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 
     
      :21scala> val rdd2=sc.parallelize(4 to 8)rdd2: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[16] at parallelize at 
      
       :21//存在重复元素scala> rdd1.union(rdd2).collectres13: Array[Int] = Array(1, 2, 3, 4, 5, 4, 5, 6, 7, 8)
      
     

（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).collectres14: 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 
     
      :21scala> val rdd2=sc.parallelize(4 to 8)rdd2: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[1] at parallelize at 
      
       :21scala> rdd1.union(rdd2).distinct.collectres0: Array[Int] = Array(6, 1, 7, 8, 2, 3, 4, 5)
      
     

（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 
     
      :21scala> val rdd2=sc.parallelize(4 to 8)rdd2: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[1] at parallelize at 
      
       :21scala> rdd1.union(rdd2).map((_,1)).groupByKey.collectres2: 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(_+_).collectres4: Array[(Int, Int)] = Array((6,1), (1,1), (7,1), (8,1), (2,1), (3,1), (4,2), (5,2))

（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 
     
      :21scala> data.sortByKey(true).collectres10: Array[(Int, Int)] = Array((1,3), (1,2), (1,4), (2,3), (2,4), (7,9))
     

（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 
     
      :21scala> val rdd2=sc.parallelize(Array((1,3)))rdd2: org.apache.spark.rdd.RDD[(Int, Int)] = ParallelCollectionRDD[32] at parallelize at 
      
       :21scala> rdd1.join(rdd2).collectres13: Array[(Int, (Int, Int))] = Array((1,(3,3)), (1,(2,3)))
      
     

scala> rdd1.leftOuterJoin(rdd2).collectres15: Array[(Int, (Int, Option[Int]))] = Array((1,(3,Some(3))), (1,(2,Some(3))))

scala> rdd1.rightOuterJoin(rdd2).collectres16: Array[(Int, (Option[Int], Int))] = Array((1,(Some(3),3)), (1,(Some(2),3)))

（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 
     
      :21scala> val rdd2=sc.parallelize(Array((1,3)))rdd2: org.apache.spark.rdd.RDD[(Int, Int)] = ParallelCollectionRDD[32] at parallelize at 
      
       :21scala> rdd1.cogroup(rdd2).collectres17: Array[(Int, (Iterable[Int], Iterable[Int]))] = Array((1,(CompactBuffer(3, 2),CompactBuffer(3))))scala> rdd1.groupWith(rdd2).collectres18: Array[(Int, (Iterable[Int], Iterable[Int]))] = Array((1,(CompactBuffer(2, 3),CompactBuffer(3))))
      
     

（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 
     
      :21scala> val rdd2=sc.parallelize(Array(5,6))rdd2: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[53] at parallelize at 
      
       :21scala> rdd1.cartesian(rdd2).collectres21: Array[(Int, Int)] = Array((1,5), (1,6), (2,5), (2,6), (3,5), (4,5), (3,6), (4,6))
      
     

（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 
     
      :21scala> val rdd2=rdd1.coalesce(2)rdd2: org.apache.spark.rdd.RDD[Int] = CoalescedRDD[56] at coalesce at 
      
       :23
      
     

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) }