return value of {....}

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 at petzel.at<mailto:valentin at 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 at hotmail.com<mailto:akshay_e4 at 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 at petzel.at<mailto:valentin at petzel.at>>
> *Sent:* Monday, January 9, 2023 9:18 PM
> *To:* akshay kulkarni <akshay_e4 at hotmail.com<mailto:akshay_e4 at hotmail.com>>
> *Cc:* R help Mailing list <r-help at r-project.org<mailto:r-help at 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 at hotmail.com<mailto:akshay_e4 at 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 at r-project.org<mailto:R-help at 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]]

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