1、groupByKey

1）def groupByKey(): RDD[(K, Iterable[V])] 2）def groupByKey(numPartitions: Int): RDD[(K, Iterable[V])] 3）def groupByKey(partitioner: Partitioner): RDD[(K, Iterable[V])]

該函式用於將RDD[K,V]中每個K對應的V值，合併到一個集合Iterable[V]中，引數numPartitions指分割槽數，partitioner指分割槽函式。

scala> var rdd1 = sc.makeRDD(Array(("A",0),("A",2),("B",1),("B",2),("C",1)))
rdd1: org.apache.spark.rdd.RDD[(String, Int)] = ParallelCollectionRDD[89] at makeRDD at :21
 
scala> rdd1.groupByKey().collect
res81: Array[(String, Iterable[Int])] = Array((A,CompactBuffer(0, 2)), (B,CompactBuffer(2, 1)), (C,CompactBuffer(1)))
 

2、reduceByKey

1）def reduceByKey(func: (V, V) => V): RDD[(K, V)] 2）def reduceByKey(func: (V, V) => V, numPartitions: Int): RDD[(K, V)] 3）def reduceByKey(partitioner: Partitioner, func: (V, V) => V): RDD[(K, V)]

該函式將RDD[K,V]中的每個K對應的V值根據對映函式來計算。引數numPartitions指分割槽數，partitioner指分割槽函式。

scala> var rdd1 = sc.makeRDD(Array(("A",0),("A",2),("B",1),("B",2),("C",1)))
rdd1: org.apache.spark.rdd.RDD[(String, Int)] = ParallelCollectionRDD[91] at makeRDD at :21
 
scala> rdd1.partitions.size
res82: Int = 15
 
scala> var rdd2 = rdd1.reduceByKey((x,y) => x + y)
rdd2: org.apache.spark.rdd.RDD[(String, Int)] = ShuffledRDD[94] at reduceByKey at :23
 
scala> rdd2.collect
res85: Array[(String, Int)] = Array((A,2), (B,3), (C,1))
 
scala> rdd2.partitions.size
res86: Int = 15
 
scala> var rdd2 = rdd1.reduceByKey(new org.apache.spark.HashPartitioner(2),(x,y) => x + y)
rdd2: org.apache.spark.rdd.RDD[(String, Int)] = ShuffledRDD[95] at reduceByKey at :23
 
scala> rdd2.collect
res87: Array[(String, Int)] = Array((B,3), (A,2), (C,1))
 
scala> rdd2.partitions.size
res88: Int = 2
 
 

3、reduceByKeyLocally：def reduceByKeyLocally(func: (V, V) => V): Map[K, V]

該函式將RDD[K,V]中每個K對應的V值根據對映函式來運算，運算結果對映到一個Map[K,V]中，而不是RDD[K,V]。

scala> var rdd1 = sc.makeRDD(Array(("A",0),("A",2),("B",1),("B",2),("C",1)))
rdd1: org.apache.spark.rdd.RDD[(String, Int)] = ParallelCollectionRDD[91] at makeRDD at :21
 
scala> rdd1.reduceByKeyLocally((x,y) => x + y)
res90: scala.collection.Map[String,Int] = Map(B -> 3, A -> 2, C -> 1)