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Clojure:地图中的聚合和计数

来源:互联网 收集:自由互联 发布时间:2021-06-22
我想这个问题可以作为入门级的clojure问题.我基本上有多次麻烦地处理一个clojure地图并提取不同类型的数据. 给定这样的映射,我试图基于多个嵌套键来计算条目: [ { "a": "X", "b": "M", "c
我想这个问题可以作为入门级的clojure问题.我基本上有多次麻烦地处理一个clojure地图并提取不同类型的数据.

给定这样的映射,我试图基于多个嵌套键来计算条目:

[
  {
    "a": "X",
    "b": "M",
    "c": 188
  },
  {
    "a": "Y",
    "b": "M",
    "c": 165
  },
  {
    "a": "Y",
    "b": "M",
    "c": 313
  },
  {
    "a": "Y",
    "b": "P",
    "c": 188
  }
]

首先,我想通过a键值对条目进行分组:

{
  "X" : [
    {
      "b": "M",
      "c": 188
    }
  ],
  "Y" : [
    {
      "b": "M",
      "c": 165
    },
    {
      "b": "M",
      "c": 313
    },
    {
      "b": "P",
      "c": 188
    }
  ]
}

其次,我想假设b键的值是重复的,并忽略其余的键:

{
  "X" : [
    {
      "b": "M"
    }
  ],
  "Y" : [
    {
      "b": "M"
    }
    {
      "b": "P"
    }
  ]
}

然后,只需计算b键的所有实例:

{
  "X" : 1,
  "Y" : 2
}

当我通过monger获取数据时,我定义了:

(defn db-query
  ([coll-name]
     (with-open [conn (mg/connect)]
       (doall (mc/find-maps (mg/get-db conn db-name) coll-name))))

然后点击路障:

(defn get-sums [request]
  (->> (db-query "data")
       (group-by :a)
       (into {})
        keys))

我怎么能从这里继续?

这是一种天真的方法,我相信有更好的方法,但它可能是你需要弄清楚的.

(into {}
  (map       

    ; f       
    (fn [ [k vs] ] ;[k `unique count`]
      [k (count (into #{} (map #(get % "b") vs)))]) 

    ; coll
    (group-by #(get % "a") DATA))) ; "a"s as keys
;user=> {"X" 1, "Y" 2}

说明:

; I am using your literal data as DATA, just removed the , and ;
(def DATA [{...

(group-by #(get % "a") DATA) ; groups by "a" as keys
; so I get a map {"X":[{},...] "Y":[{},{},{},...]}

; then I map over each [k v] pair where
; k is the map key and
; vs are the grouped maps in a vector
(fn [ [k vs] ] 
      ; here `k` is e.g. "Y" and `vs` are the maps {a _, b, _, c _}

      ; now `(map #(get % "b") vs)` gets me all the b values
      ; `into set` makes them uniqe
      ; `count` counts them
      ; finally I return a vector with the same name `k`,
      ;   but the value is the counted `b`s
      [k (count (into #{} (map #(get % "b") vs)))]) 

; at the end I just put the result `[ ["Y" 2] ["X" 1] ]` `into` a map {}
; so you get a map
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