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typeof ( substitute(expression(),..) ) -- musings

2 messages · Martin Maechler, Peter Dalgaard

Original
Martin Maechler
# 
[Relates to  math text / normal text mixing in plots ..]

Peter, I knew you'd know the answer quickly...

Anyway, one needs to be somewhat careful here [R version  pre 0.64.2]:

    typeof(print( e1 <-            expression(a + b))) # expression 
    typeof(print(se1 <- substitute(expression(a + b), list(a = 1)))) # language

    length(e1) # == 1
    length(se1)# == 2

    text.e <- function(x,y,e) {
        text(x,y,paste(deparse(substitute(e)),": "),adj=c(1,0))
	text(x,y,e, adj=0)
    }
    plot(1,type="n", axes= F)
    text.e(1,1.4, e1); text.e(1,1.3, se1); text.e(1,1.2, se1[[2]])

Which shows that you can't use the result of `substitute' directly, but
rather the 2nd element [[which is *still* not an expression !!]]

This is all +- S-plus (3.4) compatible 
     {{some details are NOT compatible; e.g. 
	     as.expression(se1[[2]])
       gives `what you expect' in R, but not in S-plus 3.4 or 5.0r3
     }}

  e1 is an expression	whereas
 se1 is a  call (aka "language" in R).

----
Is this documented anywhere? [blue book; language reference; ?? ]

Martin
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Peter Dalgaard
# 
Martin Maechler <maechler@stat.math.ethz.ch> writes:
..and the first thing to be careful with is the version, expressions
behave slightly different in 0.65!
Yes. I forgot. se1 is not an expression object but a call to the
expression() - which evaluates to an expression object, i.e. one needs 

se2 <- eval(substitute(expression(a + b), list(a = 1)))

or you can ditch the expression() bit and use

sc2 <- substitute(a + b, list(a = 1))

which is 1 + b, mode "call" and will work with *most* graphics
operations, but there's a caveat: Sometimes language objects get
evaluated when you don't want them to. The notorious example is the
do.call construction in boxplot/bxp.
It will be... (It may be in the blue book, but my fading memory of it
says that it is not all that clear on this particular subject.)