Clojure将多个地图合并为一个地图

Question

我有以下地图列表

({"child.search" {:roles #{"ROLE_ADM_UNSUBSCRIBE_SUBSCRIPTION" "ROLE_ADM_SEARCH_SUBSCRIPTION" "ROLE_ADM_VIEW_SUBSCRIPTION"}},
  "child.cc.search" {:roles #{"ROLE_ADM_CC_SEARCH_SUBSCRIPTION" "ROLE_ADM_CC_VIEW_SUBSCRIPTION"}}}
  {"child.abusereport" {:roles #{"ROLE_ADM_ABUSE_RPT"}},
   "child.manualfiltering" {:roles #{"ROLE_ADM_MANUAL_FILTERING_RPT"}}}
  {"child.assigned.advertisement" {:roles #{"ROLE_ADM_CREATE_ADVERTISING"}},
   "child.manage.advertisement" {:roles #{"ROLE_ADM_LIST_ADVERTISING"}}})

我需要有一个像下面这样的地图。

{"child.search" {:roles #{"ROLE_ADM_UNSUBSCRIBE_SUBSCRIPTION" "ROLE_ADM_SEARCH_SUBSCRIPTION" "ROLE_ADM_VIEW_SUBSCRIPTION"}}
"child.cc.search" {:roles #{"ROLE_ADM_CC_SEARCH_SUBSCRIPTION" "ROLE_ADM_CC_VIEW_SUBSCRIPTION"}}
"child.abusereport" {:roles #{"ROLE_ADM_ABUSE_RPT"}}
"child.manualfiltering" {:roles #{"ROLE_ADM_MANUAL_FILTERING_RPT"}}
"child.assigned.advertisement" {:roles #{"ROLE_ADM_CREATE_ADVERTISING"}}
"child.manage.advertisement" {:roles #{"ROLE_ADM_LIST_ADVERTISING"}}}

我该怎么做？

Answer 1

您可以使用“ into ”函数，并提供空映射{}作为第一个参数：

(into {} map-list)

这是我的REPL会话的输出; 我将你的代码复制到两个vars'map-list'和'single-map'（Clojure 1.3.0）：

(def map-list '({"child.search" {:roles #{"ROLE_ADM_UNSUBSCRIBE_SUBSCRIPTION" "ROLE_ADM_SEARCH_SUBSCRIPTION" "ROLE_ADM_VIEW_SUBSCRIPTION"}},
                 "child.cc.search" {:roles #{"ROLE_ADM_CC_SEARCH_SUBSCRIPTION" "ROLE_ADM_CC_VIEW_SUBSCRIPTION"}}}
                {"child.abusereport" {:roles #{"ROLE_ADM_ABUSE_RPT"}},
                 "child.manualfiltering" {:roles #{"ROLE_ADM_MANUAL_FILTERING_RPT"}}}
                {"child.assigned.advertisement" {:roles #{"ROLE_ADM_CREATE_ADVERTISING"}},
                 "child.manage.advertisement" {:roles #{"ROLE_ADM_LIST_ADVERTISING"}}}))

#'user/map-list
user=>
(def single-map {"child.search" {:roles #{"ROLE_ADM_UNSUBSCRIBE_SUBSCRIPTION" "ROLE_ADM_SEARCH_SUBSCRIPTION" "ROLE_ADM_VIEW_SUBSCRIPTION"}}
                 "child.cc.search" {:roles #{"ROLE_ADM_CC_SEARCH_SUBSCRIPTION" "ROLE_ADM_CC_VIEW_SUBSCRIPTION"}}
                 "child.abusereport" {:roles #{"ROLE_ADM_ABUSE_RPT"}}
                 "child.manualfiltering" {:roles #{"ROLE_ADM_MANUAL_FILTERING_RPT"}}
                 "child.assigned.advertisement" {:roles #{"ROLE_ADM_CREATE_ADVERTISING"}}
                 "child.manage.advertisement" {:roles #{"ROLE_ADM_LIST_ADVERTISING"}}})
#'user/single-map

user=>
;; Check to see if we have the desired result
(= (into {} map-list)
   single-map)
true

Answer 2

我可能会使用merge和apply

(def map-list (list {:a 1, :b 2} {:c 3, :d 4}))
(apply merge map-list) ;; returns {:a 1, :b 2, :c 3, :d 4}

Answer 3

试试这个

(def your-list '({"child.search" {:roles #{"ROLE_ADM_UNSUBSCRIBE_SUBSCRIPTION" "ROLE_ADM_SEARCH_SUBSCRIPTION" "ROLE_ADM_VIEW_SUBSCRIPTION"}},
  "child.cc.search" {:roles #{"ROLE_ADM_CC_SEARCH_SUBSCRIPTION" "ROLE_ADM_CC_VIEW_SUBSCRIPTION"}}}
  {"child.abusereport" {:roles #{"ROLE_ADM_ABUSE_RPT"}},
   "child.manualfiltering" {:roles #{"ROLE_ADM_MANUAL_FILTERING_RPT"}}}
  {"child.assigned.advertisement" {:roles #{"ROLE_ADM_CREATE_ADVERTISING"}},
   "child.manage.advertisement" {:roles #{"ROLE_ADM_LIST_ADVERTISING"}}}))

(reduce conj your-list)

Answer 4

这个答案澄清了“进入”和“结合”在这种情况下相互关联的方式，以便为您的问题提供可接受的解决方案。 也就是说，它站在前面答案的肩膀上..

您有一个格式列表：

( {key1 value1} 
  {key2 value2} ) 

并且您想要一个已提取键/值对的地图，如下所示：

{key1 value1, key2 value2} 

那么问题在于，您基本上希望在单个地图中将每个连续的k / v对“联合”到最后一个。

上面的第一个解决方案使用了。 如果我们看看clojuredoc进入，我们会看到：

“返回一个新的coll，它包含to-coll和所有项目
来自柯尔联合。“

另一个类似的答案是使用

(reduce conj '({key1 value1} {key2 value2}))

显然，这个解决方案是针对这个问题拼写出上面的“into”定义：reduce函数累积了第n +（第n + 1）个键值对的连接的每个应用程序，这样它实现了into的定义（至少，为了这个问题的目的）。