Re: [R] Wrong environment when evaluating and expression?
Thanks Bill! That is very useful. Is the str.language function in
any package (findFn(str.language) came up empty)? It certainly
helped me, not only to understand this particular problem, but in
trying to wrap my head around language objects (which I only very
poorly grasp) in general.
Josh
On Tue, Jul 5, 2011 at 11:48 AM, William Dunlap wdun...@tibco.com wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Joshua Wiley
Sent: Monday, July 04, 2011 1:12 AM
To: r-help@r-project.org
Subject: [R] Wrong environment when evaluating and expression?
Hi All,
I have constructed two expressions (e1 e2). I can see that they are
not identical, but I cannot figure out how they differ.
###
dat - mtcars
e1 - expression(with(data = dat, lm(mpg ~ hp)))
e2 - as.expression(substitute(with(data = dat, lm(f)),
list(f = mpg ~ hp)))
str(e1)
str(e2)
all.equal(e1, e2)
identical(e1, e2) # false
With the appended str.language function you can see the difference
between e1 and e2. It displays
`name` class(length)
of each component of a recursive object, along with a short text summary
of
it after a colon.
str.language(e1)
`e1` expression(1): expression(with(data = da...
`` call(3): with(data = dat, lm(mpg ~...
`` name(1): with
`data` name(1): dat
`` call(2): lm(mpg ~ hp)
`` name(1): lm
`` call(3): mpg ~ hp
`` name(1): ~
`` name(1): mpg
`` name(1): hp
str.language(e2)
`e2` expression(1): expression(with(data = da...
`` call(3): with(data = dat, lm(mpg ~...
`` name(1): with
`data` name(1): dat
`` call(2): lm(mpg ~ hp)
`` name(1): lm
`` formula(3): mpg ~ hp
`` name(1): ~
`` name(1): mpg
`` name(1): hp
`Attributes of ` list(2): structure(list(class = f...
`class` character(1): formula
`.Environment` environment(5): R_GlobalEnv dat e1 e2 s...
It is a bug in all.equal() that it ignores attributes of formulae.
E.g.,
all.equal(y~x, terms(y~x))
[1] TRUE
identical(y~x, terms(y~x))
[1] FALSE
Here is str.language
str.language -
function (object, ..., level = 0, name = deparse(substitute(object)),
attributes = TRUE)
{
abbr - function(string, maxlen = 25) {
if (length(string) 1 || nchar(string) maxlen)
paste(substring(string[1], 1, maxlen), ..., sep = )
else string
}
myDeparse - function(object) {
if (!is.environment(object)) {
deparse(object)
}
else {
ename - environmentName(object)
if (ename == )
ename - unnamed env
paste(sep = , , ename, , paste(collapse = ,
objects(object)))
}
}
cat(rep( , level), sep = )
if (is.null(name))
name -
cat(sprintf(`%s` %s(%d): %s\n, abbr(name), class(object),
length(object), abbr(myDeparse(object
a - attributes(object)
if (is.recursive(object) !is.environment(object)) {
object - as.list(object)
names - names(object)
for (i in seq_along(object)) {
str.language(object[[i]], ..., level = level + 1,
name = names[i], attributes = attributes)
}
}
if (attributes) {
a$names - NULL
if (length(a) 0) {
str.language(a, level = level + 1, name = paste(Attributes
of,
abbr(name)), attributes = attributes)
}
}
}
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
eval(e1)
eval(e2)
The context is trying to use a list of formulae to generate several
models from a multiply imputed dataset. The package I am using (mice)
has methods for with() and that is how I can (easily) get the pooled
results. Passing the formula directly does not work, so I was trying
to generate the entire call and evaluate it as if I had typed it at
the console, but I am missing something (probably rather silly).
Thanks,
Josh
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
__
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.
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
https://joshuawiley.com/
__
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.
Re: [R] Wrong environment when evaluating and expression?
No, it is not in any package. Feel free to use
it as you wish - it has no licensing restrictions.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
Sent: Wednesday, July 06, 2011 1:23 PM
To: William Dunlap
Cc: r-help@r-project.org
Subject: Re: [R] Wrong environment when evaluating and expression?
Thanks Bill! That is very useful. Is the str.language function in
any package (findFn(str.language) came up empty)? It certainly
helped me, not only to understand this particular problem, but in
trying to wrap my head around language objects (which I only very
poorly grasp) in general.
Josh
On Tue, Jul 5, 2011 at 11:48 AM, William Dunlap
wdun...@tibco.com wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Joshua Wiley
Sent: Monday, July 04, 2011 1:12 AM
To: r-help@r-project.org
Subject: [R] Wrong environment when evaluating and expression?
Hi All,
I have constructed two expressions (e1 e2). I can see
that they are
not identical, but I cannot figure out how they differ.
