kafka源码分析-consumer的分区策略

1、AbstractPartitionAssignor2、RangeAssignor3、RoundRobinAssignor4、StickyAssignor策略

consumer有三种分区策略，分别是RangeAssignor、RoundRobinAssignor和StickyAssignor，这三个策略都继承了AbstractPartitionAssignor，实现了其assign方法。该方法有两个参数：

partitionsPerTopic-每个topic的分区数量subscriptions-每个 consumerId 与其所订阅的 topic 列表的关系，可以理解成每个consumer可能被分配到的topic

public abstract Map> assign(Map partitionsPerTopic,
                                                        Map subscriptions);

先看代码实现

public class RangeAssignor extends AbstractPartitionAssignor {
 
    @Override
    public String name() {
        return "range";
    }
 
    private Map> consumersPerTopic(Map consumermetadata) {
        Map> res = new HashMap<>();
        for (Map.Entry subscriptionEntry : consumermetadata.entrySet()) {
            String consumerId = subscriptionEntry.getKey();
            for (String topic : subscriptionEntry.getValue().topics())
                put(res, topic, consumerId);
        }
        return res;
    }
 
    // partitionsPerTopic-每个topic的分区数量
    // subscriptions-每个 consumerId 与其所订阅的 topic 列表的关系
    @Override
    public Map> assign(Map partitionsPerTopic,
                                                    Map subscriptions) {
        // 将subscriptions转换成key为topic，value为consumerId的map
        Map> consumersPerTopic = consumersPerTopic(subscriptions);
 
        // 存储rebalace方案的数据结构
        Map> assignment = new HashMap<>();
        for (String memberId : subscriptions.keySet())
            assignment.put(memberId, new ArrayList());
         
        // 遍历consumersPerTopic
        for (Map.Entry> topicEntry : consumersPerTopic.entrySet()) {
            String topic = topicEntry.getKey();
            // 和这个topic相关的consumer列表
            List consumersForTopic = topicEntry.getValue();
            // 该topic的分区数
            Integer numPartitionsForTopic = partitionsPerTopic.get(topic);
            if (numPartitionsForTopic == null)
                continue;
 
            Collections.sort(consumersForTopic);
 
            // 取商
            int numPartitionsPerConsumer = numPartitionsForTopic / consumersForTopic.size();
            // 取余数
            int consumersWithExtraPartition = numPartitionsForTopic % consumersForTopic.size();
 			// 【以下关键分配步骤】
            // 生成TopicPartition列表
            List partitions = AbstractPartitionAssignor.partitions(topic, numPartitionsForTopic);
             
            for (int i = 0, n = consumersForTopic.size(); i < n; i++) {
                // 本consumer分配到的初始位置
                int start = numPartitionsPerConsumer * i + Math.min(i, consumersWithExtraPartition);
                // 长度
                int length = numPartitionsPerConsumer + (i + 1 > consumersWithExtraPartition ? 0 : 1);
                // 当consumersWithExtraPartition 不是0时，优先给前面的consumer多分配一个partition，能整除部分各consumer均分
                assignment.get(consumersForTopic.get(i)).addAll(partitions.subList(start, start + length));
            }
        }
        return assignment;
    }
 
}

重点关注源码的【关键分配步骤】，该步骤是将所有topic的分区平均分配给每个consumer，对于不能平均分配的分配给前consumersWithExtraPartition个consumer，也就是前consumersWithExtraPartition个consumer分配到的分区数会比后面的多一个。
【例1】
假设有一个topic有7个partition，有三个consumer都订阅了该topic，那么通过RangeAssignor的分配方案为：

consumer0：start=0，length=3，分配到的partition为：p0、p1、p2consumer1：start=3，length=2，分配到的partition为：p3、p4consumer2：start=5，length=2，分配到的partition为：p5、p6

【例2】
在看一个多topic分配的例子，假设consumer0订阅了topic0、topic1、topic2，consumer1订阅了topic0、topic1，consumer2订阅了topic2，topic0、topic1、topic2分别有3、2、1个partition，那么分配的结果是：

