Scala 函数返回Spark中的空列表
下面是获取压缩文件中文件名列表的代码Scala 函数返回Spark中的空列表,scala,apache-spark,functional-programming,scala-collections,Scala,Apache Spark,Functional Programming,Scala Collections,下面是获取压缩文件中文件名列表的代码 def getListOfFilesInRepo(zipFileRDD : RDD[(String,PortableDataStream)]) : (List[String]) = { val zipInputStream = zipFileRDD.values.map(x => new ZipInputStream(x.open)) val filesInZip = new ArrayBuffer[String]() var
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)
}
我以以下方式执行上述代码:
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()
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二进制文件rdd[17]位于:25的二进制文件处
scala>getListOfFilesInRepo(zipRDD)
889
0
res12:List[String]=List()
为什么我没有得到889,反而得到了0?之所以发生这种情况,是因为
filesInZip
不在工作人员之间共享foreach
对filesInZip
的本地副本进行操作,当它完成时,该副本将被丢弃并进行垃圾收集。如果要保留结果,应使用转换(很可能是flatMap
)并返回收集的聚合值
def listFiles(stream: PortableDataStream): TraversableOnce[String] = ???
zipInputStream.flatMap(listFiles)
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