[R] Wrong environment when evaluating and expression?

[William Dunlap]

> -----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/
> 
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