Evaluating lazily 'f<-' ?
Hello Andrew,
I try now to understand the evaluation of the expression:
e = expression(r(x) <- 1)
# parameter named "value" seems to be required;
'r<-' = function(x, value) {print("R");}
eval(e, list(x=2))
# [1] "R"
# both versions work
'r<-' = function(value, x) {print("R");}
eval(e, list(x=2))
# [1] "R"
### the Expression
e[[1]][[1]] # "<-", not "r<-"
e[[1]][[2]] # "r(x)"
The evaluation of "e" somehow calls "r<-", but evaluates also the
argument of r(...). I am still investigating what is actually happening.
Sincerely,
Leonard
On 9/13/2021 9:15 PM, Andrew Simmons wrote:
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 at syonic.eu
<mailto:leo.mada at 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 at r-project.org <mailto:r-help at 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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