java - 为什么这个 Spark 代码抛出 java.io.NotSerializableException

Question

我想在 RDD 的转换中访问伴生对象的方法。为什么以下不起作用:

import org.apache.spark.rdd.RDD
import spark.implicits._
import org.apache.spark.sql.{Encoder, Encoders}

class Abc {
    def transform(x: RDD[Int]): RDD[Double] = { x.map(Abc.fn) }
}

object Abc {
  def fn(x: Int): Double = { x.toDouble }
}

implicit def abcEncoder: Encoder[Abc] = Encoders.kryo[Abc]

new Abc().transform(sc.parallelize(1 to 10)).collect

上面的代码抛出一个java.io.NotSerializableException:

org.apache.spark.SparkException: Task not serializable
  at org.apache.spark.util.ClosureCleaner$.ensureSerializable(ClosureCleaner.scala:298)
  at org.apache.spark.util.ClosureCleaner$.org$apache$spark$util$ClosureCleaner$$clean(ClosureCleaner.scala:288)
  at org.apache.spark.util.ClosureCleaner$.clean(ClosureCleaner.scala:108)
  at org.apache.spark.SparkContext.clean(SparkContext.scala:2094)
  at org.apache.spark.rdd.RDD$$anonfun$map$1.apply(RDD.scala:370)
  at org.apache.spark.rdd.RDD$$anonfun$map$1.apply(RDD.scala:369)
  at org.apache.spark.rdd.RDDOperationScope$.withScope(RDDOperationScope.scala:151)
  at org.apache.spark.rdd.RDDOperationScope$.withScope(RDDOperationScope.scala:112)
  at org.apache.spark.rdd.RDD.withScope(RDD.scala:362)
  at org.apache.spark.rdd.RDD.map(RDD.scala:369)
  at Abc.transform(<console>:19)
  ... 47 elided
Caused by: java.io.NotSerializableException: Abc
Serialization stack:
        - object not serializable (class: Abc, value: Abc@4f598dfb)
        - field (class: Abc$$anonfun$transform$1, name: $outer, type: class Abc)
        - object (class Abc$$anonfun$transform$1, <function1>)
  at org.apache.spark.serializer.SerializationDebugger$.improveException(SerializationDebugger.scala:40)
  at org.apache.spark.serializer.JavaSerializationStream.writeObject(JavaSerializer.scala:46)
  at org.apache.spark.serializer.JavaSerializerInstance.serialize(JavaSerializer.scala:100)
  at org.apache.spark.util.ClosureCleaner$.ensureSerializable(ClosureCleaner.scala:295)
  ... 57 more

即使为类 Abc 定义一个 Encoder 在这里也无济于事。但更重要的问题是，为什么要尝试对 Abc 类的对象进行序列化？我的第一个想法是伴随对象是该类的单例对象，因此可能会尝试对其进行序列化。但似乎并非如此，因为当我从另一个类(class)调用 Abc.fn 时:

class Xyz {
    def transform(x: RDD[Int]): RDD[Double] = { x.map(Abc.fn) }
}

implicit def xyzEncoder: Encoder[Xyz] = Encoders.kryo[Xyz]

new Xyz().transform(sc.parallelize(1 to 10)).collect

我得到一个 java.io.NotSerializableException: Xyz

Answer 1

这是一篇讨论 Apache Spark 中“可序列化”对象与“不可序列化对象”的精彩文章:

Using Non-Serializable Objects in Apache Spark, Nicola Ferraro

这篇文章提供了几个建议:

• 在您的具体案例中发生了什么

• 一些替代方案，因此您的对象不需要“可序列化”