Clojure，合并两个地图数组答案

【问题标题】：Clojure, merging two array of mapsClojure，合并两个地图数组
【发布时间】：2012-04-18 12:17:28
【问题描述】：

我有两个地图数组

第一个是[{:a 1 :b 2 :d 6} {:a 2 :b 2} {:a 7 :b 7}]

第二个是[{:a 3 :c 3 :e 9 :y 7} {:a 2 :b 6 :c 8}]

取决于a 的值，即如果它在第二个数组中匹配，则“第二个地图”应该与“第一个地图”合并，并且生成的地图数组应该是

资源应该是[{:a 1 :b 2 :d 6} {:a 2 :b 6 :c 8} {:a 7 :b 7} {:a 3 :c 3 :e 9 :y 7}]

谁能帮我解决这个问题。提前致谢。

【问题讨论】：

我不太确定算法是什么，换句话说，我不太确定你在比较什么来获得“Res”。
我正在比较键“a”的值。如果第一个映射中的“a”的值与第二个数组中任何映射中的“a”的值匹配，则执行此操作（合并 {1st map with 'a' value = 1} {2nd map with 'a' value = 1 }) 并将这个结果映射添加到最终映射中

标签： clojure

【解决方案1】：

给你：

user> (def xs [{:a 1 :b 2 :d 6} {:a 2 :b 2} {:a 7 :b 7}])
#'user/xs
user> (def ys  [{:a 3 :c 3 :e 9 :y 7} {:a 2 :b 6 :c 8}])
#'user/ys
user> (for [[a ms] (group-by :a (concat xs ys))] (apply merge ms))
({:a 1, :b 2, :d 6} {:a 2, :c 8, :b 6} {:a 7, :b 7} {:y 7, :a 3, :c 3, :e 9})

【讨论】：

【解决方案2】：

这个数据结构在我看来非常笨拙，但这是我的看法：

(defn key-by-a [coll]
  "Convert a list of maps to a map of maps keyed by their vals at :a"
  (apply hash-map (mapcat (juxt :a identity) coll)))

(defn merge-map-lists [l1 l2]
  (->> [l1 l2]
    (map key-by-a)
    (apply merge-with merge)
    (vals)))

它不做的一件事是维护输入列表的顺序，但由于不清楚哪个列表决定（两者可能具有不同顺序的相同键），所以我把它省略了。

【讨论】：

【解决方案3】：

也许 clojure.set/join 是你想要的：

这是clojure.set/join的文档：

user=> (def animals #{{:name "betsy" :owner "brian" :kind "cow"}
                      {:name "jake"  :owner "brian" :kind "horse"}
                      {:name "josie" :owner "dawn"  :kind "cow"}})

user=> (def personalities #{{:kind "cow" :personality "stoic"}
                            {:kind "horse" :personality "skittish"}})
#'user/personalities
user=> (join animals personalities)

#{{:owner "dawn",  :name "josie", :kind "cow",   :personality "stoic"}
  {:owner "brian", :name "betsy", :kind "cow",   :personality "stoic"}
  {:owner "brian", :name "jake",  :kind "horse", :personality "skittish"}}

user=> (join animals personalities)
#{{:kind "horse", :owner "brian", :name "jake", :species "cow", :personality "stoic"}
  {:kind "cow", :owner "dawn", :name "josie", :species "cow", :personality "stoic"}
  {:kind "horse", :owner "brian", :name "jake", :species "horse", :personality "skittish"}
  {:kind "cow", :owner "brian", :name "betsy", :species "cow", :personality "stoic"}
  {:kind "cow", :owner "dawn", :name "josie", :species "horse", :personality "skittish"}
  {:kind "cow", :owner "brian", :name "betsy", :species "horse", :personality "skittish"}}


;; Notice that "Jake" is both a horse and a cow in the first line. That's 
;; likely not what you want. You can tell `join` to only produce output 
;; where the `:kind` value is the same as the `:species` value like this:

user=> (join animals personalities {:kind :species})
#{{:kind "cow", :owner "dawn", :name "josie", :species "cow", :personality "stoic"}
  {:kind "horse", :owner "brian", :name "jake", :species "horse", :personality "skittish"}
  {:kind "cow", :owner "brian", :name "betsy", :species "cow", :personality "stoic"}}

【讨论】：