spark编程模型(十五)之RDD键值转换操作(Transformation Operation)——cogroup、join

cogroup

  • 参数为1个RDD

    • def cogroup[W](other: RDD[(K, W)]): RDD[(K, (Iterable[V], Iterable[W]))]
    • def cogroup[W](other: RDD[(K, W)], numPartitions: Int): RDD[(K, (Iterable[V], Iterable[W]))]
    • def cogroup[W](other: RDD[(K, W)], partitioner: Partitioner): RDD[(K, (Iterable[V], Iterable[W]))]
  • 参数为2个RDD

    • def cogroup[W1, W2](other1: RDD[(K, W1)], other2: RDD[(K, W2)]): RDD[(K, (Iterable[V], Iterable[W1], Iterable[W2]))]
    • def cogroup[W1, W2](other1: RDD[(K, W1)], other2: RDD[(K, W2)], numPartitions: Int): RDD[(K, (Iterable[V], Iterable[W1], Iterable[W2]))]
    • def cogroup[W1, W2](other1: RDD[(K, W1)], other2: RDD[(K, W2)], partitioner: Partitioner): RDD[(K, (Iterable[V], Iterable[W1], Iterable[W2]))]
  • 参数为3个RDD

    • def cogroup[W1, W2, W3](other1: RDD[(K, W1)], other2: RDD[(K, W2)], other3: RDD[(K, W3)]): RDD[(K, (Iterable[V], Iterable[W1], Iterable[W2], Iterable[W3]))]
    • def cogroup[W1, W2, W3](other1: RDD[(K, W1)], other2: RDD[(K, W2)], other3: RDD[(K, W3)], numPartitions: Int): RDD[(K, (Iterable[V], Iterable[W1], Iterable[W2], Iterable[W3]))]
    • def cogroup[W1, W2, W3](other1: RDD[(K, W1)], other2: RDD[(K, W2)], other3: RDD[(K, W3)], partitioner: Partitioner): RDD[(K, (Iterable[V], Iterable[W1], Iterable[W2], Iterable[W3]))]
  • cogroup相当于SQL中的全外关联full outer join,返回左右RDD中的记录,关联不上的为空

  • 参数numPartitions用于指定结果的分区数

  • 参数partitioner用于指定分区函数

    var rdd1 = sc.makeRDD(Array(("A","1"),("B","2"),("C","3")),2)
    var rdd2 = sc.makeRDD(Array(("A","a"),("C","c"),("D","d")),2)

    scala> var rdd3 = rdd1.cogroup(rdd2)
    rdd3: org.apache.spark.rdd.RDD[(String, (Iterable[String], Iterable[String]))] = MapPartitionsRDD[12] at cogroup at :25

    scala> rdd3.partitions.size
    res3: Int = 2

    scala> rdd3.collect
    res1: Array[(String, (Iterable[String], Iterable[String]))] = Array(
    (B,(CompactBuffer(2),CompactBuffer())),
    (D,(CompactBuffer(),CompactBuffer(d))),
    (A,(CompactBuffer(1),CompactBuffer(a))),
    (C,(CompactBuffer(3),CompactBuffer(c)))
    )

    scala> var rdd4 = rdd1.cogroup(rdd2,3)
    rdd4: org.apache.spark.rdd.RDD[(String, (Iterable[String], Iterable[String]))] = MapPartitionsRDD[14] at cogroup at :25

    scala> rdd4.partitions.size
    res5: Int = 3

    scala> rdd4.collect
    res6: Array[(String, (Iterable[String], Iterable[String]))] = Array(
    (B,(CompactBuffer(2),CompactBuffer())),
    (C,(CompactBuffer(3),CompactBuffer(c))),
    (A,(CompactBuffer(1),CompactBuffer(a))),
    (D,(CompactBuffer(),CompactBuffer(d))))

    var rdd1 = sc.makeRDD(Array(("A","1"),("B","2"),("C","3")),2)
    var rdd2 = sc.makeRDD(Array(("A","a"),("C","c"),("D","d")),2)
    var rdd3 = sc.makeRDD(Array(("A","A"),("E","E")),2)

    scala> var rdd4 = rdd1.cogroup(rdd2,rdd3)
    rdd4: org.apache.spark.rdd.RDD[(String, (Iterable[String], Iterable[String], Iterable[String]))] =
    MapPartitionsRDD[17] at cogroup at :27

    scala> rdd4.partitions.size
    res7: Int = 2

    scala> rdd4.collect
    res9: Array[(String, (Iterable[String], Iterable[String], Iterable[String]))] = Array(
    (B,(CompactBuffer(2),CompactBuffer(),CompactBuffer())),
    (D,(CompactBuffer(),CompactBuffer(d),CompactBuffer())),
    (A,(CompactBuffer(1),CompactBuffer(a),CompactBuffer(A))),
    (C,(CompactBuffer(3),CompactBuffer(c),CompactBuffer())),
    (E,(CompactBuffer(),CompactBuffer(),CompactBuffer(E))))

join

  • 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 join[W](other: RDD[(K, W)], partitioner: Partitioner): RDD[(K, (V, W))]

  • join相当于SQL中的内关联join,只返回两个RDD根据K可以关联上的结果,join只能用于两个RDD之间的关联,如果要多个RDD关联,多关联几次即可

  • 参数numPartitions用于指定结果的分区数

  • 参数partitioner用于指定分区函数

    var rdd1 = sc.makeRDD(Array(("A","1"),("B","2"),("C","3")),2)
    var rdd2 = sc.makeRDD(Array(("A","a"),("C","c"),("D","d")),2)

    scala> rdd1.join(rdd2).collect
    res10: Array[(String, (String, String))] = Array((A,(1,a)), (C,(3,c)))

posted @ 2018-08-11 01:26  oldsix666  阅读(152)  评论(0)    收藏  举报