# [R] substitute, eval and hastables

Uwe Ligges ligges at statistik.uni-dortmund.de
Wed Jan 29 13:14:02 CET 2003

```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, a\$nn, 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))
> 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))
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]]
a[[nn]]

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

Uwe Ligges

```