Thanks again Bill; I agree that substitute is overkill here.
As an aside, for cases where someone may be tempted to use substitute(),
it seems quote() might be a safer alternative; compare
> lapply(list(1), function(y) c(quote(y), substitute(y)))
[[1]]
[[1]][[1]]
y
[[1]][[2]]
X[[i]]
versus in R < 3.2,
> lapply(list(1), function(y) c(quote(y), substitute(y)))
[[1]]
[[1]][[1]]
y
[[1]][[2]]
X[[1L]]
in any case, the lesson seems to be that quote and substitute are not
interchangeable, even though for example
> (function() identical(quote({a}), substitute({a})))()
[1] TRUE
On 07/29/2017 09:39 AM, William Dunlap wrote:
Functions, like your loader(), that use substitute to let users
confound things and their names, should give the user a way to avoid
the use of substitute. E.g., library() has the 'character.only'
argument; if TRUE then the package argument is treated as an ordinary
argument and not passed through substitute().
myLoader <- function(package, quietly = TRUE) {
wrapper <- if (quietly) suppressPackageStartupMessages else `{`
wrapper(library(package = package, character.only=TRUE))
}
> lapply(c("MASS","boot"), myLoader, quietly=FALSE)
[[1]]
[1] "MASS" "splines" "pryr" "stats" "graphics" "grDevices"
[7] "utils" "datasets" "methods" "base"
[[2]]
[1] "boot" "MASS" "splines" "pryr" "stats"
"graphics"
[7] "grDevices" "utils" "datasets" "methods" "base"
"Non-standard" evaluation (using substitute(), formulas, promises, the
rlang or lazyeval packages, etc.) has it uses but I wouldn't use it
for such a function as your loader().
Bill Dunlap
TIBCO Software
wdunlap tibco.com <http://tibco.com>
On Fri, Jul 28, 2017 at 8:20 PM, Benjamin Tyner <bty...@gmail.com
<mailto:bty...@gmail.com>> wrote:
Thanks Bill. I think my confusion may have been in part due to my
conflating two distinct meanings of the term "evaluate"; the help
for force says it "forces the evaluation of a function argument"
whereas the help for eval says it "evaluates the ... argument ...
and returns the computed value". I found it helpful to compare:
> lapply(list(a=1,b=2,c=3), function(x){ force(substitute(x)) })
$a
X[[i]]
$b
X[[i]]
$c
X[[i]]
versus
> lapply(list(a=1,b=2,c=3), function(x){ eval(substitute(x)) })
Error in eval(substitute(x)) : object 'X' not found
Now for the context my question arose in: given a function
loader <- function(package, quietly = TRUE) {
wrapper <- if (quietly) suppressPackageStartupMessages else `{`
expr <- substitute(wrapper(library(package = package)))
eval(expr)
}
prior to R version 3.2, one could do things like
lapply(c("MASS", "boot"), loader)
but not anymore (which is fine; I agree that one should not depend
on lapply's implementation details).
Regards,
Ben
On 07/28/2017 06:53 PM, William Dunlap wrote:
1: substitute(), when given an argument to a function (which
will be a promise) gives you the unevaluated expression given
as the argument:
> L <- list(a=1, b=2, c=3)
> str(lapply(L, function(x) substitute(x)))
List of 3
$ a: language X[[i]]
$ b: language X[[i]]
$ c: language X[[i]]
The 'X' and 'i' are in a frame constructed by lapply and you
are not really supposed to depend on the precise form of those
expressions.
2: An evaluated promise is still a promise: it has the
'evaled' field set to TRUE and the 'value' field set to the
result of evaluating 'code' in 'env'.
> f <- function(x, force) {
if (force) force(x)
if (pryr::is_promise(x)) promise_info(x)
else "not a promise"
}
> str(f(log(-1), force=FALSE))
List of 4
$ code : language log(-1)
$ env :<environment: R_GlobalEnv>
$ evaled: logi FALSE
$ value : NULL
> str(f(log(-1), force=TRUE))
List of 4
$ code : language log(-1)
$ env : NULL
$ evaled: logi TRUE
$ value : num NaN
Warning message:
In log(-1) : NaNs produced
Can you give a concrete example of what you are try to accomplish?
Bill Dunlap
TIBCO Software
wdunlap tibco.com <http://tibco.com> <http://tibco.com>
On Fri, Jul 28, 2017 at 3:04 PM, Benjamin Tyner
<bty...@gmail.com <mailto:bty...@gmail.com>
<mailto:bty...@gmail.com <mailto:bty...@gmail.com>>> wrote:
Hi,
I thought I understood the change to lapply semantics
resulting
from this,
https://bugs.r-project.org/bugzilla/show_bug.cgi?id=16093
<https://bugs.r-project.org/bugzilla/show_bug.cgi?id=16093>
<https://bugs.r-project.org/bugzilla/show_bug.cgi?id=16093
<https://bugs.r-project.org/bugzilla/show_bug.cgi?id=16093>>
However, would someone care to explain why this does not work?
> L <- list(a=1, b=2, c=3)
> str(lapply(L, function(x){ y <- substitute(x); force(x);
eval(y) }))
Error in eval(y) : object 'X' not found
Basically, my primary goal is to achieve the same result as,
> str(lapply(L, function(x){ eval.parent(substitute(x)) }))
List of 3
$ a: num 1
$ b: num 2
$ c: num 3
but without having to resort to eval.parent as that seems
to rely
on an implementation detail of lapply.
My secondary goal is to understand why force(x) does not
actually
force the promise here,
> str(lapply(L, function(x){ force(x);
pryr::is_promise(x) }))
List of 3
$ a: logi TRUE
$ b: logi TRUE
$ c: logi TRUE
,
Regards
Ben
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