F#:如何正确枚举多个文件?

F#: How to enumerate through multiple files correctly?

我有一堆几个 MiB 大小的文件,非常简单:

按字典顺序排序时,它们包含按所需顺序排列的所有值。

我无法将所有内容作为 float listfloat array 保存在内存中,因此我需要一个 float seq 在实际访问时遍历必要的文件。通过序列的部分实际上使用 GetEnumerator() 以命令式方式执行,因为我不希望任何资源泄漏并希望正确关闭所有文件。

我的第一个函数式方法是:

let readFile file = 
    let rec readReader (maybeReader : BinaryReader option) = 
        match maybeReader with
        | None -> 
            let openFile() = 
                printfn "Opening the file"
                new BinaryReader(new FileStream(file, FileMode.Open, FileAccess.Read, FileShare.Read))
                |> Some
                |> readReader
            seq { yield! openFile() }
        | Some reader when reader.BaseStream.Position >= reader.BaseStream.Length -> 
            printfn "Closing the file"
            reader.Dispose()
            Seq.empty
        | Some reader -> 
            reader.BaseStream.Position |> printfn "Reading from position %d"
            let bytesToRead = Math.Min(1048576L, reader.BaseStream.Length - reader.BaseStream.Position) |> int
            let bytes = reader.ReadBytes bytesToRead
            let doubles = Array.zeroCreate<float> (bytesToRead / 8)
            Buffer.BlockCopy(bytes, 0, doubles, 0, bytesToRead)
            seq { 
                yield! doubles
                yield! readReader maybeReader
            }
    readReader None

然后,当我有一个包含所有文件的 string list 时,我可以这样说:

let values = files |> Seq.collect readFile
use ve = values.GetEnumerator()
// Do stuff that only gets partial data from one file

但是,这只会在 reader 结束时关闭文件(查看函数时很清楚)。因此,作为第二种方法,我强制执行了文件枚举:

type FileEnumerator(file : string) = 
    let reader = new BinaryReader(new FileStream(file, FileMode.Open, FileAccess.Read, FileShare.Read))
    let mutable _current : float = Double.NaN
    do file |> printfn "Enumerator active for %s"

    interface IDisposable with
        member this.Dispose() = 
            reader.Dispose()
            file |> printfn "Enumerator disposed for %s"

    interface IEnumerator with
        member this.Current = _current :> obj
        member this.Reset() = reader.BaseStream.Position <- 0L
        member this.MoveNext() = 
            let stream = reader.BaseStream
            if stream.Position >= stream.Length then false
            else 
                _current <- reader.ReadDouble()
                true

    interface IEnumerator<float> with
        member this.Current = _current

type FileEnumerable(file : string) = 

    interface IEnumerable with
        member this.GetEnumerator() = new FileEnumerator(file) :> IEnumerator

    interface IEnumerable<float> with
        member this.GetEnumerator() = new FileEnumerator(file) :> IEnumerator<float>

let readFile' file = new FileEnumerable(file) :> float seq

现在,当我说

let values = files |> Seq.collect readFile'
use ve = values.GetEnumerator()
// do stuff with the enumerator

正确处理枚举器会冒泡到我的命令式枚举器。

虽然这是我想要实现的一个可行的解决方案(我可以像第一个函数式方法一样通过分块阅读它来加快它的速度,但为了简洁我没有在这里这样做)我想知道是否有一个真正的避免枚举器中可变状态的功能方法。

当您说使用 GetEnumerator() 将防止资源泄漏并允许正确关闭所有文件时,我不太明白您的意思。以下是我对此的尝试(出于演示目的忽略块复制部分),我认为它会导致文件正确关闭。

let eof (br : BinaryReader) = 
  br.BaseStream.Position = br.BaseStream.Length  

let readFileAsFloats filePath = 
    seq{
        use file = new FileStream(filePath, FileMode.Open, FileAccess.Read, FileShare.Read)
        use reader = new BinaryReader(file)
        while (not (eof reader)) do
            yield reader.ReadDouble()
    }

let readFilesAsFloats filePaths = 
    filePaths |> Seq.collect readFileAsFloats

let floats = readFilesAsFloats ["D:\floatFile1.txt"; "D:\floatFile2.txt"]

这是你的想法吗?