MySQL位图索引如何解决用户画像问题
这篇文章主要介绍了MySQL位图索引如何解决用户画像问题,具有一定借鉴价值,感兴趣的朋友可以参考下,希望大家阅读完这篇文章之后大有收获,下面让小编带着大家一起了解一下。
在甘泉等地区,都构建了全面的区域性战略布局,加强发展的系统性、市场前瞻性、产品创新能力,以专注、极致的服务理念,为客户提供成都网站建设、网站设计 网站设计制作定制网站建设,公司网站建设,企业网站建设,成都品牌网站建设,成都全网营销推广,外贸营销网站建设,甘泉网站建设费用合理。
用户画像的原始表,有一亿记录,100多个维度(100多个列),比如年龄,性别,爱好,是否有车,是否有房什么的.
测试环境800w数据,大概在5G左右
需要解决的问题一 :在100列中任选N列,过滤查询,执行时间小于一秒。实际上N一般在5到10
即类似
select * from 画像表 where 性别=‘男’ and 年龄 between 20 and 30 and 有车='yes' and 有房='yes' and 已婚='no'
问题二:全体数据的随意聚合,执行时间小于5秒
比如
select 年龄,性别,count(*) from 画像表 group by 年龄,性别
在数据库解决这个问题有一些麻烦,传统建索引优化的方式不起作用了。
100多个列随意选择几列查询,索引不可能提前建出这么多.
先看测试数据
CREATE TABLE `o_huaxiang_big` ( `id` bigint(20) NOT NULL AUTO_INCREMENT, `user_id` bigint(20) DEFAULT NULL, `umc_sex` varchar(20) DEFAULT NULL, `age` varchar(30) DEFAULT NULL, PRIMARY KEY (`id`) ) ;
处理这个问题,我自然想到模拟一个位图.
一般画像数据有几种类型
1.数值类型
2.日期类型
3.日期时间类型
4.字符串类型
其中 日期和字符串类型可以作为离散值,
日期时间类型也可以转化为日期类型,作为离散值处理。
数值类型比较麻烦,需要人为介入判断是否是离散值,如果不是还需要划分范围。
总之,所有的值都要映射为离散值
然后以上图前5个数据为例,将离散值映射为位图
男 0 0 0 0 1
未知 1 0 0 1 0
女 0 1 1 0 0
一个bigint 是8字节的,为了取整,我存放60个记录的位信息。
然后建位图表如下
CREATE TABLE `bitmap20` ( `table_name` varchar(32) NOT NULL DEFAULT '' comment '位图表记录的原始表名称', `column_name` varchar(32) NOT NULL DEFAULT '' comment '列名称', `min_id` int(11) DEFAULT NULL comment '起始ID', `max_id` int(11) DEFAULT NULL comment '终止ID', `gid` int(11) NOT NULL DEFAULT '0' comment '分组ID,每组1200记录' , `grouped` varchar(32) NOT NULL DEFAULT '' comment '离散值', `total` bigint(21) NOT NULL DEFAULT '0' comment '总数', `c20` bigint(20) NOT NULL DEFAULT '0', `c19` bigint(20) NOT NULL DEFAULT '0', `c18` bigint(20) NOT NULL DEFAULT '0', `c17` bigint(20) NOT NULL DEFAULT '0', `c16` bigint(20) NOT NULL DEFAULT '0', `c15` bigint(20) NOT NULL DEFAULT '0', `c14` bigint(20) NOT NULL DEFAULT '0', `c13` bigint(20) NOT NULL DEFAULT '0', `c12` bigint(20) NOT NULL DEFAULT '0', `c11` bigint(20) NOT NULL DEFAULT '0', `c10` bigint(20) NOT NULL DEFAULT '0', `c9` bigint(20) NOT NULL DEFAULT '0', `c8` bigint(20) NOT NULL DEFAULT '0', `c7` bigint(20) NOT NULL DEFAULT '0', `c6` bigint(20) NOT NULL DEFAULT '0', `c5` bigint(20) NOT NULL DEFAULT '0', `c4` bigint(20) NOT NULL DEFAULT '0', `c3` bigint(20) NOT NULL DEFAULT '0', `c2` bigint(20) NOT NULL DEFAULT '0', `c1` bigint(20) NOT NULL DEFAULT '0', PRIMARY KEY (`column_name`,`gid`,`grouped`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 ROW_FORMAT=COMPRESSED comment '位图表';
c1-c20,一共20个bigint类型的字段,每个bigint记录60个位信息。
也就是位图表每行存储1200个原始记录的位图信息,并且位图表启用了压缩。
测试环境
4C 8G内存(innodb buffer 2G) SSD硬盘
800万原始画像数据,占用硬盘5G
初始化位图表
insert into bitmap20 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 << (MOD(id, 60))) bitmap FROM o_huaxiang_big o GROUP BY g1200 , g60 , umc_sex ) v1 group by g1200,grouped ) v2; insert into bitmap20 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 << (MOD(id, 60))) bitmap FROM o_huaxiang_big o GROUP BY g1200 , g60 , age ) v1 group by g1200,grouped ) v2;
性别和年龄的初始化分别耗时36秒和49秒
两个维度的索引占用磁盘40M
聚合查询,800万数据耗时1.7秒.因为是CPU密集型操作,IO非常小,所以可以通过多线程再优化.
