sparksql使用collect_list自定义排序的实现方式

原始数据如下:
+---+-----+----+
|id |name |type|
+---+-----+----+
|1  |name1|p   |
|2  |name2|p   |
|3  |name3|p   |
|1  |x1   |q   |
|2  |x2   |q   |
|3  |x3   |q   |
+---+-----+----+
目标数据如下:
+----+---------------------+
|type|value_list           |
+----+---------------------+
|p   |[name3, name2, name1]|
|q   |[x3, x2, x1]         |
+----+---------------------+

spark-shell
val df=Seq((1,"name1","p"),(2,"name2","p"),(3,"name3","p"),(1,"x1","q"),(2,"x2","q"),(3,"x3","q")).toDF("id","name","type")
df.show(false)
1.使用开窗函数

df.createOrReplaceTempView("test")
spark.sql("select type,max(c) as c1 from (select type,concat_ws('&',collect_list(trim(name)) over(partition by type order by id desc)) as c  from test) as x group by type ")
因为使用开窗函数本身会使用比较多的资源,
这种方式在大数据量下性能会比较慢,所以尝试下面的操作。
2.使用struct和sort_array(array,asc?)的方式来进行,效率高些:
val df3=spark.sql("select type, concat_ws('&',sort_array(collect_list(struct(id,name)),false).name) as c from test group by type ")
df3.show(false)
3.udf的方式
import org.apache.spark.sql.functions._
import org.apache.spark.sql._
val sortUdf = udf((rows: Seq[Row]) => {
  rows.map { case Row(id:Int, value:String) => (id, value) }
    .sortBy { case (id, value) => -id } //id if asc
    .map { case (id, value) => value }
})

val grouped = df.groupBy(col("type")).agg(collect_list(struct("id", "name")) as "id_name")
val r1 = grouped.select(col("type"), sortUdf(col("id_name")).alias("value_list"))
r1.show(false)


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posted on 2021-05-21 11:03  tneduts  阅读(325)  评论(0编辑  收藏  举报

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