Add word-count exercise

[mbertheau]

No description provided.

[lexi-lambda]

Here's my implementation of this problem, which I think is more Racket-y.

(define (word-count word)
  (for/fold ([result (hash)])
            ([word (in-list (string-split word))])
    (hash-set result word (add1 (hash-ref result word 0)))))

This code uses immutable hash tables. Mutation is usually considered unidiomatic in Racket, so hash-set returns a new hash table instead of updating the existing one. It also uses add1 instead of (+ 1 ...), passes 0 as the default value to hash-ref, and uses for/fold for the accumulation of result.

[mbertheau]

Nice. Feels much more idiomatic! :)