F＃

Question

我们假设我有一系列的序列，例如，

{1, 2, 3}, {1, 2, 3}, {1, 2, 3} 

什么是转动或压缩这个序列，所以我反而有最好的方式，

{1, 1, 1}, {2, 2, 2}, {3, 3, 3} 

有没有这样做，而不诉诸底层IEnumerator<_>类型的操纵的理解呢？

要澄清，这些是seq<seq<int>>对象。每个序列（内部和外部）可以有任意数量的项目。

Answer 1

如果您要寻找Seq语义解决方案，您将不得不一直保持懒惰。

let zip seq = seq 
      |> Seq.collect(fun s -> s |> Seq.mapi(fun i e -> (i, e))) //wrap with index 
      |> Seq.groupBy(fst) //group by index 
      |> Seq.map(fun (i, s) -> s |> Seq.map snd) //unwrap 

测试：

let seq = Enumerable.Repeat((seq [1; 2; 3]), 3) //don't want to while(true) yield. bleh. 
printfn "%A" (zip seq) 

输出：

seq [seq [1; 1; 1]; seq [2; 2; 2]; seq [3; 3; 3]] 

Answer 2

这似乎非常不雅，但它得到正确的答案：

(seq [(1, 2, 3); (1, 2, 3); (1, 2, 3);]) 
|> Seq.fold (fun (sa,sb,sc) (a,b,c) ->a::sa,b::sb,c::sc) ([],[],[]) 
|> fun (a,b,c) -> a::b::c::[] 

Answer 3

它看起来像矩阵转置。

let data = 
    seq [ 
     seq [1; 2; 3] 
     seq [1; 2; 3] 
     seq [1; 2; 3] 
    ] 

let rec transpose = function 
    | (_::_)::_ as M -> List.map List.head M :: transpose (List.map List.tail M) 
    | _ -> [] 

// I don't claim it is very elegant, but no doubt it is readable 
let result = 
    data 
    |> List.ofSeq 
    |> List.map List.ofSeq 
    |> transpose 
    |> Seq.ofList 
    |> Seq.map Seq.ofList 

或者，您可以采用相同的方法seq，感谢this answer一个优雅的活动模式：

let (|SeqEmpty|SeqCons|) (xs: 'a seq) = 
    if Seq.isEmpty xs then SeqEmpty 
    else SeqCons(Seq.head xs, Seq.skip 1 xs) 

let rec transposeSeq = function 
    | SeqCons(SeqCons(_,_),_) as M -> 
     Seq.append 
      (Seq.singleton (Seq.map Seq.head M)) 
      (transposeSeq (Seq.map (Seq.skip 1) M)) 
    | _ -> Seq.empty 

let resultSeq = data |> transposeSeq 

又见this answer的技术细节和两个引用：到的PowerPack的Microsoft.FSharp.Math.Matrix和涉及可变数据的另一种方法。

Answer 4

这是相同的答案@Asti，只是清理了一点：

[[1;2;3]; [1;2;3]; [1;2;3]] 
    |> Seq.collect Seq.indexed 
    |> Seq.groupBy fst 
    |> Seq.map (snd >> Seq.map snd);;