分区是不是免费的,无论是生成和探测(左,右),双方需要进行分区做分区加盟。每个分区都需要一个交换计划片段作为孩子,并且每个都会产生网络传输。但是,如果构建侧很小,那么每个节点都可以拥有它的副本(即广播),然后利用左侧未分区的探测构建侧散列表,而不在探测器侧引入额外的子交换。事实上,广播所需的交换特别昂贵,因为每个发送者都需要发送给N个接收者。
什么是“足够小”来执行广播连接?它取决于许多因素,但最明显也很重要的是,构建端哈希表应该适合内存。
下面是一个示例方案,其中加入的策略是BROADCAST:
[localhost:21000] > explain select * from alltypes t1 join alltypessmall t2 on t1.id = t2.id;
Query: explain select * from alltypes t1 join alltypessmall t2 on t1.id = t2.id
+-----------------------------------------------------------+
| Explain String |
+-----------------------------------------------------------+
| Estimated Per-Host Requirements: Memory=160.01MB VCores=2 |
| |
| 04:EXCHANGE [UNPARTITIONED] |
| | |
| 02:HASH JOIN [INNER JOIN, BROADCAST] |
| | hash predicates: t1.id = t2.id |
| | |
| |--03:EXCHANGE [BROADCAST] |
| | | |
| | 01:SCAN HDFS [functional.alltypessmall t2] |
| | partitions=4/4 files=4 size=6.32KB |
| | |
| 00:SCAN HDFS [functional.alltypes t1] |
| partitions=24/24 files=24 size=478.45KB |
+-----------------------------------------------------------+
而且这里是连接策略划分样本:
Query: explain select * from tpch.lineitem t1 join tpch.lineitem t2 on t1.l_orderkey = t2.l_orderkey
+-----------------------------------------------------------+
| Explain String |
+-----------------------------------------------------------+
| Estimated Per-Host Requirements: Memory=815.44MB VCores=2 |
| |
| 05:EXCHANGE [UNPARTITIONED] |
| | |
| 02:HASH JOIN [INNER JOIN, PARTITIONED] |
| | hash predicates: t1.l_orderkey = t2.l_orderkey |
| | |
| |--04:EXCHANGE [HASH(t2.l_orderkey)] |
| | | |
| | 01:SCAN HDFS [tpch.lineitem t2] |
| | partitions=1/1 files=1 size=718.94MB |
| | |
| 03:EXCHANGE [HASH(t1.l_orderkey)] |
| | |
| 00:SCAN HDFS [tpch.lineitem t1] |
| partitions=1/1 files=1 size=718.94MB |
+-----------------------------------------------------------+
Fetched 16 row(s) in 0.03s
注意,后者计划有一个额外的交换。这意味着有一个额外的扫描计划片段(ID 00)。