spark源码分析

RDD之前之前的分区为3个，然后每个分区都要取出60个样例数据，不足60的全部取出，总共取出180个数据，然后这180个数据进行排序，然后用这个180个数据再均匀分出3个分区，然后开始遍历数据分别进入不同的分区，然后再分区内排序

 /**
   * Persist this RDD with the default storage level (`MEMORY_ONLY`).
   */
  def persist(): this.type = persist(StorageLevel.MEMORY_ONLY)

  /**
   * Persist this RDD with the default storage level (`MEMORY_ONLY`).
   */
  def cache(): this.type = persist()

 /**
   * Return a new RDD containing the distinct elements in this RDD.
   */
  def distinct(numPartitions: Int)(implicit ord: Ordering[T] = null): RDD[T] = withScope {
    map(x => (x, null)).reduceByKey((x, y) => x, numPartitions).map(_._1)
  }

/**
   * Return a new RDD containing the distinct elements in this RDD.
   */
  def distinct(): RDD[T] = withScope {
    distinct(partitions.length)
  }

/**
   * 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.
   */
  def repartition(numPartitions: Int)(implicit ord: Ordering[T] = null): RDD[T] = withScope {
    coalesce(numPartitions, shuffle = true)
  }

 /**
   * Return a new RDD that is reduced into `numPartitions` partitions.
   *
   * 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).
   *
   * @note With shuffle = true, you can actually coalesce to a larger number
   * of partitions. This is useful if you have a small number of partitions,
   * say 100, potentially with a few partitions being abnormally large. Calling
   * coalesce(1000, shuffle = true) will result in 1000 partitions with the
   * data distributed using a hash partitioner. The optional partition coalescer
   * passed in must be serializable.
   */
  def coalesce(numPartitions: Int, shuffle: Boolean = false,
               partitionCoalescer: Option[PartitionCoalescer] = Option.empty)
              (implicit ord: Ordering[T] = null)
      : RDD[T] = withScope {
    require(numPartitions > 0, s"Number of partitions ($numPartitions) must be positive.")
    if (shuffle) {
      /** Distributes elements evenly across output partitions, starting from a random partition. */
      val distributePartition = (index: Int, items: Iterator[T]) => {
        var position = (new Random(index)).nextInt(numPartitions)
        items.map { t =>
          // Note that the hash code of the key will just be the key itself. The HashPartitioner
          // will mod it with the number of total partitions.
          position = position + 1
          (position, t)
        }
      } : Iterator[(Int, T)]

      // include a shuffle step so that our upstream tasks are still distributed
      new CoalescedRDD(
        new ShuffledRDD[Int, T, T](mapPartitionsWithIndex(distributePartition),
        new HashPartitioner(numPartitions)),
        numPartitions,
        partitionCoalescer).values
    } else {
      new CoalescedRDD(this, numPartitions, partitionCoalescer)
    }
  }

  /**
   * Return the union of this RDD and another one. Any identical elements will appear multiple
   * times (use `.distinct()` to eliminate them).
   */
  def union(other: RDD[T]): RDD[T] = withScope {
    sc.union(this, other)
  }

  /**
   * Return the union of this RDD and another one. Any identical elements will appear multiple
   * times (use `.distinct()` to eliminate them).
   */
  def ++(other: RDD[T]): RDD[T] = withScope {
    this.union(other)
  }

  /**
   * Return the union of this RDD and another one. Any identical elements will appear multiple
   * times (use `.distinct()` to eliminate them).
   */
  def union(other: RDD[T]): RDD[T] = withScope {
    sc.union(this, other)
  }

  /**
   * Return the union of this RDD and another one. Any identical elements will appear multiple
   * times (use `.distinct()` to eliminate them).
   */
  def ++(other: RDD[T]): RDD[T] = withScope {
    this.union(other)
  }

  /**
   * Return this RDD sorted by the given key function.
   */
  def sortBy[K](
      f: (T) => K,
      ascending: Boolean = true,
      numPartitions: Int = this.partitions.length)
      (implicit ord: Ordering[K], ctag: ClassTag[K]): RDD[T] = withScope {
    this.keyBy[K](f)
        .sortByKey(ascending, numPartitions)
        .values
  }

该函数是将RDD中每一个分区中各个元素合并成一个Array，这样每一个分区就只有一个数组元素。

  /**
   * Return an RDD created by coalescing all elements within each partition into an array.
   */
  def glom(): RDD[Array[T]] = withScope {
    new MapPartitionsRDD[Array[T], T](this, (context, pid, iter) => Iterator(iter.toArray))
  }

/**
   * Return an RDD of grouped items. Each group consists of a key and a sequence of elements
   * mapping to that key. The ordering of elements within each group is not guaranteed, and
   * may even differ each time the resulting RDD is evaluated.
   *
   * @note This operation may be very expensive. If you are grouping in order to perform an
   * aggregation (such as a sum or average) over each key, using `PairRDDFunctions.aggregateByKey`
   * or `PairRDDFunctions.reduceByKey` will provide much better performance.
   */
  def groupBy[K](f: T => K, p: Partitioner)(implicit kt: ClassTag[K], ord: Ordering[K] = null)
      : RDD[(K, Iterable[T])] = withScope {
    val cleanF = sc.clean(f)
    this.map(t => (cleanF(t), t)).groupByKey(p)
  }

/**
   * Return an array that contains all of the elements in this RDD.
   *
   * @note This method should only be used if the resulting array is expected to be small, as
   * all the data is loaded into the driver's memory.
   */
  def collect(): Array[T] = withScope {
    val results = sc.runJob(this, (iter: Iterator[T]) => iter.toArray)
    Array.concat(results: _*)
  }