[R] substitute question

[Gabor Grothendieck]

Tony, Thomas.  Thanks for your help.  Your comments were very
useful.

Unfortunately, my next step gives me a new round of problems.

The following is the same as the last example except that instead
of hard coding the function into the expression I wanted to 
pass it to the expression.  It seems like one has to do a double
substitute to get this effect but I am having problems getting this
to work.

In the code below we first show that the keep.source option has been
set to FALSE so that the source attribute does not mislead us.

Then we do a double substitute.  The outer substitute just creates
the substitute that was in my previous question.  This outer
substitute produces z, a call object.  So far its as expected.

We then do ze <- eval(z) but we get a function object right away.  I was
expecting that we get an expression object.  Even worse, the function does not
have  a  replaced with  b -- even though this did work in the previous example 
that I posted.  

What's wrong?


> options()$keep.source
[1] FALSE
> f <- function(){a+1}
> z <- substitute(substitute(f,list(a=quote(b))),list(f=f))
> class(z)
[1] "call"
> as.list(z)
[[1]]
substitute

[[2]]
function () 
{
    a + 1
}

[[3]]
list(a = quote(b))

> ze <- eval(z)
> class(ze)
[1] "function"
> ze
function () 
{
    a + 1
}
> attr(ze,"source")
NULL
> 

---

Date:   Thu, 18 Mar 2004 09:00:02 -0800 (PST) 
From:   Thomas Lumley <tlumley at u.washington.edu>
To:   Gabor Grothendieck <ggrothendieck at myway.com> 
Cc:   <R-help at stat.math.ethz.ch> 
Subject:   RE: [R] substitute question 

 
On Wed, 17 Mar 2004, Gabor Grothendieck wrote:

>
>
> I left out the brackets in my last email but the problem
> (a reappears after have been substituted out) still remains:
>
> > z <- substitute( function(){a+1}, list(a=quote(b)) )
> > z
> function() {
> b + 1
> }
> > eval(z)
> function(){a+1}


Interesting.

Appearances are misleading, however:
> z <- substitute( function(){a+1}, list(a=quote(b)) )
> z
function() {
b + 1
}
> f<-eval(z)
> f()
Error in f() : Object "b" not found
> f
function(){a+1}
> attr(f,"source")<-NULL
> f
function ()
{
b + 1
}

So it isn't that eval(z) has a+1 inside, it just has a "source" attribute
with a+1.

Looking more carefully at z
> as.list(z)
[[1]]
`function`

[[2]]
NULL

[[3]]
{
b + 1
}

[[4]]
[1] "function(){a+1}"

so the original construction of z has kept the source (not, however, as a
"source" attribute).

There is method to our madness here. It is impossible (or at least too
complicated) to keep comments in the right place as a function is
parsed and deparsed. In the old days, comments would occasionally move
around, sometimes in very misleading ways (IIRC with if(){}else{} cases)

Now we keep a copy of the source code with functions created interactively
or with source(), and drop the comments on parsing. This is controlled by
options("keep.source").

If you do a lot of substitute()-style programming you may want
options(keep.source=FALSE).

     -thomas

PS: There are, of course, interesting possibilities for creative abuse of
the source attribute....

---

Date:   Thu, 18 Mar 2004 08:41:54 -0700 
From:   Tony Plate <tplate at blackmesacapital.com>
To:   <ggrothendieck at myway.com>, <R-help at stat.math.ethz.ch> 
Subject:   RE: [R] substitute question 

 
This is because of the saved attribute "source" on z (that doesn't get 
printed out before evaluating z, because z is not yet then a function).

To complete your example:

> z <- substitute( function(){a+1}, list(a=quote(b)) )
> z
function() {
b + 1
}
> eval(z)
function(){a+1}
> ze <- eval(z)
> attributes(ze)
$source
[1] "function(){a+1}"

> attr(ze, "source") <- NULL
> ze
function ()
{
b + 1
}
>

I previously wrote on this topic:

Date: Fri, 24 Oct 2003 09:42:55 -0600
To: R-help at stat.math.ethz.ch, Peter Dalgaard <p.dalgaard at biostat.ku.dk>
From: Tony Plate <tplate at blackmesacapital.com>
Subject: Re: [R] what's going on here with substitute() ?
Mime-Version: 1.0
Content-Type: text/plain; charset="us-ascii"; format=flowed

Peter, thank you for the explanation. This is indeed what is happening.

Might I suggest the following passage for inclusion in the help page for 
"function", and possibly also "body", in the DETAILS or WARNING section:

"Note that the text of the original function definition is saved as an 
attribute "source" on the function, and this is printed out when the 
function is printed. Hence, if the function body is changed in some way 
other than by assigning a value via body() (which removes the "source" 
attribute), the printed form of the function may not be the same as the 
actual function body."

Something along these lines could also go in the help for "eval", though if 
it were only there it might be very difficult to find if one were trying to 
look up puzzling behavior of a function.

Here is a transcript that shows what is happening, with another suggestion 
following it.

> eval(substitute(this.is.R <- function() X, 
list(X=!is.null(options("CRAN")[[1]]))))
> this.is.R
function() X
> body(this.is.R)
[1] TRUE
> attributes(this.is.R)
$source
[1] "function() X"
> attributes(this.is.R) <- NULL
> this.is.R
function ()
TRUE
> # the "source" attribute comes from function definition:
> attributes(function() X)
$source
[1] "function() X"
> # and seems to be added by "eval":
> attr(eval(parse(text="function() TRUE")[[1]]), "source")
[1] "function() TRUE"
>

> # we can assign bogus "source"
> attr(this.is.R, "source") <- "a totally bogus function body"
> this.is.R
a totally bogus function body
> # assigning to body() removes "source"
> body(this.is.R) <- list(666)
> this.is.R
function ()
666
> attr(this.is.R, "source")
NULL
>

An even better approach might be something that gave a warning on printing 
if the parsed "source" attribute was not identical to the language object 
being printed. This would probably belong in the code for "case LANGSXP:" 
in the function PrintValueRec in main/print.c (if it were written in R, I 
could contribute a patch, but right now I don't have time to try to 
understand the C there.) R code to do the test could be something like this:

> f <- this.is.R
> identical(f, eval(parse(text=attr(f, "source"))[[1]]))
[1] FALSE
> f <- function() TRUE
> identical(f, eval(parse(text=attr(f, "source"))[[1]]))
[1] TRUE
>

-- Tony Plate


At Wednesday 08:09 PM 3/17/2004, Gabor Grothendieck wrote:


>I left out the brackets in my last email but the problem
>(a reappears after have been substituted out) still remains:
>
> > z <- substitute( function(){a+1}, list(a=quote(b)) )
> > z
>function() {
> b + 1
>}
> > eval(z)
>function(){a+1}
>
>
>
>---
>Date: Wed, 17 Mar 2004 20:10:43 -0500 (EST)
>From: Gabor Grothendieck <ggrothendieck at myway.com>
>[ Add to Address Book | Block Address | Report as Spam ]
>To: <R-help at stat.math.ethz.ch>
>Subject: [R] substitute question
>
>
>
>
>
>
>Consider the following example:
>
># substitute a with b in the indicated function. Seems to work.
> > z <- substitute( function()a+1, list(a=quote(b)) )
> > z
>function() b + 1
>
># z is an object of class call so use eval
># to turn it into an object of class expression; however,
># when z is evaluated, the variable a returns.
> > eval(z)
>function()a+1
>
>Why did a suddenly reappear again after it had already been replaced?