List of indexes

[cameyo]

How to create a list of indexes of all the elements of a generic list?

Example:

Input: (setq lst '(1 (2 (3 4)) (5 6)))

(lst 0)

1

(lst 1)

(2 (3 4))

(lst 1 0)

2

(lst 1 1)

(3 4)

(lst 1 1 0)

3

(lst 1 1 1)

4

(lst 2)

(5 6)

(lst 2 0)

5

(lst 2 1)

6



Output: List of indexes of lst

((0) (1) (1 0) (1 1) (1 1 0) (1 1 1) (2) (2 0) (2 1))

or

(0 1 (1 0) (1 1) (1 1 0) (1 1 1) 2 (2 0) (2 1))



p.s. can't post formatted code on forum (Internal Server Error)

[fdb]

Something like below? (not very elegant, I know)

Code Select Expand

(define (index-list lst)
  (setq mylist '())
  (define (h-index-list lst agg)
    (dolist (x lst)
      (setq mv (append agg (list $idx)))
      (push mv mylist -1)
      (if (list? x)
        (h-index-list x mv)))
    mylist)
  (h-index-list lst '()))

[rickyboy]

Here's a version that uses recursive calls in a classic way (think, SICP) where even the loop is handled by recursive call.  (Nota bene: This is not a good newLISP implementation because newLISP doesn't turn tail calls into loops; fdb's implementation is the better one for newLISP.)

Code Select Expand


(define (get-indices L (child 0) (parents '()) (result '()))
  (if (empty? L)
      result
      (list? (L 0))
      (get-indices (1 L) (+ 1 child) parents
                   (append (snoc result (snoc parents child))
                           (get-indices (L 0) 0 (snoc parents child))))
      (get-indices (1 L) (+ 1 child) parents
                   (snoc result (snoc parents child)))))


You will need this utility function, snoc, which acts like cons but does the reverse (it says it in the name lol :) : it takes the element argument and puts it at the end of the list.  (Note also that the arguments are reversed as compared with cons.)

Code Select Expand


(define (snoc xs x) (push x xs -1))


[rickyboy]

You could also "factor out" the repeated code, but it may not make the code more readable.

Code Select Expand


(define (get-indices L (child 0) (parents '()) (result '()))
  (if (empty? L)
      result
      (get-indices (1 L) (+ 1 child) parents
        (append
          (snoc result (snoc parents child))
          (if (list? (L 0))
              (get-indices (L 0) 0 (snoc parents child))
              '())))))


[cameyo]

Thank you guys

Very nice solutions

@rickyboy: "Nota bene" is Italian :-)

[cameyo]

Faster solution:

Code Select Expand

(setq a '(1 (2 (3 4)) (5 6)))
;Indexes
(ref-all nil a (fn (x) true))
((0) (1) (1 0) (1 1) (1 1 0) (1 1 1) (2) (2 0) (2 1))
;Elements
(ref-all nil a (fn (x) true) true)
(1 (2 (3 4)) 2 (3 4) 3 4 (5 6) 5 6)