[R] evaluating expressions with sub expressions

[Bert Gunter]

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
>
> ______________________________________________
> R-help at r-project.org mailing list
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> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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