static void MultipleFilesToSingleFile(string dirPath, string filePattern, string destFile)
{
    string[] fileAry = Directory.GetFiles(dirPath, filePattern);
    Console.WriteLine("Total File Count : " + fileAry.Length);
    using (TextWriter tw = new StreamWriter(destFile, true))
    {
        foreach (string filePath in fileAry)
        {
            using (TextReader tr = new StreamReader(filePath))
            {
                tw.WriteLine(tr.ReadToEnd());
                tr.Close();
                tr.Dispose();
            }
            Console.WriteLine("File Processed : " + filePath);
        }
        tw.Close();
        tw.Dispose();
    }
}

一般答案

为什么不直接使用Stream.CopyTo(Stream destination)方法？

private static void CombineMultipleFilesIntoSingleFile(string inputDirectoryPath, string inputFileNamePattern, string outputFilePath)
{
    string[] inputFilePaths = Directory.GetFiles(inputDirectoryPath, inputFileNamePattern);
    Console.WriteLine("Number of files: {0}.", inputFilePaths.Length);
    using (var outputStream = File.Create(outputFilePath))
    {
        foreach (var inputFilePath in inputFilePaths)
        {
            using (var inputStream = File.OpenRead(inputFilePath))
            {
                // Buffer size can be passed as the second argument.
                inputStream.CopyTo(outputStream);
            }
            Console.WriteLine("The file {0} has been processed.", inputFilePath);
        }
    }
}

缓冲区大小调整

请注意，上述方法已重载。

有两种方法重载：

1. CopyTo(Stream destination) .
2. CopyTo(Stream destination, int bufferSize) .

第二种方法重载通过 bufferSize 参数提供缓冲区大小调整。

一种选择是利用 copy 命令，让它做擅长的事情。

像这样：

static void MultipleFilesToSingleFile(string dirPath, string filePattern, string destFile)
{
    var cmd = new ProcessStartInfo("cmd.exe", 
        String.Format("/c copy {0} {1}", filePattern, destFile));
    cmd.WorkingDirectory = dirPath;
    cmd.UseShellExecute = false;
    Process.Start(cmd);
}

我会使用BlockingCollection来读取，以便您可以同时读取和写入。
显然应该写入单独的物理磁盘以避免硬件争用。此代码将保持顺序。
读取将比写入更快，因此不需要并行读取。
同样，由于读取速度会更快，因此限制集合的大小，因此读取不会比写入更远。
在写入电流的同时并行读取单次的简单任务存在文件大小不同的问题 - 写入小文件比读取大文件更快。

我使用此模式在 T1 上读取和解析文本，然后在 T2 上插入到 SQL。

public void WriteFiles()
{
    using (BlockingCollection<string> bc = new BlockingCollection<string>(10))
    {
        // play with 10 if you have several small files then a big file
        // write can get ahead of read if not enough are queued
        TextWriter tw = new StreamWriter(@"c:'temp'alltext.text", true);
        // clearly you want to write to a different phyical disk 
        // ideally write to solid state even if you move the files to regular disk when done
        // Spin up a Task to populate the BlockingCollection
        using (Task t1 = Task.Factory.StartNew(() =>
        {
            string dir = @"c:'temp'";
            string fileText;      
            int minSize = 100000; // play with this
            StringBuilder sb = new StringBuilder(minSize);
            string[] fileAry = Directory.GetFiles(dir, @"*.txt");
            foreach (string fi in fileAry)
            {
                Debug.WriteLine("Add " + fi);
                fileText = File.ReadAllText(fi);
                //bc.Add(fi);  for testing just add filepath
                if (fileText.Length > minSize)
                {
                    if (sb.Length > 0)
                    { 
                       bc.Add(sb.ToString());
                       sb.Clear();
                    }
                    bc.Add(fileText);  // could be really big so don't hit sb
                }
                else
                {
                    sb.Append(fileText);
                    if (sb.Length > minSize)
                    {
                        bc.Add(sb.ToString());
                        sb.Clear();
                    }
                }
            }
            if (sb.Length > 0)
            {
                bc.Add(sb.ToString());
                sb.Clear();
            }
            bc.CompleteAdding();
        }))
        {
            // Spin up a Task to consume the BlockingCollection
            using (Task t2 = Task.Factory.StartNew(() =>
            {
                string text;
                try
                {
                    while (true)
                    {
                        text = bc.Take();
                        Debug.WriteLine("Take " + text);
                        tw.WriteLine(text);                  
                    }
                }
                catch (InvalidOperationException)
                {
                    // An InvalidOperationException means that Take() was called on a completed collection
                    Debug.WriteLine("That's All!");
                    tw.Close();
                    tw.Dispose();
                }
            }))
                Task.WaitAll(t1, t2);
        }
    }
}

阻塞集合类

谢尔盖-布鲁诺夫发布的用于合并 2GB 文件的尝试解决方案。系统为此工作占用了大约 2 GB 的 RAM。我进行了一些更改以进行更多优化，现在合并 350GB 文件需要 2MB RAM。

private static void CombineMultipleFilesIntoSingleFile(string inputDirectoryPath, string inputFileNamePattern, string outputFilePath)
        {
            string[] inputFilePaths = Directory.GetFiles(inputDirectoryPath, inputFileNamePattern);
            Console.WriteLine("Number of files: {0}.", inputFilePaths.Length);
            foreach (var inputFilePath in inputFilePaths)
            {
                using (var outputStream = File.AppendText(outputFilePath))
                {
                    // Buffer size can be passed as the second argument.
                    outputStream.WriteLine(File.ReadAllText(inputFilePath));
                    Console.WriteLine("The file {0} has been processed.", inputFilePath);
                }
            }
        }

您可以执行以下几项操作：

• 根据我的经验，默认缓冲区大小可以增加，并显着提高到大约 120K，我怀疑在所有流上设置一个大缓冲区将是最简单、最明显的性能助推器：

  new System.IO.FileStream("File.txt", System.IO.FileMode.Open, System.IO.FileAccess.Read, System.IO.FileShare.Read, 150000);
  

• 使用 Stream 类，而不是StreamReader类。

• 将内容读入大型缓冲区，立即将它们转储到输出流中 - 这将加快小文件操作的速度。
• 不需要多余的关闭/处置：您有using语句。

    // Binary File Copy
    public static void mergeFiles(string strFileIn1, string strFileIn2, string strFileOut, out string strError)
    {
        strError = String.Empty;
        try
        {
            using (FileStream streamIn1 = File.OpenRead(strFileIn1))
            using (FileStream streamIn2 = File.OpenRead(strFileIn2))
            using (FileStream writeStream = File.OpenWrite(strFileOut))
            {
                BinaryReader reader = new BinaryReader(streamIn1);
                BinaryWriter writer = new BinaryWriter(writeStream);
                // create a buffer to hold the bytes. Might be bigger.
                byte[] buffer = new Byte[1024];
                int bytesRead;
                // while the read method returns bytes keep writing them to the output stream
                while ((bytesRead =
                        streamIn1.Read(buffer, 0, 1024)) > 0)
                {
                    writeStream.Write(buffer, 0, bytesRead);
                }
                while ((bytesRead =
                        streamIn2.Read(buffer, 0, 1024)) > 0)
                {
                    writeStream.Write(buffer, 0, bytesRead);
                }
            }
        }
        catch (Exception ex)
        {
            strError = ex.Message;
        }
    }