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Spark算子执行流程详解之三

2017-03-03 09:31:22      个评论      
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10.aggregate

用与聚合RDD中的元素,先使用seqOp将RDD中每个分区中的T类型元素聚合成U类型,再使用combOp将之前每个分区聚合后的U类型聚合成U类型,特别注意seqOp和combOp都会使用zeroValue的值,zeroValue的类型为U,

def aggregate[U: ClassTag](zeroValue:U)(seqOp: (U,T) => U, combOp: (U,U) => U): U = withScope {
// Clone the zero value since we will also be serializing it as part of tasks
var jobResult = Utils.clone(zeroValue, sc.env.serializer.newInstance())
val cleanSeqOp = sc.clean(seqOp)
val cleanCombOp = sc.clean(combOp)

// zeroValue即初始值,aggregatePartition是在excutor上执行的

val aggregatePartition = (it:Iterator[T]) => it.aggregate(zeroValue)(cleanSeqOp, cleanCombOp)

// jobResult即初始值,其合并每个分区的结果是在driver端执行的
val mergeResult = (index: Int, taskResult:U) => jobResult = combOp(jobResult, taskResult)

sc.runJob(this, aggregatePartition, mergeResult)
jobResult
}

例如:

var rdd1 = sc.makeRDD(1 to 10,2)

##第一个分区中包含5,4,3,2,1

##第二个分区中包含10,9,8,7,6

scala> rdd1.aggregate(1)(

| {(x : Int,y : Int) => x + y},

| {(a : Int,b : Int) => a + b}

| )

res17: Int = 58

为什么是58呢?且看下面的执行流程:

\

11.fold

简化的aggregate,将aggregate中的seqOp和combOp使用同一个函数op。

/**
* Aggregate the elements of each partition, and then the results for all the partitions, using a
* given associative and commutative function and a neutral "zero value". The function
* op(t1, t2) is allowed to modify t1 and return it as its result value to avoid object
* allocation; however, it should not modify t2.
*
* This behaves somewhat differently from fold operations implemented for non-distributed
* collections in functional languages like Scala. This fold operation may be applied to
* partitions individually, and then fold those results into the final result, rather than
* apply the fold to each element sequentially in some defined ordering. For functions
* that are not commutative, the result may differ from that of a fold applied to a
* non-distributed collection.
*/

def fold(zeroValue: T)(op: (T, T) => T): T= withScope {
// Clone the zero value since we will also be serializing it as part of tasks
var jobResult = Utils.clone(zeroValue, sc.env.closureSerializer.newInstance())
val cleanOp = sc.clean(op)

//先在excutor上针对分区进行一次fold操作
val foldPartition = (iter: Iterator[T]) => iter.fold(zeroValue)(cleanOp)

//然后在driver端合并每个分区上的结果
val mergeResult = (index: Int, taskResult:T) => jobResult = op(jobResult, taskResult)
sc.runJob(this, foldPartition, mergeResult)
jobResult

}

例如可以将aggregate小节里面的例子操作转化为fold操作:

scala>var rdd1 = sc.makeRDD(1 to 10,2)

##第一个分区中包含5,4,3,2,1

##第二个分区中包含10,9,8,7,6

scala> rdd1.fold(1)(

| (x,y) => x + y

| )

res19: Int = 58

##结果同上面使用aggregate的第一个例子一样,即:

scala> rdd1.aggregate(1)(

| {(x,y) => x + y},

| {(a,b) => a + b}

| )

res20: Int = 58

 

12.treeAggregate

分层进行aggregate,由于aggregate的时候其分区的结算结果是传输到driver端再进行合并的,如果分区比较多,计算结果返回的数据量比较大的话,那么driver端需要缓存大量的中间结果,这样就会加大driver端的计算能力,因此treeAggregate把分区计算结果的合并仍旧放在excutor端进行,将结果在excutor端不断合并缩小返回driver的数据量,最后再driver端进行最后一次合并。

/**
* Aggregates the elements of this RDD in a multi-level tree pattern.
*
*
@param depth suggested depth of the tree (default: 2)
*
@see [[org.apache.spark.rdd.RDD#aggregate]]
*/

def treeAggregate[U: ClassTag](zeroValue:U)(
seqOp: (U, T) =>U,
combOp: (U, U) =>U,
depth: Int = 2): U = withScope {
require(depth >= 1, s"Depth must be greater than or equal to 1 but got$depth.")
if (partitions.length == 0) {
Utils.clone(zeroValue, context.env.closureSerializer.newInstance())
} else {
val cleanSeqOp = context.clean(seqOp)
val cleanCombOp = context.clean(combOp)

//针对初始分区的聚合函数
val aggregatePartition =
(it: Iterator[T]) => it.aggregate(zeroValue)(cleanSeqOp, cleanCombOp)

//针对初始的各分区先进行部分聚合
var partiallyAggregated = mapPartitions(it =>Iterator(aggregatePartition(it)))
var numPartitions = partiallyAggregated.partitions.length

//根据传入的depth计算出需要迭代计算的程度
val scale = math.max(math.ceil(math.pow(numPartitions,1.0 / depth)).toInt, 2)
// If creating an extra level doesn't help reduce
// the wall-clock time, we stop tree aggregation.

