解释直方图信息
解释直方图信息
适用于：
Oracle Database - Enterprise Edition - Version 7.3.0.0 and later
Oracle Database - Standard Edition - Version 7.3.0.0 and later
Oracle Database - Personal Edition - Version 7.3.0.0 and later
Information in this document applies to any platform.
目的：
直方图信息是怎么被存储的,是怎么被解释的。
范围：
其他有用的直方图参考：
Document 1445372.1 Histograms: An Overview (10g and Above)
细节：
直方图是一种机制，该机制用来存储 列数据（column data）的详细信息。该数据被CBO使用，用来决定一个查询语句最优化的访问路径（access path）.
没有直方图时，优化器依靠的所有信息是：一个列的高值和低值，该列的不同值个数，该列的空值个数，该table的记录总数。
（实际上列的高值和低值是以raw 格式存储的，因此不是特别有用），其他的信息可以从dictionary views中查询到。
没有列的统计信息时，优化器假设数据是均衡分布的，对于等值谓词，生成以一个选择率(column selectivity)，该选择率是如下计算的：1/NVD(Number of Distinct Values)
有直方图时，你可以访问行数据的更多分布信息。
当一个列的数据分布不均衡时（即：列的数据分布 高度倾斜--数据分布倾斜的很厉害），Oracle 可以存储列的直方图以给出更好的选择率.这会产生比使用标准的统计信息（high and low values plus Number of Distinct Values）更好的执行计划
就具体实现而言（In terms of implementation）,我们可以选择 将 每个不同值和该值的记录数存放在一起，对于值很少的记录数是有效的，此时，'width balanced' histograms 被使用。
随着不同值数量的增长，存储数据的数量变得过高，我们需要使用一个不同的方法来存储直方图数据。此时，我们可以选择 height balanced histograms.
使用如上两种方法，列直方图提供了一个有效和集中的方法来展现数据分布。当建立直方图时，存储的信息依靠“不同值的数量是否小于等于bucket（默认75个，最大254个）的数量”进行不同的解释。
如果不同值的数量小于等于直方图bucket的数量（bucket最多254个），那么 Frequency Histogram 被建立
如果不同值的数量大于直方图bucket的数量， Height Balanced Histogram 被建立。
Frequency Histogram
Frequency Histogram 使用bucket来记录每一个不同值的记录个数
Height Balanced Histogram
Height Balanced Histogram 通过把数据分割到不同bucket中来实现。每个bucket 包括相同数量的列值。每个bucket中的最高值（or END_POINT）和最低值被记录在零号bucket中。
一旦数据被存储于bucket中，我们可以识别两个类型的data value--- Non-popular values and popular values
Non-popular values--are those that do not occur multiple times as end points.不会出现多次
Popular values--occur multiple times as end points.会出现多次。
We can use Popular and Non-Popular Values to provide use with various statistics.Since we know how many values there are in a bucket we can use this information to estimate the number of rows in total that are covered by Popular and Non-Popular values.
?The selectivity for popular values can be obtained by calculation the proportion of bucket endpoints filled by that popular value.
?The selectivity for non popular values can now be calculated as 1/number non-popular bucket endpoints, so we can now be more accurate about selectivities than the original 1/NDV, because we have removed the popular values from the equation.
How histograms are used
直方图被用来得到column predicate 更好的selectivity 估算
Where there are fewer distinct values than buckets, the selectivity is simply calculated as we have accurate row information for each value. For the case where we have more distinct values than buckets, the following outlines how these selectivities are obtained.
Equality Predicate Selectivity calculated from:
?Popular Value:
Number of buckets for value / Total Number of buckets
?Non-Popular Value:
Density see:
Document 43041.1 Query Optimizer: What is Density?
Less than < (Same principle applies for > & >= )
?All Values:
Buckets with endpoints < value / Total No. of buckets
Histogram Examples
Table TAB1 SQL> desc tab1 Name Null? Type ------------------------------- -------- ---- A NUMBER(6) B NUMBER(6)
Column A contains unique values from 1 to 10000.
