evaluating expressions with sub expressions

-----Original Message-----
From: r-help-bounces at r-project.org 
[mailto:r-help-bounces at r-project.org] On Behalf Of Bert Gunter
Sent: Friday, January 29, 2010 1:38 PM
To: 'Gabor Grothendieck'; 'Jennifer Young'
Cc: r-help at r-project.org
Subject: Re: [R] evaluating expressions with sub expressions

Folks:

Stripped to its essentials, Jennifer's request seemed simple: 
substitute a
subexpression as a named variable for a variable name in an 
expression, also
expressed as a named variable. A simple example is:

e <- expression(0,a*b)
z1 <- quote(1/t) ## explained below

The task is to "substitute" the expression in z1, "1/t", for "b" in e,
yielding the substituted expression as the result.

Gabor provided a solution, but it seemed to me like trying to 
swat a fly
with a baseball bat -- a lot of machinery for what should be a more
straightforward task. Of course, just because I think it **should be**
straightforward does not mean it actually is. But I fooled 
around a bit
(guided by Gabor's approach and an old Programmer's Niche 
column of Bill
Venables) and came up with:

f <- lapply(e,function(x){do.call(substitute,list(x,list(b=z1)))})
f

[[1]]
[1] 0

[[2]]
a * (1/t)

## f is a list. Turn it back into an expression
f <- as.expression(f)
## check that this works as intended
f

expression(0, a * (1/t))

a <- 2
t <- 3
eval(f)

[1] 0.6666667

Now you'll note that to do this I explicitly used quote() to 
produce the
variable holding the subexpression to be substituted. You may 
ask, why not
use expression() instead, as in 

z2 <- expression(1/t)

This doesn't work:

f <- lapply(e,function(x){do.call(substitute,list(x,list(b=z2)))})
f

[[1]]
[1] 0

[[2]]
a * expression(1/t)

f <- as.expression(f)

## Yielding ...

f

expression(0, a * expression(1/t)) #### Not what we want! 
## And sure enough ...

eval(f)

Error in a * expression(1/t) : non-numeric argument to binary operator

I think I understand why the z <- expression() approach does 
not work; but I
do not understand why the z <- quote() approach does! The 
mode of the return
from both of these is "call", but they are different (because 
identical()
tells me so). Could someone perhaps elaborate on this a bit 
more? And is
there a yet simpler and more straightforward way to do the 
above than what I
proposed?

Cheers,

Bert Gunter
Genentech Nonclinical Statistics


-----Original Message-----
From: r-help-bounces at r-project.org 
[mailto:r-help-bounces at r-project.org] On
Behalf Of Gabor Grothendieck
Sent: Friday, January 29, 2010 11:01 AM
To: Jennifer Young
Cc: r-help at r-project.org
Subject: Re: [R] evaluating expressions with sub expressions

The following recursively walks the expression tree.  The esub
function is from this page (you may wish to read that entire thread):
http://tolstoy.newcastle.edu.au/R/help/04/03/1245.html

esub <- function(expr, sublist) do.call("substitute", 
list(expr, sublist))

proc <- function(e, env = parent.frame()) {
   for(nm in all.vars(e)) {
      if (exists(nm, env) && is.language(g <- get(nm, env))) {
         if (is.expression(g)) g <- g[[1]]
            g <- Recall(g, env)
            L <- list(g)
            names(L) <- nm
             e <- esub(e, L)
	  }
        }
     e
}

mat <- expression(0, f1*s1*g1)
g1 <- expression(1/Tm)
vals <- data.frame(f1=1, s1=.5, Tm=2)
e <- sapply(mat, proc)
sapply(e, eval, vals)

The last line should give:

sapply(e, eval, vals)

[1] 0.00 0.25


On Fri, Jan 29, 2010 at 11:51 AM, Jennifer Young
<Jennifer.Young at math.mcmaster.ca> wrote:

Hallo

I'm having trouble figuring out how to evaluate an 

expression when one of

the variables in the expression is defined separately as a 

sub expression.

Here's a simplified example

mat <- expression(0, f1*s1*g1) ?# vector of formulae
g1 <- expression(1/Tm) ? ? ? ? ?# expansion of the definition of g1
vals <- data.frame(f1=1, s1=.5, Tm=2) # one set of possible 

values for

variables

before adding this sub expression I was using the following 

to evaluate
"mat"

sapply(mat, eval, vals)

Obviously I could manually substitute in 1/Tm for each g1 in the
definition of "mat", but the actual expression vector is 

much longer, and

the sub expression more complicated. Also, the 

subexpression is often

adjusted for different scenarios. ?Is there a simple way of 

changing this

or redefining "mat" so that I can define "g1" like a macro 

to be used in

the expression vector.

Thanks!
Jennifer