scala - 数据帧空值在 UDF 后转换为 0。为什么？

Question

访问数据框 Row 值时如何处理 Null？ Null pointer Exception 真的需要手动处理吗？必须有更好的解决方案。

case class FirstThing(id:Int, thing:String, other:Option[Double])

val df = Seq(FirstThing(1, "first", None), FirstThing(1, "second", Some(2)), FirstThing(1, "third", Some(3))).toDS
df.show

val list = df.groupBy("id").agg(collect_list(struct("thing", "other")).alias("mylist"))
list.show(false)

NPE 失败:

val xxxx = udf((t:Seq[Row])=> t.map(elem => elem.getDouble(1)))
list.withColumn("aa", xxxx(col("mylist"))).show(false)

这奇怪地给出了 0:

val xxxx = udf((t:Seq[Row])=> t.map(elem => elem.getAs[Double]("other")))
list.withColumn("aa", xxxx(col("mylist"))).show(false)

+---+-----------------------------------------+---------------+
|id |mylist                                   |aa             |
+---+-----------------------------------------+---------------+
|1  |[[first,null], [second,2.0], [third,3.0]]|[0.0, 2.0, 3.0]|
+---+-----------------------------------------+---------------+

遗憾的是，这种适用于数据框/数据集的方法也失败了:

val xxxx = udf((t:Seq[Row])=> t.map(elem => elem.getAs[Option[Double]]("other")))
list.withColumn("aa", xxxx(col("mylist"))).show(false)

ClassCastException: java.lang.Double cannot be cast to scala.Option

Answer 1

使用 getAs[Double] 并将其包装在 Option 中将给出预期的结果:

val xxxx = udf((t: Seq[Row])=> t.map(elem => Option(elem.getAs[Double]("other"))))
list.withColumn("aa", xxxx($"mylist")).show(false)

+---+-----------------------------------------+----------------+
|id |mylist                                   |aa              |
+---+-----------------------------------------+----------------+
|1  |[[first,null], [second,2.0], [third,3.0]]|[null, 2.0, 3.0]|
+---+-----------------------------------------+----------------+

getAs[Option[Double]] 不起作用的原因可能是数据框架构不了解列具有选项。 udf 之前的架构:

root
 |-- id: integer (nullable = false)
 |-- mylist: array (nullable = true)
 |    |-- element: struct (containsNull = true)
 |    |    |-- thing: string (nullable = true)
 |    |    |-- other: double (nullable = true)