[英]Removing from a list of a tuples that contains an empty element in the second projection F#
我需要學習對 Pair 類型進行模式匹配的正確方法:
let pairToBeFiltered = Ok ([(1,[]);(2,[3;4]);(5,[6;7;8]);(9,[]);(10,[])])
let filterEmpty (pair: int * int list) =
match pair with
| (x,y) when y <> [] -> (x,y) //This gives error because of incomplete pattern matching!
let filtering = List.map(filterEmpty) pairToBeFiltered
所需的 output:
Ok([(2,[3;4]);(5,[6;7;8])])
三個版本:
(* using match statement *)
module Version1 =
let pairsToBeFiltered : Result<_, string> =
Ok [(1,[]);(2,[3;4]);(5,[6;7;8]);(9,[]);(10,[])]
let myWhere (pair : int * List<int>) =
match pair with
| _, [] -> false
| _, _ -> true
let myFilter l0 = l0 |> Result.map (List.filter myWhere)
let result = pairsToBeFiltered |> myFilter
(* using lambda functions and List.isEmpty *)
module Version2 =
let pairsToBeFiltered : Result<_, string> =
Ok [(1,[]);(2,[3;4]);(5,[6;7;8]);(9,[]);(10,[])]
let myFilter l0 =
l0
|> Result.map (fun l1 ->
l1 |> List.filter (fun (_, l2) ->
l2 |> List.isEmpty |> not))
let result = pairsToBeFiltered |> myFilter
(* shortening Version2 (point free style - take care, can be confusing) *)
module Version3 =
let pairsToBeFiltered : Result<_, string> =
Ok [(1,[]);(2,[3;4]);(5,[6;7;8]);(9,[]);(10,[])]
let myFilter = Result.map (List.filter (snd >> List.isEmpty >> not))
let result = pairsToBeFiltered |> myFilter
這應該這樣做:
let pairsToBeFiltered = Ok ([(1,[]);(2,[3;4]);(5,[6;7;8]);(9,[]);(10,[])])
let filterEmpty pairs =
List.where (fun (_, y) -> y <> []) pairs // pattern match on pair occurs here
let filtering : Result<_, string> =
pairsToBeFiltered
|> Result.map filterEmpty
printfn "%A" filtering // Ok [(2, [3; 4]); (5, [6; 7; 8])]
這里有很多問題:
filterEmpty
以便它處理整個列表,而不是一對。 這是我們使用模式匹配應用過濾 function, List.where
的地方。 (在您的代碼中,請注意帶有match
表達式的List.map
不會過濾任何內容。)Result
中,因此您需要通過Result.map
將其解包以進行處理。 (由於您沒有指定'TError
類型,我假設使用string
來安撫編譯器。)
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