similar to: using nls for gamma distribution (a,b,d)

Dear all i want to fit the self start model in nls. i have two question. i have a function, (asfr ~ I(((a*b)/c))+ ((c/age)^3/2)+ exp((-b^2)*(c/age)+(age/c)-2) i am wondering how to build the selfstart model. there is lost of example, (i.e. SSgompertz, SSmicman, SSweibull, etc). my question is, how to derive the function of parameters. and also which model to use for get the initials values. In the

Hello Everyone,   Learning about R and its wonderful array of functions. If it's not obvious, I usually try to find out what a function stands for. I think this helps me remember better.   One function that has me stumped is "sink." Can anyone tell me if this stands for something?   Thanks,   Paul         __________________________________________________ [[alternative HTML

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,

I could use some help understanding how nls parses the formula argument to a model.frame and estimates the model. I am trying to utilize the functionality of the nls formula argument to modify garchFit() to handle other variables in the mean equation besides just an arma(u,v) specification. My nonlinear model is y<-nls(t~a*sin(w*2*pi/365*id+p)+b*id+int,data=t1,

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

Hello, i have a problem with the function nls(). This are my data in "k": V1 V2 [1,] 0 0.367 [2,] 85 0.296 [3,] 122 0.260 [4,] 192 0.244 [5,] 275 0.175 [6,] 421 0.140 [7,] 603 0.093 [8,] 831 0.068 [9,] 1140 0.043 With the nls()-function i want to fit following formula whereas a,b, and c are variables: y~1/(a*x^2+b*x+c) With the standardalgorithm

I have found some "issues" (bugs?) with nls confidence intervals ... some with the relatively new "port" algorithm, others more general (but possibly in the "well, don't do that" category). I have corresponded some with Prof. Ripley about them, but I thought I would just report how far I've gotten in case anyone else has thoughts. (I'm finding the code

Hi everyone, I'm trying to perform a bi exponential Fit with the package NLS. the plinear algorithm seems to be a good choice see: p<-3000 q<-1000 a<--0.03 b<--0.02 t<-seq(0:144);t y<-p*exp(a*t) + q*exp(b*t)+rnorm(t,sd=0.3*(p* exp(a*t) + q*exp(b*t))) fittA <- nls(y~cbind(exp(a*t), exp(b*t)), algorithm="plinear",start=list(a=-.1, b=-0.2), data=list(y=y, t=t),

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 have the following vector which is created from 3 distinct distribution (three components) of gamma: x=c(rgamma(30,shape=.2,scale=14),rgamma(30,shape=12,scale=10),rgamma(30,shape=5,scale=6)) I want to plot the density curve of X, in a way that it shows a distinct 3 curves that represent each component. How can I do that? I tried this but doesn't work: lines(density(x)) Please

Hello, I am trying to do a nonlinear model using the "nls" command in R software. The data I am using is as follows: A<-c(7.132000,8.668667,9.880667,8.168000,10.863333,10.381333,11.059333,7.589333,4.716667,4.268667,7.265333,10.309333,8.456667,13.359333,8.624000,13.571333,12.523333,4.084667 ,NaN,NaN)

hello all i havnt had a chance to read through the references provided for the "nls" function (since the libraries are closed now). can anyone shed some light on how the "plinear" algorithm works? also, how are the fitted values obtained? also, WHAT DOES THE ".lin" below REPRESENT? thanking you in advance ###################################### i have a quick

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

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: >

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'm trying to fit a function y=k*l^(m*x) to some data points, with reasonable starting value estimates (I think). I keep getting "singular matrix 'a' in solve". This is the code: ox <- c(-600,-300,-200,1,100,200) ir <- c(1,2.5,4,9,14,20) model <- nls(ir ~ k*l^(m*ox),start=list(k=10,l=3,m=0.004),algorithm="plinear") summary(model) plot(ox,ir) testox <-

The documentation for nls says the following about the starting values: start: a named list or named numeric vector of starting estimates. Since R 2.4.0, when 'start' is missing, a very cheap guess for 'start' is tried (if 'algorithm != "plinear"'). It may be a good idea to document that when algorithm = "port", if start is a named

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:

This data are kilojoules of energy that are consumed in starving fish over a time period (Days). The KJ reach a lower asymptote and level off and I would like to use a non-linear plot to show this leveling off. The data are noisy and the sample sizes not the largest. I have tried selfstarting weibull curves and tried the following, both end with errors. Days<-c(12, 12, 12, 12, 22, 22, 22,

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 <-