前文的问题

第一版:
http://blog.itpub.net/29254281/viewspace-2157111/

第二版:
http://blog.itpub.net/29254281/viewspace-2157209/

第二版用时 33秒左右.

在原来的基础上,稍加改进,即可提升三分之一的性能.

select query_time,d,max(ts) ts from (
select t2.query_time,ts,rn,round(rn/total,10) percent,
case
when 0.71>=round(rn/total,10) then 0.71
when 0.81>=round(rn/total,10) then 0.81
when 0.91>=round(rn/total,10) then 0.91
end d
from (
select query_time,ts,
case when @gid=query_time then @rn:=@rn+1 when @gid:=query_time then @rn:=1 end rn
from (
select * from t ,(select @gid:='',@rn:=0) vars order by query_time,ts
) t1
) t2 inner join (
select query_time,count(*) total from t group by query_time
) t3 on(t2.query_time=t3.query_time)
where round(rn/total,10)>=0.71
) t6
where d is not null
group by query_time,d

where round(rn/total,10)>=0.71
即用定义的最小的百分位数进行过滤后,再group by

此时查询时间可以低至 20.531 s

当然，这个SQL还有进一步提升的空间

计算某个百分位数的位置,有如下的公式:
loc=1+（n-1）*p，n是元素数，p是分位点。loc大小介于1和n之间

那么SQL可以进行如下优化

select t5.query_time,t5.ts,t2.v from (

    select query_time,total,v, floor(1+(total-1)*v) rn

    from (

         select query_time,count(*) total from t group by query_time

    ) t3, (select 0.71 v,1 seq union all select 0.81,2 union all select 0.91,3) t4

)

t2 inner join (

    select

    query_time,

    case when @gid=query_time then @rn:=@rn+1 when @gid:=query_time then @rn:=1 end rn,

    ts

    from (

        select * from t ,(select @gid:='',@rn:=0) vars order by query_time,ts

    ) t1

) t5 on (t2.query_time=t5.query_time and t2.rn=t5.rn )

除了本身简化了SQL复杂度,查询时间也低至 15秒左右

ITPub博客

MySQL百分位数计算(再优化版)

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