evaluating expressions with sub expressions

as.list(quote(a*b))

e <- as.list(expression(a*b))
e

mode(e[[1]])

as.list(e[[1]])

Folks:

Stripped to its essentials, Jennifer's request seemed simple: substitute a
subexpression as a named variable for a variable name in an expression,

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

do not understand why the z <- quote() approach does! The mode of the

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

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]

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

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