f# - 需要返回 seq 而不是 seq> 吗？

Question

以下函数files返回seq<seq<R>> 。如何让它返回seq<R>相反？

type R = { .... }
let files = seqOfStrs |> Seq.choose(fun s ->
    match s with
    | Helper.ParseRegex "(\w+) xxxxx" month ->
        let currentMonth =  .....
        if currentMonth = month.[0] then
            doc.LoadHtml(s)
            Some (
                doc.DucumentNode.SelectNodes("....")
                |> Seq.map(fun tr ->
                    { ..... } ) //R. Some code return record type R. Omitted
            )
        else
            printfn "Expect %s found %s." currentMonth month.[0]
            None
    | _ ->
        printfn "No '(Month) Payment Data On Line' prompt."
        None

Answer 1

您的代码片段不完整，因此我们无法为您提供完整的答案。但是:

• 您的代码正在使用 Seq.choose，并且您将返回 None 或 Some 以及值集合。然后你会得到一系列序列...

• 您可以使用 Seq.collect 来展平序列，并将 None 替换为空序列，将 Some 替换为序列.

沿着这些思路(未经测试):

let files = seqOfStrs |> Seq.collect (fun s ->
    match s with
    | Helper.ParseRegex "(\w+) xxxxx" month ->
        let currentMonth =  .....
        if currentMonth = month.[0] then
            doc.LoadHtml(s)
            doc.DucumentNode.SelectNodes("....")
            |> Seq.map(fun tr ->
                { ..... } ) //R. Some code return record type R. Omitted
        else
            printfn "Expect %s found %s." currentMonth month.[0]
            Seq.empty
    | _ ->
        printfn "No '(Month) Payment Data On Line' prompt."
        Seq.empty )

其他选项，例如将 Seq.concat 或 Seq.collect id 添加到管道末尾显然也可以工作。