similar to: nls, the four parameter logisitc equation, and prediction band

The error msg is telling you that R cannot evaluate the loss function, so you should not expect answers. You might try examining the data -- Are there NA or Inf entries? Or prepare a dataframe with just X and Y, sort by X and graph. Then check the nls computations by sampling, say, every 100 X's to give you a dataset with about 160 observations. If that doesn't work, it is at least

Greetings, We are performing a meta-analysis of mink pup survival data versus chemical concentration. We have modeled percent survival successfully using nls as shown below and the plot. What we need to do is construct a confidence interval on the concentration at which we get 50% survival (aka the EC50, although we may want other percent survivals in the future). My first question is, what seems

Shalini Raghavan 3M Pharmaceuticals Research Building 270-03-A-10, 3M Center St. Paul, MN 55144 E-mail: sraghavan at mmm.com Tel: 651-736-2575 Fax: 651-733-5096 ----- Forwarded by Shalini Raghavan/US-Corporate/3M/US on 08/16/2004 11:25 AM ----- Shalini

Hi, I've been trying to work with the gnls() function in the "nlme" package. My decision to use gnls() was so that I could fit varPower and such to some of the data. However, in working with a small dataset, I've found that the results given by gnls() don't seem to make any sense and they differ substantially from those produced by nls(). I suspect that I am just

Dear R users,   I am struggling to fit expo-linear equation to my data using "nls" function. I am always getting error message as i highlighted below in yellow color:     ### Theexpo-linear equation which i am interested to fit my data:       response_variable =  (c/r)*log(1+exp(r*(Day-tt))), where "Day" is time-variable   ## my response variable   rl <-

I am trying to fit a curve to a cumulative mortality curve (logistic) where y is the cumulative proportion of mortalities, and t is the time in hours (see below). Asym. at 0 and 1 > y [1] 0.00000000 0.04853859 0.08303777 0.15201970 0.40995074 0.46444992 0.62862069 0.95885057 1.00000000 [10] 1.00000000 1.00000000 > t [1] 0 13 20 24 37 42 48 61 72 86 90 I tried to find starting values for

Folks I have modified an existing function to calculate 'ec/ld/lc' 50 values and their associated Fieller's confidence limits. It is based on EC50.calc (writtien by John Bailer) - but also borrows from the dose.p (MASS) function. My goal was to make the original EC50.calc function flexible with respect to 1) probability at which to calculate the expected dose, and 2) the link

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

Hi, I posted a question (bellow) a few weeks ago and had a reply (thanks Christian) that partly solves the problem, but I still would like to be able to restrict some of the independent variables in a nls model to be always >0, (is there a way to do it)?? Thanks, Angel >From: "Christian Ritz" <ritz at dina.kvl.dk> >To: "Angel -" <angel_lul at

Moved from r-help: On Thu, 14 Aug 2003 09:08:26 -0700, Spencer Graves <spencer.graves@pdf.com> wrote : >p.s. The following command in S-Plus 6.1 seems to work fine but >produces an error in R 1.7.1: > >nls(y~a, data=tstDf, start=list(a=1)) >Error in nlsModel(formula, mf, start) : singular gradient matrix at >initial parameter estimates This looks like a bug in

Hi Everyone, I am trying to use NLS to fit a dataset using a Kappa function, but I am having problems. Depending on the start values that I provide, I get either: Error in numericDeriv(form[[3L]], names(ind), env) : Missing value or an infinity produced when evaluating the model Or Error in nls(FldFatRate ~ funct3(MeanDepth_m, h, k, z, a), data = data1, : singular gradient I think these

Hi, I am rather new to R, but both myself and another much more experience user cannot figure this out. I have a collegue who has a 800+ nonlinear regressions to run for seed germination (different species, treatments, etc.) over time. I created a looping structure to extract the parameters from each regression; I will then use the parameters themselves for further analysis. I would like to

Hi, df <- read.table("data.txt", header=T); library(nls); fm <- nls(y ~ a*(x+d)^(-b), df, start=list(a=max(df->y,na.rm=T)/2,b=1,d=0)); I was using the following routine which was giving Singular Gradient, Error in numericDeriv(form[[3]], names(ind), env) : Missing value or an Infinity produced when evaluating the model errors. I also tried the

Dear R-Help, I have a generated file that looks like the following: ----- Begin file ----- # # Output File # float Version 2002.700000000000 int Numdays 31 int NumOFEs 1 # # Hillslope-specific variables # char HillVarNames[ 3 ] {Days In Simulation} {Hillslope: Precipitation (mm)}

Dear R community, I have currently some problems with non linear regression analysis in R. My data correspond to the degradation kinetic of a pollutant in two different soil A and B, x data are time in day and y data are pollutant concentration in soil. In a first time, I want to fit the data for the soil A by using the Ct = C0*exp(-k*Tpst) with Ct the concentration of pollutant at time t, C0

Dear list, I am running some linear regressions through lapply, >lapply(c('EMAX','EC50','KOUT','GAMMA'),function(x)confint(lm(get(x)~RR0,dataset2))) I got results like [[1]] 2.5 % 97.5 % (Intercept) 0.6595789212 0.8821691261 RR0 -0.0001801771 0.0001489083 [[2]] 2.5 % 97.5 % (Intercept) -63.83694930

Hi, I am a non-expert user of R. I am essaying the fit of two different functions to my data, but I receive two different error messages. I suppose I have two different problems here... But, of which nature? In the first instance I did try with some different starting values for the parameters, but without success. If anyone could suggest a sensible way to proceed to solve these I would be

I'm running into problems trying to use the nls function to fit the some data. I'm invoking nls using nls(s~k/(a+r)^b, start=list(k=1, a=13, b=0.59)) but I get errors indicating that the step has been reduced below the minimum step size or an inifinity is generated in numericDeriv. I've tried to use a variety of starting values for a, b, k but get similar errors. Is there

Hello, I want to fit a Gompertz model for tree diameter growth that depends on a 4 levels edaphic factor (?Drain?) and I don?t manage to introduce the factor variable in the formula. Dinc is the annual diameter increment and D is the Diameter. >treestab > Dinc D Drain [1,] 0.03 26.10 2 [2,] 0.04 13.05 1 [3,] 0.00 24.83 1 [4,] 0.00 15.92 4

Hello. I am using the nonlinear least squares function (nls). The function that I am trying to fit seems to be very sensitive to the starting values and, if these are not chosen properly, the nls function stops and gives an error message: Error in numericDeriv(form[[3]], names(ind), env) : Missing value or an Infinity produced when evaluating the model In addition: Warning