[R] computing on expressions

[Vadim Ogranovich]

The following seems to work too:

expr2 <- expression(a>0)
expr2 <- as.expression( bquote(.(expr2[[1]]) & b>0 ) )

Here as.expression converts the 'call' type returned by bquote to 'expression' type.

Thanks again Dunkan.

Vadim

-----Original Message-----
From: Duncan Murdoch [mailto:murdoch at stats.uwo.ca]
Sent: Friday, October 03, 2008 11:41 AM
To: Vadim Organovich
Cc: r-help at r-project.org
Subject: Re: [R] computing on expressions

On 10/3/2008 11:48 AM, Vadim Organovich wrote:
> Dear R-users,
>
> Suppose I have an expression:
>
> expr = expression(a>0)
>
> and now I want to modify it to expression(a>0 & b>0). The following doesn't work:
>
> expr = expression(expr & b>0)
>
> What would be a good way of doing this?

I'd avoid using parse because it doesn't generalize well.  Here's a way
to do it without that.  But first, some background:

expression() returns a special kind of list that contains language
objects.  So expr[[1]] is initially  a>0, and you want to change it to
a>0 & b>0.  I think you don't really want expr & b>0, because that would
be    expression(a>0) & b>0  which is nonsense.

The bquote() function is probably the easiest way to make substitutions.
  The idea is that you write out the answer you want within bquote(),
using .() around things you want substituted.  (You can also use the
substitute() function, but it's a little trickier.)

Putting these together, this gives you the final result you want:

expr <- expression(a>0)
expr[[1]] <- bquote( .(expr[[1]]) & b>0 )

The following lines almost give you what you want:

expr2 <- expression(a>0)
expr2 <- bquote( expression( .(expr2[[1]]) & b>0 ) )

but not quite:  expr2 becomes an unevaluated call to the expression()
function, not an actual expression (though they print the same).  You
need this for the equivalent to the first version:

expr2 <- expression(a>0)
expr2 <- eval( bquote( expression( .(expr2[[1]]) & b>0 ) ) )

Duncan Murdoch

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