[R] return value of {....}

[Bill Dunlap]

R's
   { expr1; expr2; expr3}
acts much like C's
   ( expr1, expr2, expr3)

E.g.,

$ cat a.c
#include <stdio.h>

int main(int argc, char* argv[])
{
    double y = 10 ;
    double x = (printf("Starting... "), y = y + 100, y * 20);
    printf("Done: x=%g, y=%g\n", x, y);
    return 0;
}
$ gcc -Wall a.c
$ ./a.out
Starting... Done: x=2200, y=110

I don't like that syntax (e.g., commas between expressions instead of the
usual semicolons after expressions).  Perhaps John Chambers et all didn't
either.

-Bill

On Fri, Jan 13, 2023 at 2:28 AM Valentin Petzel <valentin using petzel.at> wrote:

> Hello Akshay,
>
> R is quite inspired by LISP, where this is a common thing. It is not in
> fact that {...} returned something, rather any expression evalulates to
> some value, and for a compound statement that is the last evaluated
> expression.
>
> {...} might be seen as similar to LISPs (begin ...).
>
> Now this is a very different thing compared to {...} in something like C,
> even if it looks or behaves similarly. But in R {...} is in fact an
> expression and thus has evaluate to some value. This also comes with some
> nice benefits.
>
> You do not need to use {...} for anything that is a single statement. But
> you can in each possible place use {...} to turn multiple statements into
> one.
>
> Now think about a statement like this
>
> f <- function(n) {
> x <- runif(n)
> x**2
> }
>
> Then we can do
>
> y <- f(10)
>
> Now, you suggested way would look like this:
>
> f <- function(n) {
> x <- runif(n)
> y <- x**2
> }
>
> And we'd need to do something like:
>
> f(10)
> y <- somehow_get_last_env_of_f$y
>
> So having a compound statement evaluate to a value clearly has a benefit.
>
> Best Regards,
> Valentin
>
> 09.01.2023 18:05:58 akshay kulkarni <akshay_e4 using hotmail.com>:
>
> > Dear Valentin,
> >                           But why should {....} "return" a value? It
> could just as well evaluate all the expressions and store the resulting
> objects in whatever environment the interpreter chooses, and then it would
> be left to the user to manipulate any object he chooses. Don't you think
> returning the last, or any value, is redundant? We are living in the
> 21st century world, and the R-core team might,I suppose, have a definite
> reason for"returning" the last value. Any comments?
> >
> > Thanking you,
> > Yours sincerely,
> > AKSHAY M KULKARNI
> >
> > ----------------------------------------
> > *From:* Valentin Petzel <valentin using petzel.at>
> > *Sent:* Monday, January 9, 2023 9:18 PM
> > *To:* akshay kulkarni <akshay_e4 using hotmail.com>
> > *Cc:* R help Mailing list <r-help using r-project.org>
> > *Subject:* Re: [R] return value of {....}
> >
> > Hello Akshai,
> >
> > I think you are confusing {...} with local({...}). This one will
> evaluate the expression in a separate environment, returning the last
> expression.
> >
> > {...} simply evaluates multiple expressions as one and returns the
> result of the last line, but it still evaluates each expression.
> >
> > Assignment returns the assigned value, so we can chain assignments like
> this
> >
> > a <- 1 + (b <- 2)
> >
> > conveniently.
> >
> > So when is {...} useful? Well, anyplace where you want to execute
> complex stuff in a function argument. E.g. you might do:
> >
> > data %>% group_by(x) %>% summarise(y = {if(x[1] > 10) sum(y) else
> mean(y)})
> >
> > Regards,
> > Valentin Petzel
> >
> > 09.01.2023 15:47:53 akshay kulkarni <akshay_e4 using hotmail.com>:
> >
> >> Dear members,
> >>                              I have the following code:
> >>
> >>> TB <- {x <- 3;y <- 5}
> >>> TB
> >> [1] 5
> >>
> >> It is consistent with the documentation: For {, the result of the last
> expression evaluated. This has the visibility of the last evaluation.
> >>
> >> But both x AND y are created, but the "return value" is y. How can this
> be advantageous for solving practical problems? Specifically, consider the
> following code:
> >>
> >> F <- function(X) {  expr; expr2; { expr5; expr7}; expr8;expr10}
> >>
> >> Both expr5 and expr7 are created, and are accessible by the code
> outside of the nested braces right? But the "return value" of the nested
> braces is expr7. So doesn't this mean that only expr7 should be accessible?
> Please help me entangle this (of course the return value of F is expr10,
> and all the other objects created by the preceding expressions are deleted.
> But expr5 is not, after the control passes outside of the nested braces!)
> >>
> >> Thanking you,
> >> Yours sincerely,
> >> AKSHAY M KULKARNI
> >>
> >>     [[alternative HTML version deleted]]
> >>
> >> ______________________________________________
> >> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >> 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.
>
>         [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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.
>

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