similar to: nls NAs

Why fails nls with "singular gradient" here? I post a minimal example on the bottom and would be very happy if someone could help me. Kind regards, ########### # define some constants smallc <- 0.0001 t <- seq(0,1,0.001) t0 <- 0.5 tau1 <- 0.02 # generate yy(t) yy <- 1/2 * ( 1- tanh((t - t0)/smallc) * exp(-t / tau1) ) + rnorm(length(t))*0.01 # show the curve

SAS has DUD (Does not Use Derivatives)/Secant Method for nonlinear regression, does R offer this option for nonlinear regression? I have read the helpfile for nls() and could not find such option, any suggestion? Thanks, Derek [[alternative HTML version deleted]]

Hi all, Like a lot of people I noticed that I get different results when I use nls in R compared to the exponential fit in excel. A bit annoying because often the R^2 is higher in excel but when I'm reading the different topics on this forum I kind of understand that using R is better than excel? (I don't really understand how the difference occurs, but I understand that there is a

Dear all, I am using the nlsLM function to fit a Lorentzian function to my experimental data. The LM algorithm should allow to specify limits, but the upper limit appears not to work as expected in my code. The parameter 'w', which is peak width at half maximuim always hits the upper limit if the limit is specified. I would expect the value to be in-between the upper and lower limit with

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:

Dear friends, This is the dataset I am currently working with: >dput(mod14data2_random) structure(list(index = c(14L, 27L, 37L, 33L, 34L, 16L, 7L, 1L, 39L, 36L, 40L, 19L, 28L, 38L, 32L), y = c(0.44, 0.4, 0.4, 0.4, 0.4, 0.43, 0.46, 0.49, 0.41, 0.41, 0.38, 0.42, 0.41, 0.4, 0.4 ), x = c(16, 24, 32, 30, 30, 16, 12, 8, 36, 32, 36, 20, 26, 34, 28)), row.names = c(NA, -15L), class =

Hi I?m trying to fit a nonlinear model to a derivative of the logistic function y = a/(1+exp((b-x)/c)) (this is the parametrization for the SSlogis function with nls) The derivative calculated with D function is: > logis<- expression(a/(1+exp((b-x)/c))) > D(logis, "x") a * (exp((b - x)/c) * (1/c))/(1 + exp((b - x)/c))^2 So I enter this expression in the nls function:

Hi All, I have a situation where I want an 'if' variable to be parameterized. It's entirely possible that the way I'm trying to do this is wrong, especially given the error message I get that indicates I can't do this using an 'if' statement. Essentially, I have data where I think a relationship enters when a variable (here Pwd) is below some value (z). I don't

Hi,all: I met a problem of nls. My data: x y 60 0.8 80 6.5 100 20.5 120 45.9 I want to fit exp curve of data. My code: > nls(y ~ exp(a + b*x)+d,start=list(a=0,b=0,d=1)) Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial parameter estimates I can't find out the reason for the error. Any suggesions are welcome. Many thanks. [[alternative HTML

Previously, I've posted queries about this, and thanks to postings and messages in response have recently had some success, to the extent that there is now a package called nlmrt on the R-forge project https://r-forge.r-project.org/R/?group_id=395 for solving nonlinear least squares problems that include small or zero residual problems via a Marquardt method using a call that mirrors the nls()

I have a question regarding robust nonlinear regression with nlrob. I would like to place lower bounds on the parameters, but when I call nlrob with limits it returns the following error: "Error in psi(resid/Scale, ...) : unused argument(s) (lower = list(Asym = 1, mid = 1, scal = 1))" After consulting the documentation I noticed that upper and lower are not listed as parameter in

Hi- I am trying to fit a log function to my data, with the ultimate goal of finding the second derivative of the function. However, I am stalled on the first step of fitting a curve. When I use the following code: FG2.model<-(nls((CO2~log(a*Time)+b), start=setNames(coef(lm(CO2 ~ log(Time), data=FG2)), c("a", "b")),data=FG2)) I get the following error: Error in

Hi all, I am attempting to apply a nonlinear model developed using nls to a new dataset and assess the fit of that model. At the moment, I am using the fitted model from my fit dataset as the starting point for an nls fit for my test dataset (see below). I would like to be able to view the t-statistic and p-values for each of the iterations using the trace function, but have not yet worked out

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 all, I hope you are doing well? I?m currently using the lm() function from the package stats to fit linear multifactor regressions. Unfortunately, I didn?t yet find a way to fit linear multifactor regressions with constraint coefficients? I would like the slope coefficients to be all inside an interval, let?s say, between 0 and 1. Further, if possible, the slope coefficients should add up to

Hi all I have a set of data whose scatter plot shows a very nice power relationship. My problem is when I fit a Power Trend Line in an Excel spreadsheet, I get the model y= 44.23x^2.06 with an R square value of 0.72. Now, if I input the same data into R and use model< -nls(y~ a*x^b , trace=TRUE, data= my_data, start = c(a=40, b=2)) I get a solution with a = 246.29 and b = 1.51. I have tried

Hi Gabor, Thanks a lot for your help! I tried to implement your nonlinear least squares solver on my data set. I was just wondering about the argument start. If I would like to force all my coefficients to be inside an interval, let?s say, between 0 and 1, what kind of starting values are normally recommended for the start argument (e.g. Using a 4 factor model with b1, b2, b3 and b4, I tried

This is a quadratic programming problem that you can solve using either a quadratic programming solver with constraints or a general nonlinear solver with constraints. See https://cran.r-project.org/web/views/Optimization.html for more info on what is available. Here is an example using a nonlinear least squares solver and non-negative bound constraints. The constraint that the coefficients sum

Hello, I have used the arima.sim function to generate a lot of time series, but to day I got som results that I didn't quite understand. Generating two time series z0 and z1 as eps <- rnorm(n, sd=0.03) z0 <- arima.sim(list(ar=c(0.9)), n=n, innov=eps) and z1 <- arima.sim(list(ar=c(0.9)), n=n, sd=0.03), I would expect z0 and z1 to be qualitatively similar. However, with n=10 the

The nls2 package can be used to get starting values. On Thu, Apr 28, 2016 at 8:42 AM, Aleksandrovic, Aljosa (Pfaeffikon) <Aljosa.Aleksandrovic at man.com> wrote: > Hi Gabor, > > Thanks a lot for your help! > > I tried to implement your nonlinear least squares solver on my data set. I was just wondering about the argument start. If I would like to force all my coefficients to