在我的Clojure webapp中,我有各种模型命名空间,其中的函数将地图作为一个agrument,并以某种方式将该映射插入到数据库中。在插入之前,我希望能够从地图中取出所需的键。
这方面的一个基本例子是:
(let [msg-keys [:title :body]
msg {:title "Hello" :body "This is an example" :somekey "asdf" :someotherkey "asdf"}]
(select-keys msg msg-keys))
;; => {:title "Hello" :body "This is an example"}
当地图有点复杂并且我想选择一组特定的嵌套键时, select-keys
不是一个选项:
(let [person {:name {:first "john" :john "smith"} :age 40 :weight 155 :something {:a "a" :b "b" :c "c" :d "d"}}]
(some-select-key-fn person [:name [:first] :something [:a :b]]))
;; => {:name {:first "john"} :something {:a "a" :b "b"}}
有没有办法用核心功能做到这一点?有没有办法纯粹用解构来做这件事?
答案 0 :(得分:3)
此主题为discussed in the Clojure Google Group以及一些解决方案。
解构可能是最接近"核心"功能,如果您的问题是静态的并且地图具有所有预期的键(从而避免nil
),则可能是一个很好的解决方案。它可能看起来像:
(let [person {:name {:first "john" :john "smith"} :age 40 :weight 155 :something {:a "a" :b "b" :c "c" :d "d"}}
{{:keys [first]} :name {:keys [a b]} :something} person]
{:name {:first first} :something {:a a :b b}})
;; => {:name {:first "john"}, :something {:a "a", :b "b"}}
以下是适用于您的示例地图的Clojure Google Group主题中的解决方案调查。它们各自对如何指定要选择的嵌套键有不同的看法。
以下是Christophe Grand' s solution:
(defprotocol Selector
(-select [s m]))
(defn select [m selectors-coll]
(reduce conj {} (map #(-select % m) selectors-coll)))
(extend-protocol Selector
clojure.lang.Keyword
(-select [k m]
(find m k))
clojure.lang.APersistentMap
(-select [sm m]
(into {}
(for [[k s] sm]
[k (select (get m k) s)]))))
使用它需要稍微修改一下语法:
(let [person {:name {:first "john" :john "smith"} :age 40 :weight 155 :something {:a "a" :b "b" :c "c" :d "d"}}]
(select person [{:name [:first] :something [:a :b]}]))
;; => {:something {:b "b", :a "a"}, :name {:first "john"}}
这是Moritz Ulrich的解决方案(他警告说,它不能在seqs作为键的地图上工作):
(defn select-in [m keyseq]
(loop [acc {} [k & ks] (seq keyseq)]
(if k
(recur
(if (sequential? k)
(let [[k ks] k]
(assoc acc k
(select-in (get m k) ks)))
(assoc acc k (get m k)))
ks)
acc)))
使用它需要另一种略微修改的语法:
(let [person {:name {:first "john" :john "smith"} :age 40 :weight 155 :something {:a "a" :b "b" :c "c" :d "d"}}]
(select-in person [[:name [:first]] [:something [:a :b]]]))
;; => {:something {:b "b", :a "a"}, :name {:first "john"}}
这是Jay Fields的解决方案:
(defn select-nested-keys [m top-level-keys & {:as pairs}]
(reduce #(update-in %1 (first %2) select-keys (last %2)) (select-keys m top-level-keys) pairs))
它使用不同的语法:
(let [person {:name {:first "john" :john "smith"} :age 40 :weight 155 :something {:a "a" :b "b" :c "c" :d "d"}}]
(select-nested-keys person [:name :something] [:name] [:first] [:something] [:a :b]))
;; => {:something {:b "b", :a "a"}, :name {:first "john"}}
这是Baishampayan Ghose的solution:
(defprotocol ^:private IExpandable
(^:private expand [this]))
(extend-protocol IExpandable
clojure.lang.Keyword
(expand [k] {k ::all})
clojure.lang.IPersistentVector
(expand [v] (if (empty? v)
{}
(apply merge (map expand v))))
clojure.lang.IPersistentMap
(expand [m]
(assert (= (count (keys m)) 1) "Number of keys in a selector map can't be more than 1.")
(let [[k v] (-> m first ((juxt key val)))]
{k (expand v)}))
nil
(expand [_] {}))
(defn ^:private extract* [m selectors expand?]
(let [sels (if expand? (expand selectors) selectors)]
(reduce-kv (fn [res k v]
(if (= v ::all)
(assoc res k (m k))
(assoc res k (extract* (m k) v false))))
{} sels)))
(defn extract
"Like select-keys, but can select nested keys.
Examples -
(extract [{:b {:c [:d]}} :g] {:a 1 :b {:c {:d 1 :e 2}} :g 42 :xxx 11})
;=> {:g 42, :b {:c {:d 1}}}
(extract [:g] {:a 1 :b {:c {:d 1 :e 2}} :g 42 :xxx 11})
;=> {:g 42}
(extract [{:b [:c]} :xxx] {:a 1 :b {:c {:d 1 :e 2}} :g 42 :xxx 11})
;=> {:b {:c {:d 1, :e 2}}, :xxx 11}
Also see - exclude"
[selectors m]
(extract* m selectors true))
它使用另一种语法(参数相反):
(let [person {:name {:first "john" :john "smith"} :age 40 :weight 155 :something {:a "a" :b "b" :c "c" :d "d"}}]
(extract [{:name [:first]} {:something [:a :b]}] person))
;; => {:name {:first "john"}, :something {:a "a", :b "b"}}
答案 1 :(得分:0)
您最好的选择可能是在结构的每个嵌套部分使用选择键。
(-> person
(select-keys [:name :something])
(update-in [:name] select-keys [:first])
(update-in [:something] select-keys [:a :b]))
您当然可以使用上面的通用版本来实现您在函数中建议的语法(使用reduce
而不是->
形式,最有可能的是,每个嵌套的递归调用选择键)。解构无济于事,它使绑定嵌套数据变得方便,但对于构造值并不是那么有用。
以下是我如何使用reduce
和递归:
(defn simplify
[m skel]
(if-let [kvs (not-empty (partition 2 skel))]
(reduce (fn [m [k nested]]
(if nested
(update-in m [k] simplify nested)
m))
(select-keys m (map first kvs))
kvs)
m))
请注意,您提出的参数格式不方便,所以我稍微改了一下
user=> (simplify {:name {:first "john" :john "smith"}
:age 40
:weight 155
:something {:a "a" :b "b" :c "c" :d "d"}}
[:name [:first nil] :something [:a nil :b nil]])
{:something {:b "b", :a "a"}, :name {:first "john"}}
您提出的语法需要更复杂的实现