Question

我已经尝试了在StackOverflow上找到的这个问题的所有解决方案但是，尽管如此，我还是无法解决它。我有一个“MainObj”对象，它实例化一个“Recommendation”对象。当我调用“recommendedProducts”方法时，我总是会收到错误。以下是该方法的代码：

def recommendationProducts(item: Int): Unit = {

val aMatrix = new DoubleMatrix(Array(1.0, 2.0, 3.0))

def cosineSimilarity(vec1: DoubleMatrix, vec2: DoubleMatrix): Double = {
  vec1.dot(vec2) / (vec1.norm2() * vec2.norm2())
}

val itemFactor = model.productFeatures.lookup(item).head
val itemVector = new DoubleMatrix(itemFactor)

//Here is where I get the error:
val sims = model.productFeatures.map { case (id, factor) =>
  val factorVector = new DoubleMatrix(factor)
  val sim = cosineSimilarity(factorVector, itemVector)
  (id, sim)
}

val sortedSims = sims.top(10)(Ordering.by[(Int, Double), Double] {
  case (id, similarity) => similarity
})

println("\nTop 10 products:")
sortedSims.map(x => (x._1, x._2)).foreach(println)

这是错误：

Exception in thread "main" 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$$anonfun$map$1.apply(RDD.scala:369)
at org.apache.spark.rdd.RDDOperationScope$.withScope(RDDOperationScope.scala:151)
at org.apache.spark.rdd.RDDOperationScope$.withScope(RDDOperationScope.scala:112)
at org.apache.spark.rdd.RDD.withScope(RDD.scala:362)
at org.apache.spark.rdd.RDD.map(RDD.scala:369)
at RecommendationObj.recommendationProducts(RecommendationObj.scala:269)
at MainObj$.analisiIUNGO(MainObj.scala:257)
at MainObj$.menu(MainObj.scala:54)
at MainObj$.main(MainObj.scala:37)
at MainObj.main(MainObj.scala)
Caused by: java.io.NotSerializableException: org.apache.spark.SparkContext
Serialization stack:
- object not serializable (class: org.apache.spark.SparkContext, value: org.apache.spark.SparkContext@7c2312fa)
- field (class: RecommendationObj, name: sc, type: class org.apache.spark.SparkContext)
- object (class MainObj$$anon$1, MainObj$$anon$1@615bad16)
- field (class: RecommendationObj$$anonfun$37, name: $outer, type: class RecommendationObj)
- object (class RecommendationObj$$anonfun$37, <function1>)
at org.apache.spark.serializer.SerializationDebugger$.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)
... 14 more

我尝试添加： 1）“将Serializable”（Scala）扩展到我的班级 2）“extends extends java.io.Serializable”到我的Class 3）“@transient”到某些部分 4）在这个类中获取模型（和其他特性）（现在我从另一个对象中获取它们，然后将它们传递给我的类，就像参数一样）

我该如何解决？我变得疯了！提前谢谢！

Answer 1

关键在这里：

 field (class: RecommendationObj, name: sc, type: class org.apache.spark.SparkContext)

所以你有一个名为sc的类型为SparkContext的字段。 Spark想要序列化类，所以他也尝试序列化所有字段。

你应该：

使用@transient注释并检查是否为null，然后重新创建
不使用字段中的SparkContext，而是将其放入方法的参数中。但请记住，你永远不应该在map，flatMap等中使用SparkContext里面的闭包。

Spark任务不可序列化

1 个答案: