Question

在Clojure中计算嵌套地图地图的某些属性的惯用方法是什么？

鉴于以下数据结构：

(def x {
    :0 {:attrs {:attributes {:dontcare "something"
                             :1 {:attrs {:abc "some value"}}}}}
    :1 {:attrs {:attributes {:dontcare "something"
                             :1 {:attrs {:abc "some value"}}}}}
    :9 {:attrs {:attributes {:dontcare "something"
                             :5 {:attrs {:xyz "some value"}}}}}})

我如何产生所需的输出：

(= (count-attributes x) {:abc 2, :xyz 1})

这是我迄今为止的最大努力：

(defn count-attributes 
 [input]
 (let [result (for [[_ {{attributes :attributes} :attrs}] x
                 :let [v (into {} (remove (comp not :attrs) (vals attributes)))]]
              (:attrs v))]
      (frequencies result)))

产生以下内容：

{{:abc "some value"} 2, {:xyz "some value"} 1}

Answer 1

我喜欢使用线程构建这样的函数，因此步骤更容易阅读

user> (->> x 
           vals                    ; first throw out the keys
           (map #(get-in % [:attrs :attributes]))  ; get the nested maps
           (map vals)              ; again throw out the keys
           (map #(filter map? %))  ; throw out the "something" ones. 
           flatten                 ; we no longer need the sequence sequences
           (map vals)              ; and again we don't care about the keys
           flatten                 ; the map put them back into a list of lists
           frequencies)            ; and then count them.
{{:abc "some value"} 2, {:xyz "some value"} 1}

(remove (comp not :attrs)与select-keys很相似 for [[_ {{attributes :attributes} :attrs}]提醒我get-in

Answer 2

我发现tree-seq对这些情况非常有用：

(frequencies (filter #(and (map? %) (not-any? map? (vals %))) (tree-seq map? vals x)))

从Clojure中的嵌套映射中收集数据

2 个答案: