Spark中的shuffle算子

最简单的方法，不用查文件

It is actually extremely easy to find this out, without the documentation. For any of these functions just create an RDD and call to debug string, here is one example you can do the rest on ur own.

scala> val a  = sc.parallelize(Array(1,2,3)).distinct
scala> a.toDebugString
MappedRDD[5] at distinct at <console>:12 (1 partitions)
  MapPartitionsRDD[4] at distinct at <console>:12 (1 partitions)
    **ShuffledRDD[3] at distinct at <console>:12 (1 partitions)**
      MapPartitionsRDD[2] at distinct at <console>:12 (1 partitions)
        MappedRDD[1] at distinct at <console>:12 (1 partitions)
          ParallelCollectionRDD[0] at parallelize at <console>:12 (1 partitions)

So as you can see distinct creates a shuffle. It is also particularly important to find out this way rather than docs because there are situations where a shuffle will be required or not required for a certain function. For example join usually requires a shuffle but if you join two RDD's that branch from the same RDD spark can sometimes elide the shuffle.

当然，也可列一些常用的出来：

去重
def distinct()
def distinct(numPartitions: Int)

聚合
def reduceByKey(func: (V, V) => V, numPartitions: Int): RDD[(K, V)]
def reduceByKey(partitioner: Partitioner, func: (V, V) => V): RDD[(K, V)]
def groupBy[K](f: T => K, p: Partitioner):RDD[(K, Iterable[V])]
def groupByKey(partitioner: Partitioner):RDD[(K, Iterable[V])]
def aggregateByKey[U: ClassTag](zeroValue: U, partitioner: Partitioner): RDD[(K, U)]
def aggregateByKey[U: ClassTag](zeroValue: U, numPartitions: Int): RDD[(K, U)]
def combineByKey[C](createCombiner: V => C, mergeValue: (C, V) => C, mergeCombiners: (C, C) => C): RDD[(K, C)]
def combineByKey[C](createCombiner: V => C, mergeValue: (C, V) => C, mergeCombiners: (C, C) => C, numPartitions: Int): RDD[(K, C)]
def combineByKey[C](createCombiner: V => C, mergeValue: (C, V) => C, mergeCombiners: (C, C) => C, partitioner: Partitioner, mapSideCombine: Boolean = true, serializer: Serializer = null): RDD[(K, C)]

排序
def sortByKey(ascending: Boolean = true, numPartitions: Int = self.partitions.length): RDD[(K, V)]
def sortBy[K](f: (T) => K, ascending: Boolean = true, numPartitions: Int = this.partitions.length)(implicit ord: Ordering[K], ctag: ClassTag[K]): RDD[T]

重分区
def coalesce(numPartitions: Int, shuffle: Boolean = false, partitionCoalescer: Option[PartitionCoalescer] = Option.empty)
def repartition(numPartitions: Int)(implicit ord: Ordering[T] = null)

集合或者表操作
def intersection(other: RDD[T]): RDD[T]
def intersection(other: RDD[T], partitioner: Partitioner)(implicit ord: Ordering[T] = null): RDD[T]
def intersection(other: RDD[T], numPartitions: Int): RDD[T]
def subtract(other: RDD[T], numPartitions: Int): RDD[T]
def subtract(other: RDD[T], p: Partitioner)(implicit ord: Ordering[T] = null): RDD[T]
def subtractByKey[W: ClassTag](other: RDD[(K, W)]): RDD[(K, V)]
def subtractByKey[W: ClassTag](other: RDD[(K, W)], numPartitions: Int): RDD[(K, V)]
def subtractByKey[W: ClassTag](other: RDD[(K, W)], p: Partitioner): RDD[(K, V)]
def join[W](other: RDD[(K, W)], partitioner: Partitioner): RDD[(K, (V, W))]
def join[W](other: RDD[(K, W)]): RDD[(K, (V, W))]
def join[W](other: RDD[(K, W)], numPartitions: Int): RDD[(K, (V, W))]
def leftOuterJoin[W](other: RDD[(K, W)]): RDD[(K, (V, Option[W]))]