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...@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]]
>
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