similar to: Error in NLS example in the documentation

Hi R experts, I'm trying to use nls() for a piecewise linear regression with the first slope constrained to 0. There are 10 data points and when it does converge the second slope is almost always over estimated for some reason. I have many sets of these 10-point datasets that I need to do. The following segment of code is an example, and sorry for the overly precise numbers, they are just

Hi, I have a small dataset that I'm fitting a segmented regression using nls on. I get a step below minimum factor error, which I presume is because residual sum of square is still "not small enough" when steps in the parameter space is already below specified/default value. However, when I look at the trace, the convergence seems to have been reached. I initially thought I might

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?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 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

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

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

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. I am trying to do a non-linear fit using the 'nls' command. The data that I'm using is as follows pH k 1 3.79 34.21 2 4.14 25.85 3 4.38 20.45 4 4.57 15.61 5 4.74 12.42 6 4.92 9.64 7 5.11 7.30 8 5.35 5.15 9 5.67 3.24 with a transformation of pH to H <- 10^-pH When using the nls command for a set of parameters - a, b and c, I receive two sets of errors: >

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:

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 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 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

Greets, I'm trying to learn to use nls and was running the example code for an exponential model: > x <- -(1:100)/10 > y <- 100 + 10 * exp(x / 2) + rnorm(x)/10 > nlmod <- nls(y ~ Const + A * exp(B * x)) Error in B * x : non-numeric argument to binary operator In addition: Warning message: In nls(y ~ Const + A * exp(B * x)) : No starting values specified for some

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

Dear All, A couple of questions about the nls package. 1. I'm trying to run a nonlinear least squares regression but the routine gives me the following error message: step factor 0.000488281 reduced below `minFactor' of 0.000976563 even though I previously wrote the following command: nls.control(minFactor = 1/4096), which should set the minFactor to a lower level than the default

I  am trying to learn nls using a simple simulation. I assumed that the binomial prob varies linearly as 0.2 + 0.3*x in  x {0,1}, and the objective is to recover the known parameters a=0.2, b=0.3 ..data frame d has 1000 rows... d$x<-runif(0,1)               d$y<-rbinom(1000,1,0.2+0.3*d$x)  table(d$y,cut(d$x,breaks=5));   (-0.000585,0.199] (0.199,0.399] (0.399,0.599] (0.599,0.799]

Subject: nls, step factor 0.000488281 reduced below 'minFactor' of 0.000976563 Hi there, I'm trying to conduct nls regression using roughly the below code: nls1 <- nls(y ~ a*(1-exp(-b*x^c)), start=list(a=a1,b=b1,c=c1)) I checked my start values by plotting the relationship etc. but I kept getting an error message saying maximum iterations exceeded. I have tried changing these

When I run gnls I get the error: Error in nls(y ~ cbind(1, 1/(1 + exp((xmid - x)/exp(lscal)))), data = xy, : step factor 0.000488281 reduced below 'minFactor' of 0.000976563 My first thought was to decrease minFactor but gnlsControl does not contain minFactor nor nlsMinFactor (see below). It does however contain nlsMaxIter and nlsTol which I assume are the analogs of