按F中的元组分组#

一种解决方案:

let tuples = [("A",12); ("A",10); ("B",1);  ("C",2); ("C",1)]
tuples 
|> Seq.groupBy fst 
|> Seq.map (fun (key, values) -> (key, values |> Seq.sumBy snd))

编辑: …或不带管道:

let tuples = [("A",12); ("A",10); ("B",1);  ("C",2); ("C",1)]
Seq.map (fun (key, group) -> key, Seq.sumBy snd group)
        (Seq.groupBy fst tuples)

为了扩展Johan的答案,我倾向于做这类事情,所以做了下面的广义函数。

let group_fold key value fold acc seq =
    seq |> Seq.groupBy key 
        |> Seq.map (fun (key, seq) -> (key, seq |> Seq.map value |> Seq.fold fold acc))

它适用于您的tuple case,如下所示

let tuples = [("A",12); ("A",10); ("B",1);  ("C",2); ("C",1)]

let regular = group_fold fst snd (+) 0 tuples 
let piped = tuples  |> group_fold fst snd (+) 0

但也可以与其他seqence一起使用,比如字符串列表

let strings = ["A12"; "A10"; "B1";  "C2"; "C1"]

let regular = group_fold (fun (x : string) -> x.[0]) (fun (x : string) -> int x.[1..]) (+) 0 strings 
let piped = strings  |> group_fold (fun x -> x.[0]) (fun x -> int x.[1..]) (+) 0