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

Alex V picture Alex V · Apr 8, 2013 · Viewed 16.5k times · Source

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 the function?

Answer

Óscar López picture Óscar López · Apr 8, 2013

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!