Hi Bert, I am indeed creating a mathematical expression, but ?plotmath doesn't cover how to do such a vectorized substitution.
Best, Wolfgang -----Original Message----- From: Bert Gunter [mailto:bgunter.4...@gmail.com] Sent: Tuesday, 26 March, 2019 15:52 To: Viechtbauer, Wolfgang (SP) Cc: r-help mailing list Subject: Re: [R] Substitution in expressions I believe you're going about this the wrong way. You seem to want mathematical expressions. Fot this, see ?plotmath. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Tue, Mar 26, 2019 at 6:28 AM Viechtbauer, Wolfgang (SP) <wolfgang.viechtba...@maastrichtuniversity.nl> wrote: Hi All, I am trying to create a vector of expressions, where the elements in the expressions are contained in other vectors (i.e., they should be substituted). I made some attempts with substitute() and bquote(), but couldn't get this to work. My solution so far is: base <- 1:5 expo <- c(2,2,3,3,4) exvec <- as.expression(unname(mapply(function(x,y) bquote(.(x)^.(y)), base, expo))) plot(NA, NA, xlim=c(0,6), ylim=c(0,2)) text(1:5, 1, exvec) Any ideas how I could get this to work with substitute() and/or bquote()? Best, Wolfgang ______________________________________________ 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.
