Is is possible to do something like,
x = function(n,v) paste("<rel name=\"",quote(n),"\" value=\"",quote(v),"\"/>",sep="")
so that x(y,1) produces,
"<rel name=\"y\" value=\"1\"/>"
of course this doesn't work and instead produces,
"<rel name=\"n\" value=\"v\"/>"
Also I have a nagging feeling that this kind of operation has a technical name, anybody know what it is?
Essentially, it would be nice if I didn't have do x("y","1").
You're looking for substitute:
x = function(n,v) paste("<rel name=\"",substitute(n),"\" value=\"",
substitute(v),"\"/>",sep="")
x(y,1)
#[1] "<rel name=\"y\" value=\"1\"/>"
Or if you're going to have more complex expressions, deparse(substitute(:
x = function(n,v) paste("<rel name=\"",deparse(substitute(n)),"\" value=\"",
deparse(substitute(v)),"\"/>",sep="")
x(y + 2, 3)
#[1] "<rel name=\"y + 2\" value=\"3\"/>"
quote doesn't handle semicolons either.
; signals end of expression, but you haven't closed your parentheses yet; in other words - that will never work, no matter what x isYou could use deparse(substitute() or match.call. Note I have used sprintf as I find it easier to decipher than paste in these situations.
xx <- function(n,v){
x <- sapply(as.list(match.call())[-1],deparse)
sprintf(fmt ='<rel name=\"%s\" value=\"%s\">',x['n'],x['v'])}
xx(y,2)
## [1] "<rel name=\"y\" value=\"2\">"
xx(y, fun(x,b,v))
## [1] "<rel name=\"y\" value=\"fun(x, b, v)\">"
Note that x(y,fun(p;d)) will not parse as it is not a valid R expression (it won't get past the language intepreter to even start
