Good morning Erin, eval(parse(text = "pexp(3.2,rate=1)"))
seems to work But the general rule applies: library(fortunes) fortune("parse()") Best, Michael On Tue, Nov 22, 2011 at 1:23 PM, Erin Hodgess <erinm.hodg...@gmail.com> wrote: > Dear R People: > > Hope you're having a nice day. > > Here is a character vector: > >> yz > [1] "pexp(3.2,rate=1)" >> str(yz) > chr "pexp(3.2,rate=1)" >> > And I would like to evaluate that vector. > > I tried: >> eval(as.expression(yz)) > [1] "pexp(3.2,rate=1)" >> > > But that doesn't work. > > Any suggestions would be most welcome. I have a feeling that it's > quite simple and that I'm having a forest vs. trees issue. > > Thanks, > Erin > > > -- > Erin Hodgess > Associate Professor > Department of Computer and Mathematical Sciences > University of Houston - Downtown > mailto: erinm.hodg...@gmail.com > > ______________________________________________ > R-help@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. ______________________________________________ R-help@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.