###
dat - mtcars
e1 - expression(with(data = dat, lm(mpg ~ hp)))
e2 - as.expression(substitute(with(data = dat, lm(f)),
list(f = mpg ~ hp)))
str(e1)
str(e2)
all.equal(e1, e2)
identical(e1, e2) # false
With the appended str.language function you can see the difference
between e1 and e2. It displays
`name` class(length)
of each component of a recursive object, along with a short
text summary
of
it after a colon.
str.language(e1)
`e1` expression(1): expression(with(data = da...
`` call(3): with(data = dat, lm(mpg ~...
`` name(1): with
`data` name(1): dat
`` call(2): lm(mpg ~ hp)
`` name(1): lm
`` call(3): mpg ~ hp
`` name(1): ~
`` name(1): mpg
`` name(1): hp
str.language(e2)
`e2` expression(1): expression(with(data = da...
`` call(3): with(data = dat, lm(mpg ~...
`` name(1): with
`data` name(1): dat
`` call(2): lm(mpg ~ hp)
`` name(1): lm
`` formula(3): mpg ~ hp
`` name(1): ~
`` name(1): mpg
`` name(1): hp
`Attributes of ` list(2): structure(list(class = f...
`class` character(1): formula
`.Environment` environment(5): R_GlobalEnv dat e1 e2 s...
It is a bug in all.equal() that it ignores attributes of formulae.
E.g.,
all.equal(y~x, terms(y~x))
[1] TRUE
identical(y~x, terms(y~x))
[1] FALSE
Here is str.language
str.language -
function (object, ..., level = 0, name =
deparse(substitute(object)),
attributes = TRUE)
{
abbr - function(string, maxlen = 25) {
if (length(string) 1 || nchar(string) maxlen)
paste(substring(string[1], 1, maxlen), ..., sep = )
else string
}
myDeparse - function(object) {
if (!is.environment(object)) {
deparse(object)
}
else {
ename - environmentName(object)
if (ename == )
ename - unnamed env
paste(sep = , , ename, , paste(collapse = ,
objects(object)))
}
}
cat(rep( , level), sep = )
if (is.null(name))
name -
cat(sprintf(`%s` %s(%d): %s\n, abbr(name), class(object),
length(object), abbr(myDeparse(object
a - attributes(object)
if (is.recursive(object) !is.environment(object)) {
object - as.list(object)
names - names(object)
for (i in seq_along(object)) {
str.language(object[[i]], ..., level = level + 1,
name = names[i], attributes = attributes)
}
}
if (attributes) {
a$names - NULL
if (length(a) 0) {
str.language(a, level = level + 1, name =
paste(Attributes
of,
abbr(name)), attributes = attributes)
}
}
}
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
eval(e1)
eval(e2)
The context is trying to use a list of formulae to generate several
models from a multiply imputed dataset. The package I am
using (mice)
has methods for with() and that is how I can (easily) get
the pooled
results. Passing the formula directly does not work, so I
was trying
to generate the entire call and evaluate it as if I had typed it at
the console, but I am missing something (probably rather silly).
Thanks,
Josh
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
__
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
Re: [R] Wrong environment when evaluating and expression?
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Joshua Wiley
Sent: Monday, July 04, 2011 1:12 AM
To: r-help@r-project.org
Subject: [R] Wrong environment when evaluating and expression?
Hi All,
I have constructed two expressions (e1 e2). I can see that they are
not identical, but I cannot figure out how they differ.
###
dat - mtcars
e1 - expression(with(data = dat, lm(mpg ~ hp)))
e2 - as.expression(substitute(with(data = dat, lm(f)),
list(f = mpg ~ hp)))
str(e1)
str(e2)
all.equal(e1, e2)
identical(e1, e2) # false
With the appended str.language function you can see the difference
between e1 and e2. It displays
`name` class(length)
of each component of a recursive object, along with a short text summary
of
it after a colon.
str.language(e1)
`e1` expression(1): expression(with(data = da...
`` call(3): with(data = dat, lm(mpg ~...
`` name(1): with
`data` name(1): dat
`` call(2): lm(mpg ~ hp)
`` name(1): lm
`` call(3): mpg ~ hp
`` name(1): ~
`` name(1): mpg
`` name(1): hp
str.language(e2)
`e2` expression(1): expression(with(data = da...
`` call(3): with(data = dat, lm(mpg ~...
`` name(1): with
`data` name(1): dat
`` call(2): lm(mpg ~ hp)
`` name(1): lm
`` formula(3): mpg ~ hp
`` name(1): ~
`` name(1): mpg
`` name(1): hp
`Attributes of ` list(2): structure(list(class = f...
`class` character(1): formula
`.Environment` environment(5): R_GlobalEnv dat e1 e2 s...
It is a bug in all.equal() that it ignores attributes of formulae.
