similar to: fitting models with gnls

Displaying 20 results from an estimated 500 matches similar to: "fitting models with gnls"

2001 Jun 01
1
nls works but not gnls
This works fine: fit42<-nls(Vfs~SSlogis(Months,Asym.Int+Asym.Group*Groupdum,xmid,scal), data=df, start=c(Asym.Int=22,Asym.Group=5,xmid=2,scal=6), na.action=na.omit) But this, identical except using gnls, doesn't converge: fit43<-gnls(Vfs~SSlogis(Months,Asym.Int+Asym.Group*Groupdum,xmid,scal), data=df, start=c(Asym.Int=22,Asym.Group=5,xmid=2,scal=6), na.action=na.omit) Error in gnls(Vfs
2006 May 24
1
problem-nlme
Hi, I have great problems with my work in R. I look for to model the growth of fish. I have "Longitudinal data", a serie of repeated measures for each individual. Using the corresponding packages "nlme" in R. I treat to fit to the data different growth functions, wich were entered by me. Unfortunately for no it was arrived at the convergence, several error messages appeared. I
2006 Jan 27
1
how calculation degrees freedom
Hi, I' m having a hard time understanding the computation of degrees of freedom when runing nlme() on the following model: > formula(my data.gd) dLt ~ Lt | ID TasavB<- function(Lt, Linf, K) (K*(Linf-Lt)) my model.nlme <- nlme (dLt ~ TasavB(Lt, Linf, K), data = my data.gd, fixed = list(Linf ~ 1, K ~ 1), start = list(fixed = c(70, 0.4)), na.action= na.include,
2007 Jun 14
1
R programming question
Dear All., I've created R-code for which a user will be asked to choose between 2 analyses. I've written one function for each type of analysis. Within each function, the users is prompted to enter information. An example is: cat("Enter value for lower Linf :\n") L1<-scan(n=1) cat("Enter value for upper Linf :\n") L2<-scan(n=1)
2012 Sep 04
3
Comparing Von Bertalanffy Growth Curves
I am trying to compare Vbert growth curves from several years of fish data. I am following the code provided by: http://www.ncfaculty.net/dogle/fishR/gnrlex/VonBertalanffy/VonBertalanffy.pdf. Specifically the section on VBGM Comparisons between groups. ? This code is pretty cut and dry. I am able to run it perfectly with the "fake" data that is provided. But when I run it with my own
2008 Aug 01
1
Confidence intervals with nls()
I have data that looks like O.lengthO.age 176 1 179 1 182 1 ... 493 5 494 5 514 5 606 5 462 6 491 6 537 6 553 6 432 7 522 7 625 8 661 8 687 10 704 10 615 12 (truncated) with a simple VonB growth model from within nls(): plot(O.length~O.age, data=OS) Oto = nls(O.length~Linf*(1-exp(-k*(O.age-t0))), data=OS, start=list(Linf=1000, k=0.1, t0=0.1), trace=TRUE) mod <- seq(0, 12)
2008 Jun 10
3
newbie nls question
I'm tyring to fit a relatively simple nls model to some data, but keep coming up against the same error (code follows): Oto=nls(Otolith ~ Linf*(1-exp(-k(AGE-to))), data = ages, start = list(Linf=1000, k=0.1, to=0.1), trace = TRUE) The error message I keep getting is "Error in eval(expr, envir, enclos) : could not find function "k"". I've used this
2008 Sep 02
2
nls.control()
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,
2003 Jun 20
1
Power Law Exponents
I am having difficulty with the calculation of the power law exponent for set of nodes within a graph. Specifically, I am interested in the distribution of in-degree and out-degree among communities of web pages where the web pages are the nodes of the graph and the hyperlinks the edges. According to the literature, the distribution of incoming and outgoing links obeys a power law distribution
2010 Sep 29
2
fitting model to resampled data
I apologize if this comes across as confusing. I will try to explain my situation as best I can. I have R bootstrapping my growth data for fish. It's resampling my database of age and length data and then produces several new datasets for me. In this case, it's resampling my data to create three new datasets of age and length data. Here is my code with my original data called
2003 Jun 02
1
Help - Curvature measures of nonlinearity
