Question

鉴于此：

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

是否有标准功能来实现这一目标？

[
    ("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，我会尝试自己动手：）

Answer 1

傻了我，我没注意到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"])]

Answer 2

对于O（1）查找：

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

F＃相当于toLookup

2 个答案: