Recursion Elixir map_减少引用最后一个元素

在与列表中的最后一个元素进行比较时，如何循环（递归地或使用

reduce

）一个

映射的
列表

例如，假设我有一个这样的地图列表：

    datetime = Timex.beginning_of_day(Timex.now)
    data = [%{a: 0, cluster: 0, time: datetime},
            %{a: 1, cluster: 0, time: Timex.shift(datetime, minutes: 3)},
            %{a: 2, cluster: 0, time: Timex.shift(datetime, minutes: 6)},
            %{a: 3, cluster: 0, time: Timex.shift(datetime, minutes: 9)},
            %{a: 4, cluster: 1, time: Timex.shift(datetime, minutes: 12)},
            %{a: 5, cluster: 1, time: Timex.shift(datetime, minutes: 15)},
            %{a: 6, cluster: 0, time: Timex.shift(datetime, minutes: 18)},
            %{a: 7, cluster: 0, time: Timex.shift(datetime, minutes: 21)},
            %{a: 8, cluster: 0, time: Timex.shift(datetime, minutes: 23)},
            %{a: 9, cluster: 2, time: Timex.shift(datetime, minutes: 26)},
            %{a: 10, cluster: 2, time: Timex.shift(datetime, minutes: 29)},
            %{a: 11, cluster: 2, time: Timex.shift(datetime, minutes: 32)},
            %{a: 12, cluster: 1, time: Timex.shift(datetime, minutes: 35)},
            %{a: 13, cluster: 1, time: Timex.shift(datetime, minutes: 38)}]

            [%{a: 0, cluster: 0, time: datetime},
            %{a: 1, cluster: 0, time: Timex.shift(datetime, minutes: 3)},
            %{a: 2, cluster: 0, time: Timex.shift(datetime, minutes: 6)},
            %{a: 3, cluster: 0, time: Timex.shift(datetime, minutes: 9)},
            %{a: 4, cluster: 1, time: Timex.shift(datetime, minutes: 12)},
            %{a: 5, cluster: 1, time: Timex.shift(datetime, minutes: 15)},
            %{a: 6, cluster: 2, time: Timex.shift(datetime, minutes: 18)},
            %{a: 7, cluster: 2, time: Timex.shift(datetime, minutes: 21)},
            %{a: 8, cluster: 2, time: Timex.shift(datetime, minutes: 23)},
            %{a: 9, cluster: 3, time: Timex.shift(datetime, minutes: 26)},
            %{a: 10, cluster: 3, time: Timex.shift(datetime, minutes: 29)},
            %{a: 11, cluster: 3, time: Timex.shift(datetime, minutes: 32)},
            %{a: 12, cluster: 4, time: Timex.shift(datetime, minutes: 35)},
            %{a: 13, cluster: 4, time: Timex.shift(datetime, minutes: 38)}]

我想修改每个映射的
cluster
属性，以表示它所处的组（相对于它所处的顺序）

如果
集群
ID没有重复，那么使用
groupby
是很好的选择

我想在他们更改组时对他们进行分组，结果如下：

    datetime = Timex.beginning_of_day(Timex.now)
    data = [%{a: 0, cluster: 0, time: datetime},
            %{a: 1, cluster: 0, time: Timex.shift(datetime, minutes: 3)},
            %{a: 2, cluster: 0, time: Timex.shift(datetime, minutes: 6)},
            %{a: 3, cluster: 0, time: Timex.shift(datetime, minutes: 9)},
            %{a: 4, cluster: 1, time: Timex.shift(datetime, minutes: 12)},
            %{a: 5, cluster: 1, time: Timex.shift(datetime, minutes: 15)},
            %{a: 6, cluster: 0, time: Timex.shift(datetime, minutes: 18)},
            %{a: 7, cluster: 0, time: Timex.shift(datetime, minutes: 21)},
            %{a: 8, cluster: 0, time: Timex.shift(datetime, minutes: 23)},
            %{a: 9, cluster: 2, time: Timex.shift(datetime, minutes: 26)},
            %{a: 10, cluster: 2, time: Timex.shift(datetime, minutes: 29)},
            %{a: 11, cluster: 2, time: Timex.shift(datetime, minutes: 32)},
            %{a: 12, cluster: 1, time: Timex.shift(datetime, minutes: 35)},
            %{a: 13, cluster: 1, time: Timex.shift(datetime, minutes: 38)}]

            [%{a: 0, cluster: 0, time: datetime},
            %{a: 1, cluster: 0, time: Timex.shift(datetime, minutes: 3)},
            %{a: 2, cluster: 0, time: Timex.shift(datetime, minutes: 6)},
            %{a: 3, cluster: 0, time: Timex.shift(datetime, minutes: 9)},
            %{a: 4, cluster: 1, time: Timex.shift(datetime, minutes: 12)},
            %{a: 5, cluster: 1, time: Timex.shift(datetime, minutes: 15)},
            %{a: 6, cluster: 2, time: Timex.shift(datetime, minutes: 18)},
            %{a: 7, cluster: 2, time: Timex.shift(datetime, minutes: 21)},
            %{a: 8, cluster: 2, time: Timex.shift(datetime, minutes: 23)},
            %{a: 9, cluster: 3, time: Timex.shift(datetime, minutes: 26)},
            %{a: 10, cluster: 3, time: Timex.shift(datetime, minutes: 29)},
            %{a: 11, cluster: 3, time: Timex.shift(datetime, minutes: 32)},
            %{a: 12, cluster: 4, time: Timex.shift(datetime, minutes: 35)},
            %{a: 13, cluster: 4, time: Timex.shift(datetime, minutes: 38)}]

