[R] Evaluating lazily 'f<-' ?

[Andrew Simmons]

R's parser doesn't work the way you're expecting it to. When doing an
assignment like:


padding(right(df)) <- 1


it is broken into small stages. The guide "R Language Definition" claims
that the above would be equivalent to:


`<-`(df, `padding<-`(df, value = `right<-`(padding(df), value = 1)))


but that is not correct, and you can tell by using `substitute` as you were
above. There isn't a way to do what you want with the syntax you provided,
you'll have to do something different. You could add a `which` argument to
each style function, and maybe put the code for `match.arg` in a separate
function:


match.which <- function (which)
match.arg(which, c("bottom", "left", "top", "right"), several.ok = TRUE)


padding <- function (x, which)
{
    which <- match.which(which)
    # more code
}


border <- function (x, which)
{
    which <- match.which(which)
    # more code
}


some_other_style <- function (x, which)
{
    which <- match.which(which)
    # more code
}


I hope this helps.

On Mon, Sep 13, 2021 at 12:17 PM Leonard Mada <leo.mada using syonic.eu> wrote:

> Hello Andrew,
>
>
> this could work. I will think about it.
>
> But I was thinking more generically. Suppose we have a series of functions:
> padding(), border(), some_other_style();
> Each of these functions has the parameter "right" (or the group of
> parameters c("right", ...)).
>
>
> Then I could design a function right(FUN) that assigns the value to this
> parameter and evaluates the function FUN().
>
>
> There are a few ways to do this:
> 1.) Other parameters as ...
> right(FUN, value, ...) = value; and then pass "..." to FUN.
> right(value, FUN, ...) = value; # or is this the syntax? (TODO: explore)
>
> 2.) Another way:
> right(FUN(...other parameters already specified...)) = value;
> I wanted to explore this 2nd option: but avoid evaluating FUN, unless the
> parameter "right" is injected into the call.
>
> 3.) Option 3:
> The option you mentioned.
>
>
> Independent of the method: there are still weird/unexplained behaviours
> when I try the initial code (see the latest mail with the improved code).
>
>
> Sincerely,
>
>
> Leonard
>
>
> On 9/13/2021 6:45 PM, Andrew Simmons wrote:
>
> I think you're trying to do something like:
>
> `padding<-` <- function (x, which, value)
> {
>     which <- match.arg(which, c("bottom", "left", "top", "right"),
> several.ok = TRUE)
>     # code to pad to each side here
> }
>
> Then you could use it like
>
> df <- data.frame(x=1:5, y = sample(1:5, 5))
> padding(df, "right") <- 1
>
> Does that work as expected for you?
>
> On Mon, Sep 13, 2021, 11:28 Leonard Mada via R-help <r-help using r-project.org>
> wrote:
>
>> I try to clarify the code:
>>
>>
>> ###
>> right = function(x, val) {print("Right");};
>> padding = function(x, right, left, top, bottom) {print("Padding");};
>> 'padding<-' = function(x, ...) {print("Padding = ");};
>> df = data.frame(x=1:5, y = sample(1:5, 5)); # anything
>>
>> ### Does NOT work as expected
>> 'right<-' = function(x, value) {
>>      print("This line should be the first printed!")
>>      print("But ERROR: x was already evaluated, which printed
>> \"Padding\"");
>>      x = substitute(x); # x was already evaluated before substitute();
>>      return("Nothing"); # do not now what the behaviour should be?
>> }
>>
>> right(padding(df)) = 1;
>>
>> ### Output:
>>
>> [1] "Padding"
>> [1] "This line should be the first printed!"
>> [1] "But ERROR: x was already evaluated, which printed \"Padding\""
