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15.8.10.3 Estimating ANALYZE TABLE Complexity for InnoDB Tables
ANALYZE TABLE
complexity for
InnoDB
tables is dependent on:
The number of pages sampled, as defined by
innodb_stats_persistent_sample_pages
.The number of indexed columns in a table
The number of partitions. If a table has no partitions, the number of partitions is considered to be 1.
Using these parameters, an approximate formula for estimating
ANALYZE TABLE
complexity would
be:
The value of
innodb_stats_persistent_sample_pages
* number of indexed columns in a table * the number of
partitions
Typically, the greater the resulting value, the greater the
execution time for ANALYZE TABLE
.
innodb_stats_persistent_sample_pages
defines the number of pages sampled at a global level. To set
the number of pages sampled for an individual table, use the
STATS_SAMPLE_PAGES
option with
CREATE TABLE
or
ALTER TABLE
. For more
information, see Section 15.8.10.1, “Configuring Persistent Optimizer Statistics Parameters”.
If
innodb_stats_persistent=OFF
,
the number of pages sampled is defined by
innodb_stats_transient_sample_pages
.
See Section 15.8.10.2, “Configuring Non-Persistent Optimizer Statistics Parameters” for
additional information.
For a more in-depth approach to estimating ANALYZE
TABLE
complexity, consider the following example.
In Big
O notation, ANALYZE TABLE
complexity is described as:
O(n_sample
* (n_cols_in_uniq_i
+ n_cols_in_non_uniq_i
+ n_cols_in_pk * (1 + n_non_uniq_i))
* n_part)
where:
n_sample
is the number of pages sampled (defined byinnodb_stats_persistent_sample_pages
)n_cols_in_uniq_i
is total number of all columns in all unique indexes (not counting the primary key columns)n_cols_in_non_uniq_i
is the total number of all columns in all nonunique indexesn_cols_in_pk
is the number of columns in the primary key (if a primary key is not defined,InnoDB
creates a single column primary key internally)n_non_uniq_i
is the number of nonunique indexes in the tablen_part
is the number of partitions. If no partitions are defined, the table is considered to be a single partition.
Now, consider the following table (table t
),
which has a primary key (2 columns), a unique index (2 columns),
and two nonunique indexes (two columns each):
For the column and index data required by the algorithm
described above, query the
mysql.innodb_index_stats
persistent index
statistics table for table t
. The
n_diff_pfx%
statistics show the columns that
are counted for each index. For example, columns
a
and b
are counted for
the primary key index. For the nonunique indexes, the primary
key columns (a,b) are counted in addition to the user defined
columns.
For additional information about the InnoDB
persistent statistics tables, see
Section 15.8.10.1, “Configuring Persistent Optimizer Statistics Parameters”
- +------------+--------------+------------------+
- | index_name | stat_name | stat_description |
- +------------+--------------+------------------+
- | PRIMARY | n_diff_pfx01 | a |
- | PRIMARY | n_diff_pfx02 | a,b |
- | i1uniq | n_diff_pfx01 | c |
- | i1uniq | n_diff_pfx02 | c,d |
- | i2nonuniq | n_diff_pfx01 | e |
- | i2nonuniq | n_diff_pfx02 | e,f |
- | i2nonuniq | n_diff_pfx03 | e,f,a |
- | i2nonuniq | n_diff_pfx04 | e,f,a,b |
- | i3nonuniq | n_diff_pfx01 | g |
- | i3nonuniq | n_diff_pfx02 | g,h |
- | i3nonuniq | n_diff_pfx03 | g,h,a |
- | i3nonuniq | n_diff_pfx04 | g,h,a,b |
- +------------+--------------+------------------+
Based on the index statistics data shown above and the table definition, the following values can be determined:
n_cols_in_uniq_i
, the total number of all columns in all unique indexes not counting the primary key columns, is 2 (c
andd
)n_cols_in_non_uniq_i
, the total number of all columns in all nonunique indexes, is 4 (e
,f
,g
andh
)n_cols_in_pk
, the number of columns in the primary key, is 2 (a
andb
)n_non_uniq_i
, the number of nonunique indexes in the table, is 2 (i2nonuniq
andi3nonuniq
))n_part
, the number of partitions, is 1.
You can now calculate
innodb_stats_persistent_sample_pages
* (2 + 4
+ 2 * (1 + 2)) * 1 to determine the number of leaf pages that
are scanned. With
innodb_stats_persistent_sample_pages
set to
the default value of 20
, and with a default
page size of 16 KiB
(innodb_page_size
=16384), you
can then estimate that 20 * 12 * 16384 bytes
are read for table t
, or about 4
MiB
.
All 4 MiB
may not be read from disk, as
some leaf pages may already be cached in the buffer pool.
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Document créé le 26/06/2006, dernière modification le 26/10/2018
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