How to create a list of lists in scheme?

Question

I may have missed this in the R5RS document but how do I create a list of lists in (Chicken) Scheme? I want to be able to take a list, a, invoke (list-ref a b), assign the result to c, and then invoke (list-ref c d), where b and d are index values.

Edit: For clarification, suppose I have these lists:

(define citrus (list "oranges" "limes"))
(define apples (list "macintosh" "rome" "delicious"))

And then I want to create a list called fruit with the citrus and apples as list entries.

Answer 1

If you want to make a list containing those lists, just call list with them as arguments:

(define fruit (list citrus apples))

(list-ref (list-ref fruit 0) 1)
=> "lime"

Answer 2

Here's how you create a list of lists:

(list (list 1 2) (list 3 4))   

Or even simpler:

'((1 2) (3 4))

Now, if you already have the other sublists defined as separate lists, put them inside an outer list calling list again on them:

(define the-first  (list 1 2))
(define the-second (list 3 4))
(define list-of-lists (list the-first the-second))
list-of-lists
=> '((1 2) (3 4)) 

To access a position given two indexes, do this - remember, indexes are zero-based:

(define lst '((1 2) (3 4)))
(list-ref (list-ref lst 1) 0)
=> 3

So, the first example in the question would look like this:

(define a '((1 2) (3 4)))
(define b 1)
(define c (list-ref a b))
(define d 0)
(list-ref c d)
=> 3

And the second example (after the edit) would look like this:

(define citrus (list "oranges" "limes"))
(define apples (list "macintosh" "rome" "delicious"))
(define fruit (list citrus apples)) ; here's the list of lists

Now, to access an element first we have to pass the index of the outermost list (let's say we want apples, which are at index 1 in the outermost list) and then the index of the innermost list (let's say we want a macintosh, which is at index 0 in the apples sublist):

(list-ref (list-ref fruit 1) 0)
=> "macintosh"