def coalesce(numPartitions: Int, shuffle: Boolean = false, partitionCoalescer: Option[PartitionCoalescer] = Option.empty)(implicit ord: Ordering[(K, V)] = null): RDD[(K, V)]Return a new RDD that is reduced into
numPartitionspartitions.This results in a narrow dependency, e.g. if you go from 1000 partitions to 100 partitions, there will not be a shuffle, instead each of the 100 new partitions will claim 10 of the current partitions. If a larger number of partitions is requested, it will stay at the current number of partitions.
However, if you're doing a drastic coalesce, e.g. to numPartitions = 1, this may result in your computation taking place on fewer nodes than you like (e.g. one node in the case of numPartitions = 1). To avoid this, you can pass shuffle = true. This will add a shuffle step, but means the current upstream partitions will be executed in parallel (per whatever the current partitioning is).
译文：
返回一个经过简化到numPartitions个分区的新RDD。这会导致一个窄依赖，例如：你将1000个分区转换成100个分区，这个过程不会发生shuffle，相反如果10个分区转换成100个分区将会发生shuffle。然而如果你想大幅度合并分区，例如合并成一个分区，这会导致你的计算在少数几个集群节点上计算（言外之意：并行度不够）。为了避免这种情况，你可以将第二个shuffle参数传递一个true，这样会在重新分区过程中多一步shuffle，这意味着上游的分区可以并行运行。
注意：
第二个参数shuffle=true，将会产生多于之前的分区数目，例如你有一个个数较少的分区，假如是100，调用coalesce(1000, shuffle = true)将会使用一个 HashPartitioner产生1000个分区分布在集群节点上。这个（对于提高并行度）是非常有用的。
def repartition(numPartitions: Int)(implicit ord: Ordering[(K, V)] = null): RDD[(K, V)]Return a new RDD that has exactly numPartitions partitions.
Can increase or decrease the level of parallelism in this RDD. Internally, this uses a shuffle to redistribute data.
If you are decreasing the number of partitions in this RDD, consider using
coalesce, which can avoid performing a shuffle.TODO Fix the Shuffle+Repartition data loss issue described in SPARK-23207.
译文：
返回一个恰好有numPartitions个分区的RDD，可以增加或者减少此RDD的并行度。内部，这将使用shuffle重新分布数据，如果你减少分区数，考虑使用coalesce，这样可以避免执行shuffle
参考:https://blog.csdn.net/dax1n/article/details/53431373
coalesce与repartition：重分区
（*）都是重分区
（*）区别：coalesce 默认不会进行shuffle(false)
repartition 就会进行shuffle
（*）举例：
val rdd1 = sc.parallelize(List(1,2,3,4,5,6,7,8,9), 2)
查看分区个数：rdd1.partitions.length
重新分区: val rdd2 = rdd1.repartition(3)
val rdd3 = rdd1.coalesce(3,false) ---> 分区数：2
val rdd4 = rdd1.coalesce(3,true) ---> 分区数：3
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