SparkContext在配对对象中不可序列化

Question

我目前正在尝试扩展使用Scala和Spark的机器学习应用程序.我正在使用我在Github上发现的Dieterich Lawson先前项目的结构

https://github.com/dieterichlawson/admm

该项目基本上使用SparkContext构建训练样本块的RDD,然后对每个集合执行局部计算(例如求解线性系统).

我遵循相同的方案,但对于我的本地计算,我需要在每个训练样本块上执行L-BFGS算法.为了做到这一点,我想使用mlLib中的L-BFGS算法,该算法具有以下特征.

runLBFGS(RDD<scala.Tuple2<Object,Vector>> data, Gradient gradient, 
         Updater updater, int numCorrections, double convergenceTol, 
         int maxNumIterations, double regParam, Vector initialWeights)

如上所述,该方法将训练样本的RDD [Object,Vector]作为输入.问题是在每个工作者本地我不再保留数据的RDD结构.因此,我正在尝试在矩阵的每个块上使用SparkContext的并行化功能.但是当我这样做时,我得到了一个序列化器异常.(确切的异常消息在问题的最后).

这是我如何处理SparkContext的详细解释.

首先,在主应用程序中,它用于打开文本文件,它在LogRegressionXUpdate类的工厂中使用:

val A = sc.textFile("ds1.csv")
A.checkpoint
val f = LogRegressionXUpdate.fromTextFile(A,params.rho,1024,sc)

在应用程序中,LogRegressionXUpdate类实现如下

class LogRegressionXUpdate(val training: RDD[(Double, NV)],
                           val rho: Double) extends Function1[BDV[Double],Double] with Prox  with Serializable{

def prox(x: BDV[Double], rho: Double): BDV[Double] = {
    val numCorrections = 10
    val convergenceTol = 1e-4
    val maxNumIterations = 20
    val regParam = 0.1
    val (weights, loss) = LBFGS.runLBFGS(
        training,
        new GradientForLogRegADMM(rho,fromBreeze(x)),
        new SimpleUpdater(),
        numCorrections,
        convergenceTol,
        maxNumIterations,
        regParam,
        fromBreeze(x))
    toBreeze(weights.toArray).toDenseVector
}

def apply(x: BDV[Double]): Double = {
    Math.pow(1,2.0)
}

}

使用以下配套对象:

object LogRegressionXUpdate {
    def fromTextFile(file: RDD[String], rho: Double, blockHeight: Int = 1024, @transient sc: SparkContext): RDF[LogRegressionXUpdate] = {
        val fns = new BlockMatrix(file, blockHeight).blocks.
        map(X => new LogRegressionXUpdate(sc.parallelize((X(*,::).map(fila => (fila(-1),fromBreeze(fila(0 to -2))))).toArray),rho))
        new RDF[LogRegressionXUpdate](fns, 0L)
    }
}

这个构造函数导致序列化错误,虽然我并不是真的需要SparkContext来在本地构建每个RDD.我已经搜索了这个问题的解决方案,添加@transient并没有解决它.然后,我的问题是:是否真的可以构建这些"第二层RDD",或者我被迫执行L-BFGS算法的非分布式版本.提前致谢!

错误日志:

Exception in thread "main" org.apache.spark.SparkException: Task not serializable
at org.apache.spark.util.ClosureCleaner$.ensureSerializable(ClosureCleaner.scala:315)
at org.apache.spark.util.ClosureCleaner$.org$apache$spark$util$ClosureCleaner$$clean(ClosureCleaner.scala:305)
at org.apache.spark.util.ClosureCleaner$.clean(ClosureCleaner.scala:132)
at org.apache.spark.SparkContext.clean(SparkContext.scala:1891)
at org.apache.spark.rdd.RDD$$anonfun$map$1.apply(RDD.scala:294)
at org.apache.spark.rdd.RDD$$anonfun$map$1.apply(RDD.scala:293)
at org.apache.spark.rdd.RDDOperationScope$.withScope(RDDOperationScope.scala:148)
at org.apache.spark.rdd.RDDOperationScope$.withScope(RDDOperationScope.scala:109)
at org.apache.spark.rdd.RDD.withScope(RDD.scala:286)
at org.apache.spark.rdd.RDD.map(RDD.scala:293)
at admm.functions.LogRegressionXUpdate$.fromTextFile(LogRegressionXUpdate.scala:70)
at admm.examples.Lasso$.run(Lasso.scala:96)
at admm.examples.Lasso$$anonfun$main$1.apply(Lasso.scala:70)
at admm.examples.Lasso$$anonfun$main$1.apply(Lasso.scala:69)
at scala.Option.map(Option.scala:145)
at admm.examples.Lasso$.main(Lasso.scala:69)
at admm.examples.Lasso.main(Lasso.scala)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:497)
at com.intellij.rt.execution.application.AppMain.main(AppMain.java:140)
Caused by: java.io.NotSerializableException: org.apache.spark.SparkContext
Serialization stack:
- object not serializable (class: org.apache.spark.SparkContext, value: org.apache.spark.SparkContext@20576557)
- field (class: admm.functions.LogRegressionXUpdate$$anonfun$1, name: sc$1, type: class org.apache.spark.SparkContext)
- object (class admm.functions.LogRegressionXUpdate$$anonfun$1, <function1>)
at org.apache.spark.serializer.SerializationDebugger$.improveException(SerializationDebugger.scala:40)
at org.apache.spark.serializer.JavaSerializationStream.writeObject(JavaSerializer.scala:47)
at org.apache.spark.serializer.JavaSerializerInstance.serialize(JavaSerializer.scala:81)
at org.apache.spark.util.ClosureCleaner$.ensureSerializable(ClosureCleaner.scala:312)
... 21 more

Answer 1

RDD只应从驱动程序访问s.每当你打电话的时候

myRDD.map(someObject.someMethod)

spark序列化计算所需的任何内容,someMethod并将其发送给工作人员.在那里,该方法被反序列化,然后它独立地在每个分区上运行.

但是,您尝试使用本身使用spark 的方法:您尝试创建新的RDD.但是,这是不可能的,因为它们只能在驱动程序中创建.您看到的错误是spark尝试序列化spark上下文本身,因为每个块的计算都需要它.有关序列化的更多信息可以在这个问题的第一个答案中找到.

"......虽然我不是真的需要SparkContext来在本地构建每个RDD" - 实际上这正是你在调用时所做的sc.parallelize.底线 - 您需要找到(或写入)L-BFGS的本地实现.