将列表转换为元组列表

Question

我有一个这样的列表：

[
    ["5b71d7e458c37fa04a7ce768", "5b3f77502dfe0deeb8912b42", "1538077790705827"],
     ["5b71d7e458c37fa04a7ce768","5b3f77502dfe0deeb8912b42","1538078530667847"],
     ["5b71d7e458c37fa04a7ce768","5b3f77502dfe0deeb8912b42","1538077778390908"],
     ["5b71d7e458c37fa04a7ce768","5bad45b1e990057961313822","1538082492283531"]
]

我想将其转换为如下列表：

[ 
  {"5b3f77502dfe0deeb8912b42",
   [{"5b71d7e458c37fa04a7ce768","5b3f77502dfe0deeb8912b42", "1538077790705827"},
    {"5b71d7e458c37fa04a7ce768","5b3f77502dfe0deeb8912b42", "1538078530667847"},
    {"5b71d7e458c37fa04a7ce768", "5b3f77502dfe0deeb8912b42" ,"1538077778390908"}
   ]},

{"5bad45b1e990057961313822",
   [{"5b71d7e458c37fa04a7ce768","5bad45b1e990057961313822","1538082492283531"}
   ]}
]

所以关键是从原始列表中的第二项创建元组

[“5b71d7e458c37fa04a7ce768”，“5b3f77502dfe0deeb8912b42”，“1538077790705827”]，

Answer 1

首先，Enum.group_by列表的第二个元素。然后，将生成的映射值的每个值映射到List.to_tuple。Enum.map与映射一起使用时，会自动将每个键/值对转换为元组。

list = [
  ["5b71d7e458c37fa04a7ce768", "5b3f77502dfe0deeb8912b42", "1538077790705827"],
  ["5b71d7e458c37fa04a7ce768", "5b3f77502dfe0deeb8912b42", "1538078530667847"],
  ["5b71d7e458c37fa04a7ce768", "5b3f77502dfe0deeb8912b42", "1538077778390908"],
  ["5b71d7e458c37fa04a7ce768", "5bad45b1e990057961313822", "1538082492283531"]
]

list
|> Enum.group_by(&Enum.at(&1, 1))
|> Enum.map(fn {k, v} -> {k, Enum.map(v, &List.to_tuple/1)} end)
|> IO.inspect

输出：

[
  {"5b3f77502dfe0deeb8912b42",
   [
     {"5b71d7e458c37fa04a7ce768", "5b3f77502dfe0deeb8912b42",
      "1538077790705827"},
     {"5b71d7e458c37fa04a7ce768", "5b3f77502dfe0deeb8912b42",
      "1538078530667847"},
     {"5b71d7e458c37fa04a7ce768", "5b3f77502dfe0deeb8912b42",
      "1538077778390908"}
   ]},
  {"5bad45b1e990057961313822",
   [
     {"5b71d7e458c37fa04a7ce768", "5bad45b1e990057961313822",
      "1538082492283531"}
   ]}
]