Spark 任务无法使用滞后窗口函数进行序列化

我注意到，在 DataFrame 上使用 Window 函数后，如果我使用函数调用 map()，Spark 将返回“任务不可序列化”异常 这是我的代码：

val hc:org.apache.spark.sql.hive.HiveContext =
    new org.apache.spark.sql.hive.HiveContext(sc)

import hc.implicits._
import org.apache.spark.sql.expressions.Window
import org.apache.spark.sql.functions._

def f() : String = "test"
case class P(name: String, surname: String)
val lag_result: org.apache.spark.sql.Column = 
    lag($"name",1).over(Window.partitionBy($"surname"))
val lista: List[P] = List(P("N1","S1"), P("N2","S2"), P("N2","S2"))
val data_frame: org.apache.spark.sql.DataFrame = 
    hc.createDataFrame(sc.parallelize(lista))

df.withColumn("lag_result", lag_result).map(x => f)

// This works
// df.withColumn("lag_result", lag_result).map{ case x =>
//     def f():String = "test";f}.collect

这是堆栈跟踪：

org.apache.spark.SparkException：任务无法序列化 org.apache.spark.util.ClosureCleaner$.ensureSerialized(ClosureCleaner.scala:304) 在 org.apache.spark.util.ClosureCleaner$.org$apache$spark$util$ClosureCleaner$$clean(ClosureCleaner.scala:294) 在 org.apache.spark.util.ClosureCleaner$.clean(ClosureCleaner.scala:122) 在 org.apache.spark.SparkContext.clean(SparkContext.scala:2055) 处 org.apache.spark.rdd.RDD$$anonfun$map$1.apply(RDD.scala:324) 在 org.apache.spark.rdd.RDD$$anonfun$map$1.apply(RDD.scala:323) 在... 以及更多原因：java.io.NotSerializedException： org.apache.spark.sql.Column 序列化堆栈：

• 对象不可序列化（类：org.apache.spark.sql.Column，值：'lag（name，1，null）windowspecdefinition（surname，UnspecifiedFrame））
• 字段（类：$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$ iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC$$iwC, 名称：lag_result，类型：class org.apache.spark.sql.Column) ... 和 更多的

lag回报o.a.s.sql.Column这是不可序列化的。同样的事情也适用于WindowSpec。在交互模式下，这些对象可以作为闭包的一部分包含在内map:

scala> import org.apache.spark.sql.expressions.Window
import org.apache.spark.sql.expressions.Window

scala> val df = Seq(("foo", 1), ("bar", 2)).toDF("x", "y")
df: org.apache.spark.sql.DataFrame = [x: string, y: int]

scala> val w = Window.partitionBy("x").orderBy("y")
w: org.apache.spark.sql.expressions.WindowSpec = org.apache.spark.sql.expressions.WindowSpec@307a0097

scala> val lag_y = lag(col("y"), 1).over(w)
lag_y: org.apache.spark.sql.Column = 'lag(y,1,null) windowspecdefinition(x,y ASC,UnspecifiedFrame)

scala> def f(x: Any) = x.toString
f: (x: Any)String

scala> df.select(lag_y).map(f _).first
org.apache.spark.SparkException: Task not serializable
    at org.apache.spark.util.ClosureCleaner$.ensureSerializable(ClosureCleaner.scala:304)
...
Caused by: java.io.NotSerializableException: org.apache.spark.sql.expressions.WindowSpec
Serialization stack:
    - object not serializable (class: org.apache.spark.sql.expressions.WindowSpec, value: org.apache.spark.sql.expressions.WindowSpec@307a0097)

一个简单的解决方案是将两者标记为瞬态：

scala> @transient val w = Window.partitionBy("x").orderBy("y")
w: org.apache.spark.sql.expressions.WindowSpec = org.apache.spark.sql.expressions.WindowSpec@7dda1470

scala> @transient val lag_y = lag(col("y"), 1).over(w)
lag_y: org.apache.spark.sql.Column = 'lag(y,1,null) windowspecdefinition(x,y ASC,UnspecifiedFrame)

scala> df.select(lag_y).map(f _).first
res1: String = [null]