similar to: standard errors in predict.nls ()

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 =

Following an example on p 111 in 'Nonlinear Regression with R' by Ritz & Streibig, I have been fitting nls models using square brackets with the grouping variable inside. In their book is this example, in which 'state' is a factor indicating whether a treatment has been used or not: > Puromycin.m1 <- nls(rate ~ Vm[state] * + conc/(K[state] + conc), data = Puromycin,

The option se.fit in predict.nls is currently ignored. Is there any other function available to calculate the error in the predictions? Thanks, Manuel ______________________________________________ LLama Gratis a cualquier PC del Mundo. Llamadas a fijos y m??viles desde 1 c??ntimo por minuto.

Hi all I have fitted a model usinf nls function to these data: > x [1] 1 0 0 4 3 5 12 10 12 100 100 100 > y [1] 1.281055090 1.563609934 0.001570796 2.291579783 0.841891853 [6] 6.553951324 14.243274230 14.519899320 15.066473610 21.728809880 [11] 18.553054450 23.722637370 The model fitted is: modellogis<-nls(y~SSlogis(x,a,b,c)) It runs OK. Then I calculate

Hi all, I need to use the parameter estimates of nls() for further analysis. I know how to do in S+, e.g. nls(...)$parameters. In R, the attributes(nls(...)) does not have parameters, how would one get the parameter values out of nls()? Thanks in advance. Nancy

Dear all, Is there a way to retrieve standard errors from nls models? The help page tells me that arguments such as se.fit are ignored... Many thanks and best wishes Christoph -- Dr. rer.nat. Christoph Scherber University of Goettingen DNPW, Agroecology Waldweg 26 D-37073 Goettingen Germany phone +49 (0)551 39 8807 fax +49 (0)551 39 8806 Homepage http://www.gwdg.de/~cscherb1

Hi all, I would like to predict some values for an nls regression function (functional response model Rogers type II). This is an asymptotic function of which I would like to predict the asymptotic value I estimated the paramters with nls, but can't seem to get predictions for values of m choice...... This is my script: RogersII_N <-

All - I'm having an issue with trying to plot a model derived from nls() onto a simple plot.? I have included a sample data set and the code that I've been using. ?? year month day?????? date location mileage? cost gallon????? cpg ? mpg????????? x 2009???? 1?? 4?? 1/4/2009????? BZN? 124585 19.39? 14.37 1.349339 10.71677 2009-01-04 2009???? 1? 15? 1/15/2009????? BZN? 124888? 23.2? 16.12

I'm drawing a fitted normal distribution over a histogram. The use case is trivial (fitting normal distributions on densities) but I want to extend it to other fitting scenarios. What has stumped me so far is how to take the list that is returned by nls() and use it for curve(). I realize that I can easily do all of this with a few intermediate steps for any specific case. But I had expected

Colleagues, I am using "nls" successfully (2.11.1, OS X) but I am having difficulties retrieving part of the output - residual sum of squares. I have assigned the output to FIT: > > FIT > Nonlinear regression model > model: NEWY ~ PMESOR + PAMPLITUDE * cos(2 * pi * (NEWX - POFFSET)/PERIOD) > data: parent.frame() > PMESOR PAMPLITUDE POFFSET >

I haven't followed your example closely, but can't you use the predict() method for this? To draw a curve, the function that will be used in curve() sets up a newdata dataframe and passes it to predict(fit, newdata= ...) to get predictions at those locations. Duncan Murdoch On 17/10/2020 5:27 a.m., Boris Steipe wrote: > I'm drawing a fitted normal distribution over a

Hello all, I'm trying to establish some confidence intervals on predictions I am making using >predict(nls(...)) and predict.nls (unfortunately) does not utilize the se.fit option. A little more background is that I am trying to match the output with older SAS routines to maintain consistency. Because predict.nls does not provide se's for individual predictions, I have been using a

Do I have the right impression that it's currently not possible to produce confidence intervals for the nls predictions using R? I had a course were we used SAS PROC nlin and there you could get intervals for the parameters and the prediction but I do not have access to SAS. Would it be difficult to implement, I tried to dig into the help pages of nls, vcov and nlsModel but I could not

I am trying to find a confidence band for a fitted non-linear curve. I see that the predict.nls function has an interval argument, but a previous post indicates that this argument has not been implemented. Is this still true? I have tried various ways to extract the interval information from the model object without success. My code is: Model.predict <- predict(My.nls.model,

Hello Is there any function to calculate White's standard errors in R in an NLS regression. The sandwich and car package do it but they need an lm object to calculate the error's. Does anyone have idea how to do it for an NLS object ? Regards The woods are lovely, dark and deep But I have promises to keep And miles before I go to sleep And miles before I go to sleep ----- [[alternative

I've tried to predict the values from a new data.frame using the nls.predict function and keep getting the error message: Error in if (sum(wrong) == 1) stop(gettextf("variable '%s' was fitted with class \"%s\" but class \"%s\" was supplied", : missing value where TRUE/FALSE needed I first thought that it was becuase there may have been something

hello, can anyone help with this: ########################################################### ###data: measurments (response = trans) run several times at the same predictor value level (press) por<-data.frame(list(structure(list(run = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L), .Label = c("1", "3", "4"), class = "factor"), press

Dear List, my exploration of R goes on... And I REALLY enjoy it ! (thanks to all guRus). Yesterday a colleague ask me for fitting some data presented as a data.frame, something like : >a x X158.7 X150.0 ... 1 -0.25 506 183.1 2 -0.75 633 210.7 3 -1.25 674 220.3 4 -1.50 NA 244.6 5 -1.75 742 261.2 6 -2.25 787 269.1 7 -2.50 NA 283.5 8

Hi: I have usually used the GROWTH() excel function to do this but now want to see if I can do this with R. I want to predict values into the future, possibly with the predict.arima Function. I have the following weekly fish weight averages: weight <- c("2.1","2.4","2.8","3.6","4.1","5.2","6.3") week <-

Hello. I am trying to fit an asymptotic relationship (nonlinear) to some ecological data, and am having problems. I am interested in the upper bound on the data (i.e. if there is an upper limit to 'y' across a range of 'x'). As such, I am using the nonlinear quantile regression package (nlrq) to fit a michaelis mention type model. The errors I get (which are dependant on