On 06/11/2021 11:32 p.m., Deepayan Sarkar wrote:
On Sun, Nov 7, 2021 at 6:05 AM Rolf Turner <r.tur...@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.


Rolf said these were automatically produced functions. Those don't always deparse properly, because manipulating expressions can produce things that can never be produced by the parser. I'm not sure this happened in this case. You'd need to examine the parse trees of d1 and d2 to see.

There's also a possibility that the srcref attached to them is lying, and we're not seeing the deparsed versions of the functions. Printing removeSource(d1) and removeSource(d2) should reveal that.

Duncan Murdoch

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