Evaluating lazily 'f<-' ?
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