从Clojure中的嵌套映射收集数据

Clojure中计算嵌套映射的某些属性的惯用方法是什么

给定以下数据结构：

(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"}}}}}})

如何生成所需的输出：

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

这是我迄今为止最大的努力：

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

这将产生以下结果：

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

---

我喜欢用线程构建这样的函数，这样步骤更容易阅读

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} 

（删除（comp-not:attrs）

非常类似于

选择键

---

对于[[[{{attributes:attributes}:attrs}]

让我想起了

进入
我发现
树序列对于这些情况非常有用：

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

我认为
tree-seq
的答案比Marz的“Specter”库要好得多，这可能对以下方面有用：