Substitution in expressions

Vectorize(function(x,y) as.expression(bquote(.(x)^.(y))))(base,expo)

On 26 Mar 2019, at 15:04 , Viechtbauer, Wolfgang (SP) <wolfgang.viechtbauer at maastrichtuniversity.nl> wrote:

Apologies for not being clearer. The code does what I want, but I was wondering if there is a simpler way of doing this, using substitute()/bquote() directly without the mapply().

Best,
Wolfgang

-----Original Message-----
From: peter dalgaard [mailto:pdalgd at gmail.com] 
Sent: Tuesday, 26 March, 2019 14:42
To: Viechtbauer, Wolfgang (SP)
Cc: r-help mailing list
Subject: Re: [R] Substitution in expressions

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 at 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