clojure - 将转换应用于数据映射的正确方法？

Question

我正在尝试想出一种将特定转换函数应用于数据 map 的好方法。

以 map 为例:

{:wrapper {:firstName "foo" 
           :lastName "bar"
           :addressLine1 "line1"
           :addressLine2 "line2"
           :birthDate {:iso "1930-03-12"}}}

并将其转换为:

{:name "foo bar"
 :address "line1 /n line2"
 :age 86}

我也希望转换以相反的方式工作，尽管我不介意编写单独的转换。

到目前为止，我已经尝试编写转换函数列表:(伪)

(-> start-map
    transform-name
    transform-address
    transform-age)

每个变换都采用[start-map {accumulator-map}]。我还尝试编写一个映射，其中包含转换后的映射的键以及转换函数(和参数)作为它们的值。我觉得我错过了一个技巧。

Answer 1

变换:

(require '[clj-time.core :as t])
(require '[clj-time.format :as f])

(def data {:wrapper {:firstName "foo" 
                     :lastName "bar"
                     :addressLine1 "line1"
                     :addressLine2 "line2"
                     :birthDate {:iso "1930-03-12"}}})

(def transformed
  {:name (str (get-in data [:wrapper :firstName])
              " "
              (get-in data [:wrapper :lastName]))
   :address (str (get-in data [:wrapper :addressLine1])
                 "\n"
                 (get-in data [:wrapper :addressLine2]))
   :age (t/in-years (t/interval (f/parse
                                 (get-in data [:wrapper :birthDate :iso] data))
                                (t/now)))})

逆变换。请注意，日期失去了精度。

(require '[clojure.string :as str])

(def untransformed
  (let [[firstname lastname] (str/split
                              (:name transformed)
                              #" ")
        [addressline1 addressline2] (str/split
                                     (:address transformed)
                                     #"\n")]
    {:wrapper
     {:firstName firstname
      :lastName lastname
      :addressLine1 addressline1
      :addressLine2 addressline2
      :birthDate
      {:iso (f/unparse
             (f/formatters :date)
             (t/minus (t/now)
                      (t/years (:age transformed))))}}}))