函数返回星火空List [英] Function returns an empty List in Spark

问题描述

下面是code为获得一个压缩文件的文件名列表

Below is code for getting list of file Names in a zipped file

def getListOfFilesInRepo(zipFileRDD : RDD[(String,PortableDataStream)]) : (List[String]) = {
    val zipInputStream = zipFileRDD.values.map(x => new ZipInputStream(x.open))
    val filesInZip =  new ArrayBuffer[String]()
    var ze : Option[ZipEntry] = None
    zipInputStream.foreach(stream =>{
      do{
        ze = Option(stream.getNextEntry);
        ze.foreach{ze =>
          if(ze.getName.endsWith("java") && !ze.isDirectory()){
            var fileName:String = ze.getName.substring(ze.getName.lastIndexOf("/")+1,ze.getName.indexOf(".java"))
            filesInZip += fileName
          }
        }
        stream.closeEntry()
      } while(ze.isDefined)
      println(filesInZip.toList.length) // print 889 (correct)
    })
    println(filesInZip.toList.length) // print 0 (WHY..?)
    (filesInZip.toList)
  }

我在下面的方式执行上面code：

I execute above code in the following manner :

scala> val zipFileRDD = sc.binaryFiles("./handsOn/repo~apache~storm~14135470~false~Java~master~2210.zip")
zipFileRDD: org.apache.spark.rdd.RDD[(String, org.apache.spark.input.PortableDataStream)] = ./handsOn/repo~apache~storm~14135470~false~Java~master~2210.zip BinaryFileRDD[17] at binaryFiles at <console>:25

scala> getListOfFilesInRepo(zipRDD)
889
0
res12: List[String] = List()

为什么我没有得到889，而是让0？

Why i am not getting 889 and instead getting 0?

推荐答案

这是因为 filesInZip 不是工人之间共享。 的foreach 运行在 filesInZip 的本地副本，当它完成这个副本简单地抛弃，垃圾收集。如果你想保留的结果，你应该使用变换（最有可能是 flatMap ），并返回收集汇总值。

It happens because filesInZip is not shared between workers. foreach operates on a local copy of filesInZip and when it finishes this copy is simply discarded and garbage collected. If you want to keep the results you should use transformation (most likely a flatMap) and return collected aggregated values.

def listFiles(stream: PortableDataStream): TraversableOnce[String] = ???

zipInputStream.flatMap(listFiles)

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