Column B contains 10 distinct values.
The value '5' occurs 9991 times.
Values '1, 2, 3, 4, 9996, 9997, 9998, 9999, 10000' occur only once.
i.e.
select distinct B , count(*)
from HTAB1
group by B
order by B
;
B COUNT(*)
---------- ----------
1 1
2 1
3 1
4 1
5 9991
9996 1
9997 1
9998 1
9999 1
10000 1
10 rows selected.
There is an index on Column B.
Statistics are gathered without Histograms using:
exec DBMS_STATS.GATHER_TABLE_STATS (NULL,'HTAB1', method_opt => 'FOR ALL COLUMNS SIZE 1');
Setup:
drop table HTAB1;
create table HTAB1 (a number, b number);
Insert into HTAB1 ( A,B) values ( 1,1);
Insert into HTAB1 ( A,B) values ( 2,2);
Insert into HTAB1 ( A,B) values ( 3,3);
Insert into HTAB1 ( A,B) values ( 4,4);
Insert into HTAB1 ( A,B) values ( 9996,9996);
Insert into HTAB1 ( A,B) values ( 9997,9997);
Insert into HTAB1 ( A,B) values ( 9998,9998);
Insert into HTAB1 ( A,B) values ( 9999,9999);
Insert into HTAB1 ( A,B) values ( 10000,10000);
commit;
begin
for i in 5 .. 9995 loop
Insert into HTAB1 ( A,B)
values ( i,5);
if (mod(i,100) = 0) then
commit;
end if;
end loop;
commit;
end;
/
commit;
create index HTAB1_B on HTAB1(b);
exec DBMS_STATS.GATHER_TABLE_STATS (NULL,'HTAB1', method_opt => 'FOR ALL COLUMNS SIZE 1');
alter session set OPTIMIZER_DYNAMIC_SAMPLING = 0;
Function to convert raw data in to numeric data:
create or replace function raw_to_number(my_input raw)
return number
as
my_output number;
begin
dbms_stats.convert_raw_value(my_input,my_output);
return my_output;
end;
/
This results in statistics as follows:
column COLUMN_NAME format a5 heading COL
column NUM_DISTINCT format 99990
column LOW_VALUE format 99990
column HIGH_VALUE format 99990
column DENSITY format 99990
column NUM_NULLS format 99990
column NUM_BUCKETS format 99990
column SAMPLE_SIZE format 99990
select COLUMN_NAME,NUM_DISTINCT,raw_to_number(LOW_VALUE) Low,raw_to_number(HIGH_VALUE) High,DENSITY,NUM_NULLS,
NUM_BUCKETS,LAST_ANALYZED,SAMPLE_SIZE,HISTOGRAM
from user_tab_columns
where table_name = 'HTAB1';
COL NUM_DISTINCT LOW HIGH DENSITY NUM_NULLS NUM_BUCKETS LAST_ANALYZED SAMPLE_SIZE HISTOGRAM
----- ------------ ---------- ---------- ------- --------- ----------- -------------------- ----------- ---------------
A 10000 1 10000 0 0 1 31-jan-2013 09:32:08 10000 NONE
B 10 1 10000 0 0 1 31-jan-2013 09:32:08 10000 NONE
select lpad(TABLE_NAME,10) TAB, lpad(COLUMN_NAME, 10) COL,
ENDPOINT_NUMBER, ENDPOINT_VALUE
from user_histograms
where table_name='HTAB1'
order by COL, ENDPOINT_NUMBER;
TAB COL ENDPOINT_NUMBER ENDPOINT_VALUE
---------- ---------- --------------- --------------
HTAB1 A 0 1
HTAB1 A 1 10000
HTAB1 B 0 1
HTAB1 B 1 10000
In the above you can see that the statistics gathering has not created a histogram. There is a single bucket and high and a low ENDPOINT_NUMBER for each column value ( you will always get 2 entries in USER_HISTOGRAMS for each column, for the high and low values respectively).
