Clojure函数定义：路径树

我想定义一个函数

f

，它返回一棵树（作为嵌套映射），给定通过该树的路径序列

e、 g

假设所有向量都表示源于同一根的路径

如何在一次过程中从功能上定义此函数

---

我对与之相反的行为感兴趣。

我会使用以下地图：

(def paths [[:a :b :c]
            [:a :b :d]
            [:b :d]
            [:b :e]
            [:b :e :a]])

(defn update-with-path
  [tree-map path-vec]
  (assoc-in tree-map path-vec nil) )

(defn build-tree [paths]
  (reduce
    update-with-path
    {} paths )
  )

(deftest t-global
  (is= {:a nil} (update-with-path {} [:a]))
  (is= {:a {:b nil}} (update-with-path {} [:a :b]))
  (is= {:a {:b {:c nil}}} (update-with-path {} [:a :b :c]))

  (is= (build-tree paths)
    { :a
      { :b
        { :c nil
          :d nil}}
      :b
      {
        :d nil
        :e
          { :a nil }}}
    )
  )

(ns xyz...
 (:require
    [tupelo.core :as t]  ))
(t/refer-tupelo)

(def t1 (build-tree paths))
(def cum (atom []))
(defn walker [path tree]
  (if (nil? tree)
    (swap! cum t/append path)
    (doseq [key (keys tree)]
      (walker (t/append path key) (t/grab key tree)))))
(walker [] t1)
(spyx @cum)

 => [[:a :b :c] [:a :b :d] [:b :d] [:b :e :a]]

在assoc中使用地图很容易。所有叶值的值均为

nil

，而非叶值的值均为map。此外，使用嵌套映射结构意味着我们不必担心重复

您可以找到所有叶节点，如下所示：

(def paths [[:a :b :c]
            [:a :b :d]
            [:b :d]
            [:b :e]
            [:b :e :a]])

(defn update-with-path
  [tree-map path-vec]
  (assoc-in tree-map path-vec nil) )

(defn build-tree [paths]
  (reduce
    update-with-path
    {} paths )
  )

(deftest t-global
  (is= {:a nil} (update-with-path {} [:a]))
  (is= {:a {:b nil}} (update-with-path {} [:a :b]))
  (is= {:a {:b {:c nil}}} (update-with-path {} [:a :b :c]))

  (is= (build-tree paths)
    { :a
      { :b
        { :c nil
          :d nil}}
      :b
      {
        :d nil
        :e
          { :a nil }}}
    )
  )

(ns xyz...
 (:require
    [tupelo.core :as t]  ))
(t/refer-tupelo)

(def t1 (build-tree paths))
(def cum (atom []))
(defn walker [path tree]
  (if (nil? tree)
    (swap! cum t/append path)
    (doseq [key (keys tree)]
      (walker (t/append path key) (t/grab key tree)))))
(walker [] t1)
(spyx @cum)

 => [[:a :b :c] [:a :b :d] [:b :d] [:b :e :a]]

然后修改树创建代码，创建一棵带有向量叶的新树：

(defn update-with-path-2
  [tree-map path-vec]
  (assoc-in tree-map path-vec path-vec))

(defn build-tree-2 [paths]
  (reduce
    update-with-path-2
    {} paths)
  )
(spyx (build-tree-2 @cum))

 => {:a {:b {:c [:a :b :c], 
             :d [:a :b :d]}}, 
     :b {:d [:b :d], 
         :e {:a [:b :e :a]}}}

---

此代码使用。

如何修改此定义以包含结果映射中每个路径叶的原始路径向量（例如，[:a:b:c]来代替

nil

？在

用路径更新
的定义中，将nil替换为对
路径向量的引用？对不起，@alan，我无意中点击了[enter]很早。我现在正在准备例子。