[R] all.vars for nested expressions

Eik Vettorazzi E.Vettorazzi at uke.de
Mon Apr 29 16:45:58 CEST 2013

Hi Felix,
I thought, this could be an easy task for substitute, and the following
works as expected:

# [1] "tp" "fn"

But (of course)
does not yield the desired result, and I can't figure out, how to set up
as.name, bquote, eval, deparse etc to do the task properly.

Instead, my approach is a recursive call to all.vars

  #check if there is an object with name x
  if(exists(x)) lapply(all.vars(get(x)),xall.help) else x}

   if (!is.character(x)) x<-paste(substitute(x))
   #for convenience put in a single vecotr
   #xall.help returns a 'parsed tree'

a <- expression(fn+tp)
sen <- expression(tp/a)

# [1] "tp" "n1" "n2"


Am 29.04.2013 13:33, schrieb flxms:
> Dear R fellows,
> Assume I define
> a <- expression(fn+tp)
> sen <- expression(tp/a)
> Now I'd like to know, which variables are necessary for calculating sen
> all.vars(sen)
> This results in a vector c(tp,a). But I'd like all.vars to evaluate the
> sen-object down to the ground level, which would result in a vector
> c(tp,fn) (because a was defined as fn+tp). In other words, I'd like
> all.vars to expand the a-object (and all other downstream objects). I am
> looking for a solution, that works with much more levels. This is just a
> very simple example.
> I'd appreciate any suggestions how to do that very much!
> Thanks in advance,
> Felix
> 	[[alternative HTML version deleted]]
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Eik Vettorazzi
Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf

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