在Spark中配置函数/ lambda序列化

Question

如何配置Spark以将KryoSerializer用于lambda？还是我在Spark中发现了错误？我们对其他地方的数据序列化没有问题，只是这些lambda使用默认值而不是Kryo。

这里是代码：

JavaPairRDD<String, IonValue> rdd; // provided
IonSexp filterExpression; // provided
Function<Tuple2<String, IonValue>, Boolean> filterFunc = record -> myCustomFilter(filterExpression, record);
rdd = rdd.filter(filterFunc);

抛出异常：

org.apache.spark.SparkException: Task not serializable
    at org.apache.spark.util.ClosureCleaner$.ensureSerializable(ClosureCleaner.scala:403)
    at org.apache.spark.util.ClosureCleaner$.org$apache$spark$util$ClosureCleaner$$clean(ClosureCleaner.scala:393)
    at org.apache.spark.util.ClosureCleaner$.clean(ClosureCleaner.scala:162)
    at org.apache.spark.SparkContext.clean(SparkContext.scala:2326)
    at org.apache.spark.rdd.RDD$$anonfun$filter$1.apply(RDD.scala:388)
    at org.apache.spark.rdd.RDD$$anonfun$filter$1.apply(RDD.scala:387)
    at org.apache.spark.rdd.RDDOperationScope$.withScope(RDDOperationScope.scala:151)
    at org.apache.spark.rdd.RDDOperationScope$.withScope(RDDOperationScope.scala:112)
    at org.apache.spark.rdd.RDD.withScope(RDD.scala:363)
    at org.apache.spark.rdd.RDD.filter(RDD.scala:387)
    at org.apache.spark.api.java.JavaPairRDD.filter(JavaPairRDD.scala:99)
    at com.example.SomeClass.process(SomeClass.java:ABC)
    {more stuff}
Caused by: java.io.NotSerializableException: com.amazon.ion.impl.lite.IonSexpLite
Serialization stack:
    - object not serializable (class: com.amazon.ion.impl.lite.IonSexpLite, value: (and (equals (literal 1) (path marketplace_id)) (equals (literal 351) (path product gl_product_group))))
    - element of array (index: 1)
    - array (class [Ljava.lang.Object;, size 2)
    - field (class: java.lang.invoke.SerializedLambda, name: capturedArgs, type: class [Ljava.lang.Object;)
    - object (class java.lang.invoke.SerializedLambda, SerializedLambda[capturingClass=class com.example.SomeClass, functionalInterfaceMethod=org/apache/spark/api/java/function/Function.call:(Ljava/lang/Object;)Ljava/lang/Object;, implementation=invokeSpecial com/example/SomeClass.lambda$process$8f20a2d2$1:(Lcom/amazon/ion/IonSexp;Lscala/Tuple2;)Ljava/lang/Boolean;, instantiatedMethodType=(Lscala/Tuple2;)Ljava/lang/Boolean;, numCaptured=2])
    - writeReplace data (class: java.lang.invoke.SerializedLambda)
    - object (class com.example.SomeClass$$Lambda$36/263969036, com.example.SomeClass$$Lambda$36/263969036@31880efa)
    - field (class: org.apache.spark.api.java.JavaPairRDD$$anonfun$filter$1, name: f$1, type: interface org.apache.spark.api.java.function.Function)
    - object (class org.apache.spark.api.java.JavaPairRDD$$anonfun$filter$1, <function1>)
    at org.apache.spark.serializer.SerializationDebugger$.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:400)
    ... 18 more

在这种情况下，有问题的filterExpression是Ion S-Expression对象，它没有实现java.io.Serializable。我们正在使用Kryo序列化器，并已对其进行注册和配置，以便可以对其进行序列化。

初始化火花配置时的代码：

sparkConf = new SparkConf().setAppName("SomeAppName").setMaster("MasterLivesHere")
        .set("spark.serializer", KryoSerializer.class.getCanonicalName())
        .set("spark.kryo.registrator", KryoRegistrator.class.getCanonicalName())
        .set("spark.kryo.registrationRequired", "false");

我们的注册人代码：

kryo.register(com.amazon.ion.IonSexp.class);
kryo.register(Class.forName("com.amazon.ion.impl.lite.IonSexpLite"));

如果我尝试使用以下代码手动序列化该lambda，则>]

SerializationUtils.serialize(filterFunc);
由于filterExpression不可序列化，因此失败，并出现与预期相同的错误。但是，下面的代码有效：

sparkContext.env().serializer().newInstance().serialize(filterFunc, ClassTag$.MODULE$.apply(filterFunc.getClass()));
同样，这是预期的，因为我们的Kryo设置能够处理这些对象。

所以我的问题/困惑是，当我们明确配置了Lambda以使用Kryo时，为什么Spark会尝试使用org.apache.spark.serializer.JavaSerializer序列化该lambda？

如何配置Spark以将KryoSerializer用于lambda？还是我在Spark中发现了错误？我们对其他地方的数据序列化没有任何问题，只是这些lambda使用默认的...

Answer 1

经过更多的挖掘，事实确实是有一个不同的序列化器用于闭包。由于Kryo的错误，关闭序列化器被硬编码为默认值。