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MySQL位图索引如何解决用户画像问题

发表于:2025-01-20 作者:千家信息网编辑
千家信息网最后更新 2025年01月20日,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。用户画像的原始表,有一亿记录,1
千家信息网最后更新 2025年01月20日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。

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)) )  cfrom (        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

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