[R] Evaluating lazily 'f<-' ?

[Andrew Simmons]

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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