Very interesting. But what explicitly happens with ***source()***
in Splus???
cheers,
Rolf
On 08/06/13 11:33, William Dunlap wrote:
foo <- function(x) {
sin(42)
x^2
}
foo(3)
[1] 9
The value of sin(42) is never seen.
This is true in both R and S+. Only the return value of a function is available
for autoprinting and sin(42) is not the return value. Perhaps you are
remembering
the case where the last subexpression in the function is an assignment, in which
case S+ autoprints its value but R does not.
R> f <- function(x) { y <- x + 1 }
R> f(10)
R> .Last.value
[1] 11
S+> f <- function(x) { y <- x + 1 }
Warning messages:
Last expression in function is an assignment
(You probably wanted to return the left-hand side)
in: y <- x + 1
S+> f(10)
[1] 11
S+> .Last.value
[1] 11
Autoprinting is implemented quite differently in R and S+. I think R attaches
the autoprint flag to the returned object while S+ assigns .Auto.print<-TRUE
in the frame of the caller (yuck). That leads to differences like the following
R> f <- function(x) invisible(x+1)
R> g <- function(x) 10 * x
R> g(f(23)) # g() and f() are evaluated in same frame
[1] 240
R> .Last.value
[1] 240
S+> f <- function(x) invisible(x+1)
S+> g <- function(x) 10 * x
S+> g(f(23)) # g() and f() are evaluated in same frame
S+> .Last.value
[1] 240
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-----Original Message-----
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of Rolf Turner
Sent: Friday, June 07, 2013 4:17 PM
To: Duncan Murdoch
Cc: r-help@r-project.org; Scott Raynaud
Subject: Re: [R] SPlus script
On 07/06/13 23:05, Duncan Murdoch wrote:
On 13-06-06 6:22 PM, Rolf Turner wrote:
On 07/06/13 03:19, Scott Raynaud wrote:
I actually had tried placing arguments in the call but it didn't
work. However, I did
not think about writing it to a variable and printing. That seems
to have done the
trick. Funny, I don't remember having to do that before, but that's
not surprising.
If I remember correctly --- haven't used Splus for decades --- this is a
difference
between Splus and R.
In R the output of a function is returned *invisibly* if that function
is called
from within another function. And source() is one such other function.
Actually this depends on the caller. source() does return its results
invisibly, but many other functions don't.
From FAQ 7.16:
If you type '1+1' or 'summary(glm(y~x+z, family=binomial))' at the
command line the returned value is automatically printed (unless it is
|invisible()|), but in other circumstances, such as in a |source()|d
file or ***inside a function*** it isn't printed unless you
specifically print it.
(Emphasis added.)
I think that you have misinterpreted what I wrote. Many (most?) functions
*return* their results (values) visibly. But if you put an expression
into the code
of that function (an expression which is not part of the returned value)
you never
see the result of evaluating that expression.
E.g.:
foo <- function(x) {
sin(42)
x^2
}
foo(3)
[1] 9
The value of sin(42) is never seen.
The main point however is that IIRC Splus is different from R in respect
of whether the values of (un-assigned) expressions inside source are
visible. In R they are invisible; in Splus I *believe* (vaguely recall)
that
they are visible. I cannot check this since I have no access to Splus.
I do wish someone would confirm (or deny, as the case may be) that
my recollections about Splus are correct.
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