理解Spark RDD中的aggregate函数-JobPlus

def aggregate[U](zeroValue: U)(seqOp: (U, T) ⇒ U, combOp: (U, U) ⇒ U)(implicit arg0: ClassTag[U]): U

Aggregate the elements of each partition, and then the results for all the partitions, using given combine functions and a neutral “zero value”. This function can return a different result type, U, than the type of this RDD, T. Thus, we need one operation for merging a T into an U and one operation for merging two U’s, as in scala.TraversableOnce. Both of these functions are allowed to modify and return their first argument instead of creating a new U to avoid memory allocation. seqOp操作会聚合各分区中的元素，然后combOp操作把所有分区的聚合结果再次聚合，两个操作的初始值都是zeroValue. seqOp的操作是遍历分区中的所有元素(T)，第一个T跟zeroValue做操作，结果再作为与第二个T做操作的zeroValue，直到遍历完整个分区。combOp操作是把各分区聚合的结果，再聚合。aggregate函数返回一个跟RDD不同类型的值。因此，需要一个操作seqOp来把分区中的元素T合并成一个U，另外一个操作combOp把所有U聚合。

zeroValue
the initial value for the accumulated result of each partition for the seqOp operator, and also the initial value for the combine results from different partitions for the combOp operator - this will typically be the neutral element (e.g. Nil for list concatenation or 0 for summation)
seqOp
an operator used to accumulate results within a partition
combOp
an associative operator used to combine results from different partitions

举个例子。假如List(1,2,3,4,5,6,7,8,9,10)，对List求平均数，使用aggregate可以这样操作。

C:\Windows\System32>scala
Welcome to Scala 2.11.8 (Java HotSpot(TM) Client VM, Java 1.8.0_91).
Type in expressions for evaluation. Or try :help.

scala> val rdd = List(1,2,3,4,5,6,7,8,9)
rdd: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9)

scala> rdd.par.aggregate((0,0))(

(acc,number) => (acc._1 + number, acc._2 + 1),

(par1,par2) => (par1._1 + par2._1, par1._2 + par2._2)

)
res0: (Int, Int) = (45,9)

scala> res0._1 / res0._2
res1: Int = 5

过程大概这样：

首先，初始值是(0,0)，这个值在后面2步会用到。

然后，(acc,number) => (acc._1 + number, acc._2 + 1)，number即是函数定义中的T，这里即是List中的元素。所以acc._1 + number, acc._2 + 1的过程如下。

1. 0+1, 0+1

2. 1+2, 1+1

3. 3+3, 2+1

4. 6+4, 3+1

5. 10+5, 4+1

6. 15+6, 5+1

7. 21+7, 6+1

8. 28+8, 7+1

9. 36+9, 8+1

结果即是(45,9)。这里演示的是单线程计算过程，实际Spark执行中是分布式计算，可能会把List分成多个分区，假如3个，p1(1,2,3,4)，p2(5,6,7,8)，p3(9)，经过计算各分区的的结果（10,4），（26,4），（9,1），这样，执行(par1,par2) => (par1._1 + par2._1, par1._2 + par2._2)就是（10+26+9,4+4+1）即（45,9）.再求平均值就简单了。

def aggregate[U](zeroValue: U)(seqOp: (U, T) ⇒ U, combOp: (U, U) ⇒ U)(implicit arg0: ClassTag[U]): U Aggregate the elements of each partition, and then the results for all the partitions, using given combine functions and a neutral “zero value”. This function can return a different result type, U, than the type of this RDD, T. Thus, we need one operation for merging a T into an U and one operation for merging two U’s, as in scala.TraversableOnce. Both of these functions are allowed to modify and return their first argument instead of creating a new U to avoid memory allocation.   seqOp操作会聚合各分区中的元素，然后combOp操作把所有分区的聚合结果再次聚合，两个操作的初始值都是zeroValue.   seqOp的操作是遍历分区中的所有元素(T)，第一个T跟zeroValue做操作，结果再作为与第二个T做操作的zeroValue，直到遍历完整个分区。combOp操作是把各分区聚合的结果，再聚合。aggregate函数返回一个跟RDD不同类型的值。因此，需要一个操作seqOp来把分区中的元素T合并成一个U，另外一个操作combOp把所有U聚合。<ul><li>zeroValue</li><li>the initial value for the accumulated result of each partition for the seqOp operator, and also the initial value for the combine results from different partitions for the combOp operator - this will typically be the neutral element (e.g. Nil for list concatenation or 0 for summation)</li><li>seqOp</li><li>an operator used to accumulate results within a partition</li><li>combOp</li><li>an associative operator used to combine results from different partitions</li></ul>举个例子。假如List(1,2,3,4,5,6,7,8,9,10)，对List求平均数，使用aggregate可以这样操作。C:\Windows\System32>scala Welcome to Scala 2.11.8 (Java HotSpot(TM) Client VM, Java 1.8.0_91). Type in expressions for evaluation. Or try :help. scala> val rdd = List(1,2,3,4,5,6,7,8,9) rdd: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9)scala> rdd.par.aggregate((0,0))((acc,number) => (acc._1 + number, acc._2 + 1),(par1,par2) => (par1._1 + par2._1, par1._2 + par2._2)) res0: (Int, Int) = (45,9)scala> res0._1 / res0._2 res1: Int = 5 过程大概这样：首先，初始值是(0,0)，这个值在后面2步会用到。然后，(acc,number) => (acc._1 + number, acc._2 + 1)，number即是函数定义中的T，这里即是List中的元素。所以acc._1 + number, acc._2 + 1的过程如下。1.   0+1,  0+12.  1+2,  1+13.  3+3,  2+14.  6+4,  3+15.  10+5,  4+16.  15+6,  5+17.  21+7,  6+18.  28+8,  7+19.  36+9,  8+1 结果即是(45,9)。这里演示的是单线程计算过程，实际Spark执行中是分布式计算，可能会把List分成多个分区，假如3个，p1(1,2,3,4)，p2(5,6,7,8)，p3(9)，经过计算各分区的的结果（10,4），（26,4），（9,1），这样，执行(par1,par2) => (par1._1 + par2._1, par1._2 + par2._2)就是（10+26+9,4+4+1）即（45,9）.再求平均值就简单了。