similar to: 4 parameter logistic model

Dear sirs It seems like there is a bug in `profile.nls` with `algorithm = "plinear"` when a matrix is supplied on the right hand side. Here is the bug and a potential fix ##### # example where profile.nls does not work with `plinear` but does with # `default` require(graphics) set.seed(1) DNase1 <- subset(DNase, Run == 1) x <- rnorm(nrow(DNase1)) f1 <- nls(density ~ b1/(1 +

Full_Name: Madeleine Yeh Version: 1.9.1 OS: AIX 5.2 Submission from: (NULL) (151.121.225.1) After compiling R-1.9.1 on AIX 5.2 using the IBM cc compiler, I ran the checks. One of them failed. Here is the output from running the check solo. root@svweb:/fsapps/test/build/R/1.9.1/R-1.9.1/tests/Examples: ># ../../bin/R --vanilla < stats-Ex.R R : Copyright 2004, The R

Hi, this following problem is a S-Plus problem, I know many guys here are also experts in S-plus, so I am posting here, too. Thanks I encountered a weird problem of fitting nls inside a loop, it works well in R, but not in S-plus. The code is: >

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

Hello, nls library provides 6 self-starting models, among them: SSfp, a four parameters logistic function. Its self-starting procedure involves several steps. One of these steps is: pars <- as.vector(coef(nls(y ~ cbind(1, 1/(1 + exp((xmid - x)/exp(lscal)))), data = xydata, start = list(lscal = 0), algorithm = "plinear"))) which assumes an initial value of lscal equal to 0. If lscal

Dear R users I'm a new user of R and I have a basic question about the 4-parameter logistic model. According to the information from Pinheiro & Bates the model is: y(x)=theta1+(theta2-theta1)/(1+exp((theta3-x)/theta4)) == y(x)=A+(B-A)/(1+exp((xmid-input)/scal)) from the graph in page 518 of the book of the same authors (mixed models in S) theta 1 corresponds to the horizontal asymptote

Howdee, I am able to fit a 4-parameter logistic growth curve to a dataset which comprise many individuals (using R v. 2.3.1). Yet, if I want to obtain the parameters for each individual (i.e., for each 'id') using nlsList, then I obtain an Error message which I have trouble interpreting. Any advice as to how I can solve this problem? Thanks for your time, Marc > reg <-nls(mass ~

Dear all, I am trying to fit a non linear regression model to time series data. If I do this: reg.logis = nls(myVar~SSlogis(myTime,Asym,xmid,scal)) I get this error message (translated to English from French): Erreur in nls(y ~ 1/(1 + exp((xmid - x)/scal)), data = xy, start = list(xmid = aux[1], : le pas 0.000488281 became inferior to 'minFactor' of 0.000976562 I then tried to set

Hi, I am not sure this is a bug but I can repeat it, The functions and data are below. I know this is nasty data, and it is very questionable whether a 4pl model is appropriate, but it is data fed to an automated tool and I would have hoped for an error. Does this repeat for anyone else? My details: > version _ platform i686-pc-linux-gnu

Dear all Please help me with correct syntax of predict.nlme. I would like to predict from nlme object for new data. I used predict(fit.nlme6, data=newdata) but I have always got fitted values, no matter how I changed newdata. I have > summary(fit.nlme6) Nonlinear mixed-effects model fit by maximum likelihood Model: konverze ~ SSfpl(tepl, A, B, xmid, scal) Data: limity.gr AIC

Hello, How do I specify the formula and random effects without a startup object ? I thought it would be a mixture of nls and lme. after trying very hard, I ask for help on using nlme. Can someone hint me to some examples? I constructed a try using the example from nls: #variables are density, conc and Run #all works fine with nls DNase1 <- subset(DNase, Run == 1 ) fm2DNase1 <- nls(

Hi! When checking validity of a model for a large number of experimental data I thought it to be interesting to check the information provided by the summary method programmatically. Still I could not find out which method to use to get to those data. Example (not my real world data, but to show the point): [BEGIN] > DNase1 <- subset(DNase, Run == 1) > fm1DNase1 <- nls(density ~

Hi, I would like to be able to access the calculated results from the nls package. Using the example in R, fm3DNase1 we can reurn certain parts of the calculations: > coef(fm3DNase1) Asym xmid scal 2.345179 1.483089 1.041454 > resid(fm3DNase1) [1] -0.0136806237 -0.0126806237 0.0089488569 0.0119488569 -0.0025803222 [6] 0.0064196778 0.0026723396 -0.0003276604

Hi, I am trying to use F test or Chi-square test to test if 2 5-parameter (A, B, xmid, scal and H) logistic curves are parallel based on residual sum of squares. What's usually done is to first fit the 2 curves using a constraint (or global) model where all parameters are kept the same except for "xmid"; then fit 2 independent curves using unconstraint models where all 5 parameters

Hi R help, I am trying to determine how nls() generates a function based on the self-starting SSlogis and what the formula for the function would be. I've scoured the help site, and other literature to try and figure this out but I still am unsure if I am correct in what I am coming up with. ************************************************************************** dat <-

Dear all, I am studying a bit the various support functions that exist for extracting information from fitted model objects. From the help files it is not completely clear to me whether the number returned by nobs() should be the same as the "nobs" attribute of the object returned by logLik(). If so, then there is a slight inconsistency in the methods for 'nls' objects with

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

Dear list, I have encountered a problem with predict.nls (Windows XP, R.2.5.0), but I am not sure if it is a bug... On the nls man page, an example is: DNase1 <- subset(DNase, Run == 1) fm2DNase1 <- nls(density ~ 1/(1 + exp((xmid - log(conc))/scal)), data = DNase1, start = list(xmid = 0, scal = 1)) alg = "plinear", trace =

Dear R-help subscribers Would you tell me how to apply SSfpl with binary data as below? Unfortunately, there is not the EXAMPLE in help(SSfpl) for binary data but for quantitative data(Chick). V1: dose V2: log-transformed dose V3: response (rate) V1 V2 V3 1 0.775 -0.2548922 0.1666667 2 5.000 1.6094379 0.8148148 3 10.000 2.3025851 0.5000000 4 20.000 2.9957323

Dear all, I encounter some discrepancies when comparing the deviance of a weighted and unweigthed model with the AIC values. A general example (from 'nls'): DNase1 <- subset(DNase, Run == 1) fm1DNase1 <- nls(density ~ SSlogis(log(conc), Asym, xmid, scal), DNase1) This is the unweighted fit, in the code of 'nls' one can see that 'nls' generates a vector