Removing elements in a nested R list by name - r

Question

I have a nested element like this
> x <- list(a=list(from="me", id="xyz"), b=list(comment=list(list(message="blabla", id="abc"), list(message="humbug", id="jkl"))), id="123")
> str(x)
List of 3
$ a :List of 2
..$ from: chr "me"
..$ id : chr "xyz"
$ b :List of 1
..$ comment:List of 2
.. ..$ :List of 2
.. .. ..$ message: chr "blabla"
.. .. ..$ id : chr "abc"
.. ..$ :List of 2
.. .. ..$ message: chr "humbug"
.. .. ..$ id : chr "jkl"
$ id: chr "123"
How can I remove all the elements with name id in all levels of the list? i.e. the expected output is
> str(x)
List of 2
$ a:List of 1
..$ from: chr "me"
$ b:List of 1
..$ comment:List of 2
.. ..$ :List of 1
.. .. ..$ message: chr "blabla"
.. ..$ :List of 1
.. .. ..$ message: chr "humbug"
Solutions using rlist package would be particularly welcome, but I'm happy with anything that works.

Answer 1

Recursion is also how I did it:
# recursive function to remove name from all levels of list
stripname <- function(x, name) {
thisdepth <- depth(x)
if (thisdepth == 0) {
return(x)
} else if (length(nameIndex <- which(names(x) == name))) {
x <- x[-nameIndex]
}
return(lapply(x, stripname, name))
}
# function to find depth of a list element
# see http://stackoverflow.com/questions/13432863/determine-level-of-nesting-in-r
depth <- function(this, thisdepth=0){
if (!is.list(this)) {
return(thisdepth)
} else{
return(max(unlist(lapply(this,depth,thisdepth=thisdepth+1))))
}
}
str(stripname(x, "id"))
## List of 2
## $ a:List of 1
## ..$ from: chr "me"
## $ b:List of 1
## ..$ comment:List of 2
## .. ..$ :List of 1
## .. ..$ :List of 1
## .. .. ..$ message: chr "blabla"
## .. .. ..$ message: chr "humbug"

Answer 2

Try a recursive function in the veins of
f <- function(i)
lapply(i, function(x)
if (is.list(x)) {
if(!is.null(names(x))) f(x[names(x)!="id"]) else f(x)
} else x
)
str(f(x[names(x)!="id"]))
# List of 2
# $ a:List of 1
# ..$ from: chr "me"
# $ b:List of 1
# ..$ comment:List of 2
# .. ..$ :List of 1
# .. .. ..$ message: chr "blabla"
# .. ..$ :List of 1
# .. .. ..$ message: chr "humbug"

Answer 3

This is an old question, but this can also be done quite conveniently with rrapply() in the rrapply-package (revisit of base rapply()):
rrapply::rrapply(
x, ## nested list
condition = \(x, .xname) .xname != "id", ## filter condition
how = "prune" ## how to structure result
) |>
str()
#> List of 2
#> $ a:List of 1
#> ..$ from: chr "me"
#> $ b:List of 1
#> ..$ comment:List of 2
#> .. ..$ :List of 1
#> .. .. ..$ message: chr "blabla"
#> .. ..$ :List of 1
#> .. .. ..$ message: chr "humbug"