如何在R中制作一个无限递归列表：混淆[和[[

Question

Edit: this question is silly, I confused [ and [[ (thanks @josilber), but I can't delete it.

How can one make an infinitely recursive list, L == L[1], in R ?
A terrible way (for an R novice) is to look at dataset fmri in package astsa :

data( fmri, package="astsa" )  # a list with fmri[1] == fmri[1][1] ... ??

print_recursive = function( x ){  # x x[1] x[1][1] ... while list
    for( j in 1:5 ){
        cat( class(x), len(x), names(x), "\n" )
        str(x)
        cat( "\n" )

        x = x[1]  # <-- wrong, should be x[[1]]
        if( class(x) != "list" )  break
    }
    x
}

x = print_recursive( fmri )

Answer 1

The answer to "How can one make an infinitely recursive list, L == L[1], in R" is that it is impossible, because you would need an infinite amount of memory to store a list of infinite recursive depth.

That being said, you can build a recursive list of a specified depth with a simple for loop, at each iteration creating a new list that stores the old list as one of its elements:

l <- list()
depth <- 50
for (k in seq(depth-1)) {
  l <- list(l)
}

You could write a recursive function to check the depth of a recursive list:

recursive.depth <- function(l) {
  if (!is.list(l)) 0
  else if (length(l) == 0) 1
  else 1+max(sapply(l, recursive.depth))
}
recursive.depth(l)
# [1] 50
recursive.depth(fmri)
# [1] 1

Getting back to the example from the question, the list you have is actually not recursive at all (it's just a list of matrices). The reason you think it is recursive is that indexing l[1] subsets the list (aka it returns the list with just its first element). For instance, consider the following very simple list:

(l <- list(2))
# [[1]]
# [1] 2

No matter how many times I subset with the [ notation, I will get back the exact same list:

l[1]
# [[1]]
# [1] 2
l[1][1]
# [[1]]
# [1] 2

Running this list l through your print_recursive function will also result in an infinite loop. If you wanted to actually extract the first element of the list instead of subsetting the list itself, you should use the [[ notation (eg l[[1]] ).