Dear colleagues, Von Bertalanffy model is commonly adjust to data on fish length (TL) and age (AGE) TL= Linf*(1-exp(-K*(AGE-t0)). Linf, K and t0 are parameters of the model. One main goal of the growth study is the comparison of growth parameter estimates between sexes of the same species, or estimates from different populations. The realibility statistical tests normally applied are highly
2010 Sep 30
1
getting the output after bootstraping
Thanks to the help of people from this forum I was able to bootstrap my data and then apply a model to it. Thanks for all your help. Everything worked out well, but I am having a difficult time getting the new parameter values. I bootstrapped the data 300 times and I want to get the 300 sets of parameter estimates and plot them in Excel. Here is my code:
2012 Mar 10
1
Draw values from multiple data sets as inputs to a Monte-Carlo function; then apply across entire matrix
Hi all, I am trying to implement a Monte-Carlo simulation for each cell in a spatial matrix (using mcd2 package) . I have figured out how to conduct the simulation using data from a single location (where I manually input distribution parameters into the R code), but am having trouble (a) adjusting the code to pull input variables from my various data sets and then (b) applying the entire
2004 Jul 27
1
Integration with "adapt"
Hi all, I need to calculate a multidimensional integration on R. I am using the command "adapt" (from library adapt), although sometimes I get the following error message: Ifail=2, lenwrk was too small. -- fix adapt() ! Check the returned relerr! in: adapt(3, linf, lsup, functn = Integrando1) I guess it happens because the domain of integration is too small, although I tried a
2012 Sep 10
0
More help need on Von Bertalanffy Growth Curves
Howdy, Last week I got some great help on why I was getting an error code when trying to run this model, thanks everyone!  I was able to get the code up and running beautifully for several data sets.  Now I am getting different errors with this new data set.  I can't figure out why, I have more data points with this species, and it is ordered exactly the same as the other species I have been
2006 Jan 26
0
degrees freedom in nlme
I'm having hard time understanding the computation of degrees of freedom when runing nlme () on the following model: > formula(my data.gd) dLt ~ Lt | ID TasavB<- function(Lt, Linf, K) (K*(Linf-Lt)) my model.nlme <- nlme (dLt ~ TasavB(Lt, Linf, K), data = my data.gd, fixed = list(Linf ~ 1, K ~ 1), start = list(fixed = c(70, 0.4)), na.action= na.include,
2009 Sep 14
1
Error: C stack usage is too close to the limit
R-help, I 'm trying to optimize a model to data using log-likelihoods but I encounter the following error message: > l= c(49.4, 57.7,64.8,70.9,78.7,86.6,88.3,91.6,99,115) > t=3:12 > fn <- function(params, l=l, t=t) { Linf <- params[1] k <- params[2] t0 <- params[3] sigma <- params[4]
2008 Feb 26
0
NLS -- multiplicative errors and group comparison
Hello, I am attempting to fit a non-linear model (Von Bertalanffy growth model) to fish length-at-age data with the purpose of determining if any of the three parameters differ between male and female fish. I believe that I can successfully accomplish this goal assuming an additive error structure (illustrated in section 1 below). My trouble begins when I attempt this analysis using a model
2006 Feb 02
1
Significance of degrees of freedom in nlme
Dear Dr. Bates, Thank you very much for your response. I had consulted the algorithm described in Pinheiro and Bates. However, what I don't understand (among other things) is why my two parameters appear to be estimated at different grouping levels (based on the DF values). Affect this different values of DF at the estimates parameters? The estimates fixed effects were get at the same level of
2006 Aug 07
2
finding x values to meet a y
Hi, I'd like to find which values of x will give me a y. In other words, in the example of a gaussian curve, I want to find the values of x that will give me a density, let's say, of 0.02. curve(((1/(sqrt(2*pi)*10))*exp(-((x-50)^2)/(2*10^2))),xlim=c(0,100)) Thanks for any help, Antonio Olinto ------------------------------------------------- WebMail Bignet - O seu provedor do