Re: [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

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