Question

我有一个spark scala DataFrame，它有四个值：（id，day，val，order）。我想用列创建一个新的DataFrame：（id，day，value_list：List（val1，val2，...，valn））其中val1到valn按asc顺序值排序。

例如：

(50, 113, 1, 1), 
(50, 113, 1, 3), 
(50, 113, 2, 2), 
(51, 114, 1, 2), 
(51, 114, 2, 1), 
(51, 113, 1, 1)

会变成：

((51,113),List(1))
((51,114),List(2, 1)
((50,113),List(1, 2, 1))

我很接近，但在将数据汇总到列表后不知道该怎么办。我不知道如何通过int命令对每个值列表进行spark命令：

import org.apache.spark.sql.Row

val testList = List((50, 113, 1, 1), (50, 113, 1, 3), (50, 113, 2, 2), (51, 114, 1, 2), (51, 114, 2, 1), (51, 113, 1, 1))
val testDF = sqlContext.sparkContext.parallelize(testList).toDF("id1", "id2", "val", "order")

val rDD1 = testDF.map{case Row(key1: Int, key2: Int, val1: Int, val2: Int)  => ((key1, key2), List((val1, val2)))}
val rDD2 = rDD1.reduceByKey{case (x, y) =>  x ++ y}

输出如下：

((51,113),List((1,1)))
((51,114),List((1,2), (2,1)))
((50,113),List((1,3), (1,1), (2,2)))

下一步是制作：

((51,113),List((1,1)))
((51,114),List((2,1), (1,2)))
((50,113),List((1,1), (2,2), (1,3)))

Answer 1

您只需要映射RDD并使用sortBy：

scala> val df = Seq((50, 113, 1, 1), (50, 113, 1, 3), (50, 113, 2, 2), (51, 114, 1, 2), (51, 114, 2, 1), (51, 113, 1, 1)).toDF("id1", "id2", "val", "order")
df: org.apache.spark.sql.DataFrame = [id1: int, id2: int, val: int, order: int]

scala> import org.apache.spark.sql.Row
import org.apache.spark.sql.Row

scala> val rDD1 = df.map{case Row(key1: Int, key2: Int, val1: Int, val2: Int)  => ((key1, key2), List((val1, val2)))}
rDD1: org.apache.spark.rdd.RDD[((Int, Int), List[(Int, Int)])] = MapPartitionsRDD[10] at map at <console>:28

scala> val rDD2 = rDD1.reduceByKey{case (x, y) =>  x ++ y}
rDD2: org.apache.spark.rdd.RDD[((Int, Int), List[(Int, Int)])] = ShuffledRDD[11] at reduceByKey at <console>:30

scala> val rDD3 = rDD2.map(x => (x._1, x._2.sortBy(_._2)))
rDD3: org.apache.spark.rdd.RDD[((Int, Int), List[(Int, Int)])] = MapPartitionsRDD[12] at map at <console>:32

scala> rDD3.collect.foreach(println)
((51,113),List((1,1)))
((50,113),List((1,1), (2,2), (1,3)))
((51,114),List((2,1), (1,2)))

Answer 2

testDF.groupBy("id1","id2").agg(collect_list($"val")).show
+---+---+-----------------+                                                     
|id1|id2|collect_list(val)|
+---+---+-----------------+
| 51|113|              [1]|
| 51|114|           [1, 2]|
| 50|113|        [1, 1, 2]|
+---+---+-----------------+

Spark Scala：将DataFrame列值聚合到有序列表中

2 个答案: