從數組的seq中創建元組。

Question

基本上我希望利用Seq.Windowed的輸出，返回數組的序列，並把它變成元組

的順序，所以我想利用這個

[[|1;2;3|];[|4;5;6|]] 

，並把它變成

[(1,2,3);(4,5,6)] 

在此先感謝。

Answer 1

> let x = [[|1;2;3|];[|4;5;6|]];; 

val x : int [] list = [[|1; 2; 3|]; [|4; 5; 6|]] 

> let y = [for [|a; b; c|] in x do yield (a, b, c)];; 

    let y = [for [|a; b; c|] in x do yield (a, b, c)];; 
    ----------------------------^ 

stdin(6,29): warning FS0025: Incomplete pattern matches on this expression. 
For example, the value '[|_; _; _; _|]' may indicate a case not covered by 
the pattern(s). 

val y : (int * int * int) list = [(1, 2, 3); (4, 5, 6)] 

如果您可以保證所有陣列具有相同的形狀，則可以忽略上述警告。如果警告真的困擾你，你可以寫：

> x |> List.map (function [|a;b;c|] -> a, b, c | _ -> failwith "Invalid array length");; 
val it : (int * int * int) list = [(1, 2, 3); (4, 5, 6)] 

Answer 2

我不確定它是否是一個typeo或不是你的數據與窗口不匹配。

let firstThreeToTuple (a : _[]) = (a.[0], a.[1], a.[2]) 

seq {1 .. 6} 
|> Seq.windowed 3 
|> Seq.map firstThreeToTuple 
|> Seq.iter (printfn "%A") 

(1, 2, 3) 
(2, 3, 4) 
(3, 4, 5) 
(4, 5, 6) 

如果你想有一個函數，它接受一個序列，並扒其成陣列序列您可以使用此代碼從另一個question。

let chunks n (sequence: seq<_>) = 
    let fold_fce (i, s) value = 
     if i < n then (i+1, Seq.append s (Seq.singleton value)) 
       else ( 1, Seq.singleton value) 
    in sequence 
    |> Seq.scan (fold_fce) (0, Seq.empty) 
    |> Seq.filter (fun (i,_) -> i = n) 
    |> Seq.map (Seq.to_array << snd) 

然後你可以通過firstThreeToTuple運行結果。

seq {1 .. 6} 
|> chunks 3 
|> Seq.map firstThreeToTuple 
|> Seq.iter (printfn "%A") 

(1, 2, 3) 
(4, 5, 6)