Here a different version, without the sequence abstraction. I realise
its idiomatically obsolete, but thought you guys might enjoy it.
(defn sift
"Sift. Old Skool."
[key-pred in-s]
(reduce
(fn [coll elem]
(if (key-pred elem)
(conj coll [(str elem) []])
(conj
(pop coll)
[(first (last coll))
(conj (last (last coll)) (str elem))])))
[]
in-s))
On Jan 25, 10:21 am, Christophe Grand <[email protected]> wrote:
> Hi Patrick,
>
> On Sun, Jan 24, 2010 at 9:58 PM, CuppoJava <[email protected]> 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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