select t1p,t2p,sum(total) from ( select t1.grouped t1p, t2.grouped t2p, bit_count(t1.c1&t2.c1) + bit_count(t1.c2&t2.c2) + bit_count(t1.c3&t2.c3) + bit_count(t1.c4&t2.c4) + bit_count(t1.c5&t2.c5) + bit_count(t1.c6&t2.c6) + bit_count(t1.c7&t2.c7) + bit_count(t1.c8&t2.c8) + bit_count(t1.c9&t2.c9) + bit_count(t1.c10&t2.c10) + bit_count(t1.c11&t2.c11) + bit_count(t1.c12&t2.c12) + bit_count(t1.c13&t2.c13) + bit_count(t1.c14&t2.c14) + bit_count(t1.c15&t2.c15) + bit_count(t1.c16&t2.c16) + bit_count(t1.c17&t2.c17) + bit_count(t1.c18&t2.c18) + bit_count(t1.c19&t2.c19) + bit_count(t1.c20&t2.c20) total from bitmap20 t1 inner join bitmap20 t2 on(t1.gid=t2.gid) where t1.column_name='umc_sex' and t2.column_name='age' ) t3 where total>0 group by t1p,t2p
还有一个问题是过滤
select max_id , concat( concat(right(c20,1),left(c20,59)) , concat(right(c19,1),left(c19,59)) , concat(right(c18,1),left(c18,59)) , concat(right(c17,1),left(c17,59)) , concat(right(c16,1),left(c16,59)) , concat(right(c15,1),left(c15,59)) , concat(right(c14,1),left(c14,59)) , concat(right(c13,1),left(c13,59)) , concat(right(c12,1),left(c12,59)) , concat(right(c11,1),left(c11,59)) , concat(right(c10,1),left(c10,59)) , concat(right(c9,1),left(c9,59)) , concat(right(c8,1),left(c8,59)) , concat(right(c7,1),left(c7,59)) , concat(right(c6,1),left(c6,59)) , concat(right(c5,1),left(c5,59)) , concat(right(c4,1),left(c4,59)) , concat(right(c3,1),left(c3,59)) , concat(right(c2,1),left(c2,59)) , concat(right(c1,1),left(c1,59)) ) c from ( select gid,min_id,max_id, lpad(conv(bit_and(c20),10,2),60,'0') c20, lpad(conv(bit_and(c19),10,2),60,'0') c19, lpad(conv(bit_and(c18),10,2),60,'0') c18, lpad(conv(bit_and(c17),10,2),60,'0') c17, lpad(conv(bit_and(c16),10,2),60,'0') c16, lpad(conv(bit_and(c15),10,2),60,'0') c15, lpad(conv(bit_and(c14),10,2),60,'0') c14, lpad(conv(bit_and(c13),10,2),60,'0') c13, lpad(conv(bit_and(c12),10,2),60,'0') c12, lpad(conv(bit_and(c11),10,2),60,'0') c11, lpad(conv(bit_and(c10),10,2),60,'0') c10, lpad(conv(bit_and(c9),10,2),60,'0') c9, lpad(conv(bit_and(c8),10,2),60,'0') c8, lpad(conv(bit_and(c7),10,2),60,'0') c7, lpad(conv(bit_and(c6),10,2),60,'0') c6, lpad(conv(bit_and(c5),10,2),60,'0') c5, lpad(conv(bit_and(c4),10,2),60,'0') c4, lpad(conv(bit_and(c3),10,2),60,'0') c3, lpad(conv(bit_and(c2),10,2),60,'0') c2, lpad(conv(bit_and(c1),10,2),60,'0') c1 from bitmap20 where ( (column_name='umc_sex' and grouped='未知') or (column_name='age' and grouped='117')) group by gid,min_id,max_id having count(distinct column_name)=2 ) v1
用max_id 减去 ‘1’在c字符串的位置,就是原始的ID
感谢你能够认真阅读完这篇文章,希望小编分享的“MySQL位图索引如何解决用户画像问题”这篇文章对大家有帮助,同时也希望大家多多支持创新互联,关注创新互联行业资讯频道,更多相关知识等着你来学习!
网站标题:MySQL位图索引如何解决用户画像问题
文章链接:http://scyanting.com/article/gecsdc.html