Find the index of element in list

Question

I need to get index of element in list in scheme. For example:

(... 2 '(2 3 4 5))

0

(... 4 '(2 3 4 5))

2

Can someone help?

Answer 1

Somthing like this

(define list-index
        (lambda (e lst)
                (if (null? lst)
                        -1
                        (if (eq? (car lst) e)
                                0
                                (if (= (list-index e (cdr lst)) -1) 
                                        -1
                                        (+ 1 (list-index e (cdr lst))))))))

Answer 2

The following is the clearest solution I could come up with:

(define (get-list-index l el)
    (if (null? l)
        -1
        (if (= (car l) el)
            0
            (let ((result (get-list-index (cdr l) el)))
                (if (= result -1)
                    -1
                    (1+ result))))))

This solution is largely the same as merriav's except that I've added a let at the end so that the recursive call is not unnecessarily repeated (in written code or execution).

The accepted solution does not seem to account for an empty list, or a list that does not contain the element being sought after.

Answer 3

You could implement index using reverse, member, length, and cdr as follows:

(define (index a b)
  (let [(tail (member a (reverse b)))]
    (and tail (length (cdr tail))))