kafka consumer 分区reblance算法

Posted sanmutongzi

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了kafka consumer 分区reblance算法相关的知识,希望对你有一定的参考价值。

转载请注明原创地址 http://www.cnblogs.com/dongxiao-yang/p/6238029.html

     最近需要详细研究下kafka reblance过程中分区计算的算法细节,网上搜了部分说法,感觉比较晦涩且不太易懂,还是自己抠源码比较简便一点。

kafka reblance计算部分代码如下:

class RangeAssignor() extends PartitionAssignor with Logging {

  def assign(ctx: AssignmentContext) = {
    val valueFactory = (topic: String) => new mutable.HashMap[TopicAndPartition, ConsumerThreadId]
    val partitionAssignment =
      new Pool[String, mutable.Map[TopicAndPartition, ConsumerThreadId]](Some(valueFactory))
    for (topic <- ctx.myTopicThreadIds.keySet) {
      val curConsumers = ctx.consumersForTopic(topic)
      val curPartitions: Seq[Int] = ctx.partitionsForTopic(topic)

      val nPartsPerConsumer = curPartitions.size / curConsumers.size
      val nConsumersWithExtraPart = curPartitions.size % curConsumers.size

      info("Consumer " + ctx.consumerId + " rebalancing the following partitions: " + curPartitions +
        " for topic " + topic + " with consumers: " + curConsumers)

      for (consumerThreadId <- curConsumers) {
        val myConsumerPosition = curConsumers.indexOf(consumerThreadId)
        assert(myConsumerPosition >= 0)
        val startPart = nPartsPerConsumer * myConsumerPosition + myConsumerPosition.min(nConsumersWithExtraPart)
        val nParts = nPartsPerConsumer + (if (myConsumerPosition + 1 > nConsumersWithExtraPart) 0 else 1)

        /**
         *   Range-partition the sorted partitions to consumers for better locality.
         *  The first few consumers pick up an extra partition, if any.
         */
        if (nParts <= 0)
          warn("No broker partitions consumed by consumer thread " + consumerThreadId + " for topic " + topic)
        else {
          for (i <- startPart until startPart + nParts) {
            val partition = curPartitions(i)
            info(consumerThreadId + " attempting to claim partition " + partition)
            // record the partition ownership decision
            val assignmentForConsumer = partitionAssignment.getAndMaybePut(consumerThreadId.consumer)
            assignmentForConsumer += (TopicAndPartition(topic, partition) -> consumerThreadId)
          }
        }
      }
    }

 

  def getPartitionsForTopics(topics: Seq[String]): mutable.Map[String, Seq[Int]] = {
    getPartitionAssignmentForTopics(topics).map { topicAndPartitionMap =>
      val topic = topicAndPartitionMap._1
      val partitionMap = topicAndPartitionMap._2
      debug("partition assignment of /brokers/topics/%s is %s".format(topic, partitionMap))
      (topic -> partitionMap.keys.toSeq.sortWith((s,t) => s < t))
    }
  }

 

  def getConsumersPerTopic(group: String, excludeInternalTopics: Boolean) : mutable.Map[String, List[ConsumerThreadId]] = {
    val dirs = new ZKGroupDirs(group)
    val consumers = getChildrenParentMayNotExist(dirs.consumerRegistryDir)
    val consumersPerTopicMap = new mutable.HashMap[String, List[ConsumerThreadId]]
    for (consumer <- consumers) {
      val topicCount = TopicCount.constructTopicCount(group, consumer, this, excludeInternalTopics)
      for ((topic, consumerThreadIdSet) <- topicCount.getConsumerThreadIdsPerTopic) {
        for (consumerThreadId <- consumerThreadIdSet)
          consumersPerTopicMap.get(topic) match {
            case Some(curConsumers) => consumersPerTopicMap.put(topic, consumerThreadId :: curConsumers)
            case _ => consumersPerTopicMap.put(topic, List(consumerThreadId))
          }
      }
    }
    for ( (topic, consumerList) <- consumersPerTopicMap )
      consumersPerTopicMap.put(topic, consumerList.sortWith((s,t) => s < t))
    consumersPerTopicMap
  }

 

 

计算过程主要由上述高亮代码部分实现,举例说明,一个拥有十个分区的topic,相同group拥有三个consumerid为aaa,ccc,bbb的消费者

1 由后两段代码可知,获取consumerid列表和partition分区列表都是已经排好序的,所以

curConsumers=(aaa,bbb,ccc)

curPartitions=(0,1,2,3,4,5,6,7,8,9)

2

nPartsPerConsumer=10/3  =3

nConsumersWithExtraPart=10%3  =1

3 假设当前客户端id为aaa

myConsumerPosition= curConsumers.indexof(aaa) =0

4 计算分区范围

startPart= 3*0+0.min(1) = 0

nParts = 3+(if (0 + 1 > 1) 0 else 1)=3+1=4

所以aaa对应的分区号为[0,4),即0,1,2,3前面四个分区

同理可得bbb对应myConsumerPosition=1,对应分区4,5,6中间三个分区

ccc对应myConsumerPosition=2,对应7,8,9最后三个分区。

 

以上是关于kafka consumer 分区reblance算法的主要内容,如果未能解决你的问题,请参考以下文章

Kafka暂停消费--consumer.pause()

kafkakafka 指定分区消费 不会触发 reblance

kafka consumer offset机制

kafka可以修改分区不

如何确定Kafka的分区数key和consumer线程数

如何确定Kafka的分区数key和consumer线程数