[R] Substitution in expressions
peter dalgaard
pd@|gd @end|ng |rom gm@||@com
Tue Mar 26 14:41:51 CET 2019
Er, I'm confused.
You post some code, the code does something. In which sense is this not what you want?
This is slightly more direct:
> mapply(function(x,y) as.expression(bquote(.(x)^.(y))), base, expo)
expression(1L^2, 2L^2, 3L^3, 4L^3, 5L^4)
but I sense that you are looking for something else?
-pd
> On 26 Mar 2019, at 14:27 , Viechtbauer, Wolfgang (SP) <wolfgang.viechtbauer using 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
>
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--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
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