Here's my take:
(defn partition' [coll]
(let [coll' (partition-by keyword? coll)]
(reduce merge {}
(map vector (flatten (take-nth 2 coll')) (take-nth 2 (rest
coll'))))))
Note that this breaks if some keys point to empty sets
On Wednesday, February 20, 2013 6:50:42 AM UTC-5, Stefan Kamphausen wrote:
>
> Hi,
>
> given a vector of the form
>
>
>
> [:key1 1 2 3 :key2 4 :key3 5 6 7]
>
> I wand to create a map collecting the items behind each keyword as a vector
> like this:
>
> {:key1 [1 2 3]
> :key2 [4]
> :key3 [5 6 7]}
>
> I have already written two functions which achieve this, but neither of them
> "feels good" and I am interested in more elegant and idiomatic solutions.
>
> (defn keycollect-partition [coll]
> (->> coll
> (partition-by keyword?) ; bundle at key
> (partition 2) ; kw + its arg
> (map (fn [[[sec] arg]] [sec (vec arg)])) ; destruct the mess
> (into {}))) ; make it a map
>
> (defn keycollect-reduce [coll]
> (apply zipmap
> (reduce
> (fn [ac x]
> (if (keyword? x)
> [(conj (first ac) x) (conj (second ac) [])]
> (update-in ac [1 (dec (count (first ac)))] conj x))) [[] []]
> coll)))
>
> Added complexity: My original vector does not yet contain keywords, but I
> construct them in a map regexp-matching each item and creating a keyword from
> the first group of the match.
>
> Any pointers or ideas appreciated.
>
> Regards,
> Stefan
>
>
>
>
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