Evaluating lazily 'f<-' ?
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> 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>
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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