托卢库普的F#等价物

鉴于此：

[
    ("A","A122");
    ("A","A123");
    ("B","B122");
    ("B","B123");
    ("C","C122");
]

有没有一个标准的函数来获取这个

[
    ("A",["A122";"A123"]);
    ("B",["B122";"B123"]);
    ("C",["C122"])
]

我想到了Seq.distinctBy、List.partition、Set、Map，但它们似乎都不是我想要的

---

谢谢。。。在我等待的时候，我会试着自己滚：）

傻我，我没有注意到Seq.groupBy

[
    ("A","A122");
    ("A","A123");
    ("B","B122");
    ("B","B123");
    ("C","C122");
]
 |> Seq.groupBy (fun (a, b) -> a)
 |> Seq.map (fun (a, b) -> (a, Seq.map snd b))

输出：

seq
[("A", seq ["A122"; "A123"]); ("B", seq ["B122"; "B123"]);
 ("C", seq ["C122"])]

对于O（1）查找：

---

如果希望获得ILookup的高效O（logn）查找性能，您可能希望使用Map.ofSeq创建映射

[
    ("A","A122");
    ("A","A123");
    ("B","B122");
    ("B","B123");
    ("C","C122");
]
|> Seq.groupBy fst
|> dict