java - 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。

代码如下:

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

异常抛出:

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 是一个 Ion S-Expression对象，它没有实现 java.io.Serializable。我们正在使用 Kryo 序列化程序并已注册和配置它，以便它可以很好地序列化它。

初始化spark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");

注册器中的代码:

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

如果我尝试使用下面的代码手动序列化该 lambda

SerializationUtils.serialize(filterFunc);

正如预期的那样，它失败并出现相同的错误，因为 filterExpression 不可序列化。但是，下面的代码有效:

sparkContext.env().serializer().newInstance().serialize(filterFunc, ClassTag$.MODULE$.apply(filterFunc.getClass()));

这再次符合预期，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，为什么 Spark 尝试使用 org.apache.spark.serializer.JavaSerializer 序列化该 lambda，而我们已经明确地将其配置为使用 Kryo？

Answer 1

经过更多的挖掘后发现确实有一个不同的序列化器被用于闭包。由于 Kryo 的错误，闭包序列化程序被硬编码为默认序列化程序。

这个答案很好地解释了它:https://stackoverflow.com/a/40261550/2158122

不过，我能够使用广播解决我的特殊问题。

这是我的代码现在的样子:

JavaSparkContext sparkContext; // provided
JavaPairRDD<String, IonValue> rdd; // provided
IonSexp filterExpression; // provided

Broadcast<IonSexp> filterExprBroadcast = sparkContext.broadcast(filterExpression);
rdd = rdd.filter(record -> myCustomFilter(filterExprBroadcast.value(), record));
filterExprBroadcast.destroy(false); // Only do this after an action is executed

广播的处理方式与 RDD 类似，因此它确实使用了已配置的 Kryo 序列化器。