000 tree seedlings and trying to fit non-linear Weibull growth curves through the data of each plant. Since they differ a lot in their shape, initial parameters cannot be set for all plants. That’s why I use the self-starting function SSweibull(). However, I often got two error messages:

data(Boston, package='MASS') y <- Boston$nox x <- Boston$dis nls(y~ A + B * exp(C * x), start=list(A=1, B=1, C=1)) Error in nls(y ~ A + B * exp(C * x), start = list(A = 1, B = 1, C = 1), : step factor 0.000488281 reduced below 'minFactor' of 0.000976562 I don't know how to fix this error. I think my problem is that I set the wrong start. Could somebody help please?

Hello there, I am trying to use R function NLS to analyze my data and one of the examples in the documentation is - ## the nls() internal cheap guess for starting values can be sufficient: x <- -(1:100)/10 y <- 100 + 10 * exp(x / 2) + rnorm(x)/10 nlmod <- nls(y ~ Const + A * exp(B * x), trace=TRUE) plot(x,y, main = "nls(*), data, true function and fit, n=100") curve(100 +

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'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 <-

Dear R-helpers, I'm using R 2.5.0 under Windows and am trying to use nlsList from nlme 3.1-80 with the selfstart function for the four parametric logistic function. My first test went well, but now I'm trying to do some more sophisticated things and it does not work anymore. I simulate my data from a five parametric logistic function like this:

Dear Bert, Thank you so much for your kind and valuable feedback. I tried finding the starting values using the approach you mentioned, then did the following to fit the nonlinear regression model: nlregmod2 <- nls(y ~ theta1 - theta2*exp(-theta3*x), start = list(theta1 = 0.37, theta2 = exp(-1.8), theta3 =

Oh, sorry; I changed signs in the model, fitting theta0 + theta1*exp(theta2*x) So for theta0 - theta1*exp(-theta2*x) use theta1= -.exp(-1.8) and theta2 = +.055 as starting values. -- Bert On Sun, Aug 20, 2023 at 11:50?AM Paul Bernal <paulbernal07 at gmail.com> wrote: > Dear Bert, > > Thank you so much for your kind and valuable feedback. I tried finding the > starting

Hi, In the third edition of the MASS book, Chapter 8, Page 252 ~ 253. I tried applying the same technique on using a negative exponential model (as suggested in Exercise 1), and trying to get a confidence interval. What I have changed is follows: expn <- function(b0, b1, th, x) { temp <- exp(-x/th) model.func <- b0 + b1 * temp # Differentiate temp derive.negexp <- (x

Hello, I am trying to do a nonlinear model using the "nls" command in R software. The data I am using is as follows: A<-c(7.132000,8.668667,9.880667,8.168000,10.863333,10.381333,11.059333,7.589333,4.716667,4.268667,7.265333,10.309333,8.456667,13.359333,8.624000,13.571333,12.523333,4.084667 ,NaN,NaN)

nls() is using 1) only a Gauss-Newton code which is prone to some glitches 2) approximate derivatives Package nlmrt uses symbolic derivatives for expressions (you have to provide Jacobian code for R functions) and an aggressive Marquardt method to try to reduce the sum of squares. It does return more information about the problem (singular values of the final Jacobian and gradient at the proposed

A couple of weeks ago there was a question regarding apparent convergence in nls when using the SSfol selfStart model for fitting a first-order pharmacokinetic model. I can't manage to find the original message either in my archive or in the list archives but the data were 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