Dictionary Clojure-与小径同行

Dictionary Clojure-与小径同行,dictionary,data-structures,clojure,clojurescript,Dictionary,Data Structures,Clojure,Clojurescript,我正在寻找一个与中类似的函数,该函数有一个内部函数作为参数: 不是键和值,就像clojure.walk/walk函数一样 但是从顶级数据结构访问值所需的键向量 递归遍历所有数据 例如: ;; not good since it takes `[k v]` as argument instead of `[path v]`, and is not recursive. user=> (clojure.walk/walk (fn [[k v]] [k (* 10 v)]) identity

我正在寻找一个与中类似的函数,该函数有一个
内部
函数作为参数:

  • 不是键和值,就像clojure.walk/walk函数一样
  • 但是从顶级数据结构访问值所需的键向量
  • 递归遍历所有数据
例如:

;; not good since it takes `[k v]` as argument instead of `[path v]`, and is not recursive.
user=> (clojure.walk/walk (fn [[k v]] [k (* 10 v)]) identity {:a 1 :b {:c 2}})
;; {:a 10, :c 30, :b 20}

;; it should receive as arguments instead :
[[:a] 1]
[[:b :c] 2]
注:

  • 它也应该使用数组,使用键0、1、2。。。(就像在
    进入
    中一样)
  • 如果允许简化代码,我并不真正关心
    outer
    参数

目前正在学习clojure,我尝试将此作为练习。 然而,我发现直接实现它是相当棘手的,因为它是沿着树向下走的,在运行时应用内部函数

为了达到您想要的结果,我将任务分为两部分:

  • 首先将嵌套结构转换为以路径为键的字典,并将值
  • 然后将内部函数映射到外部函数,或使用外部函数进行缩减
我的实施:

;; Helper function to have vector's indexes work like for get-in
(defn- to-indexed-seqs [coll]
  (if (map? coll)
    coll
    (map vector (range) coll)))

;; Flattening the tree to a dict of (path, value) pairs that I can map over
;; user>  (flatten-path [] {:a {:k1 1 :k2 2} :b [1 2 3]})
;; {[:a :k1] 1, [:a :k2] 2, [:b 0] 1, [:b 1] 2, [:b 2] 3}
(defn- flatten-path [path step]
  (if (coll? step)
    (->> step
         to-indexed-seqs
         (map (fn [[k v]] (flatten-path (conj path k) v)))
         (into {}))
    [path step]))

;; Some final glue
(defn path-walk [f coll]
  (->> coll
      (flatten-path [])
      (map #(apply f %))))

;; user> (println (clojure.string/join "\n" (path-walk #(str %1 " - " %2) {:a {:k1 1 :k2 2} :b [1 2 3]})))
;; [:a :k1] - 1
;; [:a :k2] - 2
;; [:b 0] - 1
;; [:b 1] - 2
;; [:b 2] - 3

结果是Stuart Halloway发表了一篇文章,这篇文章可能有一些用处(它使用了一个协议,这也使得它具有可扩展性):

注意:在我的情况下,我也可以使用,因此如果您正在阅读此文章,您可能还想查看它。

还有


在您的示例中,数据不应该是
{[:a:k1]1、[:a:k2]2、[:b0]1、[:b1]2、[:b2]3}
?哦,真的!to索引seqs中有一个错误,应该是:(映射向量(范围)coll)而不是:(映射向量coll(范围))我将更新我的答案:)谢谢!我接受了您的回答,但仍然希望有一个不需要枚举路径的解决方案(我的用例是需要一些转换的数据库迁移)。
(ns user)

(def app
  "Intenal Helper"
  (fnil conj []))

(defprotocol PathSeq
  (path-seq* [form path] "Helper for path-seq"))

(extend-protocol PathSeq
  java.util.List
  (path-seq*
   [form path]
   (->> (map-indexed
         (fn [idx item]
           (path-seq* item (app path idx)))
         form)
        (mapcat identity)))

  java.util.Map
  (path-seq*
   [form path]
   (->> (map
         (fn [[k v]]
           (path-seq* v (app path k)))
         form)
        (mapcat identity)))

  java.util.Set
  (path-seq*
   [form path]
   (->> (map
         (fn [v]
           (path-seq* v (app path v)))
         form)
        (mapcat identity)))


  java.lang.Object
  (path-seq* [form path] [[form path]])

  nil
  (path-seq* [_ path] [[nil path]]))

(defn path-seq
  "Returns a sequence of paths into a form, and the elements found at
   those paths.  Each item in the sequence is a map with :path
   and :form keys. Paths are built based on collection type: lists
   by position, maps by key, and sets by value, e.g.

   (path-seq [:a [:b :c] {:d :e} #{:f}])

   ({:path [0], :form :a}
    {:path [1 0], :form :b}
    {:path [1 1], :form :c}
    {:path [2 :d], :form :e}
    {:path [3 :f], :form :f})
   "
  [form]
  (map
   #(let [[form path] %]
      {:path path :form form})
   (path-seq* form nil)))

(comment
  (path-seq [:a [:b :c] {:d :e} #{:f}])

  ;; finding nils hiding in data structures:
  (->> (path-seq [:a [:b nil] {:d :e} #{:f}])
       (filter (comp nil? :form)))

  ;; finding a nil hiding in a Datomic transaction
  (->> (path-seq {:db/id 100
                  :friends [{:firstName "John"}
                            {:firstName nil}]})
       (filter (comp nil? :form)))


  )
(def node (:b (first (root :a))))

(= node {:c 1}) ;; => true

(c/context node) ;; => [:a 0 :b]