>> [1] "Padding = " # How did this happen ???
>>
>>
>> ### Problems:
>>
>> 1.) substitute(x): did not capture the expression;
>> - the first parameter of 'right<-' was already evaluated, which is not
>> the case with '%f%';
>> Can I avoid evaluating this parameter?
>> How can I avoid to evaluate it and capture the expression: "right(...)"?
>>
>>
>> 2.) Unexpected
>> 'padding<-' was also called!
>> I did not know this. Is it feature or bug?
>> R 4.0.4
>>
>>
>> Sincerely,
>>
>>
>> Leonard
>>
>>
>> On 9/13/2021 4:45 PM, Duncan Murdoch wrote:
>> > On 13/09/2021 9:38 a.m., Leonard Mada wrote:
>> >> Hello,
>> >>
>> >>
>> >> I can include code for "padding<-"as well, but the error is before
>> that,
>> >> namely in 'right<-':
>> >>
>> >> right = function(x, val) {print("Right");};
>> >> # more options:
>> >> padding = function(x, right, left, top, bottom) {print("Padding");};
>> >> 'padding<-' = function(x, ...) {print("Padding = ");};
>> >> df = data.frame(x=1:5, y = sample(1:5, 5));
>> >>
>> >>
>> >> ### Does NOT work
>> >> 'right<-' = function(x, val) {
>> >>         print("Already evaluated and also does not use 'val'");
>> >>         x = substitute(x); # x was evaluated before
>> >> }
>> >>
>> >> right(padding(df)) = 1;
>> >
>> > It "works" (i.e. doesn't generate an error) for me, when I correct
>> > your typo:  the second argument to `right<-` should be `value`, not
>> > `val`.
>> >
>> > I'm still not clear whether it does what you want with that fix,
>> > because I don't really understand what you want.
>> >
>> > Duncan Murdoch
>> >
>> >>
>> >>
>> >> I want to capture the assignment event inside "right<-" and then call
>> >> the function padding() properly.
>> >>
>> >> I haven't thought yet if I should use:
>> >>
>> >> padding(x, right, left, ... other parameters);
>> >>
>> >> or
>> >>
>> >> padding(x, parameter) <- value;
>> >>
>> >>
>> >> It also depends if I can properly capture the unevaluated expression
>> >> inside "right<-":
>> >>
>> >> 'right<-' = function(x, val) {
>> >>
>> >> # x is automatically evaluated when using 'f<-'!
>> >>
>> >> # but not when implementing as '%f%' = function(x, y);
>> >>
>> >> }
>> >>
>> >>
>> >> Many thanks,
>> >>
>> >>
>> >> Leonard
>> >>
>> >>
>> >> On 9/13/2021 4:11 PM, Duncan Murdoch wrote:
>> >>> On 12/09/2021 10:33 a.m., Leonard Mada via R-help wrote:
>> >>>> How can I avoid evaluation?
>> >>>>
>> >>>> right = function(x, val) {print("Right");};
>> >>>> padding = function(x) {print("Padding");};
>> >>>> df = data.frame(x=1:5, y = sample(1:5, 5));
>> >>>>
>> >>>> ### OK
>> >>>> '%=%' = function(x, val) {
>> >>>>        x = substitute(x);
>> >>>> }
>> >>>> right(padding(df)) %=% 1; # but ugly
>> >>>>
>> >>>> ### Does NOT work
>> >>>> 'right<-' = function(x, val) {
>> >>>>        print("Already evaluated and also does not use 'val'");
>> >>>>        x = substitute(x); # is evaluated before
>> >>>> }
>> >>>>
>> >>>> right(padding(df)) = 1
>> >>>
>> >>> That doesn't make sense.  You don't have a `padding<-` function, and
>> >>> yet you are trying to call right<- to assign something to padding(df).
>> >>>
>> >>> I'm not sure about your real intention, but assignment functions by
>> >>> their nature need to evaluate the thing they are assigning to, since
>> >>> they are designed to modify objects, not create new ones.
>> >>>
>> >>> To create a new object, just use regular assignment.
>> >>>
>> >>> Duncan Murdoch
>> >
>>
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>

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