Apache Spark将多行连接成单行的列表 [英] Apache Spark concatenate multiple rows into list in single row
本文介绍了Apache Spark将多行连接成单行的列表的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我需要从源表中创建一个表(hive表/ spark数据框),该表将多行用户的数据存储到单行的列表中。
I need to create a table(hive table/spark dataframe) from a source table that stores data of users in multiple rows into list in single row.
User table:
Schema: userid: string | transactiondate:string | charges: string |events:array<struct<name:string,value:string>>
----|------------|-------| ---------------------------------------
123 | 2017-09-01 | 20.00 | [{"name":"chargeperiod","value":"this"}]
123 | 2017-09-01 | 30.00 | [{"name":"chargeperiod","value":"last"}]
123 | 2017-09-01 | 20.00 | [{"name":"chargeperiod","value":"recent"}]
123 | 2017-09-01 | 30.00 | [{"name":"chargeperiod","value":"0"}]
456 | 2017-09-01 | 20.00 | [{"name":"chargeperiod","value":"this"}]
456 | 2017-09-01 | 30.00 | [{"name":"chargeperiod","value":"last"}]
456 | 2017-09-01 | 20.00 | [{"name":"chargeperiod","value":"recent"}]
456 | 2017-09-01 | 30.00 | [{"name":"chargeperiod","value":"0"}]
应该是
Output table should be
userid:String | concatenatedlist :List[Row]
-------|-----------------
123 | [[2017-09-01,20.00,[{"name":"chargeperiod","value":"this"}]],[2017-09-01,30.00,[{"name":"chargeperiod","value":"last"}]],[2017-09-01,20.00,[{"name":"chargeperiod","value":"recent"}]], [2017-09-01,30.00, [{"name":"chargeperiod","value":"0"}]]]
456 | [[2017-09-01,20.00,[{"name":"chargeperiod","value":"this"}]],[2017-09-01,30.00,[{"name":"chargeperiod","value":"last"}]],[2017-09-01,20.00,[{"name":"chargeperiod","value":"recent"}]], [2017-09-01,30.00, [{"name":"chargeperiod","value":"0"}]]]
Spark版本:1.6.2
Spark version: 1.6.2
推荐答案
val rdd = sc.parallelize(Seq(("1","2017-02-01","20.00","abc"),("1","2017-02-01","30.00","abc2"),("2","2017-02-01","20.00","abc"),("2","2017-02-01","30.00","abc")))
val df = rdd.toDF("id","date","amt","array")
df.withColumn("new",concat_ws(",",$"date",$"amt",$"array")).select("id","new").groupBy("id").agg(concat_ws(",",collect_list("new")))
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