可以看到分配结果并不均匀，甚至有consumer闲置。

先看源码

public class RoundRobinAssignor extends AbstractPartitionAssignor {
 
    @Override
    public Map> assign(Map partitionsPerTopic,
                                                    Map subscriptions) {
        // 存储数据结构
        Map> assignment = new HashMap<>();
        for (String memberId : subscriptions.keySet())
            assignment.put(memberId, new ArrayList());
 		// 【关键分配步骤】
        // 将consumer排序并生成环状迭代器
        CircularIterator assigner = new CircularIterator<>(Utils.sorted(subscriptions.keySet()));
        // 遍历所有的partition
        for (TopicPartition partition : allPartitionsSorted(partitionsPerTopic, subscriptions)) {
            final String topic = partition.topic();
            // 找到订阅这个partition对应topic的消费者
            while (!subscriptions.get(assigner.peek()).topics().contains(topic))
                // 返回迭代器当前位置的元素，并将迭代器计数+1
                assigner.next();
            // 将partition分配给消费者
            assignment.get(assigner.next()).add(partition);
        }
        return assignment;
    }
 
    // 获取所有的partition，并且同一个topic的partition是相连的
    public List allPartitionsSorted(Map partitionsPerTopic,
                                                    Map subscriptions) {
        SortedSet topics = new TreeSet<>();
        for (Subscription subscription : subscriptions.values())
            topics.addAll(subscription.topics());
 
        List allPartitions = new ArrayList<>();
        for (String topic : topics) {
            Integer numPartitionsForTopic = partitionsPerTopic.get(topic);
            if (numPartitionsForTopic != null)
                allPartitions.addAll(AbstractPartitionAssignor.partitions(topic, numPartitionsForTopic));
        }
        return allPartitions;
    }
 
    @Override
    public String name() {
        return "roundrobin";
    }
 
}

重点关注【关键分配步骤】它的分配规则是遍历所有的topic-partition，对每个consumer以环状的形式进行分配，从当前位置往后找到第一个可以分配的consumer，将当前partition分配给找到的consumer，并将位置+1，继续下一个partition的分配。
看2中的【例2】使用RoundRobinAssignor的分配方案

可以看到相比于RangeAssignor分配更加均匀，但是这种方式并不能完全解决平均分配的问题，看一下下面的例子
【例3】
假设consumer0订阅了topic0、topic1、topic2，consumer1订阅了topic0、topic1，consumer2订阅了topic2，topic0、topic1、topic2分别有4、3、2个partition，那么分配的结果是：

可以看到将t2p1分配给consumer2的话，分配结果更加平均。

sticky-粘性的，非常形象的名字，StickyAssignor从0.11版本才开始引入的，主要有两个目的

目的1：分区的分配要尽可能均匀目的2：分区的分配要尽可能与上次分配的保持相同

目的1和目的2冲突时，目的1优于目的2。
StickyAssignor的代码较多，先看下面的流程图

以3中的【例3】为例来看一下StickyAssignor策略的分配结果

相比于RoundRobinAssignor，分配结果达到均衡。假设新加入了consumer3，其订阅了topic0，那么再看一下再均衡的过程：
（1）所有partition的排序结果：t0p0、t0p1、t0p2、t0p3、t1p0、t1p1、t1p1、t2p0、t2p1
（2）初始的consumer排序：consumer3、consumer2、consumer1、consumer0
（3）开始再均衡，显然t0p0是可以再分配的，重新分配后的结果及consuemr顺序是：

继续处理t0p1，显然也可以再分配，由consumer1调整到consumer3

（4）完成再分配过程，返回结果
流程图中省略了很多细节，相关内容可见下面源码分析：

public class StickyAssignor extends AbstractPartitionAssignor {
    private static final Logger log = LoggerFactory.getLogger(StickyAssignor.class);
 
    // these schemas are used for preserving consumer's previously assigned partitions
    // list and sending it as user data to the leader during a rebalance
    private static final String TOPIC_PARTITIONS_KEY_NAME = "previous_assignment";
    private static final String TOPIC_KEY_NAME = "topic";
    private static final String PARTITIONS_KEY_NAME = "partitions";
    private static final Schema TOPIC_ASSIGNMENT = new Schema(
            new Field(TOPIC_KEY_NAME, Type.STRING),
            new Field(PARTITIONS_KEY_NAME, new ArrayOf(Type.INT32)));
    private static final Schema STICKY_ASSIGNOR_USER_DATA = new Schema(
            new Field(TOPIC_PARTITIONS_KEY_NAME, new ArrayOf(TOPIC_ASSIGNMENT)));
 
    private List memberAssignment = null;
    private PartitionMovements partitionMovements;
 
    public Map> assign(Map partitionsPerTopic,
                                                    Map subscriptions) {
        Map> currentAssignment = new HashMap<>();
        partitionMovements = new PartitionMovements();
 
