If its good enough to have one level of substitution then esub in my
post (originally due to Tony Plate -- see reference in my post) is all
that is needed:
esub(mat[[2]], list(g1 = g1[[1]]))
but I think the real problem could require multiple levels of
substitution in which case repeated application of esub is needed as
you walk the expression tree which is what proc() in my post does.
For example, suppose mat[[2]] is a function of g1 which is a function
of Tm which is a function of z. Then continuing the example in the
original post this does the repeated substitution needed (which would
be followed by an eval, not shown here, as in my original post):
> Tm <- expression(z^2)
> sapply(mat, proc)
[[1]]
[1] 0
[[2]]
f1 * s1 * (1/z^2)
To answer your question, quote() produces a call object but expression
produces a call wrapped in an expression which is why there is special
handling of expression objects in the proc() function in my post.
On Fri, Jan 29, 2010 at 4:38 PM, Bert Gunter <gunter.ber...@gene.com> wrote:
> 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-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Gabor Grothendieck
> Sent: Friday, January 29, 2010 11:01 AM
> To: Jennifer Young
> Cc: r-help@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.yo...@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@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.
>>
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> 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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