Reputation: 107
I am vexed by the way R dynamically creates lists and I am hoping someone can help me understand what is going on and what to do to fix my code. My problem is that for assignments of a vector of length one, a named vector is assigned, but assignments of a vector of length greater than one, a list is assigned. My desired outcome is that a list is assigned no matter the length of the vector I am assigning. How do I achieve such a result?
For example,
types <- c("a", "b")
lst <- vector("list", length(types))
names(lst) <- types
str(lst)
List of 2
$ a: NULL
$ b: NULL
lst$a[["foo"]] <- "hi"
lst$b[["foo"]] <- c("hi", "SO")
str(lst)
List of 2
$ a: Named chr "hi"
..- attr(*, "names")= chr "foo"
$ b:List of 1
..$ foo: chr [1:2] "hi" "SO"
str(lst$a)
Named chr "hi"
- attr(*, "names")= chr "foo"
str(lst$b)
List of 1
$ foo: chr [1:2] "hi" "SO"
What I want to have as the outcome is a data structure that looks like this.
List of 2
$ a:List of 1
..$ foo: chr [1] "hi"
$ b:List of 1
..$ foo: chr [1:2] "hi" "SO"
Upvotes: 3
Views: 10042
Reputation: 89097
While I also find it surprising, it is documented in ?[[
:
Recursive (list-like) objects:
[...]
When ‘$<-’ is applied to a ‘NULL’ ‘x’, it first coerces ‘x’ to
‘list()’. This is what also happens with ‘[[<-’ if the
replacement value ‘value’ is of length greater than one: if
‘value’ has length 1 or 0, ‘x’ is first coerced to a zero-length
vector of the type of ‘value’.
To override that behavior, you could specifically create empty lists before dynamically assigning to them:
lst$a <- list()
lst$b <- list()
or like Josh suggested below, replace your lst <- vector("list", length(types))
with lst <- replicate(length(types), list())
.
Now that ‘x’
(lst$a
or lst$b
) is not ‘NULL’
but an empty list, your code should work as you expected:
lst$a[["foo"]] <- "hi"
lst$b[["foo"]] <- c("hi", "SO")
str(lst)
# List of 2
# $ a:List of 1
# ..$ foo: chr "hi"
# $ b:List of 1
# ..$ foo: chr [1:2] "hi" "SO"
Upvotes: 3
Reputation: 368629
I think you just need to create the types you want and assign them:
R> qq <- list( a=list(foo="Hi"), b=list(foo=c("Hi", "SO")))
R> qq
$a
$a$foo
[1] "Hi"
$b
$b$foo
[1] "Hi" "SO"
R>
where all your requirements are met:
R> class(qq)
[1] "list"
R> names(qq)
[1] "a" "b"
R> sapply(qq, names)
a b
"foo" "foo"
R>
Upvotes: 2