Jeffrey.Morris at sanofipasteur.com
2011-Jan-20 19:37 UTC
[R] Inverse Prediction with splines
Hello, I have fit a simple spline model to the following data. Data x y 0 1.298 2 0.605 3 0.507 4 0.399 5 0.281 6 0.203 7 0.150 8 0.101 Model Sp.1=lm(y~bs(x,df=4)) Now I wish to inverse predict the x for y=.75, say. Optimize works fine for a polynomial but I can figure out how to get the spline model into the function argument. Can anyone help me out. Thanks!! Jeff Jeff Morris Sanofi Pasteur This communication, including any attachments, is intended solely for the use of the addressee and may contain information which is privileged, confidential, exempt from disclosure under applicable law or subject to copyright. If you are not an intended recipient, any use, disclosure, distribution, reproduction, review or copying is unauthorized and may be unlawful. If you have received this transmission in error, please notify the sender immediately. Thank you. Cette communication,y compris les pieces jointes, est reservee a l'usage exclusif du destinataire et peut contenir des informations privilegiees, confidentielles, exemptees de divulgation selon la loi ou protegees par les droits de publication. Si vous n'etes pas un destinataire, toute utilisation, divulgation, distribution, reproduction, examen ou copie est non-autorisee et peut etre illegale. Si vous avez recu cette communication par erreur, veuillez aviser l'expediteur immediatement. Merci. [[alternative HTML version deleted]]
On Jan 20, 2011, at 2:37 PM, <Jeffrey.Morris at sanofipasteur.com> <Jeffrey.Morris at sanofipasteur.com > wrote:> Hello, I have fit a simple spline model to the following data. > > Data > x y > 0 1.298 > 2 0.605 > 3 0.507 > 4 0.399 > 5 0.281 > 6 0.203 > 7 0.150 > 8 0.101 > > Model > Sp.1=lm(y~bs(x,df=4)) > > Now I wish to inverse predict the x for y=.75, say. Optimize works > fine > for a polynomial but I can figure out how to get the spline model into > the function argument.Why not pass the reversed x and y vectors from the spline fit to approxfun()? > pred.Spl <- predict(Sp.1, data.frame(x=seq(0,8, by=0.01) ) ) > approxfun(x=pred.Spl, y=seq(0,8, by=0.01) )(0.75) [1] 1.447311 Looks plausible, anyway. -- David.> > Can anyone help me out. > > Thanks!! > > Jeff > > Jeff Morris > Sanofi Pasteur > > > > This communication, including any attachments, is intended solely > for the use of the addressee and may contain information which is > privileged, confidential, exempt from disclosure under applicable > law or subject to copyright. If you are not an intended recipient, > any use, disclosure, distribution, reproduction, review or copying > is unauthorized and may be unlawful. If you have received this > transmission in error, please notify the sender immediately. Thank > you. > > Cette communication,y compris les pieces jointes, est reservee a > l'usage exclusif du destinataire et peut contenir des informations > privilegiees, confidentielles, exemptees de divulgation selon la loi > ou protegees par les droits de publication. Si vous n'etes pas un > destinataire, toute utilisation, divulgation, distribution, > reproduction, examen ou copie est non-autorisee et peut etre > illegale. Si vous avez recu cette communication par erreur, veuillez > aviser l'expediteur immediatement. Merci. > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at r-project.org mailing list > stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.David Winsemius, MD West Hartford, CT
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