similar to: question about non-linear least squares in R

Displaying 20 results from an estimated 3000 matches similar to: "question about non-linear least squares in R"

2010 Apr 19
3
nls for piecewise linear regression not converging to least square
Hi R experts, I'm trying to use nls() for a piecewise linear regression with the first slope constrained to 0. There are 10 data points and when it does converge the second slope is almost always over estimated for some reason. I have many sets of these 10-point datasets that I need to do. The following segment of code is an example, and sorry for the overly precise numbers, they are just
2013 May 28
6
Ocultar componentes de una lista
Saludos a todos, Quisiera saber como ocultar algunos componentes de una lista para que cuando crees una funcion solo se muestre una parte de esa lista, y que al resto tengas acceso cuando descompongas ese objeto, por ejemplo: Tengo la siguiente data: data=data.frame(Samples=1:10,Sobs=c(12.6,22.4,27.4,30.5,32.5,34.3,36,37.4,38.9,39.9)) Cuando uso la funcion nls: A <-
2009 Jul 12
2
Nonlinear Least Squares nls() programming help
Hi, I am trying to use the nls() function to closely approximate a vector of values, colC and I'm running into trouble. I am not sure how if I am asking the program to do what I think its doing, because the same minimization in Excel's Solver does not run into problems. If anyone can tell me what is going wrong, and why I'm getting a singular convergence(7) error, please tell me. I
2005 Jan 06
1
nls - convergence problem
Dear list, I do have a problem with nls. I use the following data: >test time conc dose 0.50 5.40 1 0.75 11.10 1 1.00 8.40 1 1.25 13.80 1 1.50 15.50 1 1.75 18.00 1 2.00 17.00 1 2.50 13.90 1 3.00 11.20 1 3.50 9.90 1 4.00 4.70 1 5.00 5.00 1 6.00 1.90 1 7.00 1.90 1 9.00 1.10 1 12.00 0.95 1 14.00
2007 Apr 15
1
nls.control( ) has no influence on nls( ) !
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*128))" inside the brackets of nls, but the same error message
2007 Sep 04
1
Help: how can i build a constrained non-linear model?
Dear I have a data.frame, and want to fit a constrained non-linear model: data: x y -0.08 20.815 -0.065 19.8128 -0.05 19.1824 -0.03 18.7346 -0.015 18.3129 0.015 18.0269 0.03 18.4715 0.05 18.9517 0.065 19.4184 0.08 20.146 0 18.2947 model: y~exp(a)*(x-m)^4+exp(b)*(x-m)^2+const I try to use nls() and set start=list(a=1,b=1,c=1,m=1), but which always give me a error message that
2012 Jan 25
1
solving nls
Hi, I have some data I want to fit with a non-linear function using nls, but it won't solve. > regresjon<-nls(lcfu~lN0+log10(1-(1-10^(k*t))^m), data=cfu_data, > start=(list(lN0 = 7.6, k = -0.08, m = 2))) Error in nls(lcfu ~ lN0 + log10(1 - (1 - 10^(k * t))^m), data = cfu_data, : step factor 0.000488281 reduced below 'minFactor' of 0.000976562 Tried to increase minFactor
2007 Sep 05
3
'singular gradient matrix’ when using nls() and how to make the program skip nls( ) and run on
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)). The error message is: "number of iterations exceeded maximum of
2012 Apr 17
3
error using nls with logistic derivative
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:
2006 Aug 04
1
gnlsControl
When I run gnls I get the error: Error in nls(y ~ cbind(1, 1/(1 + exp((xmid - x)/exp(lscal)))), data = xy, : step factor 0.000488281 reduced below 'minFactor' of 0.000976563 My first thought was to decrease minFactor but gnlsControl does not contain minFactor nor nlsMinFactor (see below). It does however contain nlsMaxIter and nlsTol which I assume are the analogs of
2013 May 28
3
Ocultar componentes de una lista
Hola Javier, Creo que comprendio mal mi pregunta, quiero saber como ocultar algunos componentes de una lista producida por una funcion que uno crea, pero que solo esten ocultos y que pueda tener acceso a esos componentes cuando uso la funcion "str", la cual despliega todos los componentes de la lista. Además como hacer para que salga ordenado o de una forma preferida los componentes
2009 Jul 09
1
nls, reach limit bounds
Hi, I am trying to fit a 4p logistic to this data, using nls function. The function didn't freely converge; however, it converged if I put a lower and an upper bound (in algorithm port). Also, the b1.A parameter always takes value of the upper bound, which is very strange. Has anyone experienced about non-convergent of nls and how to deal with this kind of problem? Thank you very much.