        // 先构建出当前的分配状态：currentAssignment
        // 这个方法里用的userData是自定义信息，但是是怎么用的？？
        prepopulateCurrentAssignments(subscriptions, currentAssignment);
        // 判断是否是全新的分配,true-是
        boolean isFreshAssignment = currentAssignment.isEmpty();
 
        // 记录partirion可以分配给哪些consumer
        final Map> partition2AllPotentialConsumers = new HashMap<>();
        // 记录consumer能够被分配到哪些partition
        final Map> consumer2AllPotentialPartitions = new HashMap<>();
 
        // 初始化partition2AllPotentialConsumers，value是空List
        for (Entry entry: partitionsPerTopic.entrySet()) {
            for (int i = 0; i < entry.getValue(); ++i)
                partition2AllPotentialConsumers.put(new TopicPartition(entry.getKey(), i), new ArrayList());
        }
 
        // 遍历subscriptions
        for (Entry entry: subscriptions.entrySet()) {
            String consumer = entry.getKey();
            // 初始化consumer2AllPotentialPartitions的每个consumer
            consumer2AllPotentialPartitions.put(consumer, new ArrayList());
            for (String topic: entry.getValue().topics()) {
                for (int i = 0; i < partitionsPerTopic.get(topic); ++i) {
                    TopicPartition topicPartition = new TopicPartition(topic, i);
                    // 将这个consumer订阅的topic的所有partition存入consumer2AllPotentialPartitions
                    consumer2AllPotentialPartitions.get(consumer).add(topicPartition);
                    // 将这个topic的所有partition都记录上consumer
                    partition2AllPotentialConsumers.get(topicPartition).add(consumer);
                }
            }
 
            // 当前consumer的分配方案不存在
            if (!currentAssignment.containsKey(consumer))
                // 初始化当前consumer的分配存储结构
                currentAssignment.put(consumer, new ArrayList());
        }
        // 以上完成partition2AllPotentialConsumers和consumer2AllPotentialPartitions的初始化
 
        // 记录当前分配方案中的partition和consumer的关系
        Map currentPartitionConsumer = new HashMap<>();
        for (Map.Entry> entry: currentAssignment.entrySet())
            for (TopicPartition topicPartition: entry.getValue())
                currentPartitionConsumer.put(topicPartition, entry.getKey());
         
        // 对有效分区进行排序，以便它们在潜在的重新分配阶段以适当的顺序进行处理，从而使消费者之间的分区移动最小（因此尊重最大粘性）
        // 详细解析见下面sortPartitions的源码
        List sortedPartitions = sortPartitions(
                currentAssignment, isFreshAssignment, partition2AllPotentialConsumers, consumer2AllPotentialPartitions);
 
        // 将排序后的partition存到unassignedPartitions
        List unassignedPartitions = new ArrayList<>(sortedPartitions);
         
        // 遍历当前的分配方案，进行删除处理
        for (Iterator>> it = currentAssignment.entrySet().iterator(); it.hasNext();) {
            Map.Entry> entry = it.next();
            if (!subscriptions.containsKey(entry.getKey())) {
                // 之前的consumer不在存在，删除该consumer订阅的所有partition，并从当前分配方案中删除该consumer
                for (TopicPartition topicPartition: entry.getValue())
                    currentPartitionConsumer.remove(topicPartition);
                it.remove();
            } else {
                // otherwise (the consumer still exists)
                for (Iterator partitionIter = entry.getValue().iterator(); partitionIter.hasNext();) {
                    TopicPartition partition = partitionIter.next();
                    if (!partition2AllPotentialConsumers.containsKey(partition)) {
                        // 这个partition不再存在，从当前分配方案删除partiton，并从currentPartitionConsumer删除partition
                        partitionIter.remove();
                        currentPartitionConsumer.remove(partition);
                    } else if (!subscriptions.get(entry.getKey()).topics().contains(partition.topic())) {
                        // 该consumer不再订阅该partition对应的topic，从当前分配方案删除partiton
                        partitionIter.remove();
                    } else
                        // 该partition已经被分配过，从未分配的partition中删除partition
                        unassignedPartitions.remove(partition);
                }
            }
        }
        // at this point we have preserved all valid topic partition to consumer assignments and removed
        // all invalid topic partitions and invalid consumers. Now we need to assign unassignedPartitions
        // to consumers so that the topic partition assignments are as balanced as possible.
 
