4

我想这个问题符合入门级 clojure 问题。我基本上无法多次处理 clojure 映射并提取不同类型的数据。

给定这样的地图，我试图根据多个嵌套键计算条目：

[
  {
    "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，我想按 a-key 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-keys 的值是重复的，并忽略其余的键：

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

然后，简单地计算 b-key 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))))

然后遇到障碍：

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

我怎么能从这里继续？

4

3 回答 3

4

这是一种幼稚的方法，我相信有更好的方法，但这可能是您需要弄清楚的。

(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}

解释：

; 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
于 2016-03-21T19:53:41.243 回答
2 
(def data [{"a" "X", "b" "M", "c" 188}
       {"a" "Y", "b" "M", "c" 165}
       {"a" "Y", "b" "M", "c" 313}
       {"a" "Y", "b" "P", "c" 188}])
;; Borrowing data from @leetwinski

如果要定义数据，您可能需要考虑的一件事是使用关键字而不是字符串作为键。这带来了能够使用关键字作为函数来访问地图中的事物的好处，即(get my-map "a")变得(:a my-map)。

要获取按“a”键分组的数据：

(defn by-a-key [data] 
  (group-by #(get % "a") data))

我认为你实际上可以跳过你的第二步，如果它只是被用来让你进入你的第三步，因为它不需要这样做。在第二次阅读时，我不知道您是否只想为每个不同的“b”键保留一个元素。我假设不会，因为您没有指定如何选择要保留的内容，而且它们似乎有很大不同。

(reduce-kv 
  (fn [m k v] 
    (assoc m k 
      (count (filter #(contains? % "b") v)))) 
  {} 
  (by-a-key data))

你也可以像这样做整个事情：

(frequencies (map #(get % "a") (filter #(contains? % "b") data)))

由于您可以在分组之前按包含“b”键进行过滤，因此您可以依靠频率为您分组和计数。

于 2016-03-21T22:31:14.873 回答
1

你可以使用reduce：

(def data [{"a" "X", "b" "M", "c" 188}
           {"a" "Y", "b" "M", "c" 165}
           {"a" "Y", "b" "M", "c" 313}
           {"a" "Y", "b" "P", "c" 188}])

(def processed (reduce #(update % (%2 "a") (fnil conj #{}) (%2 "b")) 
                       {} data))

;; {"X" #{"M"}, "Y" #{"M" "P"}}
;; you create a map of "a" values to a sets of "b" values in one pass
;; and then you just create a new map with counts

(reduce-kv #(assoc %1 %2 (count %3)) {} processed)

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

因此它使用与@birdspider 的解决方案相同的逻辑，但使用较少的集合传递

在一个功能中：

(defn process [data]
  (->> data
       (reduce #(update % (%2 "a") (fnil conj #{}) (%2 "b")) {})
       (reduce-kv #(assoc %1 %2 (count %3)) {})))

 user> (process data)
 ;; {"X" 1, "Y" 2}
于 2016-03-21T20:20:07.450 回答