Spark修炼之道(进阶篇)——Spark入门到精通:第五节 Spark编程模型(二)

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Spark修炼之道(进阶篇)——Spark入门到精通:第五节 Spark编程模型(二)

周志湖 2015-09-20 23:18:00 浏览2089
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作者:周志湖
网名:摇摆少年梦
微信号:zhouzhihubeyond

本文主要内容

  1. RDD 常用Transformation函数

1. RDD 常用Transformation函数

(1)union
union将两个RDD数据集元素合并,类似两个集合的并集
union函数参数:

 /**
   * 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] 

RDD与另外一个RDD进行Union操作之后,两个数据集中的存在的重复元素
代码如下:

scala> val rdd1=sc.parallelize(1 to 5)
rdd1: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[15] at parallelize at <console>:21

scala> val rdd2=sc.parallelize(4 to 8)
rdd2: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[16] at parallelize at <console>:21
//存在重复元素
scala> rdd1.union(rdd2).collect
res13: Array[Int] = Array(1, 2, 3, 4, 5, 4, 5, 6, 7, 8)

这里写图片描述

(2)intersection
方法返回两个RDD数据集的交集
函数参数:
/**
* Return the intersection of this RDD and another one. The output will not contain any duplicate
* elements, even if the input RDDs did.
*
* Note that this method performs a shuffle internally.
*/
def intersection(other: RDD[T]): RDD[T]

使用示例:

scala> rdd1.intersection(rdd2).collect
res14: Array[Int] = Array(4, 5)

(3)distinct
distinct函数将去除重复元素
distinct函数参数:

/**
* Return a new RDD containing the distinct elements in this RDD.
*/
def distinct(): RDD[T]

scala> val rdd1=sc.parallelize(1 to 5)
rdd1: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[0] at parallelize at <console>:21

scala> val rdd2=sc.parallelize(4 to 8)
rdd2: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[1] at parallelize at <console>:21

scala> rdd1.union(rdd2).distinct.collect
res0: Array[Int] = Array(6, 1, 7, 8, 2, 3, 4, 5)

这里写图片描述

(4)groupByKey([numTasks])
输入数据为(K, V) 对, 返回的是 (K, Iterable) ,numTasks指定task数量,该参数是可选的,下面给出的是无参数的groupByKey方法
/**
* Group the values for each key in the RDD into a single sequence. Hash-partitions the
* resulting RDD with the existing partitioner/parallelism level. 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 groupByKey(): RDD[(K, Iterable[V])]

scala> val rdd1=sc.parallelize(1 to 5)
rdd1: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[0] at parallelize at <console>:21

scala> val rdd2=sc.parallelize(4 to 8)
rdd2: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[1] at parallelize at <console>:21

scala> rdd1.union(rdd2).map((_,1)).groupByKey.collect
res2: Array[(Int, Iterable[Int])] = Array((6,CompactBuffer(1)), (1,CompactBuffer(1)), (7,CompactBuffer(1)), (8,CompactBuffer(1)), (2,CompactBuffer(1)), (3,CompactBuffer(1)), (4,CompactBuffer(1, 1)), (5,CompactBuffer(1, 1)))

(5)reduceByKey(func, [numTasks])
reduceByKey函数输入数据为(K, V)对,返回的数据集结果也是(K,V)对,只不过V为经过聚合操作后的值
/**
* Merge the values for each key using an associative reduce function. This will also perform
* the merging locally on each mapper before sending results to a reducer, similarly to a
* “combiner” in MapReduce. Output will be hash-partitioned with numPartitions partitions.
*/
def reduceByKey(func: (V, V) => V, numPartitions: Int): RDD[(K, V)]

使用示例:

scala> rdd1.union(rdd2).map((_,1)).reduceByKey(_+_).collect
res4: Array[(Int, Int)] = Array((6,1), (1,1), (7,1), (8,1), (2,1), (3,1), (4,2), (5,2))

这里写图片描述

(6)sortByKey([ascending], [numTasks])
对输入的数据集按key排序
sortByKey方法定义

/**
   * Sort the RDD by key, so that each partition contains a sorted range of the elements. Calling
   * `collect` or `save` on the resulting RDD will return or output an ordered list of records
   * (in the `save` case, they will be written to multiple `part-X` files in the filesystem, in
   * order of the keys).
   */
  // TODO: this currently doesn't work on P other than Tuple2!
  def sortByKey(ascending: Boolean = true, numPartitions: Int = self.partitions.length)
      : RDD[(K, V)]

使用示例:

scala> var data = sc.parallelize(List((1,3),(1,2),(1, 4),(2,3),(7,9),(2,4)))
data: org.apache.spark.rdd.RDD[(Int, Int)] = ParallelCollectionRDD[20] at parallelize at <console>:21

scala> data.sortByKey(true).collect
res10: Array[(Int, Int)] = Array((1,3), (1,2), (1,4), (2,3), (2,4), (7,9))

