similar to: nls (PR#1532)

Full_Name: Sophie Smout Version: 1.5.0 OS: windows 2000, pc Submission from: (NULL) (138.251.190.17) I'm finding that nls iterates to the end of its iteration setting and does not jump out to report convergence even when it has clearly converged. It is also mis-reporting the number of iterations completed. These may not be unconnected... This is an example: [note - details of my computer,

I'm trying to fit a model to solve a biological problem. There are multiple independent variables, and also there are multiple responses. Each response is a function of all the independent variables, plus a set of parameters. All the responses depend on the same variables and parameters - just the form of the function changes to define each seperate response. Any ideas how I can fit

Hello, I'm using the nls function and would like to increase the number of iterations. According to the documentation as well as other postings on R-help, I've tried to do this using the "control" argument: nls(y ~ SSfpl(x, A, B, xmid, scal), data=my.data, control=nls.control(maxiter=200)) but no matter how much I increase "maxiter", I get the following error

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 R gurus, As part of finding initial values for a much more complicated fit I want to fit a function of the form y ~ a + bx + cx^d to fairly "noisy" data and have hit some problems. To demonstrate the specific R-related problem, here is an idealised data set, smaller and better fitting than reality: # idealised data set aDF <- data.frame( x= c(1.80, 9.27, 6.48, 2.61, 9.86,

//Referring to the response posted many years ago, copied below, what is the specific criterium used for singularity of the gradient matrix? Is a Singular Value Decomposition used to determine the singular values? Is it the gradient matrix condition number or some other criterion for determining singularity? // //Glenn // / / /> What does the error 'singular gradient' mean

 Hi:  Using nls how can I increase the numbers of iterations to go beyond 50.  I just want to be able to predict for the last two weeks of the year.  This is what I have:  weight_random <- runif(50,1,24)  weight <- sort(weight_random);weight weightData <- data.frame(weight,week=1:50)                          weightData plot(weight ~ week, weightData) M_model <- nls(weight ~ alpha +

Hi, I am using the nls function and it stops because the number of iterations exceeded 50, but i used the nls.control argument to allow for 500 iterations. Do you have any idea why it's not working? fm1 <- nls(npe ~ SSgompertz(npo, Asym, b2, b3), data=f,control=nls.control(maxiter=500)) Thanks for your help, Cheers, Karine.

Howdee, My question appears at #6 below: 1. I want to model the growth of each of a large number of individuals using a 4-parameter logistic growth curve. 2. nlme does not converge with the random structure that I want to use. 3. nlsList does not converge for some individuals. 4. I decided to go around nlsList using: t(sapply(split(data, list(data$id)), function(subd){coef(nls(mass ~

I'm sure this must be a nls() newbie question, but I'm stumped. I'm trying to do the example from Draper and Yang (1997). They give this snippet of S-Plus code: Specify the weight function: weight < - function(y,x1,x2,b0,b1,b2) { pred <- b0+b1*x1 + b2*x2 parms <- abs(b1*b2)^(1/3) (y-pred)/parms } Fit the model gmfit < -nls(~weight(y,x1,x2,b0,b1,b2),

Hello, I try to do a very simple nonlinear regression. The function is y = (b0 + b1*x1 + b2*x2 + b3*x3) * x4^b4 I think I do everything well, but as I set the starting value of b4 to 0 (it is the theoretically sane starting value), it converges very quickly, and to the wrong solution. Wrong in a sense, that 1) we do not expect this and 2) we do not get this on E-Views, Stata and SAS. I do not

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 Colleagues, I have x, y data (pollen and seed dispersal from oaks !) that I would like to fit with a function which look like this: p(a,b,x,y)=b/(2*pi*a?gamma(2/b))*exp(-(square_root(x?+y?)/a)power(b)) I am looking for a and b values that fit my data at best. Can someone give me hints to perform such an analysis with R ? Thanks a lot Sophie Sophie Gerber

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

Hi, I've got the following data: x<-c(1,3,5,7) y<-c(37.98,11.68,3.65,3.93) penetrationks28<-dataframe(x=x,y=y) now I need to fit a non linear function so I did: fit <- nls(y ~ I(a+b*exp(1)^(-c * x)), data = penetrationks28, start = list(a=0,b = 1,c=1), trace = T) The error message I get is: Error in nls(y ~ I(a + b * exp(1)^(-c * x)), data = penetrationks28, start = list(a =

Hi, I've noticed I get different results fitting a function to some data on my laptop to when I do it on my computer at work. Here's a code snippet of what I do: ##------------------------------------------------------------------ require(circular) ## for Bessel function I.0 ## Data: dd <- c(0.9975948929787, 0.9093316197395, 0.7838819026947, 0.9096108675003, 0.8901804089546,

Hi All. I've run into a problem with the plinear algorithm in nls that is confusing me. Assume the following reaction time data over 15 trials for a single unit. Trials are coded from 0-14 so that the intercept represents reaction time in the first trial. trl RT 0 1132.0 1 630.5 2 1371.5 3 704.0 4 488.5 5 575.5 6 613.0 7 824.5 8 509.0 9

Hi all, I am trying to use nonlinear regression (nls) to analyze some seed germination data, but am having problems with error codes. The data that I have closely matches the germination dataset included in the drc package. Here is the head of the data temp species start end germinated TotSeeds TotGerminated Prop 1 10 wheat 0 1 0 20 0 0.0 2 10 wheat

Dear list, Can anyone provide a package or code for estimating the parameters of a mixture of c (c >=2) univariate normal distributions? I've tried the algorithm provided by Venables & Ripley (1999) p 263, for the mixture of two normal, but I don't find the "ms" function in R. I've used nls instead, but I'm not sure if it works the same. The data I have is very

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