[R] Evaluating lazily 'f<-' ?

[Leonard Mada]

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
>