这里写图片描述

(7)join(otherDataset, [numTasks])
对于数据集类型为 (K, V) 及 (K, W)的RDD,join操作后返回类型为 (K, (V, W)),join函数有三种:
def join[W](other: RDD[(K, W)], partitioner: Partitioner): RDD[(K, (V, W))]
def leftOuterJoin[W](
other: RDD[(K, W)],
partitioner: Partitioner): RDD[(K, (V, Option[W]))]
def rightOuterJoin[W](other: RDD[(K, W)], partitioner: Partitioner)
: RDD[(K, (Option[V], W))]

使用示例:

scala> val rdd1=sc.parallelize(Array((1,2),(1,3))
     | )
rdd1: org.apache.spark.rdd.RDD[(Int, Int)] = ParallelCollectionRDD[24] at parallelize at <console>:21

scala> val rdd2=sc.parallelize(Array((1,3)))
rdd2: org.apache.spark.rdd.RDD[(Int, Int)] = ParallelCollectionRDD[32] at parallelize at <console>:21

scala> rdd1.join(rdd2).collect
res13: Array[(Int, (Int, Int))] = Array((1,(3,3)), (1,(2,3)))

这里写图片描述

scala> rdd1.leftOuterJoin(rdd2).collect
res15: Array[(Int, (Int, Option[Int]))] = Array((1,(3,Some(3))), (1,(2,Some(3))))

这里写图片描述

scala> rdd1.rightOuterJoin(rdd2).collect
res16: Array[(Int, (Option[Int], Int))] = Array((1,(Some(3),3)), (1,(Some(2),3)))

这里写图片描述

(8)cogroup(otherDataset, [numTasks])
如果输入的RDD类型为(K, V) 和(K, W),则返回的RDD类型为 (K, (Iterable, Iterable)) . 该操作与 groupWith等同

方法定义:
/**
* For each key k in this or other, return a resulting RDD that contains a tuple with the
* list of values for that key in this as well as other.
*/
def cogroup[W](other: RDD[(K, W)], partitioner: Partitioner)
: RDD[(K, (Iterable[V], Iterable[W]))]

scala> val rdd1=sc.parallelize(Array((1,2),(1,3))
     | )
rdd1: org.apache.spark.rdd.RDD[(Int, Int)] = ParallelCollectionRDD[24] at parallelize at <console>:21

scala> val rdd2=sc.parallelize(Array((1,3)))
rdd2: org.apache.spark.rdd.RDD[(Int, Int)] = ParallelCollectionRDD[32] at parallelize at <console>:21

scala> rdd1.cogroup(rdd2).collect
res17: Array[(Int, (Iterable[Int], Iterable[Int]))] = Array((1,(CompactBuffer(3, 2),CompactBuffer(3))))

scala> rdd1.groupWith(rdd2).collect
res18: Array[(Int, (Iterable[Int], Iterable[Int]))] = Array((1,(CompactBuffer(2, 3),CompactBuffer(3))))

这里写图片描述

(9)cartesian(otherDataset)
求两个RDD数据集间的笛卡尔积
函数定义:
/**
* Return the Cartesian product of this RDD and another one, that is, the RDD of all pairs of
* elements (a, b) where a is in this and b is in other.
*/
def cartesian[U: ClassTag](other: RDD[U]): RDD[(T, U)]

scala> val rdd1=sc.parallelize(Array(1,2,3,4))
rdd1: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[52] at parallelize at <console>:21

scala> val rdd2=sc.parallelize(Array(5,6))
rdd2: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[53] at parallelize at <console>:21

scala> rdd1.cartesian(rdd2).collect
res21: Array[(Int, Int)] = Array((1,5), (1,6), (2,5), (2,6), (3,5), (4,5), (3,6), (4,6))

这里写图片描述

(10)coalesce(numPartitions)
将RDD的分区数减至指定的numPartitions分区数

函数定义:
/**
* 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.
*
* 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.
*/
def coalesce(numPartitions: Int, shuffle: Boolean = false)(implicit ord: Ordering[T] = null)
: RDD[T]

示例代码:

scala> val rdd1=sc.parallelize(1 to 100,3)
rdd1: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[55] at parallelize at <console>:21

scala> val rdd2=rdd1.coalesce(2)
rdd2: org.apache.spark.rdd.RDD[Int] = CoalescedRDD[56] at coalesce at <console>:23

这里写图片描述

repartition(numPartitions),功能与coalesce函数相同,实质上它调用的就是coalesce函数,只不是shuffle = true,意味着可能会导致大量的网络开销。
方法定义:
/**
* 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)
}

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