71.2. 多元统计信息示例 #
71.2.1. 函数依赖
多元相关性可以通过一个非常简单的数据集来演示 — 一个包含两列的表,两列都含有相同的值:
CREATE TABLE t (a INT, b INT); INSERT INTO t SELECT i % 100, i % 100 FROM generate_series(1, 10000) s(i); ANALYZE t;
如第 14.2 节所述,规划器可以利用从 pg_class 得到的页数和行数来确定 t 的基数:
SELECT relpages, reltuples FROM pg_class WHERE relname = 't';
relpages | reltuples
----------+-----------
45 | 10000
数据分布非常简单;每一列中都只有 100 个不同的值,并且均匀分布。
下面的示例展示了对 WHERE 条件在 a 列上进行估计的结果:
EXPLAIN (ANALYZE, TIMING OFF) SELECT * FROM t WHERE a = 1;
QUERY PLAN
-------------------------------------------------------------------------------
Seq Scan on t (cost=0.00..170.00 rows=100 width=8) (actual rows=100 loops=1)
Filter: (a = 1)
Rows Removed by Filter: 9900
规划器会检查该条件,并认定此子句的选择率为 1%。将该估计与实际行数相比,可以看出估计非常准确(实际上是精确的,因为这张表非常小)。把 WHERE 条件改为使用 b 列,会生成相同的计划。但请看如果把同一条件同时应用到两列上,并用 AND 连接,会发生什么:
EXPLAIN (ANALYZE, TIMING OFF) SELECT * FROM t WHERE a = 1 AND b = 1;
QUERY PLAN
-----------------------------------------------------------------------------
Seq Scan on t (cost=0.00..195.00 rows=1 width=8) (actual rows=100 loops=1)
Filter: ((a = 1) AND (b = 1))
Rows Removed by Filter: 9900
规划器分别估计每个条件的选择率,得到与上面相同的 1% 估计值。然后它假定这些条件彼此独立,于是将它们的选择率相乘,得到最终只有 0.01% 的选择率估计。这是严重的低估,因为实际匹配这些条件的行数(100)要高出两个数量级。
这个问题可以通过创建一个统计信息对象来解决,该对象会指示 ANALYZE 在这两列上计算函数依赖的多元统计信息:
CREATE STATISTICS stts (dependencies) ON a, b FROM t;
ANALYZE t;
EXPLAIN (ANALYZE, TIMING OFF) SELECT * FROM t WHERE a = 1 AND b = 1;
QUERY PLAN
-------------------------------------------------------------------------------
Seq Scan on t (cost=0.00..195.00 rows=100 width=8) (actual rows=100 loops=1)
Filter: ((a = 1) AND (b = 1))
Rows Removed by Filter: 9900
71.2.2. 多元非重复值计数
在估计多列集合的基数时,也会出现类似的问题,例如 GROUP BY 子句会生成多少个组。当 GROUP BY 只列出一列时,非重复值数量估计(可从 HashAggregate 节点估计返回的行数看出)非常准确:
EXPLAIN (ANALYZE, TIMING OFF) SELECT COUNT(*) FROM t GROUP BY a;
QUERY PLAN
-----------------------------------------------------------------------------------------
HashAggregate (cost=195.00..196.00 rows=100 width=12) (actual rows=100 loops=1)
Group Key: a
-> Seq Scan on t (cost=0.00..145.00 rows=10000 width=4) (actual rows=10000 loops=1)
但是如果没有多元统计信息,对于在 GROUP BY 中包含两列的查询,组数估计会像下面这样差一个数量级:
EXPLAIN (ANALYZE, TIMING OFF) SELECT COUNT(*) FROM t GROUP BY a, b;
QUERY PLAN
--------------------------------------------------------------------------------------------
HashAggregate (cost=220.00..230.00 rows=1000 width=16) (actual rows=100 loops=1)
Group Key: a, b
-> Seq Scan on t (cost=0.00..145.00 rows=10000 width=8) (actual rows=10000 loops=1)
通过重新定义统计信息对象,使其包含这两列的非重复值计数,估计值就会大幅改进:
DROP STATISTICS stts;
CREATE STATISTICS stts (dependencies, ndistinct) ON a, b FROM t;
ANALYZE t;
EXPLAIN (ANALYZE, TIMING OFF) SELECT COUNT(*) FROM t GROUP BY a, b;
QUERY PLAN
--------------------------------------------------------------------------------------------
HashAggregate (cost=220.00..221.00 rows=100 width=16) (actual rows=100 loops=1)
Group Key: a, b
-> Seq Scan on t (cost=0.00..145.00 rows=10000 width=8) (actual rows=10000 loops=1)