OrderBy without Spark dataset abstraction

OrderByTimestamp.scala

import org.apache.spark.sql.SparkSession
import org.scalatest.FunSuite

case class Root(headers: Map[String, String], body: String)

class OrderByTimeStampTest extends FunSuite {

  val spark = SparkSession.builder
    .master("local[*]")
    .getOrCreate

  test("fill and order dataset") {
    val r1 = Root(Map("test" -> "a", "timestamp" -> "1"), "FirstTest")
    val r2 = Root(Map("test" -> "b", "timestamp" -> "3"), "Test")
    val r3 = Root(Map("test" -> "c", "timestamp" -> "4"), "Test")
    val r4 = Root(Map("test" -> "d", "timestamp" -> "5"), "Test")
    val r5 = Root(Map("test" -> "e", "timestamp" -> "2"), "Test")

    import spark.implicits._

    //Just to show an example, I don't use spark to read the files and store them in a dataset
    //Assume that ds is the dataset I retrieve by reading the files from HDFS
    val ds : Dataset[Root] = Seq(r1, r2, r3, r4, r5).toDS

    //SylarBenes solution
    
    val seq: Seq[Root] = ds.collect().to[Seq]
    
    seq.sortBy(x => x.headers.get("timestamp").map(_.toInt)).reverse.foreach(println(_))

  }


}