Scala中的尾递归用例解决方案

问题描述 投票:0回答:1

我正在尝试使用尾递归解决问题。用例是:

我有文件夹列表,每个文件夹都有文件列表,每个文件都有多个记录。我想对记录进行一些转换,并将它们批量写入运动学。

val listOfFolders = Folder1(File1(RF11, RF12, RF13), File2(RF21,RF22))

我想在运动学中一次写两个记录。到目前为止,我已经尝试过:

listOfFolders.map { folder =>
    val files = fetchAllFilesFromFolder(folder)
    if (files.nonEmpty) {
      sendBatch(files, Seq.empty[(ByteBuffer, String)], 2)
    } else {
      logger.info(s"No files are present in folder")
    }
  }

  @scala.annotation.tailrec
  def sendBatch(
    files: Seq[Files],
    buffer: Seq[(ByteBuffer, String)],
    numberOfRecordsToSend: Int
  ): Unit =
    files match {
      case Nil => {
        if (buffer.nonEmpty) {
          sendToKinesis(streamName, buffer) map { putDataResult =>
            val putDataList = putDataResult.getRecords.asScala.toList
            logger.info(
              s"Successfully Sent"
            )
          }
        } else {
          logger.info(s"Successfully sent")
        }
      }
      case head :: tail => {
        val fileData = readFileData()
        val byteData: Seq[(ByteBuffer, String)] = transformDataAndConvertToByteBuffer(fileData)

        val currentBatch = buffer ++ byteData
        if (currentBatch.size >= numberOfRecordsToSend) {
          sendToKinesis(streamName, buffer)  map { putRecordRes =>
            val putDataList = putRecordRes.getRecords.asScala.toList
            logger.info(
              s"Sent successfully" 
            )
          }
          sendBatch(tail, Seq.empty[(ByteBuffer, String)], 2)
        } else {
          sendBatch(tail, currentBatch, 2)
        }
      }
    }

sendToKinesis使用KCL putRecords。

上面的代码存在的问题是:

  • 从一个文件读取所有数据。因此,如果文件有5条记录将发送5条记录可以运动,但批大小为2。

    无法从地图调用尾部递归方法。

    如果-如果file1有3条记录,也应该发送2一起记录RF11,RF12,然后一起记录RF13,RF21,最后RF22。

我不想在我的代码中使用任何var。可以使用尾部记录仪解决吗?

scala scala-collections tail-recursion
1个回答
0
投票

您有两个子问题

  1. 如何发送固定大小的批次
@scala.annotation.tailrec
def sendBatch(file: Option[File], buffer: Seq[(ByteBuffer, String)], numbersOfRecrodsToSend: Int): Seq[(ByteBuffer, String)] = {
  if (buffer.length < numbersOfRecrodsToSend) {
    // case 1: too few records to be sent 
    file match {
      // case 1.1: file was not yet read
      case Some(f) => sendBatch(None, buffer ++ getByteData(f), numbersOfRecrodsToSend)
      // case 1.2: too few records, file was already read, return leftover records
      case None => buffer
    }
  } else {
    // case 2: we can send numbersOfRecrodsToSend to Kinesis
    val (toSend, newBuffer) = buffer.splitAt(numbersOfRecrodsToSend)
    sendToKinesis(streamName, toSend)
    sendBatch(file, newBuffer, numbersOfRecrodsToSend)
  }
}
  1. 如何遍历列表并发送固定大小的批次
// start with empty list of files to send and for each folder
// add it's files to the buffer and send as many records as you can
// the leftover is going to be passed to next iteration for both files and directories
val partial = listOfFolders.foldLeft(Seq.empty[(ByteBuffer, String)]) { case (acc, folder) =>
  fetchAllFilesFromFolder(folder).foldLeft(acc) { case (acc2, file) => 
    sendBatch(Some(file), acc2, numbersOfRecrodsToSend)
  }
}

// if any records have left - send them too
if (partial.nonEmpty) {
  sendToKinesis(streamName, partial)
}

希望您有主意。

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