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, se.fit=TRUE, interval = "confidence", level = 0.95) , where My.nls.model is an nls object, I was able to extract the predictions okay. Thank you for your help. Penny.
Much quicker than asking for help on the list is to read the help file (which you have been asked to do in the posting guide you hopefully read). ?predict.nls tells us: "interval A character string indicating if prediction intervals or a confidence interval on the mean responses are to be calculated. At present this argument is ignored." Best, Uwe Ligges On 07.05.2011 06:17, Penny Bilton wrote:> 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, se.fit=TRUE, interval > "confidence", level = 0.95) , > > where My.nls.model is an nls object, I was able to extract the > predictions okay. > > > Thank you for your help. > Penny. > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.
On Sat, May 7, 2011 at 12:17 AM, Penny Bilton <pennybilton at xnet.co.nz> wrote:> 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, ?se.fit=TRUE, ?interval > "confidence", ?level = 0.95) ? , > > where My.nls.model is an nls object, I was able to extract the predictions > okay. >You can get these intervals using nls2. The as.lm function has an nls method which returns the lm model tangent to an nls model and use can use predict.lm on that.> library(nls2) > fm <- nls(demand ~ SSasympOrig(Time, A, lrc), data = BOD) > predict(as.lm(fm), interval = "confidence")fit lwr upr 1 7.887451 3.701701 12.07320 2 12.524979 8.219483 16.83047 3 15.251674 11.813306 18.69004 4 16.854870 13.668094 20.04164 5 17.797489 14.026668 21.56831 6 18.677578 13.393630 23.96153> predict(as.lm(fm), interval = "prediction")fit lwr upr 1 7.887451 -0.3349547 16.10986 2 12.524979 4.2409738 20.80898 3 15.251674 7.3833942 23.11995 4 16.854870 9.0932340 24.61651 5 17.797489 9.7783530 25.81663 6 18.677578 9.8453897 27.50977 Warning message: In predict.lm(as.lm(fm), interval = "prediction") : Predictions on current data refer to _future_ responses -- Statistics & Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com
Le 07/05/2011 06:17, Penny Bilton a ?crit :> 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, se.fit=TRUE, interval = > "confidence", level = 0.95) , > > where My.nls.model is an nls object, I was able to extract the > predictions okay. > > > Thank you for your help. > Penny. > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. >You can use the bootstrap methodology using the predicted values assessed through the N replications.
How do you use bootstrap to estimate the confidence as well as the prediction intervals in nonlinear regression ? -- View this message in context: http://r.789695.n4.nabble.com/plotting-confidence-bands-from-predict-nls-tp3505012p4358572.html Sent from the R help mailing list archive at Nabble.com.
On 05/02/2012 08:10, ioanna wrote:> How do you use bootstrap to estimate the confidence as well as the prediction > intervals in nonlinear regression ?With difficulty! There is far too little here to go on, and this seems an odd question unless it is homework (why dictate a problem-strewn method of solution if this is a real problem?) The real issue is how to bootstrap nonlinear regression, and you will find that discussed in all good books on the subject, such as Venables & Ripley and Davison & Hinkley. It is not trivial and the solutions are not altogether satisfactory ....> -- > View this message in context: http://r.789695.n4.nabble.com/plotting-confidence-bands-from-predict-nls-tp3505012p4358572.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.-- Brian D. Ripley, ripley at stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595