我在做F# Wiki Book on List的練習(滾動到底部)創建一個Pair
方法。如何創建一個「Pair」函數來匹配字符串列表?
我能夠毫無問題地配對整數列表,但是爲字符串列表拋出了F#異常。對於我來說,對於像我這樣的F#初學者來說,破解這個例外意味着太神祕了。
這是我初次嘗試實施Pair
上fsi.exe
> let pair l =
- let rec loop acc = function
- | [] -> acc
- | (hd1 :: hd2 :: tl) -> loop ((hd1, hd2) :: acc) tl
- List.rev(loop [] l)
-
- printfn "%A" ([1..10] |> pair)
- printfn "%A" ([ "one"; "two"; "three"; "four"; "five" ] |> pair);;
let rec loop acc = function
-----------------------^
stdin(2,24): warning FS0025: Incomplete pattern matches on this expression.
For example, the value '[_]' will not be matched
val pair : 'a list -> ('a * 'a) list
[(1, 2); (3, 4); (5, 6); (7, 8); (9, 10)]
Microsoft.FSharp.Core.MatchFailureException:
Exception of type 'Microsoft.FSharp.Core.MatchFailureException' was thrown.
at [email protected](List`1 acc, List`1 _arg1)
at FSI_0002.pair[T](List`1 l)
at <StartupCode$FSI_0002>.$FSI_0002._main()
stopped due to error
所以Pair
不整數版本
工作和函數簽名
val pair : 'a list -> ('a * 'a) list
表示Pair
泛型列表上運行。
問題:那麼爲什麼Pair
不能在字符串列表上工作?
【答案】(我的版本)
簡單地返回累積列表else
情況下(_)的伎倆。
並且警告也被照顧。
let pair l =
let rec loop acc = function
// | [] -> acc
| (hd1 :: hd2 :: tl) -> loop ((hd1, hd2) :: acc) tl
| _ -> acc
List.rev(loop [] l)
printfn "%A" ([1..10] |> pair)
printfn "%A" ([ "one"; "two"; "three"; "four"; "five" ] |> pair)
[EDIT2]嗯,我也將張貼我的版本的Unpair
的完整性。
let unpair l = [for (a,b) in l do yield! a :: b :: []]
下面是使用的解決方案版本,針對我的有點瑕疵基準100萬個項目列表
#light
open System;
let pn l = printfn "%A" l
let duration f =
let startTime = DateTime.Now;
let returnValue = f()
let endTime = DateTime.Now;
printfn "Duration (ms): %f" (endTime - startTime).TotalMilliseconds
returnValue
let ll = [for a in 1..1000000 do yield (a)]
let tl = [for a in 1..1000000 do yield (a,a)]
let pair1 l =
let rec loop acc = function
| [] | [_] -> List.rev acc
| h1 :: h2 :: tl -> loop ((h1, h2) :: acc) tl
loop [] l
let unpair1 l =
let rec loop acc = function
| [] -> List.rev acc
| (h1, h2) :: tl -> loop (h2 :: h1 :: acc) tl
loop [] l
let pair2 l =
let rec loop acc = function
| (hd1 :: hd2 :: tl) -> loop ((hd1, hd2) :: acc) tl
| _ | [_] -> acc
List.rev(loop [] l)
let unpair2 l = [for (a,b) in l do yield! a :: b :: []]
pn(duration (fun() -> ll |> pair1))
pn(duration (fun() -> tl |> unpair1))
pn(duration (fun() -> ll |> pair2))
pn(duration (fun() -> tl |> unpair2))
基準測試結果:
Solution version
PAIR -> Duration (ms): 255.000000
UNPAIR -> Duration (ms): 840.000000
My version
PAIR -> Duration (ms): 220.000000
UNPAIR -> Duration (ms): 1624.000000
順便說一句,維基書籍列出了實施「配對」的解決方案,但我決定不作弊。 – Sung 2009-04-26 23:45:09