[英]F# match question
這是我到目前為止:
type u = {str : string} //some type that has some property str (for simplicity, only one)
type du=
| A of u
| B of u // some discriminated union that carries u with it
然后,在某個地方,我有一個du序列,我正在做distinctBy和屬性做的不同是str。 我能想出的最好的是:
Seq.distinctBy (fun d -> match d with (A u|B u) -> u.str)
代碼有效,但我不喜歡在受歧視的聯盟的a和b上匹配,並希望用某些東西替換匹配。
問題是什么? :)
編輯:
在我的情況下,被區分的聯合的a和b將總是帶有相同的類型u,一個解決方案是擺脫du並添加它的字符串形式來輸入u並簡化整個混亂,但我想保留它現在的方式,因為我打算在a和b上做匹配...
作為du的財產,這次比賽怎么樣?
type u = {str : string}
type du =
| A of u
| B of u with
member this.U =
match this with
(A u | B u) -> u
[A {str="hello"}; B {str="world"}; A {str="world"}]
|> Seq.distinctBy (fun x -> x.U.str)
//val it : seq<du> = seq [A {str = "hello";}; B {str = "world";}]
但是,我有一些想法可以更好地模擬你和你之間的關系,同時滿足你的“編輯”問題。 一種方法是簡單地使用元組:
type u = {str : string}
type du =
| A
| B
//focus on type u
[A, {str="hello"}; B, {str="world"}; A, {str="world"}]
|> Seq.distinctBy (fun (_,x) -> x.str)
//val it : seq<du * u> = seq [(A, {str = "hello";}); (B, {str = "world";})]
//focus on type du
let x = A, {str="hello"}
match x with
| A,_ -> "A"
| B,_ -> "B"
//val it : string = "A"
另一種方法是切換它並將du添加到u:
type du =
| A
| B
type u = { case : du; str : string}
//focus on type u
[{case=A; str="hello"}; {case=B; str="world"}; {case=A; str="world"}]
|> Seq.distinctBy (fun x -> x.str)
//val it : seq<u> = seq [{case = A;
// str = "hello";}; {case = B;
// str = "world";}]
//focus on type du
let x = {case=A; str="hello"}
match x with
| {case=A} -> "A"
| {case=B} -> "B"
//val it : string = "A"
你真的不能像你描述的那樣簡化它,但你可以簡化它。 謝謝@Tomas Petricek
[ A { str = "A" }; B { str = "B" }; B { str = "B" } ]
|> Seq.distinctBy (fun (A u | B u) -> u.str)
|> Seq.toArray;;
產量
[| A {str = "A";}; B {str = "B";} |]
我對F#有點新鮮,但希望這會有所幫助。 在我看來,Active Patterns可能會讓您的生活更輕松,以減少您在模式匹配中輸入的內容。 而不是使用A a | B b你可以在它的位置使用活動模式AorB。
type u = { str : string }
type du =
| A of u
| B of u
let (|AorB|) (v:du) =
match v with
| A a -> a
| B b -> b
[A { str = "A" }; B { str = "B"}; A { str = "A"}]
|> Seq.distinctBy (fun d ->
match d with
| AorB s -> s)
|> Seq.iter (fun i -> match i with AorB c -> printfn "%s" c.str)
隨着斯蒂芬的加法,最后的表達可以這樣寫出來。
[A { str = "A" }; B { str = "B"}; A { str = "A"}]
|> Seq.distinctBy (|AorB|)
|> Seq.iter (fun i -> match i with AorB c -> printfn "%s" c.str)
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