藥劑枚舉，過濾器和組

Question

什麼在藥劑中最優雅的方式來改造

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

到：

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

基本上我想遍歷輸入列表和組由前兩個元素。

Answer 1

我最好組由Enum.take(2)，然後flat_map每組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 

輸出：

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

注意，這也將工作，如果在輸入列表中的任何項目都有> 3要素;在這種情況下，它會只是Concat的他們：

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

將輸出：

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