[R] Wrong environment when evaluating and expression?

[William Dunlap]

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.psych at gmail.com] 
> Sent: Wednesday, July 06, 2011 1:23 PM
> To: William Dunlap
> Cc: r-help at 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 
> <wdunlap at tibco.com> wrote:
> >> -----Original Message-----
> >> From: r-help-bounces at r-project.org
> >> [mailto:r-help-bounces at r-project.org] On Behalf Of Joshua Wiley
> >> Sent: Monday, July 04, 2011 1:12 AM
> >> To: r-help at 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 at 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/
>