Test queries:
select * from htab1 where b=5;
select * from htab1 where b=3;
To replicate the tests you will need to disable OPTIMIZER_DYNAMIC_SAMPLING
alter session set OPTIMIZER_DYNAMIC_SAMPLING = 0;
See:
Document 336267.1 Optimizer Dynamic Sampling (OPTIMIZER_DYNAMIC_SAMPLING)Without Histograms, both queries do an INDEX RANGE SCAN because the optimizer believes that the data is uniformly distributed in column B and that each predicate with return 1/10th of the values because there are 10 distinct values:
--------------------------------------------------------------------------------------- | Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time | --------------------------------------------------------------------------------------- | 0 | SELECT STATEMENT | | 1111 | 6666 | 5 (0)| 00:00:01 | | 1 | TABLE ACCESS BY INDEX ROWID| HTAB1 | 1111 | 6666 | 5 (0)| 00:00:01 | |* 2 | INDEX RANGE SCAN | HTAB1_B | 1111 | | 3 (0)| 00:00:01 | ---------------------------------------------------------------------------------------
In fact it may be preferable to use a Full Table Scan for the select where b=5 and index lookups for the others.
Gathering Histogram Statistics
If we collect histogram statistics with the recommended settings:
exec DBMS_STATS.GATHER_TABLE_STATS (NULL,'HTAB1', method_opt => 'FOR ALL COLUMNS SIZE AUTO');
The b=5 query now does a Full Table Scan
select * from htab1 where b=5; --------------------------------------------------------------------------- | Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time | --------------------------------------------------------------------------- | 0 | SELECT STATEMENT | | 9991 | 69937 | 7 (0)| 00:00:01 | |* 1 | TABLE ACCESS FULL| HTAB1 | 9991 | 69937 | 7 (0)| 00:00:01 | ---------------------------------------------------------------------------
The query where B is 3 still uses an index:
select * from htab1 where b=3; --------------------------------------------------------------------------------------- | Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time | --------------------------------------------------------------------------------------- | 0 | SELECT STATEMENT | | 1 | 7 | 2 (0)| 00:00:01 | | 1 | TABLE ACCESS BY INDEX ROWID| HTAB1 | 1 | 7 | 2 (0)| 00:00:01 | |* 2 | INDEX RANGE SCAN | HTAB1_B | 1 | | 1 (0)| 00:00:01 | ---------------------------------------------------------------------------------------
This is because a FREQUENCY Histogram has been created:
COL NUM_DISTINCT LOW HIGH DENSITY NUM_NULLS NUM_BUCKETS LAST_ANALYZED SAMPLE_SIZE HISTOGRAM
----- ------------ ---------- ---------- ------- --------- ----------- -------------------- ----------- ---------------
A 10000 1 10000 0 0 1 31-jan-2013 09:58:01 10000 NONE
B 10 1 10000 0 0 10 31-jan-2013 09:58:01 10000 FREQUENCY
TAB COL ENDPOINT_NUMBER ENDPOINT_VALUE
---------- ---------- --------------- --------------
HTAB1 A 0 1
HTAB1 A 1 10000
HTAB1 B 1 1
HTAB1 B 2 2
HTAB1 B 3 3
HTAB1 B 4 4
HTAB1 B 9995 5
HTAB1 B 9996 9996
HTAB1 B 9997 9997
HTAB1 B 9998 9998
HTAB1 B 9999 9999
HTAB1 B 10000 10000
12 rows selected.
On Column B there are 10 buckets matching up with the 10 distinct values.
The ENDPOINT_VALUE shows the column value and the ENDPOINT_NUMBER shows the cumulative number of rows. So the number of rows for ENDPOINT_VALUE 2, it has an ENDPOINT_NUMBER 2, the previous ENDPOINT_NUMBER is 1, hence the number of rows with value 2 is 1. Another example is ENDPOINT_VALUE 5. Its ENDPOINT_NUMBER is 9995. The previous bucket ENDPOINT_NUMBER is 4, so 9995 - 4 = 9991 rows containing the value 5.
