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由于此时您只能拥有少量行,因此您只需按原样收集属性并将结果展平(Spark >= 2.4)import org.apache.spark.sql.functions.{collect_set, flatten, array_distinct}val byState = Seq( ("Canada", "America", Seq("A", "B")), ("Belgium", "Europe", Seq("Z")), ("USA", "America", Seq("A")), ("France", "Europe", Seq("Y", "X"))).toDF("country", "continent", "attributes")byState .groupBy("continent") .agg(array_distinct(flatten(collect_set($"attributes"))) as "attributes") .show+---------+----------+|continent|attributes|+---------+----------+| Europe| [Y, X, Z]|| America| [A, B]|+---------+----------+在一般情况下,事情更难处理,并且在许多情况下,如果您期望大型列表,每个组有许多重复项和许多值,则最佳解决方案*是从头开始重新计算结果,即input.groupBy($"continent").agg(collect_set($"attributes") as "attributes")一种可能的替代方法是使用Aggregatorimport org.apache.spark.sql.expressions.Aggregatorimport org.apache.spark.sql.catalyst.encoders.ExpressionEncoderimport org.apache.spark.sql.{Encoder, Encoders}import scala.collection.mutable.{Set => MSet}class MergeSets[T, U](f: T => Seq[U])(implicit enc: Encoder[Seq[U]]) extends Aggregator[T, MSet[U], Seq[U]] with Serializable { def zero = MSet.empty[U] def reduce(acc: MSet[U], x: T) = { for { v <- f(x) } acc.add(v) acc } def merge(acc1: MSet[U], acc2: MSet[U]) = { acc1 ++= acc2 } def finish(acc: MSet[U]) = acc.toSeq def bufferEncoder: Encoder[MSet[U]] = Encoders.kryo[MSet[U]] def outputEncoder: Encoder[Seq[U]] = enc}并按如下方式应用case class CountryAggregate( country: String, continent: String, attributes: Seq[String])byState .as[CountryAggregate] .groupByKey(_.continent) .agg(new MergeSets[CountryAggregate, String](_.attributes).toColumn) .toDF("continent", "attributes") .show+---------+----------+|continent|attributes|+---------+----------+| Europe| [X, Y, Z]|| America| [B, A]|+---------+----------+但这显然不是 Java 友好的选择。另请参阅如何在 groupBy 之后将值聚合到集合中?(类似,但没有唯一性约束)。* 这是因为explode可能非常昂贵,尤其是在旧 Spark 版本中,与访问 SQL 集合的外部表示相同。