[Rd] eval(parse()) within mutate() returning same value for all rows

[Duncan Murdoch]

On 29/12/2023 9:13 a.m., Mateo Obregón wrote:
> Hi all-
> 
> Looking through stackoverflow for R string combining examples, I found the
> following from 3 years ago:
> 
> <https://stackoverflow.com/questions/63881854/how-to-format-strings-using-values-from-other-column-in-r>
> 
> The top answer suggests to use eval(parse(sprintf())). I tried the suggestion
> and it did not return the expected combines strings. I thought that this might
> be an issue with some leftover values being reused, so I explicitly eval()
> with a new.env():
> 
>> library(dplyr)
>> df <- tibble(words=c("%s plus %s equals %s"),
> args=c("1,1,2","2,2,4","3,3,6"))
>> df |> mutate(combined = eval(parse(text=sprintf("sprintf('%s', %s)", words,
> args)), envir=new.env()))
> 
> # A tibble: 3 × 3
>    words                args  combined
>    <chr>                <chr> <chr>
> 1 %s plus %s equals %s 1,1,2 3 plus 3 equals 6
> 2 %s plus %s equals %s 2,2,4 3 plus 3 equals 6
> 3 %s plus %s equals %s 3,3,6 3 plus 3 equals 6
> 
> The `combined`  is not what I was expecting, as the same last eval() is
> returned for all three rows.
>   
> Am I missing something? What has changed in the past three years?
> 

I don't know if this is a change, but when `eval()` is passed an 
expression vector, it evaluates the elements in order and returns the 
value of the last one.  This is only partially documented:

"Value:  The result of evaluating the object: for an expression vector 
this is the result of evaluating the last element."


That text has been unchanged in the help page for 13 years.

Duncan Murdoch