similar to: Non linear regression using Levenberg-Marquardt method

Hi All, Is the Levenberg-Marquardt algorithm available in R. This method combines the steepest descent algorithm and Newton's method. Thanks in Advance, Dermot MacSweeney. ************************************************************** Dermot MacSweeney NMRC, Email: dsweeney at nmrc.ucc.ie Lee Maltings, Tel: +353 21 904178 Prospect Row, Fax: +353 21 270271 Cork, WWW:

Hi, It seems to me that the most suitable method in R for curve-fitting is the use of nls, which uses a Gauss-Newton (GN) algorithm, while the use of the Levenberg-Marquardt (LM) algorithm does not seem to be very stressed in R. According to this [1] by Ripley, 'Levenberg-Marquardt is hardly competitive these days' which could imply the low emphasize on LM in R. The position of LM is, to

Hi, I have a problem fitting the following Weibull Model to a set of data. The model is this one: a-b*exp(-c*x^d) If I fitted the model with CurveExpert I can find a very nice set of coefficients which create a curve very close to my data, but when I use the nls.lm function in R I can't obtain the same result. My data are these: X Y 15 13 50 13 75 9 90 4 With the commercial

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

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

Hi! I'm translating some code from Matlab to R and I found a problem. I need to translate Matlab's function 'lsqnonlin' (http://www-ccs.ucsd.edu/matlab/toolbox/optim/lsqnonlin.html) into R, and at the beginning I thought it would be the same as R's 'optim'. But then I looked at the definition of 'lsqnonlin' and I don't quite see how to make

Dear Mr Graves, Thank you very much for your response. Nobody else from this mailing list ventured to reply to me for the two weeks since I posted my question. "nlminb" and "optim" are just optimization procedures. What I need is not just optimization, but a nonlinear CURVE FITTING procedure. If there is some way to perform nonlinear curve fitting with the

Hello! I am running R-2.3.1-i386-1 on Slackware Linux 10.2. I am a former matlab user, moving to R. In matlab, via the cftool, I performed nonlinear curve fitting using the method "nonlinear least squares" with the "Trust-Region" algorithm and not using robust fitting. Is it possible to perform the same analysis in R? I read quite a lot of R documentation, but I could not find

In the course of revising a paper I have had occasion to attempt to maximize a rather complicated log likelihood using the function nlm(). This is at the demand of a referee who claims that this will work better than my proposed use of a home- grown implementation of the Levenberg-Marquardt algorithm. I have run into serious hiccups in attempting to apply nlm(). If I provide gradient and

Dear All, In the past I have often used minpack (http://bit.ly/zXVls3) relying on the Levenberg-Marquardt algorithm to perform non-linear fittings. However, I have always dealt with a function of a single variable. Is there any difference if the function depends on two variables? To fix the ideas, please consider the function f(R,N)=(a/(log(2*N))+b)*R+c*N^d, where a,b,c,d are fit parameters. For

Dear All, I'd like to fit a "kind" of logistic model to small data-set using nonlinear least-squares regression. A transcript of R-script are reproduced below. Estimated B and T (the model's coeff, herein B=-8,50 and T=5,46) seem appropriate (at least visually) but are quite diff from those obtained w/ SPSS (Levenberg-Marquardt): B=-19,56 and T=2,37. Am I doing something wrong in

The on-line documentation for par() refers to a 'gamma' argument. However, if 'par()' is invoked, there is no gamma argument. -- Art Salwin salwin@mitretek.org (202) 863-2985 (202) 863-2988 (fax) --please do not edit the information below-- Version: platform = Windows arch = x86 os = Win32 system = x86, Win32 status = major = 1 minor = 0.0 year = 2000 month = February

The on-line documentation for the 'ann' argument under par(), states "If set to 'FALSE'...". However, this parameter shows up as a numeric on my machine, not as a logical. -- Art Salwin salwin@mitretek.org (202) 863-2985 (202) 863-2988 (fax) --please do not edit the information below-- Version: platform = Windows arch = x86 os = Win32 system = x86, Win32

The documentation for 'order()' has the following sentence: `sort.list' is the same, using only one argument put allowing partial sorting. Should "put" be "but" or possibly "'partial'"? -- Art Salwin salwin@mitretek.org (202) 863-2985 (202) 863-2988 (fax) --please do not edit the information below-- Version: platform = Windows arch =

Quick question. Say I have a geographical database with counties and zip codes where counties have and belong to many zip codes. zip_codes (id, zip_code) counties (id, name) When I create the association table, the Rails way says to do the following: counties_zip_codes (county_id, zip_code_id). However, given that zip_codes.zip_code is itself a candidate key, I would much prefer to do the

I am regularly running into a problem where I can't seem to figure out how maintain correct data order when selecting data out of a dataframe. The below code shows an example of trying to pull data from a dataframe using ordered zip codes. My problem is returning the pulled data in the correct order. This is a very simple example, but it illustrates a regular problem that I am running into.

i've search the email archives, searched the documention of various map packages and done an R-site search, but have been unable to find direct resources for creating maps of the US that are colored or annotated or ... by zipcode data. For example, create a map of the US and color each zipcode region by its population using two vectors z,p containing the zipcode and population,

Is there an R function that inputs a value and a vector, and returns the percentile corresponding to the value? Something like pvalue<-function(x,vector){mean(x>=vector)} but with interpolation. Given the above function, is there a way to do the following without the loop: for(i in 1:10) y[i] <- pvalue(x[i],vector) where x and y are numeric vectors of length 10? -- Art Salwin

"An Introduction to R" notes that the function tapply() can be used to index a vector by multiple categories. But they don't show how to do this. How does one implement, for example, the example in the text in which tax accountants are indexed by both state and sex? Is there a way to use tapply() to put the vector into bins defined by the levels instead of having to exactly match

I am interested in locating a script to implement a sampling scheme that would basically make it more likely that a particular observation is chosen based on a weight associated with the observation. I am trying to select a sample of ~30 census blocks from each ZIP code area based on the proportion of women in a ZCTA living in a particular block. I want to make it more likely that a block will