我想访问RDD中转换内的伴随对象的方法。为什么下面不工作:为什么这个Spark代码抛出java.io.NotSerializableException
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
上面这段代码抛出一个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$$anonfun$map$1.apply(RDD.scala:369)
at org.apache.spark.rdd.RDDOperationScope$.withScope(RDDOperationScope.scala:151)
at org.apache.spark.rdd.RDDOperationScope$.withScope(RDDOperationScope.scala:112)
at org.apache.spark.rdd.RDD.withScope(RDD.scala:362)
at org.apache.spark.rdd.RDD.map(RDD.scala:369)
at Abc.transform(<console>:19)
... 47 elided
Caused by: java.io.NotSerializableException: Abc
Serialization stack:
- object not serializable (class: Abc, value: [email protected])
- field (class: Abc$$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
即使定义Encoder
为ABC类并不在这里帮助。但更重要的问题是,为什么Abc类的对象的系列化被尝试呢?我的第一个想法是伴侣对象是类的单例对象,所以也许会尝试序列化它。但似乎并不喜欢这样,因为当我从另一个类调用Abc.fn:
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
工作不会发生在边缘节点上;类(或对象)必须被序列化,以便数据节点可以运行它。 –
因为你实际上并没有真正定义serialise/deserialise函数,即使实现适当的接口也没有那么简单? (https://docs.oracle.com/javase/7/docs/api/java/io/NotSerializableException.html)默认情况下,sereilsiation只能访问公共设置和可获取的内容。除此之外,您需要提供自己的功能。 – Christopher