E.g.,
all.equal(y~x, terms(y~x))
[1] TRUE
identical(y~x, terms(y~x))
[1] FALSE
Here is str.language
str.language -
function (object, ..., level = 0, name = deparse(substitute(object)),
attributes = TRUE)
{
abbr - function(string, maxlen = 25) {
if (length(string) 1 || nchar(string) maxlen)
paste(substring(string[1], 1, maxlen), ..., sep = )
else string
}
myDeparse - function(object) {
if (!is.environment(object)) {
deparse(object)
}
else {
ename - environmentName(object)
if (ename == )
ename - unnamed env
paste(sep = , , ename, , paste(collapse = ,
objects(object)))
}
}
cat(rep( , level), sep = )
if (is.null(name))
name -
cat(sprintf(`%s` %s(%d): %s\n, abbr(name), class(object),
length(object), abbr(myDeparse(object
a - attributes(object)
if (is.recursive(object) !is.environment(object)) {
object - as.list(object)
names - names(object)
for (i in seq_along(object)) {
str.language(object[[i]], ..., level = level + 1,
name = names[i], attributes = attributes)
}
}
if (attributes) {
a$names - NULL
if (length(a) 0) {
str.language(a, level = level + 1, name = paste(Attributes
of,
abbr(name)), attributes = attributes)
}
}
}
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
eval(e1)
eval(e2)
The context is trying to use a list of formulae to generate several
models from a multiply imputed dataset. The package I am using (mice)
has methods for with() and that is how I can (easily) get the pooled
results. Passing the formula directly does not work, so I was trying
to generate the entire call and evaluate it as if I had typed it at
the console, but I am missing something (probably rather silly).
Thanks,
Josh
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
__
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.
__
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.
[R] Wrong environment when evaluating and expression?
Hi All, I have constructed two expressions (e1 e2). I can see that they are not identical, but I cannot figure out how they differ. ### dat - mtcars e1 - expression(with(data = dat, lm(mpg ~ hp))) e2 - as.expression(substitute(with(data = dat, lm(f)), list(f = mpg ~ hp))) str(e1) str(e2) all.equal(e1, e2) identical(e1, e2) # false eval(e1) eval(e2) The context is trying to use a list of formulae to generate several models from a multiply imputed dataset. The package I am using (mice) has methods for with() and that is how I can (easily) get the pooled results. Passing the formula directly does not work, so I was trying to generate the entire call and evaluate it as if I had typed it at the console, but I am missing something (probably rather silly). Thanks, Josh -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ __ 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.
Re: [R] Wrong environment when evaluating and expression?
On Mon, Jul 4, 2011 at 4:11 AM, Joshua Wiley jwiley.ps...@gmail.com wrote: Hi All, I have constructed two expressions (e1 e2). I can see that they are not identical, but I cannot figure out how they differ. ### dat - mtcars e1 - expression(with(data = dat, lm(mpg ~ hp))) e2 - as.expression(substitute(with(data = dat, lm(f)), list(f = mpg ~ hp))) str(e1) str(e2) all.equal(e1, e2) identical(e1, e2) # false eval(e1) eval(e2) The context is trying to use a list of formulae to generate several models from a multiply imputed dataset. The package I am using (mice) has methods for with() and that is how I can (easily) get the pooled results. Passing the formula directly does not work, so I was trying to generate the entire call and evaluate it as if I had typed it at the console, but I am missing something (probably rather silly). In e1, mpg ~ hp is a call object but in e2 its a formula with an environment: e1[[1]][[3]][[2]] mpg ~ hp e2[[1]][[3]][[2]] mpg ~ hp class(e1[[1]][[3]][[2]]) [1] call class(e2[[1]][[3]][[2]]) [1] formula environment(e2[[1]][[3]][[2]]) environment: R_GlobalEnv -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ 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.
Re: [R] Wrong environment when evaluating and expression?
Thanks Gabor, that makes sense now. In case anyone else runs into something similar, I ended up just passing a character string of the formula so it could be coerced to a formula in the correct environment. Thanks again, Josh On Mon, Jul 4, 2011 at 4:26 AM, Gabor Grothendieck ggrothendi...@gmail.com wrote: On Mon, Jul 4, 2011 at 4:11 AM, Joshua Wiley jwiley.ps...@gmail.com wrote: Hi All, I have constructed two expressions (e1 e2). I can see that they are not identical, but I cannot figure out how they differ. ### dat - mtcars e1 - expression(with(data = dat, lm(mpg ~ hp))) e2 - as.expression(substitute(with(data = dat, lm(f)), list(f = mpg ~ hp))) str(e1) str(e2) all.equal(e1, e2) identical(e1, e2) # false eval(e1) eval(e2) The context is trying to use a list of formulae to generate several models from a multiply imputed dataset. The package I am using (mice) has methods for with() and that is how I can (easily) get the pooled results. Passing the formula directly does not work, so I was trying to generate the entire call and evaluate it as if I had typed it at the console, but I am missing something (probably rather silly). In e1, mpg ~ hp is a call object but in e2 its a formula with an environment: e1[[1]][[3]][[2]] mpg ~ hp e2[[1]][[3]][[2]] mpg ~ hp class(e1[[1]][[3]][[2]]) [1] call class(e2[[1]][[3]][[2]]) [1] formula environment(e2[[1]][[3]][[2]]) environment: R_GlobalEnv -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ __ 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.
