PostgreSQL统计信息在计算选择率上的应用分析-创新互联
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一、计算选择率
单条件等值查询
测试数据生成脚本如下:
insert into t_grxx(dwbh,grbh,xm,xb,nl) select generate_series(1,100000)/10||'',generate_series(1,100000),'XM'||generate_series(1,100000), (case when (floor(random()*2)=0) then '男' else '女' end),floor(random() * 100 + 1)::int;
SQL脚本和执行计划:
testdb=# explain verbose select * from t_grxx where dwbh = '6323'; QUERY PLAN ---------------------------------------------------------------------------------------- Index Scan using idx_t_grxx_dwbh on public.t_grxx (cost=0.29..46.90 rows=30 width=24) Output: dwbh, grbh, xm, xb, nl Index Cond: ((t_grxx.dwbh)::text = '6323'::text) (3 rows) testdb=# explain verbose select * from t_grxx where dwbh = '24'; QUERY PLAN ---------------------------------------------------------------------------------------- Index Scan using idx_t_grxx_dwbh on public.t_grxx (cost=0.29..20.29 rows=10 width=24) Output: dwbh, grbh, xm, xb, nl Index Cond: ((t_grxx.dwbh)::text = '24'::text) (3 rows)
虽然都是等值查询,但执行计划中dwbh='6323'和dwbh='24'返回的行数(rows)却不一样,一个是rows=30,一个是rows=10,从生成数据的脚本来看,'6323'和'24'的rows应该是一样的,但执行计划显示的结果却不同,原因是计算选择率时'6323'出现在高频值中,因此与其他值不同.
计算过程解析
查询该列的统计信息:
testdb=# \x
Expanded display is on.
testdb=# select starelid,staattnum,stainherit,stanullfrac,stawidth,stadistinct
from pg_statistic
where starelid = 16742 and staattnum = 1;
-[ RECORD 1 ]---------
starelid | 16742
staattnum | 1
stainherit | f
stanullfrac | 0
stawidth | 4
stadistinct | -0.10015
testdb=# select staattnum,stakind1,staop1,stanumbers1,stavalues1,
stakind2,staop2,stanumbers2,stavalues2,
stakind3,staop3,stanumbers3,stavalues3
from pg_statistic
where starelid = 16742
and staattnum = 1;
-[ RECORD 1 ]----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
staattnum | 1
stakind1 | 1
staop1 | 98
stanumbers1 | {0.0003}
stavalues1 | {6323}
stakind2 | 2
staop2 | 664
stanumbers2 |
stavalues2 | {0,1092,1181,1265,1350,1443,1529,1619,171,1797,1887,1972,2058,2151,2240,2334,2423,2520,2618,271,2798,2892,2987,3076,3162,3246,3332,3421,3510,3597,3685,3777,3860,3956,4051,4136,4227,4317,4408,45,4590,4671,4760,4850,4933,5025,5120,5210,5300,5396,548,5570,5656,5747,5835,5931,6017,6109,6190,6281,6374,6465,6566,6649,6735,6830,6921,7012,7101,7192,7278,737,7455,7544,7630,7711,7801,7895,7988,8081,8167,8260,8344,8430,8520,8615,8707,8809,8901,8997,9083,918,9272,9367,9451,9538,9630,9729,982,9904,9999}
stakind3 | 3
staop3 | 664
stanumbers3 | {0.819578}
stavalues3 |条件语句是等值表达式,使用的操作符是"="(字符串等值比较,texteq/eqsel/eqjoinsel),因此使用的统计信息是高频值MCV(注意:staop1=98,这是字符串等值比较).'6323'出现在高频值中,选择率为0.0003,因此rows=100,000x0.0003=30.而'24'没有出现在高频值中,选择率=(1-0.0003)/abs(stadistinct)/Tuples=(1-0.0003)/abs(-0.10015)/100000=0.000099820269595606590000,rows=(1-0.0003)/abs(stadistinct)=10(取整).
单条件比较查询
测试脚本:
testdb=# create table t_int(c1 int,c2 varchar(20)); CREATE TABLE testdb=# testdb=# insert into t_int select generate_series(1,100000)/10,'C2'||generate_series(1,100000)/100; INSERT 0 100000 testdb=# ANALYZE t_int; ANALYZE testdb=# select oid from pg_class where relname='t_int'; oid ------- 16755 (1 row)
查询c1列的统计信息
testdb=# \x
Expanded display is on.
testdb=# select staattnum,stakind1,staop1,stanumbers1,stavalues1,
testdb-# stakind2,staop2,stanumbers2,stavalues2,
testdb-# stakind3,staop3,stanumbers3,stavalues3
testdb-# from pg_statistic
testdb-# where starelid = 16755
testdb-# and staattnum = 1;
-[ RECORD 1 ]---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
staattnum | 1
stakind1 | 1
staop1 | 96
stanumbers1 | {0.0003}
stavalues1 | {8306}
stakind2 | 2
staop2 | 97
stanumbers2 |
stavalues2 | {0,108,215,318,416,514,611,704,809,912,1015,1111,1217,1312,1410,1511,1607,1705,1805,1903,2002,2094,2189,2287,2388,2487,2592,2695,2795,2896,2998,3112,3213,3304,3408,3507,3606,3707,3798,3908,4004,4106,4205,4312,4413,4505,4606,4714,4821,4910,5014,5118,5220,5321,5418,5516,5613,5709,5807,5916,6014,6127,6235,6341,6447,6548,6648,6741,6840,6931,7032,7131,7234,7330,7433,7532,7626,7727,7827,7925,8020,8120,8217,8322,8420,8525,8630,8730,8831,8934,9032,9128,9223,9323,9425,9527,9612,9706,9804,9904,9999}
stakind3 | 3
staop3 | 97
stanumbers3 | {1}
stavalues3 |查询语句:
testdb=# explain verbose select * from t_int where c1 < 2312; QUERY PLAN ------------------------------------------------------------------- Seq Scan on public.t_int (cost=0.00..1790.00 rows=23231 width=9) Output: c1, c2 Filter: (t_int.c1 < 2312) (3 rows)
SQL使用了非等值查询(<,int4lt/scalarltsel/scalarltjoinsel),结合统计信息中MCV和直方图使用,
由于2312均小于MCV中的值,因此根据MCV得出的选择率为0.
根据直方图计算的选择率=(1-0.0003)x(23+(2312-2287-1)/(2388-2287))/100=0.2323065247,rows=100000x0.2323065247=23231(取整)
其中:
除高频值外的其他数值占比=(1-0.0003)
直方图中的总槽数=数组元素总数-1即101-1=100
2312落在第24个槽中,槽占比=(23+(2312-2287-1)/(2388-2287))/100
感谢各位的阅读，以上就是“PostgreSQL 统计信息在计算选择率上的应用分析”的内容了，经过本文的学习后，相信大家对PostgreSQL 统计信息在计算选择率上的应用分析这一问题有了更深刻的体会，具体使用情况还需要大家实践验证。这里是创新互联，小编将为大家推送更多相关知识点的文章，欢迎关注！
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