        // 根据已经分配给消费者的主题分区的数量，对消费者进行升序排序
        TreeSet sortedCurrentSubscriptions = new TreeSet<>(new SubscriptionComparator(currentAssignment));
        sortedCurrentSubscriptions.addAll(currentAssignment.keySet());
 
        balance(currentAssignment, sortedPartitions, unassignedPartitions, sortedCurrentSubscriptions,
                consumer2AllPotentialPartitions, partition2AllPotentialConsumers, currentPartitionConsumer);
        return currentAssignment;
    }
     
    // 省略一些方法
}

private List sortPartitions(Map> currentAssignment,
                                            boolean isFreshAssignment,
                                            Map> partition2AllPotentialConsumers,
                                            Map> consumer2AllPotentialPartitions) {
    List sortedPartitions = new ArrayList<>();
 
    if (!isFreshAssignment && areSubscriptionsIdentical(partition2AllPotentialConsumers, consumer2AllPotentialPartitions)) {
        // 如果不是新的分配，且每个consumer所可能分配到的partition都是一样的
 
        // 复制一份当前的分配方案
        Map> assignments = deepCopy(currentAssignment);
        // 将不属于当前consumer的partition从当前分配方案中删除
        for (Entry> entry: assignments.entrySet()) {
            List toRemove = new ArrayList<>();
            for (TopicPartition partition: entry.getValue())
                if (!partition2AllPotentialConsumers.keySet().contains(partition))
                    toRemove.add(partition);
            for (TopicPartition partition: toRemove)
                entry.getValue().remove(partition);
        }
        // SubscriptionComparator是根据两个key对应value的长度进行比较，长度相同根据key进行字符串排序
        TreeSet sortedConsumers = new TreeSet<>(new SubscriptionComparator(assignments));
        sortedConsumers.addAll(assignments.keySet());
 
        // 将partition进行排序，是按consumer的顺序倒序取出，也就是分配到更多partition的consumer
        while (!sortedConsumers.isEmpty()) {
            String consumer = sortedConsumers.pollLast();
            List remainingPartitions = assignments.get(consumer);
            if (!remainingPartitions.isEmpty()) {
                sortedPartitions.add(remainingPartitions.remove(0));
                sortedConsumers.add(consumer);
            }
        }
 
        // 将不属于consumer的partition放入到排序结果中
        for (TopicPartition partition: partition2AllPotentialConsumers.keySet()) {
            if (!sortedPartitions.contains(partition))
                sortedPartitions.add(partition);
        }
 
    } else {
        // PartitionComparator是根据两个key对应value的长度进行比较，如果长度相同则根据key进行字符串排序，在相同就根据topic的partition数排序
        TreeSet sortedAllPartitions = new TreeSet<>(new PartitionComparator(partition2AllPotentialConsumers));
        sortedAllPartitions.addAll(partition2AllPotentialConsumers.keySet());
 
        // 顺序取出partition进行排序，越少可以被consumer分配到的partition越在前面
        while (!sortedAllPartitions.isEmpty())
            sortedPartitions.add(sortedAllPartitions.pollFirst());
    }
 
    return sortedPartitions;
}

private void balance(Map> currentAssignment,
                     List sortedPartitions,
                     List unassignedPartitions,
                     TreeSet sortedCurrentSubscriptions,
                     Map> consumer2AllPotentialPartitions,
                     Map> partition2AllPotentialConsumers,
                     Map currentPartitionConsumer) {
    // 是否是初始化分配，如果最大数量partition的consumer所分配到的partition都是空的，则是新的分配
    boolean initializing = currentAssignment.get(sortedCurrentSubscriptions.last()).isEmpty();
    boolean reassignmentPerformed = false;
 
    // 分配还未分配的partition
    // 分配给最少partition的consumer，并重新将该consumer放入sortedCurrentSubscriptions,为了重新将consumer排序
    for (TopicPartition partition: unassignedPartitions) {
        // skip if there is no potential consumer for the partition
        if (partition2AllPotentialConsumers.get(partition).isEmpty())
            continue;
 
        assignPartition(partition, sortedCurrentSubscriptions, currentAssignment,
                        consumer2AllPotentialPartitions, currentPartitionConsumer);
    }
 
