[R] substitute, eval and hastables

[Uwe Ligges]

Serge Boiko wrote:
> I have the following problem. I have an automatically generated named
> list with "stringified" names:
> 
> a <- list("A"=..., "B"=..., "C"=..., )
> 
> then I want to refer to the elements of the list, stored as an vector
> of names:
> 
> nn <- c("A", "B", "C"), so that I could get list elements like
> 
> a$nn[1], a$nn[2], etc. Obviously it doesn't work. Instead I do:

a$nn is evaluated at first, which looks for a list element called "nn"!
It's the convinient form for a[["nn"]].


> nn.Exp <- substitute(expression(a$b), list(b=nn[1]))
> eval(nn.Exp)

That's an enormous *overhead*. To construct such an expression, I'd try 
(given here just for fun):

   nn.Exp <- substitute(a$b, list(b=nn[1]))
   eval(nn.Exp)


> in a result I get
> expession(a$"A") but not the value stored in the list.
> Meanwhile if I manually construct expression as expression(a$"A") and
> then evaluate it, it works fine.
> 
> How do I solve that problem? Perhaps using such a list with
> "stringified" names are not very good programming style, but this is
> convenient to store and retrieve elements if list should be filled
> automatically from numerous sources. In this way I'm trying to emulate
> a hash-table behavior. Perhaps there is a better way?
> 
> Any help is highly appreciated

Use the more general [[ ]] indexing mechanism for lists as in:

  a[[nn[1]]]
  a[[nn[2]]]

See "An Introduction to R" or any book on the S language for more 
details regarding indexing.

Uwe Ligges