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Code cleanup:

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- Move [some] engine-agnostic tests from t/selectivity.test to t/selectivity_no_engine.test
- Move Histogram::point_selectivity to sql_statistics.cc
This commit is contained in:
Sergey Petrunya
2014-03-27 13:08:00 +04:00
parent 0d67aafaa2
commit 79a8a6130b
7 changed files with 337 additions and 369 deletions
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112
sql/sql_statistics.h
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@ -241,120 +241,10 @@ public:
return sel;
}
/*
Estimate selectivity of "col=const" using a histogram
@param pos Position of the "const" between column's min_value and
max_value. This is a number in [0..1] range.
@param avg_sel Average selectivity of condition "col=const" in this table.
It is calcuated as (#non_null_values / #distinct_values).
@return
Expected condition selectivity (a number between 0 and 1)
@notes
[re_zero_length_buckets] If a bucket with zero value-length is in the
middle of the histogram, we will not have min==max. Example: suppose,
pos_value=0x12, and the histogram is:
#n #n+1 #n+2
... 0x10 0x12 0x12 0x14 ...
|
+------------- bucket with zero value-length
Here, we will get min=#n+1, max=#n+2, and use the multi-bucket formula.
The problem happens at the histogram ends. if pos_value=0, and the
histogram is:
0x00 0x10 ...
then min=0, max=0. This means pos_value is contained within bucket #0,
but on the other hand, histogram data says that the bucket has only one
value.
*/
double point_selectivity(double pos, double avg_sel)
{
double sel;
/* Find the bucket that contains the value 'pos'. */
uint min= find_bucket(pos, TRUE);
uint pos_value= (uint) (pos * prec_factor());
/* Find how many buckets this value occupies */
uint max= min;
while (max + 1 < get_width() && get_value(max + 1) == pos_value)
max++;
/*
A special case: we're looking at a single bucket, and that bucket has
zero value-length. Use the multi-bucket formula (attempt to use
single-bucket formula will cause divison by zero).
For more details see [re_zero_length_buckets] above.
*/
if (max == min && get_value(max) == ((max==0)? 0 : get_value(max-1)))
max++;
if (max > min)
{
/*
The value occupies multiple buckets. Use start_bucket ... end_bucket as
selectivity.
*/
double bucket_sel= 1.0/(get_width() + 1);
sel= bucket_sel * (max - min + 1);
}
else
{
/*
The value 'pos' fits within one single histogram bucket.
Histogram buckets have the same numbers of rows, but they cover
different ranges of values.
We assume that values are uniformly distributed across the [0..1] value
range.
*/
/*
If all buckets covered value ranges of the same size, the width of
value range would be:
*/
double avg_bucket_width= 1.0 / (get_width() + 1);
/*
Let's see what is the width of value range that our bucket is covering.
(min==max currently. they are kept in the formula just in case we
will want to extend it to handle multi-bucket case)
*/
double inv_prec_factor= (double) 1.0 / prec_factor();
double current_bucket_width=
(max + 1 == get_width() ? 1.0 : (get_value(max) * inv_prec_factor)) -
(min == 0 ? 0.0 : (get_value(min-1) * inv_prec_factor));
DBUG_ASSERT(current_bucket_width); /* We shouldn't get a one zero-width bucket */
/*
So:
- each bucket has the same #rows
- values are unformly distributed across the [min_value,max_value] domain.
If a bucket has value range that's N times bigger then average, than
each value will have to have N times fewer rows than average.
*/
sel= avg_sel * avg_bucket_width / current_bucket_width;
/*
(Q: if we just follow this proportion we may end up in a situation
where number of different values we expect to find in this bucket
exceeds the number of rows that this histogram has in a bucket. Are
we ok with this or we would want to have certain caps?)
*/
}
return sel;
}
double point_selectivity(double pos, double avg_sel);
};