Re: [Rd] WISH: eval() to preserve the visibility (now value is always visible)
On Sun, Feb 8, 2015 at 8:44 PM, Suharto Anggono Suharto Anggono via
R-devel r-devel@r-project.org wrote:
Sorry to intervene.
No, I'm very happy you intervened. You're comment is 100%
valid/correct making my wish moot.
Your explanation is very clear and nails it; one should use
eval(substitute(expr)) or evalq(expr) for what I'm trying to do.
It all came from me trying to prevent
withOptions({x - 1}, foo=1)
from printed the value, where (somewhat simplified):
withOptions - function(expr, ..., envir=parent.frame()) {
oopts - options(...)
on.exit(options(oopts))
eval(expr, envir=envir)
}
I have a few of these withNnn() functions, but for this particular one
(*) I had forgotten an expr - substitute(expr) in there, which caused
me to incorrectly blame eval(). recursive mistakeThis is very much
the same problem as you observed with my eval2() example./recursive
mistake
Thank you very much
Henrik
(*) Actually withSeeds() which is to messy to use as an example.
Argument passed to 'eval' is evaluated first.
So,
eval(x - 2)
is effectively like
{ x - 2; eval(2) } ,
which is effectively
{ x - 2; 2 } .
The result is visible.
eval(expression(x - 2))
or
eval(quote(x - 2))
or
evalq(x - 2)
gives the same effect as
x - 2 .
The result is invisible.
In function 'eval2',
res - eval(withVisible(expr), envir=envir, ...)
is effectively
res - withVisible(expr) .
---
Would it be possible to have the value of eval() preserve the
visibility of the value of the expression?
PROBLEM:
# Invisible
x - 1
# Visible
eval(x - 2)
[1] 2
TROUBLESHOOTING:
withVisible(x - 1)
$value
[1] 1
$visible
[1] FALSE
withVisible(eval(x - 2))
$value
[1] 2
$visible
[1] TRUE
WORKAROUND:
eval2 - function(expr, envir=parent.frame(), ...) {
res - eval(withVisible(expr), envir=envir, ...)
value - res$value
if (res$visible) value else invisible(value)
}
x - 1
eval(x - 2)
[1] 2
eval2(x - 3)
x
[1] 3
/Henrik
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Re: [Rd] WISH: eval() to preserve the visibility (now value is always visible)
Sorry to intervene.
Argument passed to 'eval' is evaluated first.
So,
eval(x - 2)
is effectively like
{ x - 2; eval(2) } ,
which is effectively
{ x - 2; 2 } .
The result is visible.
eval(expression(x - 2))
or
eval(quote(x - 2))
or
evalq(x - 2)
gives the same effect as
x - 2 .
The result is invisible.
In function 'eval2',
res - eval(withVisible(expr), envir=envir, ...)
is effectively
res - withVisible(expr) .
---
Would it be possible to have the value of eval() preserve the
visibility of the value of the expression?
PROBLEM:
# Invisible
x - 1
# Visible
eval(x - 2)
[1] 2
TROUBLESHOOTING:
withVisible(x - 1)
$value
[1] 1
$visible
[1] FALSE
withVisible(eval(x - 2))
$value
[1] 2
$visible
[1] TRUE
WORKAROUND:
eval2 - function(expr, envir=parent.frame(), ...) {
res - eval(withVisible(expr), envir=envir, ...)
value - res$value
if (res$visible) value else invisible(value)
}
x - 1
eval(x - 2)
[1] 2
eval2(x - 3)
x
[1] 3
/Henrik
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Re: [Rd] WISH: eval() to preserve the visibility (now value is always visible)
On 07/02/2015 12:40 PM, Henrik Bengtsson wrote:
Would it be possible to have the value of eval() preserve the
visibility of the value of the expression?
PROBLEM:
# Invisible
x - 1
# Visible
eval(x - 2)
[1] 2
TROUBLESHOOTING:
withVisible(x - 1)
$value
[1] 1
$visible
[1] FALSE
withVisible(eval(x - 2))
$value
[1] 2
$visible
[1] TRUE
WORKAROUND:
eval2 - function(expr, envir=parent.frame(), ...) {
res - eval(withVisible(expr), envir=envir, ...)
value - res$value
if (res$visible) value else invisible(value)
}
x - 1
eval(x - 2)
[1] 2
eval2(x - 3)
x
[1] 3
What's wrong with the workaround?
Duncan Murdoch
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