[ please copy me on answers, since I am not subscribed to the list ]
Dear all,
I am trying to write an R function which uses system.time
to determine which of a given list of R expressions executes
fastest. To work around the limited resolution of system.time,
I want to convert the given expressions into functions which
execute the given expressions a fixed number of times.
My current attempt is as follows:
FuncIt <- function(k, expr) {
k <- as.numeric(k)
expr <- eval.parent(substitute(expr))
eval(substitute(function() { for (funcit.i in 1:k) { expr } }))
}
This works, but seems not very robust.
My question: is there a better way of doing this?
Here are some experiments.
1) good: If I run the following using "Rscript"
test1 <- function(e1) {
e1 <- substitute(e1)
FuncIt(100, e1)
}
f <- test1(rnorm(1))
print(f)
then I get the following output:
function ()
{
for (funcit.i in 1:100) {
rnorm(1)
}
}
<environment: 0x102260c28>
This is what I want. But why do I need the extra "substitute"
in test1? I only found by experiment that this is needed.
2) bad: If I try to call FuncIt directly, it fails:
f <- FuncIt(100, rnorm(1))
print(f)
has the output:
function ()
{
for (funcit.i in 1:100) {
-0.763894772833099
}
}
<environment: 0x102265790>
This is bad, since now 'rnorm(1)' already has been
evaluated. How do I prevent this from happening,
without breaking the good case 1 above?
3) ugly: If I run the same commands in the R gui on MacOS
(R 2.15.1 released on 2012/06/22), I get different output:
> source("/Users/voss/project/statcomp/test.R")
function() { for (funcit.i in 1:k) { expr } }
<environment: 0x19cc040>
function() { for (funcit.i in 1:k) { expr } }
<environment: 0x19bc884>
This is on the same machine using (as far as I can tell) the
same R engine. So why is the output different?
Many thanks,
Jochen
--
http://seehuhn.de/
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