Question

Elixir中最优雅的转化方式

[["A","B","foo"],["A","B","bar"],["A","B","baz"],["C","D","foobar"],["C","D","bla"],["E","F","blabla"]]

成：

[["A","B","foo","bar","baz"],["C","D","foobar","bla"],["E","F","blabla"]]

基本上我想迭代输入列表并按前两个元素进行分组。

Answer 1

我按Enum.take(2)分组，然后flat_map分组Enum.drop(2)：

[["A","B","foo"],["A","B","bar"],["A","B","baz"],["C","D","foobar"],["C","D","bla"],["E","F","blabla"]]
|> Enum.group_by(&Enum.take(&1, 2))
|> Enum.map(fn {key, value} ->
  key ++ Enum.flat_map(value, &Enum.drop(&1, 2))
end)
|> IO.inspect

输出：

[["A", "B", "foo", "bar", "baz"], ["C", "D", "foobar", "bla"],
 ["E", "F", "blabla"]]

请注意，如果输入列表中的任何项目具有＆gt; 3个要素;在这种情况下，它只会连接它们：

[["A","B","foo","z","zz"],["A","B","bar"],["A","B","baz"],["C","D","foobar"],["C","D","bla"],["E","F","blabla"]]

将输出：

[["A", "B", "foo", "z", "zz", "bar", "baz"], ["C", "D", "foobar", "bla"],
 ["E", "F", "blabla"]]

Elixir Enum，过滤器和组

1 个答案: