Loops Clojure嵌套doseq循环
我是Clojure的新手,我有一个关于嵌套doseq循环的问题 我想迭代一个序列,得到一个子序列,然后得到一些键,将函数应用于所有序列元素 给定序列的结构大致如下,但包含数百本书、书架和许多图书馆:Loops Clojure嵌套doseq循环,loops,clojure,nested,Loops,Clojure,Nested,我是Clojure的新手,我有一个关于嵌套doseq循环的问题 我想迭代一个序列,得到一个子序列,然后得到一些键,将函数应用于所有序列元素 给定序列的结构大致如下,但包含数百本书、书架和许多图书馆: ([:state/libraries {6 #:library {:name "MUNICIPAL LIBRARY OF X" :id 6 :shelves {3 #:shelf {:name "GREEN SHELF" :id 3 :books {45 #:book {:id 45 :nam
([:state/libraries {6 #:library {:name "MUNICIPAL LIBRARY OF X" :id 6
:shelves {3 #:shelf {:name "GREEN SHELF" :id 3 :books
{45 #:book {:id 45 :name "NECRONOMICON" :pages {...},
{89 #:book {:id 89 :name "HOLY BIBLE" :pages {...}}}}}}}}])
这是我的密码:
(defn my-function [] (let [conn (d/connect (-> my-system :config :datomic-uri))]
(doseq [library-seq (read-string (slurp "given-sequence.edn"))]
(doseq [shelves-seq (val library-seq)]
(library/create-shelf conn {:id (:shelf/id (val shelves-seq))
:name (:shelf/name (val shelves-seq))})
(doseq [books-seq (:shelf/books (val shelves-seq))]
(library/create-book conn (:shelf/id (val shelves-seq)) {:id (:book/id (val books-seq))
:name (:book/name (val books-seq))})
)))))
问题是我想摆脱嵌套的doseq混乱,但我不知道什么是最好的方法,因为在每次迭代中键都会改变。使用重现?减少也许我的想法是完全错误的?就像Carcigenicate在评论中说的那样,假设图书馆/。。。函数只是副作用,您只需将其写入一个
doseq
(defn my-function []
(let [conn (d/connect (-> my-system :config :datomic-uri))]
(doseq [library-seq (read-string (slurp "given-sequence.edn"))
shelves-seq (val library-seq)
:let [_ (library/create-shelf conn
{:id (:shelf/id (val shelves-seq))
:name (:shelf/name (val shelves-seq))})]
books-seq (:shelf/books (val shelves-seq))]
(library/create-book conn
(:shelf/id (val shelves-seq))
{:id (:book/id (val books-seq))
:name (:book/name (val books-seq))}))))
不过,我会将“连接到数据库”与“读取文件”与“写入数据库”分开。再加上一些解构,我最终会得到一些类似于:
(defn write-to-the-db [conn given-sequence]
(doseq [library-seq given-sequence
shelves-seq (val library-seq)
:let [{shelf-id :shelf/id,
shelf-name :shelf/name
books :shelf/books} (val shelves-seq)
_ (library/create-shelf conn {:id shelf-id, :name shelf-name})]
{book-id :book/id, book-name :book/name} books]
(library/create-book conn shelf-id {:id book-id, :name book-name})))
正如Carcigenicate在评论中所说,假设图书馆/。。。函数只是副作用,您只需将其写入一个
doseq
(defn my-function []
(let [conn (d/connect (-> my-system :config :datomic-uri))]
(doseq [library-seq (read-string (slurp "given-sequence.edn"))
shelves-seq (val library-seq)
:let [_ (library/create-shelf conn
{:id (:shelf/id (val shelves-seq))
:name (:shelf/name (val shelves-seq))})]
books-seq (:shelf/books (val shelves-seq))]
(library/create-book conn
(:shelf/id (val shelves-seq))
{:id (:book/id (val books-seq))
:name (:book/name (val books-seq))}))))
不过,我会将“连接到数据库”与“读取文件”与“写入数据库”分开。再加上一些解构,我最终会得到一些类似于:
(defn write-to-the-db [conn given-sequence]
(doseq [library-seq given-sequence
shelves-seq (val library-seq)
:let [{shelf-id :shelf/id,
shelf-name :shelf/name
books :shelf/books} (val shelves-seq)
_ (library/create-shelf conn {:id shelf-id, :name shelf-name})]
{book-id :book/id, book-name :book/name} books]
(library/create-book conn shelf-id {:id book-id, :name book-name})))
library/create-*
函数完全通过副作用发挥作用?他们没有返回任何东西?如果是这样的话,嵌套的剂量可能是这里最整洁的。请注意,至少对于前两个doseq,您可以将其压缩为(doseq[库顺序…,架顺序…])
。对于每个序列,您不需要新的doseq
。它们可以像for
s那样分组。library/create-*
函数完全通过副作用发挥作用?他们没有返回任何东西?如果是这样的话,嵌套的剂量可能是这里最整洁的。请注意,至少对于前两个doseq,您可以将其压缩为(doseq[库顺序…,架顺序…])
。对于每个序列,您不需要新的doseq
。它们可以像fors那样分组。基本上就是我在头脑中散列的东西。虽然(library/create shelf
部分很不幸。我几乎会给第三个seq自己的doseq
,这样let
就不会被用来产生副作用了。是的,我同意这有点恶心。a:do
如果它存在的话会更好。基本上就是我脑海中散列的内容(库/创建书架部分很不幸。我几乎会给第三个seq自己的doseq
,这样let
就不会被用来产生副作用。是的,我同意这有点恶心。如果它存在的话,:do
会更好。