How do I find the index of an element in a list in

Question

This is trivial implement of course, but I feel there is certainly something built in to Racket that does this. Am I correct in that intuition, and if so, what is function?

Answer 1

Here's a very simple implementation:

(define (index-of l x)
  (for/or ([y l] [i (in-naturals)] #:when (equal? x y)) i))

And yes, something like this should be added to the standard library, but it's just a little tricky to do so nobody got there yet.

Note, however, that it's a feature that is very rarely useful -- since lists are usually taken as a sequence that is deconstructed using only the first/rest idiom rather than directly accessing elements. More than that, if you have a use for it and you're a newbie, then my first guess will be that you're misusing lists. Given that, the addition of such a function is likely to trip such newbies by making it more accessible. (But it will still be added, eventually.)

Answer 2

One can also use a built-in function 'member' which gives a sublist starting with the required item or #f if item does not exist in the list. Following compares the lengths of original list and the sublist returned by member:

(define (indexof n l)
  (define sl (member n l))
  (if sl 
      (- (length l)
         (length sl))
      #f))

For many situations, one may want indexes of all occurrences of item in the list. One can get a list of all indexes as follows:

(define (indexes_of1 x l)
  (let loop ((l l)
             (ol '())
             (idx 0))
    (cond
      [(empty? l) (reverse ol)]
      [(equal? (first l) x)
       (loop (rest l)
             (cons idx ol)
             (add1 idx))]
      [else
       (loop (rest l)
             ol
             (add1 idx))])))

For/list can also be used for this:

(define (indexes_of2 x l)
  (for/list ((i l)
             (n (in-naturals))
             #:when (equal? i x))
    n))

Testing:

(indexes_of1 'a '(a b c a d e a f g))
(indexes_of2 'a '(a b c a d e a f g))

Output:

'(0 3 6)
'(0 3 6)

Answer 3

Strangely, there isn't a built-in procedure in Racket for finding the 0-based index of an element in a list (the opposite procedure does exist, it's called list-ref). However, it's not hard to implement efficiently:

(define (index-of lst ele)
  (let loop ((lst lst)
             (idx 0))
    (cond ((empty? lst) #f)
          ((equal? (first lst) ele) idx)
          (else (loop (rest lst) (add1 idx))))))

But there is a similar procedure in srfi/1, it's called list-index and you can get the desired effect by passing the right parameters:

(require srfi/1)

(list-index (curry equal? 3) '(1 2 3 4 5))
=> 2

(list-index (curry equal? 6) '(1 2 3 4 5))
=> #f

UPDATE

As of Racket 6.7, index-of is now part of the standard library. Enjoy!