(use 'clojure.set)
(def data
[
{:a 1 :b 1 :c 1 :d 1}
{:a 1 :b 2 :c 1 :d 2}
{:a 1 :b 2 :c 2 :d 3}
{:a 2 :b 1 :c 1 :d 5}
{:a 2 :b 1 :c 1 :d 6}
{:a 2 :b 1 :c 1 :d 7}
{:a 2 :b 2 :c 1 :d 7}
{:a 2 :b 3 :c 1 :d 7}
]
)
(defn key-join
"join of map by key , value is distinct."
[map-list]
(let [keys (keys (first map-list))]
(into {} (for [k keys] [k (vec (set (map #(% k) map-list)))]))))
(defn group-reduce [key map-list]
(let [sdata (set map-list)
group-value (project sdata [key])]
(into {}
(for [m group-value] [(key m) (key-join (map #(dissoc % key) (select #(= (key %) (key m)) sdata)))]))))
;;other version fast than group-reduce
(defn gr [key map-list]
(let [gdata (group-by key map-list)]
(into {} (for [[k m] gdata][k (dissoc (key-join m) key)]))))
user=> (group-reduce :a data)
{1 {:c [1 2], :b [1 2], :d [1 2 3]}, 2 {:c [1], :b [1 2 3], :d [5 6 7]}}
user=> (gr :a data)
{1 {:c [1 2], :b [1 2], :d [1 2 3]}, 2 {:c [1], :b [1 2 3], :d [5 6 7]}}
在第一张地图条目中,我看到你如何到达{:b [1 2]},但不知道如何到达{:c [1 2]}。它几乎看起来应该是{:c [1 1]}。你能否凭经验陈述算法? – octopusgrabbus
由于:a是主键,前3个映射缩减为1条记录,并且在3个映射中,当其他键的不同值累积时,应该导致:b [1 2]:c [1 2]和: d [1 2 3]。嵌套/递归组是否会这样做? – user922621