鉴于您拥有dataframe作为

+----+----+----+
|colA|colB|colC|
+----+----+----+
|1   |1   |23  |
|1   |2   |63  |
|1   |3   |31  |
|2   |1   |32  |
|2   |2   |56  |
+----+----+----+

您可以通过执行以下操作Window功能

You can Window functions by doing the following

import org.apache.spark.sql.functions._
import org.apache.spark.sql.expressions._
df.withColumn("colD", collect_list("colC").over(Window.partitionBy("colA").orderBy("colB"))).show(false)

结果:

+----+----+----+------------+
|colA|colB|colC|colD        |
+----+----+----+------------+
|1   |1   |23  |[23]        |
|1   |2   |63  |[23, 63]    |
|1   |3   |31  |[23, 63, 31]|
|2   |1   |32  |[32]        |
|2   |2   |56  |[32, 56]    |
+----+----+----+------------+

collect_set的结果也与此类似.但是最后一个set中的元素顺序不会像collect_list

Similar is the result for collect_set as well. But the order of elements in the final set will not be in order as with collect_list

df.withColumn("colD", collect_set("colC").over(Window.partitionBy("colA").orderBy("colB"))).show(false)
+----+----+----+------------+
|colA|colB|colC|colD        |
+----+----+----+------------+
|1   |1   |23  |[23]        |
|1   |2   |63  |[63, 23]    |
|1   |3   |31  |[63, 31, 23]|
|2   |1   |32  |[32]        |
|2   |2   |56  |[56, 32]    |
+----+----+----+------------+

如果您按以下说明删除orderBy

If you remove orderBy as below

df.withColumn("colD", collect_list("colC").over(Window.partitionBy("colA"))).show(false)

结果应为

+----+----+----+------------+
|colA|colB|colC|colD        |
+----+----+----+------------+
|1   |1   |23  |[23, 63, 31]|
|1   |2   |63  |[23, 63, 31]|
|1   |3   |31  |[23, 63, 31]|
|2   |1   |32  |[32, 56]    |
|2   |2   |56  |[32, 56]    |
+----+----+----+------------+

我希望答案会有所帮助