Environmental oddity.

On Sun, Nov 7, 2021 at 6:05 AM Rolf Turner <r.turner at auckland.ac.nz> wrote:

I have two functions which appear to differ only in their environments.
They look like:

d1
function (x, mean = 0, sd = 1, log = FALSE)
(((x - mean)/sd)^2 - 1) * if (log) 1 else dnorm(x, mean, sd)/sd
<environment: namespace:stats>

and

d2
function (x, mean = 0, sd = 1, log = FALSE)
(((x - mean)/sd)^2 - 1) * if (log) 1 else dnorm(x, mean, sd)/sd

Typing "environment(d1)" gives

<environment: namespace:stats>

and typing "environment(d2)" gives

<environment: R_GlobalEnv>

The d2() function however gives an incorrect result:

d1(1,0,3,TRUE)
[1] -0.2962963
d2(1,0,3,TRUE)
[1] -0.8888889

It can't be as simple as that. I get the same result (as your d2) with
the following:

d <- function (x, mean = 0, sd = 1, log = FALSE) {
     (((x - mean)/sd)^2 - 1) * if (log) 1 else dnorm(x, mean, sd) / sd
}
d(1, 0, 3, TRUE)
environment(d)
environment(d) <- as.environment("package:stats")
d(1, 0, 3, TRUE)

In d2() the result of the if() statement does not get divided
by the final "sd" whereas in d1() it does (which is the desired/correct
result).

Of course the code is ridiculously kludgy (it was produced by "symbolic
differentiation").  That's not the point.  I'm just curious (idly?) as
to *why* the association of the namespace:stats environment with d1()
causes it to "do the right thing".

This sounds like a difference in precedence. The expression

if (log) 1 else dnorm(x, mean, sd) / sd

is apparently being interpreted differently as

d1: (if (log) 1 else dnorm(x, mean, sd)) / sd
d2: if (log) 1 else (dnorm(x, mean, sd)) / sd)

It's unclear how environments could affect this, so it would be very
helpful to have a reproducible example.