Frequency histograms work fine with a low number of distinct values, but when the number exceeds the maximum number of buckets, you cannot create a bucket for each value. In this case the Optimizer creates Height balanced histograms.
Height Balanced Histograms
You can demonstrate this situation by forcing the optimizer to create fewer buckets than the Number of Distinct Values. i.e. using 8 buckets for 10 Distinct Values:
exec DBMS_STATS.GATHER_TABLE_STATS (NULL,'HTAB1', method_opt => 'FOR COLUMNS B SIZE 8');
So now we have gathered a HEIGHT BALANCED HISTOGRAM for Column B:
COL NUM_DISTINCT LOW HIGH DENSITY NUM_NULLS NUM_BUCKETS LAST_ANALYZED SAMPLE_SIZE HISTOGRAM
----- ------------ ---------- ---------- ------- --------- ----------- -------------------- ----------- ---------------
A 10000 1 10000 0 0 1 31-jan-2013 09:58:01 10000 NONE
B 10 1 10000 0 0 8 31-jan-2013 09:59:09 10000 HEIGHT BALANCED
TAB COL ENDPOINT_NUMBER ENDPOINT_VALUE
---------- ---------- --------------- --------------
HTAB1 A 0 1
HTAB1 A 1 10000
HTAB1 B 0 1
HTAB1 B 7 5
HTAB1 B 8 10000
Notice that there are 8 Buckets against B now.
Oracle puts the same number of values in each bucket and records the endpoint of each bucket.
With HEIGHT BALANCED Histograms, the ENDPOINT_NUMBER is the actual bucket number and ENDPOINT_VALUE is the endpoint value of the bucket determined by the column value.
From the above, bucket 0 holds the low value for the column.
Because buckets 1-7 have the same endpoint, Oracle does not store all these rows to save space. But we have: bucket 1 with an endpoint of 5, bucket 2 with an endpoint of 5, bucket 3 with an endpoint of 5, bucket 4 with an endpoint of 5, bucket 5 with an endpoint of 5, bucket 6 with an endpoint of 5, bucket 7 with an endpoint of 5 AND bucket 8 with an endpoint of 10000 So bucket 1 contains values between 1 and 5, bucket 8 contains values between 5 and 10000.
All buckets contain the same number of values (which is why they are called height-balanced histograms), except the last bucket may have fewer values then the other buckets.
Storing Character Values in Histograms
For character columns, Oracle only stores the first 32 bytes of any string (there are also limits on numeric columns, but these are less frequently an issue since the majority of numbers are insufficiently large to encounter any problems). See:
Document 212809.1 Limitations of the Oracle Cost Based OptimizerAny predicates that contain strings greater than 32 characters will not use histogram information and the selectivity will be 1 / Number of DISTINCT Values. Data in histogram endpoints is normalized to double precision floating point arithmetic.
For Example
SQL> select * from example; A ---------- a b c d e e e e
The table contains 5 distinct values. There is one occurence of 'a', 'b', 'c' and 'd' There are 4 occurrences of 'e'. If we create a histogram: Looking in user_histograms:
TABLE COL ENDPOINT_NUMBER ENDPOINT_VALUE ---------- ----- --------------- -------------- EXAMPLE A 1 5.0365E+35 EXAMPLE A 2 5.0885E+35 EXAMPLE A 3 5.1404E+35 EXAMPLE A 4 5.1923E+35 EXAMPLE A 8 5.2442E+35
So:
ENDPOINT_VALUE 5.0365E+35 represents a 5.0885E+35 represents b 5.1404E+35 represents c 5.1923E+35 represents d 5.2442E+35 represents e
Then, if you look at the cumulative values for ENDPOINT_NUMBER, the corresponding ENDPOINT_VALUE's are correct.