[R] Evaluating lazily 'f<-' ?

[Leonard Mada]

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 <mailto: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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