Try coef( graphmodeld )
And you don't need to approximate if you use the newdata argument to the predict function. I think reading the "Introduction to R" that comes with R would help. -- Sent from my phone. Please excuse my brevity. On April 5, 2018 2:00:45 PM PDT, g l <gnuli...@gmx.com> wrote: >> Sent: Thursday, April 05, 2018 at 4:40 PM >> From: "Jeff Newmiller" <jdnew...@dcn.davis.ca.us> >> >> the coef function. >> > >For the benefit of other novices, used the following command to read >the documentation: > >?coef > >Then tried and obtained: > >> cvalue100<-coef(graphmodelp~100) >> cvalue100 >NULL > >Then looked at the model values which of course correspond to original >non-modelled values. > >graphmodelp >1 2 3 4 5 6 7 8 > >91.244636 69.457794 52.873083 40.248368 30.638107 23.322525 17.753714 >13.514590 > 9 10 11 >10.287658 7.831233 5.961339 > >This prompted to think that interpolation is required, but the function >'approx' only seems to perform constant interpolation. > >Is the correct thinking to find a function to perform interpolation, >then find/write a function to differentiate the model at a specific >value of x, to find gradient at that point? ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.
