向量映射到向量映射

Question

I would like to replace this structure,

{:a [1 2 .. 3] :b [1 2 .. 3] :c [1 2 3 ..]} 

to this.

[{:a 1 :b 1 c 1} {:a 2 :b 2 :c 2} {:a 3 :b 3 :c 3} ...] 

Number of keys is not specified.

Answer 1

I found the solution, very ugly solution, but it can explain what I need.

#(map (fn [line] 
       (zipmap (keys %) line)) 
     (partition (count (keys %)) 
                (apply interleave (vals %))))

If anyone could simplify this, it would be great.

EDIT:

#(map (fn [vs]
     (zipmap (keys %) vs))(apply map vector (vals %)))

Answer 2

(def z (sorted-map :a [1 2 3 4] :b ["a" "b" "c" "d"] :c [:i :j :k :l]))

(defn x []
  (let [k (repeat (keys z))
        v (apply map vector (vals z))]
    (map zipmap k v)))


user> (x)
({:c :i, :b "a", :a 1} {:c :j, :b "b", :a 2} {:c :k, :b "c", :a 3} {:c :l, :b "d", :a 4})

Answer 3

Here is another solution

   #(mapv 
      (partial apply merge) 
      (map val 
         (group-by (comp  second first) 
         (for [[x y] % z y] (hash-map x z)))))

;=>[{:a 1, :c 1, :b 1} {:a 2, :c 2, :b 2} {:a 3, :c 3, :b 3}]

Answer 4

(let [x  {1 [:a :b :c], 2 [:d :e :f], 3 [:g :h :i]}
      k (vec (keys x))
      r (map #(apply merge %)
             (apply map (fn [ & a]
                          (merge (map-indexed
                                  (fn [idx v] {(k idx) v}) a))) (vals x)))]
      (println r))