Question

我有两个这样的地图矢量：

1：

[{:id 54785214 :field "fff" :attributes [{:id 1 :field "ddd"}]}
{:id 1254561561 :field "fff" :attributes [{:id 1 :field "ddd"}]}
{:id 156156456 :field "fff" :attributes [{:id 1 :field "ddd"}]}
{:id 77897445 :field "fff" :attributes [{:id 1 :field "ddd"}]}
{:id 4848121 :field "fff" :attributes [{:id 1 :field "ddd"}]}
{:id 787897212 :field "fff" :attributes [{:id 1 :field "ddd"}]}
{:id 45478789 :field "fff" :attributes [{:id 1 :field "ddd"}]}
{:id 112315478 :field "fff" :attributes [{:id 1 :field "ddd"}]}
{:id 154847841 :field "fff" :attributes [{:id 1 :field "ddd"}]}
{:id 65698898 :field "fff" :attributes [{:id 1 :field "ddd"}]}]

2：

[{:id 547485214 :field "fff" :attributes [{:id 1 :field "ddd"}]}
{:id 12545561561 :field "fff" :attributes [{:id 1 :field "ddd"}]}
{:id 1561456456 :field "fff" :attributes [{:id 1 :field "ddd"}]}
{:id 778927445 :field "fff" :attributes [{:id 1 :field "ddd"}]}
{:id 4848121 :field "hhh" :attributes [{:id 1 :field "ddd"}]}
{:id 7817897212 :field "fff" :attributes [{:id 1 :field "ddd"}]}
{:id 455478789 :field "fff" :attributes [{:id 1 :field "ddd"}]}
{:id 1123615478 :field "fff" :attributes [{:id 1 :field "ddd"}]}
{:id 1541847841 :field "fff" :attributes [{:id 1 :field "ddd"}]}
{:id 656988898 :field "fff" :attributes [{:id 1 :field "ddd"}]}]

和一个函数来连接这两个向量与合并地图与相同的键（我选择：id）：

(defn join-map-sets [set-a set-b by-key]
  (map #(apply merge %) (vals (group-by by-key (concat set-a set-b)))))

将此函数应用于地图1和2后，结果不像源地图那样排序。预览结果的前3个地图：

({:id 154847841, :field "fff", :attributes [{:id 1, :field "ddd"}]}
 {:id 1541847841, :field "fff", :attributes [{:id 1, :field "ddd"}]}
 {:id 656988898, :field "fff", :attributes [{:id 1, :field "ddd"}]})

对于这个问题：如何修改我的函数以始终保留地图的顺序？就像concat一样，但是替换/合并重复项。

PS。抱歉丑陋的例子，但对于简单的例子，它没有显示问题。

Answer 1

最简单的方法是使用distinct：

(distinct (concat v1 v2))

将为您提供删除重复项的延迟序列。（v1和v2是您的向量..）

count将证明重复项被删除：

(count (distinct (concat v1 v2)))

顺便说一句，您的代码示例中没有重复的:id，请仔细查看ID编号是否完全匹配。

按选定的键加入两个地图矢量，并保留元素的顺序

1 个答案: