search for: nls

I am interested in testing two similar nls models to determine if the lines are statistically different when fitted with two different data sets; one corn, another soybean. I know I can do this in linear models by testing for interactions. See Introductory Statistics with R by Dallgaard p212-218 for an example. I have two different data...

Hello, I am running a simple nls model (which a friend ran OK) but I get the following error: Error in nls(y ~ R * (1 - (x/K)^2), data = nls.dat, start = list(R = 0.3, : object 'R_nls_iter' not found Does anyone know what the 'R_nls_iter' error is? The data are: x=1:8 ; y=c(14,19,25,34,43,56,69,76) # star...

Dear all, I would like to use the Levenberg-Marquardt algorithm for non-linear least-squares regression using function nls.lm. Can anybody help me to find a a way to compute confidence intervals on the fitted parameters as it is possible for nls (using confint.nls, which does not work for nls.lm)? Thank you for your help Michael

Hi, I am running simulations that does multiple comparisons to control. For each simulation, I need to model 7 nls functions. I loop over 7 to do the nls using try if try fails, I break out of that loop, and go to next simulation. I get warnings on nls failures, but the simulation continues to run, except when the internal call (internal to nls) of the chol2inv fails. =======================================...

...lowing: # data.frame > Raies[1:10,] Tps SolA Solb 1 0 32.97 35.92 2 0 32.01 31.35 3 1 21.73 22.03 4 1 23.73 18.53 5 2 19.68 18.28 6 2 18.56 16.79 7 3 18.79 15.61 8 3 17.60 13.43 9 4 14.83 12.76 10 4 17.33 14.91 etc... # non linear model (work well) RaiesLossA.nls<-nls(SolA~a/(1+b*Tps)^c,start=c(a=32,b=0.5,c=0.6)) RaiesLossB.nls<-nls(Solb~a/(1+b*Tps)^c,start=c(a=33,b=1.5,c=0.5)) # bootstrap library(boot) dif.param<-function(data,i){ RaiesLossA.nls<-nls(SolA[i]~a/(1+b*Tps[i])^c,start=c(a=31,b=0.5,c=0.6)) RaiesLossB.nls<-nls(Solb[i]~a/(1+b*Tp...

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*...

Hello, Using this data: http://n4.nabble.com/file/n1676330/US_Final_Values.txt US_Final_Values.txt and the following code i got the image at the end of this message: US.final.values<-read.table("c:/tmp/US_Final_Values.txt",header=T,sep=" ") US.nls.1<-nls(US.final.values$ECe~a*US.final.values$WTD^b+c,data=US.final.values,start=list(a=2.75,b=-0.95,c=0.731),trace=TRUE) f.US1<-function(x){coef(US.nls.1)["a"]*x^coef(US.nls.1)["b"]+coef(US.nls.1)["c"]} xvals.US1<-seq(min(US.final.values$WTD),max(US.final.valu...

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))....

Hello, I am trying to bootstrap a function that extracts the log-likelihood value and the nls coefficients from an nls object. I want to sample my dataset (pdd) with replacement and for each sampled dataset, I want to run nls and output the nls coefficients and the log-likelihood value. Code: x<-c(1,2,3,4,5,6,7,8,9,10) y<-c(10,11,12,15,19,23,26,28,28,30) pdd<-data.frame(x,y) i&l...

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 positi...

...tic curve, and I have an important question to ask; in particular I have some data, from which I calculate manually the CDF, and then I import them into R and try to fit: I have the x values (my original samples) and the y values (P(X<x)). To attempt the fit I've both fitdistr() and nls(), in the way you can read in the piece of code at the end of the email. Because the fit with all data doesn't work very well, I decided to take a subset of samples randomly chosen (for some random x, the correspondant y is chosen). The first big problem is that fitdistr and nls, in th...

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, version etc are at the end...

Hello I use in /etc/X11/XF86Config XkBLayout "sr" and have defined in KDE Srbija, srpski, ISO 8859-2 And allways got the messages: fixme:string:GetLocaleInfoA 'IDEFAULTMACCODEPAGE' is not defined for your language (081A). Please define it in dlls/kernel/nls/YourLanguage.nls and submit patch for inclusion into the next Wine release. fixme:keyboard:X11DRV_KEYBOARD_DetectLayout Your keyboard layout was not found! Using closest match instead (Croatian keyboard layout) for scancode mapping. Please define your layout in windows/x11drv/keyboard.c and submi...

Hello everyone, I'm trying to test the accurracy of R on the Eckerle4 dataset from NIST and I don't understand how the control option of the nls function works. I tought nls(...) was equivalent to nls(...control=nls.control()) i.e nls.control() was the default value of control, but here is the error I get : > n2=nls(V1~(b1/b2) * exp(-0.5*((V2-b3)/b2)^2),data=ecker,start=list(b1=1.5,b2=5,b3=450,control=nls.control())) Error in nlsModel...

Dear WizaRds, I would like to fit a curve to ten points with nls() for one unknown parameter gamma in the Kahnemann/ Tversky function, but somehow it won't work and I am unable to locate my mistake. p.kum <- seq(0.1,1, by=0.1) felt.prob.kum <- c(0.16, 0.23, 0.36, 0.49, 0.61, 0.71, 0.85, 0.89, 0.95, 1) ## how to find a function that fits these point...

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...

Hello all. I was wondering if the logLike.nls() and logLike.nlme() functions are still being used. Neither function seems to be available in the most recent release of R (1.9.1). The following is contained in the help file for logLik(): "classes which already have methods for this function include: 'glm', 'lm', 'nls...

Dear all Nobody responded to my previous post so far so I try with more offending subject. I just encountered a strange problem with nls formula. I tried to use nls in cycle but I was not successful. I traced the problem to some parse command. Here is an example DF<-data.frame(x=1:10, y=3*(1:10)^.5+rnorm(10)) coef(lm(log(DF[,2])~log(DF[,1]))) (Intercept) log(DF[, 1]) 0.7437320 0.6831726 # this works coef(nls(y~a*x^...

Dear list, I have encountered a problem with predict.nls (Windows XP, R.2.5.0), but I am not sure if it is a bug... On the nls man page, an example is: DNase1 <- subset(DNase, Run == 1) fm2DNase1 <- nls(density ~ 1/(1 + exp((xmid - log(conc))/scal)), data = DNase1, start = list(xmid = 0, scal = 1))...

Hi all, I need to use the parameter estimates of nls() for further analysis. I know how to do in S+, e.g. nls(...)$parameters. In R, the attributes(nls(...)) does not have parameters, how would one get the parameter values out of nls()? Thanks in advance. Nancy