我有一个RDD:
val rdd: org.apache.spark.rdd.RDD[((String, String), Double)] =
sc.parallelize(List(
(("a", "b"), 1.0),
(("a", "c"), 3.0),
(("a", "d"), 2.0)
))
我正在尝试将此RDD从类型org.apache.spark.rdd.RDD[((String, String), Double)]
转换为org.apache.spark.rdd.RDD[((String), List[Double])]
RDD中的每个键都应该是唯一的,并且它的值会被排序。
因此rdd
以上的结构将转换为:
val newRdd : [((String), List[Double])] = RDD("a" , List(1,2,3))
获取我使用的唯一键列表:
val r2 : org.apache.spark.rdd.RDD[(String, Double)] = rdd.map(m => (m._1._1 , m._2))
如何将每个键转换为包含已排序的双打列表?
整个代码:
import org.apache.spark.SparkContext;
object group {
println("Welcome to the Scala worksheet") //> Welcome to the Scala worksheet
val conf = new org.apache.spark.SparkConf()
.setMaster("local")
.setAppName("distances")
.setSparkHome("C:\\spark-1.1.0-bin-hadoop2.4\\spark-1.1.0-bin-hadoop2.4")
.set("spark.executor.memory", "1g") //> conf : org.apache.spark.SparkConf = org.apache.spark.SparkConf@1bd0dd4
val sc = new SparkContext(conf) //> 14/12/16 16:44:56 INFO spark.SecurityManager: Changing view acls to: a511381
//| ,
//| 14/12/16 16:44:56 INFO spark.SecurityManager: Changing modify acls to: a5113
//| 81,
//| 14/12/16 16:44:56 INFO spark.SecurityManager: SecurityManager: authenticatio
//| n disabled; ui acls disabled; users with view permissions: Set(a511381, ); u
//| sers with modify permissions: Set(a511381, )
//| 14/12/16 16:44:57 INFO slf4j.Slf4jLogger: Slf4jLogger started
//| 14/12/16 16:44:57 INFO Remoting: Starting remoting
//| 14/12/16 16:44:57 INFO Remoting: Remoting started; listening on addresses :[
//| akka.tcp://sparkDriver@LA342399.dmn1.fmr.com:51092]
//| 14/12/16 16:44:57 INFO Remoting: Remoting now listens on addresses: [akka.tc
//| p://sparkDriver@LA342399.dmn1.fmr.com:51092]
//| 14/12/16 16:44:57 INFO util.Utils: Successfully started service 'sparkDriver
//| ' on port 51092.
//| 14/12/16 16:44:57 INFO spark.SparkEnv: Registering MapOutputTracker
//| 14/12/16 16:44:57 INFO spark.SparkEnv:
//| Output exceeds cutoff limit.
val rdd: org.apache.spark.rdd.RDD[((String, String), Double)] =
sc.parallelize(List(
(("a", "b"), 1.0),
(("a", "c"), 3.0),
(("a", "d"), 2.0)
)) //> rdd : org.apache.spark.rdd.RDD[((String, String), Double)] = ParallelCollec
//| tionRDD[0] at parallelize at group.scala:15
val r2 : org.apache.spark.rdd.RDD[(String, Double)] = rdd.map(m => (m._1._1 , m._2))
//> r2 : org.apache.spark.rdd.RDD[(String, Double)] = MappedRDD[1] at map at gr
//| oup.scala:21
val m1 = r2.collect //> 14/12/16 16:44:59 INFO spark.SparkContext: Starting job: collect at group.sc
//| ala:23
//| 14/12/16 16:44:59 INFO scheduler.DAGScheduler: Got job 0 (collect at group.s
//| cala:23) with 1 output partitions (allowLocal=false)
//| 14/12/16 16:44:59 INFO scheduler.DAGScheduler: Final stage: Stage 0(collect
//| at group.scala:23)
//| 14/12/16 16:44:59 INFO scheduler.DAGScheduler: Parents of final stage: List(
//| )
//| 14/12/16 16:44:59 INFO scheduler.DAGScheduler: Missing parents: List()
//| 14/12/16 16:44:59 INFO scheduler.DAGScheduler: Submitting Stage 0 (MappedRDD
//| [1] at map at group.scala:21), which has no missing parents
//| 14/12/16 16:44:59 WARN util.SizeEstimator: Failed to check whether UseCompre
//| ssedOops is set; assuming yes
//| 14/12/16 16:44:59 INFO storage.MemoryStore: ensureFreeSpace(1584) called wit
//| h curMem=0, maxMem=140142182
//| 14/12/16 16:44:59 INFO storage.MemoryStore: Block broadcast_0 stored as valu
//| es in memory (estimated size 1584.0 B
//| Output exceeds cutoff limit.
m1.foreach { case (e, i) => println(e + "," + i) }
//> a,1.0
//| a,3.0
//| a,2.0
}
答案 0 :(得分:2)
嗨,使用@Imm解决方案,您的值将不会被排序,如果它发生将是一个伤亡者。 要获得排序列表,您只需添加:
val r4 = r3.mapValues(_。toList.sorted) 所以r4将有一个rdd,每个值列表将按每个键排序
我希望这会有用
答案 1 :(得分:1)
使用groupByKey
:
val r3: RDD[String, Iterable[Double]] = r2.groupByKey
如果您确实希望第二个元素是List
而不是一般Iterable
,那么您可以使用mapValues
val r4 = r3.mapValues(_.toList)
请确保import org.apache.spark.SparkContext._
位于顶部,以便这些功能可用。