Hi Patrick,
On Sun, Jan 24, 2010 at 9:58 PM, CuppoJava <patrickli_2...@hotmail.com> wrote:
> It's an elegant puzzle. Thanks Sean!
>
> Here's my take:
> (defn sift [pred? s]
> (lazy-seq
> (if (seq s)
> (let [key (first s)
> [vals remaining] (split-with #(not (pred? %)) (rest s))]
> (cons [key vals] (sift pred? remaining))))))
>
> I'm interested in what you came up with. =)
> -Patrick
Mine looks very much like yours:
(defn sift [pred coll]
(lazy-seq
(when-let [[x & xs] (seq (drop-while (complement pred) coll))]
(cons [x (take-while (complement pred) xs)] (sift pred xs)))))
except that I discard the first items when they don't match pred,
while you assume that the first item always match pred.
I prefer to use when instead of if when there's no "else" expression.
I also have a non-lazy solution:
(defn sift [pred coll]
(letfn [(alter-top [v f & args] (conj (pop v) (apply f (peek v) args)))]
(reduce #(if (pred %2)
(conj %1 [%2 []])
(alter-top %1 alter-top conj %2)) [] coll)))
Christophe
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