为此，我需要将列表中的当前项与前一项进行比较。我从下面这样的内容开始，然后停止，因为我知道它不会引用上一个项目来比较当前项目和上一个项目：

Enum.map_reduce(data, 0, fn(x, acc) -> cluster_grouping(x, acc) end)

  def cluster_grouping(x, acc) do
    cond do
      x.cluster == acc -> {Map.put(x, :cluster, acc), acc}
      x.cluster > acc -> {Map.put(x, :cluster, acc), acc + 1}
    end
  end

您需要在累加器中保留两个整数：当前生成的集群（每次更改都会增加1）和集群的最后一个原始值

datetime = Timex.beginning_of_day(Timex.now)
data = [%{a: 0, cluster: 0, time: datetime},
        %{a: 1, cluster: 0, time: Timex.shift(datetime, minutes: 3)},
        %{a: 2, cluster: 0, time: Timex.shift(datetime, minutes: 6)},
        %{a: 3, cluster: 0, time: Timex.shift(datetime, minutes: 9)},
        %{a: 4, cluster: 1, time: Timex.shift(datetime, minutes: 12)},
        %{a: 5, cluster: 1, time: Timex.shift(datetime, minutes: 15)},
        %{a: 6, cluster: 0, time: Timex.shift(datetime, minutes: 18)},
        %{a: 7, cluster: 0, time: Timex.shift(datetime, minutes: 21)},
        %{a: 8, cluster: 0, time: Timex.shift(datetime, minutes: 23)},
        %{a: 9, cluster: 2, time: Timex.shift(datetime, minutes: 26)},
        %{a: 10, cluster: 2, time: Timex.shift(datetime, minutes: 29)},
        %{a: 11, cluster: 2, time: Timex.shift(datetime, minutes: 32)},
        %{a: 12, cluster: 1, time: Timex.shift(datetime, minutes: 35)},
        %{a: 13, cluster: 1, time: Timex.shift(datetime, minutes: 38)}]

Enum.map_reduce(data, {0, 0}, fn x, {i, last} -> 
  i = if x.cluster == last, do: i, else: i + 1
  {Map.put(x, :cluster, i), {i, x.cluster}}
end)
|> elem(0)
|> IO.inspect

输出：

[
  %{a: 0, cluster: 0, time: #DateTime<2018-02-18 00:00:00Z>},
  %{a: 1, cluster: 0, time: #DateTime<2018-02-18 00:03:00Z>},
  %{a: 2, cluster: 0, time: #DateTime<2018-02-18 00:06:00Z>},
  %{a: 3, cluster: 0, time: #DateTime<2018-02-18 00:09:00Z>},
  %{a: 4, cluster: 1, time: #DateTime<2018-02-18 00:12:00Z>},
  %{a: 5, cluster: 1, time: #DateTime<2018-02-18 00:15:00Z>},
  %{a: 6, cluster: 2, time: #DateTime<2018-02-18 00:18:00Z>},
  %{a: 7, cluster: 2, time: #DateTime<2018-02-18 00:21:00Z>},
  %{a: 8, cluster: 2, time: #DateTime<2018-02-18 00:23:00Z>},
  %{a: 9, cluster: 3, time: #DateTime<2018-02-18 00:26:00Z>},
  %{a: 10, cluster: 3, time: #DateTime<2018-02-18 00:29:00Z>},
  %{a: 11, cluster: 3, time: #DateTime<2018-02-18 00:32:00Z>},
  %{a: 12, cluster: 4, time: #DateTime<2018-02-18 00:35:00Z>},
  %{a: 13, cluster: 4, time: #DateTime<2018-02-18 00:38:00Z>}
]

[
%{a:0，集群：0，时间：#DateTime}，
%{a:1，集群：0，时间：#DateTime}，
%{a:2，集群：0，时间：#DateTime}，
%{a:3，集群：0，时间：#DateTime}，
%{a:4，集群：1，时间：#DateTime}，
%{a:5，集群：1，时间：#DateTime}，
%{a:6，集群：2，时间：#DateTime}，
%{a:7，集群：2，时间：#DateTime}，
%{a:8，集群：2，时间：#DateTime}，
%{a:9，集群：3，时间：#DateTime}，
%{a:10，集群：3，时间：#DateTime}，
%{a:11，集群：3，时间：#DateTime}，
%{a:12，集群：4，时间：#DateTime}，
%{a:13，集群：4，时间：#DateTime}
]
更有效的方法是在map reducer中的函数子句上使用模式匹配：

Enum.map\u reduce（数据，{0,0}，fn
%{cluster:last}=x，{i，last}->
{%{x |簇：i}，{i，last}
%{cluster:last}=x，{i，{}->
{%{x |簇：i+1}，{i+1，last}
(完)
没门！！我很好奇。我尝试在函数的第一个参数中放入两个参数，但失败了。我不知道您可以通过
元组将多个项传递到累加器中。累加器也可以进行
映射吗？或者需要什么吗？从中可以看出，它需要
any（）
，这意味着如果需要，可以将最后一个结构完全传递给它。。。。六羟甲基三聚氰胺六甲醚。。。谢谢@Dogbert！！！是的，你可以传递任何术语。顺便说一句，我进一步简化了匿名函数中的代码。