这篇文章给大家分享的是有关如何使用Mysql位图索引解决用户画像问题的内容。小编觉得挺实用的,因此分享给大家做个参考,一起跟随小编过来看看吧。每个bigint类型包括60个记录的位信息.但是第0位表示第六十
这篇文章给大家分享的是有关如何使用Mysql位图索引解决用户画像问题的内容。小编觉得挺实用的,因此分享给大家做个参考,一起跟随小编过来看看吧。
每个bigint类型包括60个记录的位信息.
但是第0位表示第六十个记录的位
第1位至第59位表示第一至五十九的记录的位信息.
这样记录的位信息保存并不连续,
使用的时候还得把最右边的一位挪到最左边,不好理解还非常麻烦,性能也有损耗.
经过王工的改良,
使用如下sql替换,则保存的位信息就连续了
SELECT
CEIL(id / 60) g60,
CEIL(id / 1200) g1200,
age grouped,
COUNT(*) total,
BIT_OR(1 << (if(MOD(id, 60)=0,60,MOD(id, 60)))) bitmap
FROM
o_huaxiang_big_0 o
GROUP BY g1200 , g60 , age
创建位图索引的整体SQL如下
truncate table bitmap20_0;
insert into bitmap20_0
select
'o_huaxiang_big' table_name,
'umc_sex' column_name,
((g1200-1)*60)*20 min_id,
((g1200-1)*60)*20+1200 max_id,
v2.*
from (
select
g1200,
grouped,
sum(total) total,
ifnull(max(case when abs((g1200-1)*20-g60)=20 then bitmap else null end),0) c20,
ifnull(max(case when abs((g1200-1)*20-g60)=19 then bitmap else null end),0) c19,
ifnull(max(case when abs((g1200-1)*20-g60)=18 then bitmap else null end),0) c18,
ifnull(max(case when abs((g1200-1)*20-g60)=17 then bitmap else null end),0) c17,
ifnull(max(case when abs((g1200-1)*20-g60)=16 then bitmap else null end),0) c16,
ifnull(max(case when abs((g1200-1)*20-g60)=15 then bitmap else null end),0) c15,
ifnull(max(case when abs((g1200-1)*20-g60)=14 then bitmap else null end),0) c14,
ifnull(max(case when abs((g1200-1)*20-g60)=13 then bitmap else null end),0) c13,
ifnull(max(case when abs((g1200-1)*20-g60)=12 then bitmap else null end),0) c12,
ifnull(max(case when abs((g1200-1)*20-g60)=11 then bitmap else null end),0) c11,
ifnull(max(case when abs((g1200-1)*20-g60)=10 then bitmap else null end),0) c10,
ifnull(max(case when abs((g1200-1)*20-g60)=9 then bitmap else null end),0) c9,
ifnull(max(case when abs((g1200-1)*20-g60)=8 then bitmap else null end),0) c8,
ifnull(max(case when abs((g1200-1)*20-g60)=7 then bitmap else null end),0) c7,
ifnull(max(case when abs((g1200-1)*20-g60)=6 then bitmap else null end),0) c6,
ifnull(max(case when abs((g1200-1)*20-g60)=5 then bitmap else null end),0) c5,
ifnull(max(case when abs((g1200-1)*20-g60)=4 then bitmap else null end),0) c4,
ifnull(max(case when abs((g1200-1)*20-g60)=3 then bitmap else null end),0) c3,
ifnull(max(case when abs((g1200-1)*20-g60)=2 then bitmap else null end),0) c2,
ifnull(max(case when abs((g1200-1)*20-g60)=1 then bitmap else null end),0) c1
from (
SELECT
CEIL(id / 60) g60,
CEIL(id / 1200) g1200,
umc_sex grouped,
COUNT(*) total,
BIT_OR(1 << (if(MOD(id, 60)=0,60,MOD(id, 60)))) bitmap
FROM
o_huaxiang_big_0 o
GROUP BY g1200 , g60 , umc_sex
) v1 group by g1200,grouped
) v2;
insert into bitmap20_0
select
'o_huaxiang_big' table_name,
'age' column_name,
((g1200-1)*60)*20 min_id,
((g1200-1)*60)*20+1200 max_id,
v2.*
from (
select
g1200,
grouped,
sum(total) total,
ifnull(max(case when abs((g1200-1)*20-g60)=20 then bitmap else null end),0) c20,
ifnull(max(case when abs((g1200-1)*20-g60)=19 then bitmap else null end),0) c19,
ifnull(max(case when abs((g1200-1)*20-g60)=18 then bitmap else null end),0) c18,
ifnull(max(case when abs((g1200-1)*20-g60)=17 then bitmap else null end),0) c17,
ifnull(max(case when abs((g1200-1)*20-g60)=16 then bitmap else null end),0) c16,
ifnull(max(case when abs((g1200-1)*20-g60)=15 then bitmap else null end),0) c15,
ifnull(max(case when abs((g1200-1)*20-g60)=14 then bitmap else null end),0) c14,
ifnull(max(case when abs((g1200-1)*20-g60)=13 then bitmap else null end),0) c13,
ifnull(max(case when abs((g1200-1)*20-g60)=12 then bitmap else null end),0) c12,
ifnull(max(case when abs((g1200-1)*20-g60)=11 then bitmap else null end),0) c11,
ifnull(max(case when abs((g1200-1)*20-g60)=10 then bitmap else null end),0) c10,
ifnull(max(case when abs((g1200-1)*20-g60)=9 then bitmap else null end),0) c9,
ifnull(max(case when abs((g1200-1)*20-g60)=8 then bitmap else null end),0) c8,
ifnull(max(case when abs((g1200-1)*20-g60)=7 then bitmap else null end),0) c7,
ifnull(max(case when abs((g1200-1)*20-g60)=6 then bitmap else null end),0) c6,
ifnull(max(case when abs((g1200-1)*20-g60)=5 then bitmap else null end),0) c5,
ifnull(max(case when abs((g1200-1)*20-g60)=4 then bitmap else null end),0) c4,
ifnull(max(case when abs((g1200-1)*20-g60)=3 then bitmap else null end),0) c3,
ifnull(max(case when abs((g1200-1)*20-g60)=2 then bitmap else null end),0) c2,
ifnull(max(case when abs((g1200-1)*20-g60)=1 then bitmap else null end),0) c1
from (
SELECT
CEIL(id / 60) g60,
CEIL(id / 1200) g1200,
age grouped,
COUNT(*) total,
BIT_OR(1 << (if(MOD(id, 60)=0,60,MOD(id, 60)))) bitmap
FROM
o_huaxiang_big_0 o
GROUP BY g1200 , g60 , age
) v1 group by g1200,grouped
) v2;
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本文标题: 如何使用MySQL位图索引解决用户画像问题
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