[英]Pivot or zip a seq<seq<'a>> in F#
假設我有一個序列序列,例如
{1, 2, 3}, {1, 2, 3}, {1, 2, 3}
樞轉或壓縮此序列的最佳方法是什么,所以我有了
{1, 1, 1}, {2, 2, 2}, {3, 3, 3}
是否有一種可理解的方式來實現,而無需借助對基礎IEnumerator<_>
類型的操縱?
為了澄清起見,這些是seq<seq<int>>
對象。 每個序列(內部和外部)都可以具有任意數量的項目。
如果您要使用語義上為Seq的解決方案,則必須一直保持懶惰。
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
測試:
let seq = Enumerable.Repeat((seq [1; 2; 3]), 3) //don't want to while(true) yield. bleh.
printfn "%A" (zip seq)
輸出:
seq [seq [1; 1; 1]; seq [2; 2; 2]; seq [3; 3; 3]]
這似乎很不雅致,但它得到了正確的答案:
(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::[]
它看起來像矩陣轉置。
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
另外,您可以對seq
采用相同的方法,這要歸功於優雅的Active模式的答案 :
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
另請參閱此答案以獲取技術細節和兩個參考: PowerPack的Microsoft.FSharp.Math.Matrix
和涉及可變數據的另一種方法。
這和@Asti的答案是一樣的,只是清理了一下:
[[1;2;3]; [1;2;3]; [1;2;3]]
|> Seq.collect Seq.indexed
|> Seq.groupBy fst
|> Seq.map (snd >> Seq.map snd);;
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