2005 Feb 22
3
problems with nonlinear fits using nls
Hello colleagues, I am attempting to determine the nonlinear least-squares estimates of the nonlinear model parameters using nls. I have come across a common problem that R users have reported when I attempt to fit a particular 3-parameter nonlinear function to my dataset: Error in nls(r ~ tlm(a, N.fix, k, theta), data = tlm.data, start = list(a = a.st, : step factor 0.000488281
2008 Apr 10
1
(no subject)
Subject: nls, step factor 0.000488281 reduced below 'minFactor' of 0.000976563 Hi there, I'm trying to conduct nls regression using roughly the below code: nls1 <- nls(y ~ a*(1-exp(-b*x^c)), start=list(a=a1,b=b1,c=c1)) I checked my start values by plotting the relationship etc. but I kept getting an error message saying maximum iterations exceeded. I have tried changing these
2009 Jul 20
1
Argument problem in function wrapper
Dear all, we are writing a wrapper for the nls function in library stats. We are having a problem with one of the arguments (weightsArgument) which seems not to reach nls even if we explicitly assign it in the function call. We are attaching the simplest code reproducing the error and the output calling the wrapper and calling nls directly. We are using R 2.9.0 library(stats) wrappernls <-
2013 May 28
3
Ocultar componentes de una lista
Hola Eva, Gracias por la respuesta fue de mucha ayuda. Lo que me gustaria saber es por que solo algunas funciones que usan el metodo S3 son visibles en su estructura con la funcion "body". Por ejemplo si coloco body(print.ts) Me va a salir la estructura de esa funcion, sin embargo si coloco body(print.nls) Me sale el siguiente mensaje :Error in body(print.nls) :
2015 Mar 18
1
Help
Hi to All, I am fitting some models to a data using non linear least square, and whenever i run the command, parameters value have good convergence but I get the error in red as shown below. Kindly how can I fix this problem. Convergence of parameter values 0.2390121 : 0.1952981 0.9999975 1.0000000 0.03716107 : 0.1553976 0.9999910 1.0000000 0.009478433 : 0.2011017 0.9999798 1.0000000
2010 Apr 19
2
nls minimum factor error
Hi, I have a small dataset that I'm fitting a segmented regression using nls on. I get a step below minimum factor error, which I presume is because residual sum of square is still "not small enough" when steps in the parameter space is already below specified/default value. However, when I look at the trace, the convergence seems to have been reached. I initially thought I might
2010 Apr 15
2
using nls for gamma distribution (a,b,d)
Dear all i want to estimated the parameter of the gamma density(a,b,d) f(x) = (1/gamma(b)*(a^b)) * ((x-d)^(b-1)) * exp{-(x-d)/a)} for x>d f(x) = Age specific fertility rate x = age when i run this in R by usling nls() gamma.asfr <- formula(asfr ~ (((age-d)^(b-1))/((gamma(b))*(a^b)))* exp(-((age-d)/a))) gamma.asfr1 <- nls(gamma.asfr, data= asfr.aus, start = list(b = 28, a = 1, d= 0.5),
2004 Jan 14
2
Generalized least squares using "gnls" function
Hi: I have data from an assay in the form of two vectors, one is response and the other is a predictor. When I attempt to fit a 5 parameter logistic model with "nls", I get converged parameter estimates. I also get the same answers with "gnls" without specifying the "weights" argument. However, when I attempt to use the "gnls" function and try to