    // fixedPartitions记录只能被分配给唯一consumer的partition，并将这部分partition从sortedPartitions中移除
    // 到目前为止所有的所有的partition已经被分配了，剔除掉不可能被重新分配的partition
    // fixedPartitions在后面并没有什么用途，单纯的中间变量
    Set fixedPartitions = new HashSet<>();
    for (TopicPartition partition: partition2AllPotentialConsumers.keySet())
        if (!canParticipateInReassignment(partition, partition2AllPotentialConsumers))
            fixedPartitions.add(partition);
    sortedPartitions.removeAll(fixedPartitions);
 
    // 将已经分配结束的consumer从sortedCurrentSubscriptions删除，并将其对应的partition存入fixedAssignments
    Map> fixedAssignments = new HashMap<>();
    for (String consumer: consumer2AllPotentialPartitions.keySet())
        // canParticipateInReassignment:consumer是否可以被重新分配,true-可以
        // 1、没满：满即所有可能分配给这个consumer的partition已全部分配给该consumer，返回true
        // 2、满了，但是该consumer的partition有任一partition可以被分配给多个consumer，返回true
        if (!canParticipateInReassignment(consumer, currentAssignment,
                                          consumer2AllPotentialPartitions, partition2AllPotentialConsumers)) {
            sortedCurrentSubscriptions.remove(consumer);
            fixedAssignments.put(consumer, currentAssignment.remove(consumer));
        }
 
    // create a deep copy of the current assignment so we can revert to it if we do not get a more balanced assignment later
    // 备份
    Map> preBalanceAssignment = deepCopy(currentAssignment);
    Map preBalancePartitionConsumers = new HashMap<>(currentPartitionConsumer);
     
    // reassignmentPerformed-true,意味着被重新分配过
    reassignmentPerformed = performReassignments(sortedPartitions, currentAssignment, sortedCurrentSubscriptions,
            consumer2AllPotentialPartitions, partition2AllPotentialConsumers, currentPartitionConsumer);
 
    // getBalanceScore-计算平衡分数，分数越小越好，分数是正的
    // 如何计算：遍历所有consumer的partition数和其他consumer的partition数相减的绝对值累加，已经遍历过的consumer要删除（即后面的partition计算时不会在用到前面已处理过的consumer）
    if (!initializing && reassignmentPerformed && getBalanceScore(currentAssignment) >= getBalanceScore(preBalanceAssignment)) {
        // 重新分配的方案不好，重置回原来的方案
        deepCopy(preBalanceAssignment, currentAssignment);
        currentPartitionConsumer.clear();
        currentPartitionConsumer.putAll(preBalancePartitionConsumers);
    }
 
    // 将前面不需要重新分配的consumer重新放到当前分配方案和排序中
    for (Entry> entry: fixedAssignments.entrySet()) {
        String consumer = entry.getKey();
        currentAssignment.put(consumer, entry.getValue());
        sortedCurrentSubscriptions.add(consumer);
    }
 
    fixedAssignments.clear();
}

private boolean performReassignments(List reassignablePartitions,
                                     Map> currentAssignment,
                                     TreeSet sortedCurrentSubscriptions,
                                     Map> consumer2AllPotentialPartitions,
                                     Map> partition2AllPotentialConsumers,
                                     Map currentPartitionConsumer) {
    boolean reassignmentPerformed = false;
    boolean modified;
 
    // repeat reassignment until no partition can be moved to improve the balance
    do {
        modified = false;
        // reassign all reassignable partitions (starting from the partition with least potential consumers and if needed)
        // until the full list is processed or a balance is achieved
        // 重新分配所有可重新分配的分区（从潜在使用者最少的分区开始，如果需要），或者达到平衡
        Iterator partitionIterator = reassignablePartitions.iterator();
 
        // isBalanced-判断当前方案已经平衡
        // 1、当前分配方案分配到最少partition的consumer的数量大于等于最多的数量-1，返回ture
        // 2、数量比较少的consumer，没有其他consumer的partition有可能分配给该consumer，返回true
        while (partitionIterator.hasNext() && !isBalanced(currentAssignment, sortedCurrentSubscriptions, consumer2AllPotentialPartitions)) {
            TopicPartition partition = partitionIterator.next();
 
            // the partition must have at least two consumers
            if (partition2AllPotentialConsumers.get(partition).size() <= 1)
                log.error("Expected more than one potential consumer for partition '" + partition + "'");
 
            // the partition must have a current consumer
            String consumer = currentPartitionConsumer.get(partition);
            if (consumer == null)
                log.error("Expected partition '" + partition + "' to be assigned to a consumer");
 
