如何使用 scala 对 Spark 中 rdd 的每一行进行排序?

Question

我的文本文件有以下数据：

10,14,16,19,52
08,09,12,20,45
55,56,70,78,53

我想按降序对每一行进行排序。我尝试了以下代码

val file = sc.textFile("Maximum values").map(x=>x.split(","))
val sorted = file.sortBy(x=> -x(2).toInt)
sorted.collect()

我得到了以下输出

[[55, 56, 70, 78, 53], [10, 14, 16, 19, 52], [08, 09, 12, 20, 45]]

上面的结果表明，整个列表已按降序排序。但是我希望按降序对每个值进行排序

例如

[10,14,16,19,52],[08,09,12,20,45],[55,56,70,78,53]

应该是

[52,19,16,14,10],[45,20,12,09,08],[78,70,56,55,53]

请花点时间回答这个问题。提前谢谢。

Answer 1

检查一下。

val file = spark.sparkContext.textFile("in/sort.dat").map( x=> { val y = x.split(','); y.sorted.reverse.mkString(",") }  )
file.collect.foreach(println)

编辑1：不同的方法如何应用于上面的代码。

scala> val a = "10,14,16,19,52"
a: String = 10,14,16,19,52
scala> val b = a.split(',')
b: Array[String] = Array(10, 14, 16, 19, 52)
scala> b.sorted
res0: Array[String] = Array(10, 14, 16, 19, 52)
scala> b.sorted.reverse
res1: Array[String] = Array(52, 19, 16, 14, 10)
scala> b.sorted.reverse.mkString(",")
res2: String = 52,19,16,14,10
scala> b.sorted.reverse.mkString("*")
res3: String = 52*19*16*14*10
scala>

编辑2：

val file = spark.sparkContext.textFile("in/sort.dat").map( x=> { val y = x.split(',').map(_.toInt); y.sorted.reverse.mkString(",") }  )
file.collect.foreach(println)

Answer 2

这是一种方法(未经测试(

val reverseStringOrdering = Ordering[String].reverse
val file = sc.textFile("Maximum values").map(x=>x.split(",").sorted(reverseStringOrdering))
val sorted = file.sortBy(r => r, ascending = false)
sorted.collect()

Answer 3

Spark SQL 方式，

import org.apache.spark.sql.functions._
val df = Seq(
("10","14","16","19","52"),
("08","09","12","20","45"),
("55","56","70","78","53")).toDF("C1", "C2","C3","C4","C5")
df.withColumn("sortedCol", sort_array(array("C1", "C2","C3","C4","C5"), false))
.select("sortedCol")     
.show()

输出

+--------------------+
|           sortedCol|
+--------------------+
|[52, 19, 16, 14, 10]|
|[45, 20, 12, 09, 08]|
|[78, 70, 56, 55, 53]|
+--------------------+