PostgreSQL

69.2. Multivariate Statistics Examples
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69.2. Multivariate Statistics Examples

69.2.1. Functional Dependencies

Multivariate correlation can be demonstrated with a very simple data set — a table with two columns, both containing the same values:

CREATE TABLE t (a INT, b INT);
INSERT INTO t SELECT i % 100, i % 100 FROM generate_series(1, 10000) s(i);
ANALYZE t;

As explained in Section 14.2, the planner can determine cardinality of t using the number of pages and rows obtained from pg_class:

SELECT relpages, reltuples FROM pg_class WHERE relname = 't';

 relpages | reltuples
----------+-----------
       45 |     10000

The data distribution is very simple; there are only 100 distinct values in each column, uniformly distributed.

The following example shows the result of estimating a WHERE condition on the a column:

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

The planner examines the condition and determines the selectivity of this clause to be 1%. By comparing this estimate and the actual number of rows, we see that the estimate is very accurate (in fact exact, as the table is very small). Changing the WHERE condition to use the b column, an identical plan is generated. But observe what happens if we apply the same condition on both columns, combining them with AND:

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

The planner estimates the selectivity for each condition individually, arriving at the same 1% estimates as above. Then it assumes that the conditions are independent, and so it multiplies their selectivities, producing a final selectivity estimate of just 0.01%. This is a significant underestimate, as the actual number of rows matching the conditions (100) is two orders of magnitude higher.

This problem can be fixed by creating a statistics object that directs ANALYZE to calculate functional-dependency multivariate statistics on the two columns:

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

69.2.2. Multivariate N-Distinct Counts

A similar problem occurs with estimation of the cardinality of sets of multiple columns, such as the number of groups that would be generated by a GROUP BY clause. When GROUP BY lists a single column, the n-distinct estimate (which is visible as the estimated number of rows returned by the HashAggregate node) is very accurate:

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)

But without multivariate statistics, the estimate for the number of groups in a query with two columns in GROUP BY, as in the following example, is off by an order of magnitude:

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)

By redefining the statistics object to include n-distinct counts for the two columns, the estimate is much improved:

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)

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69.1. Row Estimation Examples	Home	69.3. Planner Statistics and Security

69.1. Row Estimation Examples

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69.3. Planner Statistics and Security