            // check if a better-suited consumer exist for the partition; if so, reassign it
            for (String otherConsumer: partition2AllPotentialConsumers.get(partition)) {
                if (currentAssignment.get(consumer).size() > currentAssignment.get(otherConsumer).size() + 1) {
                    // 参照下面源码解析
                    reassignPartition(partition, currentAssignment, sortedCurrentSubscriptions, currentPartitionConsumer, consumer2AllPotentialPartitions);
                    reassignmentPerformed = true;
                    modified = true;
                    break;
                }
            }
        }
    } while (modified);
 
    return reassignmentPerformed;
}

private void reassignPartition(TopicPartition partition,
                               Map> currentAssignment,
                               TreeSet sortedCurrentSubscriptions,
                               Map currentPartitionConsumer,
                               Map> consumer2AllPotentialPartitions) {
    String consumer = currentPartitionConsumer.get(partition);
 
    // sortedCurrentSubscriptions是按consumer已分配的partition数量升序，所以找到第一可以分配到该partition的consumer就是新的可分配consumer
    String newConsumer = null;
    for (String anotherConsumer: sortedCurrentSubscriptions) {
        if (consumer2AllPotentialPartitions.get(anotherConsumer).contains(partition)) {
            newConsumer = anotherConsumer;
            break;
        }
    }
 
    assert newConsumer != null;
 
    // 找到准确的需要被重新分配的partition，为了粘性分配，详情参考下面getTheActualPartitionToBeMoved源码
    TopicPartition partitionToBeMoved = partitionMovements.getTheActualPartitionToBeMoved(partition, consumer, newConsumer);
    // 将需要移动的partition进行重新分配
    // 1、将新旧consumer从排序结果中删除，为了重新存入排序
    // 2、更新partitionMovements和partitionMovementsByTopic
    // 3、更新新旧consumer的分配方案和currentPartitionConsumer
    // 4、对新旧consumer的分配结果重新排序
    processPartitionMovement(partitionToBeMoved, newConsumer, currentAssignment, sortedCurrentSubscriptions, currentPartitionConsumer);
 
    return;
}

// partitionMovementsByTopic-存储partition上一次的调整关系（由srcConsumer调整到destConsumer）
// partitionMovementsForThisTopic-存储topic维度下ConsumerPair（新旧consumer）调整的partition
private TopicPartition getTheActualPartitionToBeMoved(TopicPartition partition, String oldConsumer, String newConsumer) {
        String topic = partition.topic();
        // 该topic的所有partition从未被调整过，直接返回该partition
        if (!partitionMovementsByTopic.containsKey(topic))
            return partition;
 
        // 该partition被调整过
        if (partitionMovements.containsKey(partition)) {
            // this partition has previously moved
            // 这次调整的旧consumer一定是上次调整的新consumer
            assert oldConsumer.equals(partitionMovements.get(partition).dstMemberId);
            // 旧consumer赋值为上次调整的旧consumer
            // 这是为了下面更大限度保证粘性，比如partiiton0上次调整是A->C,这次是C->B，但是存在一个partiiton1是从B调整到A的，
            // 经过这个赋值，将会用partition1代替partition0进行调整，这样B趋于平衡，A趋于不平衡，这样下次调整就有可能将A调整到C的partiiton调整会A
            // 这个设计比较绕
            oldConsumer = partitionMovements.get(partition).srcMemberId;
        }
 
        Map> partitionMovementsForThisTopic = partitionMovementsByTopic.get(topic);
        // 新旧consumer的“反consumer对”
        ConsumerPair reversePair = new ConsumerPair(newConsumer, oldConsumer);
        // 不存在“反consumer对”，直接返回该partition
        if (!partitionMovementsForThisTopic.containsKey(reversePair))
            return partition;
        // 返回该“反consumer对”之前调整的一个partition，为了满足StickyAssignor的目的2，尽可能保证分配和上次相同
        // 举个例子，比如这次要将partition0从consumerA调整到consumerB，但是在这之前曾将partition1从consumerB调整到consumerA，那么需要用partiiton1代替partiiton0
        // 这时候partiion0和partition1都是属于consumerA的，调整partiion0和partition1都可以使分配方案趋于平衡，但是调整partition1更符合粘性策略
        return partitionMovementsForThisTopic.get(reversePair).iterator().next();
    }

本文主要对kafka的三种分区策略进行源码分析，RangeAssignor、RoundRobinAssignor都不能保证完全的均衡分配，StickyAssignor虽然实现复杂，但是相比于其他两种分配策略，均衡效果更好，而且可以减少不必要的分区调整。