while (numPartitions > scale + numPartitions / scale) {//计算迭代的程度
numPartitions /= scale
val curNumPartitions = numPartitions

//减少分区个数,合并部分分区的结果
partiallyAggregated = partiallyAggregated.mapPartitionsWithIndex {
(i, iter) => iter.map((i % curNumPartitions, _))
}.reduceByKey(new HashPartitioner(curNumPartitions), cleanCombOp).values
}

//执行最后一次reduce,返回最终结果
partiallyAggregated.reduce(cleanCombOp)
}
}

例如:

scala> def seq(a:Int,b:Int):Int={

| a+b}

seq: (a: Int, b: Int)Int

scala> def comb(a:Int,b:Int):Int={

| a+b}

comb: (a: Int, b: Int)Int

val z =sc.parallelize(List(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18),9)

scala> z.treeAggregate(0)(seq,comb,2)

res1: Int = 171

其具体的执行过程如下:

 

\

13.reduce

RDD中元素前两个传给输入函数,产生一个新的return值,新产生的return值与RDD中下一个元素(第三个元素)组成两个元素,再被传给输入函数,直到最后只有一个值为止。

/**
* Reduces the elements of this RDD using the specified commutative and
* associative binary operator.
*/

def reduce(f: (T,T) => T): T = withScope {
val cleanF = sc.clean(f)

//定义一个遍历partition的函数,这是在excutor端执行的
val
reducePartition: Iterator[T] => Option[T] = iter => {
if (iter.hasNext) {

//reduceLeft从左往后遍历
Some(iter.reduceLeft(cleanF))
} else {
None
}
}
var jobResult: Option[T] = None

//定义一个driver端处理分区计算结果的函数,这是在driver端执行的
val
mergeResult = (index: Int, taskResult: Option[T]) => {
if (taskResult.isDefined) {
jobResult = jobResult match {
case Some(value) =>Some(f(value, taskResult.get))
case None => taskResult
}
}
}
sc.runJob(this, reducePartition, mergeResult)
// Get the final result out of our Option, or throw an exception if the RDD was empty

//将结果返回
jobResult.getOrElse(throw new UnsupportedOperationException("empty collection"))
}

例如:

val c = sc.parallelize(1 to 10 , 2)

c.reduce((x, y) => x + y)//结果55

具体执行流程如下:

\

14.max

返回最大值,其排序方法对象默认的排序方法

/**
* Returns the max of this RDD as defined by the implicit Ordering[T].
*
@return the maximum element of the RDD
* */

def max()(implicitord: Ordering[T]):T = withScope {
this.reduce(ord.max)

}

其本质就是定义个排序的方法,然后调用reduce操作,实例如下:

scala>var rdd1 = sc.makeRDD(1 to 10,2)

##第一个分区中包含5,4,3,2,1

##第二个分区中包含10,9,8,7,6

scala> rdd1.max()

res19: Int = 10

其执行流程如下:

\

15.min

返回最小值,其排序方法对象默认的排序方法

/**
* Returns the min of this RDD as defined by the implicit Ordering[T].
*
@return the maximum element of the RDD
* */

def min()(implicitord: Ordering[T]):T = withScope {
this.reduce(ord.min)

}

其本质就是定义个排序的方法,然后调用reduce操作,实例如下:

scala>var rdd1 = sc.makeRDD(1 to 10,2)

##第一个分区中包含5,4,3,2,1

##第二个分区中包含10,9,8,7,6

scala> rdd1.min()

res19: Int = 1

其执行流程如下:

\

16.treeReduce

类似于treeAggregate,利用在excutor端进行多次aggregate来缩小driver的计算开销

/**
* Reduces the elements of this RDD in a multi-level tree pattern.
*
*
@param depth suggested depth of the tree (default: 2)
*
@see [[org.apache.spark.rdd.RDD#reduce]]
*/

def treeReduce(f: (T,T) => T, depth: Int =2): T = withScope {
require(depth >= 1, s"Depth must be greater than or equal to 1 but got$depth.")
val cleanF = context.clean(f)

//针对初始分区的reduce函数
val
reducePartition: Iterator[T] => Option[T] = iter => {
if (iter.hasNext) {
Some(iter.reduceLeft(cleanF))
} else {
None
}
}

//针对初始的各分区先进行部分reduce
val
partiallyReduced = mapPartitions(it =>Iterator(reducePartition(it)))
val op: (Option[T], Option[T]) => Option[T] = (c, x) => {
if (c.isDefined && x.isDefined) {
Some(cleanF(c.get, x.get))
} else if (c.isDefined) {
c
} else if (x.isDefined) {
x
} else {
None
}
}

//最终调用的还是treeAggregate方法
partiallyReduced.treeAggregate(Option.empty[T])(op, op, depth)
.getOrElse(throw new UnsupportedOperationException("empty collection"))

}

treeReduce函数先是针对每个分区利用scala的reduceLeft函数进行计算;最后,在将局部合并的RDD进行treeAggregate计算,这里的seqOp和combOp一样,初值为空。在实际应用中,可以用treeReduce来代替reduce,主要是用于单个reduce操作开销比较大,而treeReduce可以通过调整深度来控制每次reduce的规模。其具体的执行流程不再详细叙述,可以参考treeAggregate方法。


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