分析导览：分析目的

1-统计不同性别的消费频次、消费金额

2-统计不同年龄段用户的消费金额

3-统计不同月份的消费金额

4-统计多次消费的用户，第一次和最后一次消费时间的间隔

5-统计不同年龄段的用户消费金额

6-统计消费的二八法则，消费top20%的用户，贡献了多少额度分析过程

(一)文件导入

文件导入具体方法见如何在mySQL中文件导入CSV文件，这里就不在赘述。

主要步骤分为建表、导入文件，为了避免遇到很多坑，可以间我前一篇文章，这里的数据量有10w条，用命令行导入很方便。

创建的两张表的字段：

user：userid, sex,birth

orderinfo: orderid,userid,ispaid,price,paidtime

(二)用户消费者行为分析

1、统计不同月份的下单人数

select CONCAT_WS('-',left(paidtime,4),mid(paidtime,6,1)) as month_,count(distinct userid) as paidorder from test.orderinfo

where ispaid='已支付'

group by left(paidtime,4),mid(paidtime,6,1);

注：这里用到字符串拼接和选取字符串的函数，concat_ws,left,mid,还可以用substring函数。

从结果中看出，三四月份的下单人数最多，5月份竟然只有6个人。

2、统计男女的消费频次

思路是先将两个表联结，求出每个id的下单次数，再在该表基础上对男女进行分组，统计。这里有个需要注意的，我们可以统计下单的男女频次，也可以统计已支付和未支付的频次。

select a.userid,b.sex,count(1)as cs from test.orderinfo a,test.user b where a.userid=b.userid and ISPAID="已支付" and b.sex is not null

group by a.userid;

select c.sex ,round(avg(c.cs),2) as 平均次数 from (select a.userid,b.sex,count(1)as cs from test.orderinfo a,test.user b where a.userid=b.userid and ISPAID="已支付" and b.sex is not null

group by a.userid,b.sex) c group by c.sex ;

以上是一年男女平均频次，以下是以月为单位男女的平均频次

select t.month,t.sex,round(avg(t.ct),2) as 频次 from (

select left(a.paidtime,6) as month,a.userid,count(a.userid) as ct,b.sex

from test.orderinfo a,test.user b where a.ISPAID='已支付' and a.userid=b.userid

and b.sex is not null and left(a.paidtime,6) !='2016/5'

group by left(a.paidtime,6),a.userid) t

group by t.month,t.sex ;

3、统计不同年龄段的消费频次

先给不同年纪用户划分年龄段，这里用到一个新函数timestampdiff。返回时间序列的最大、最小时间差，由于我的时间是字符串类型，我用现在的时间-年份字段得到年龄。

CEIL(X) 返回不小于X的最小整数值。(天花板)

FLOOR(X) 返回不大于X的最大整数值。(地板)

这里由于我的时间是字符串，所以我用case when多条件查询，

select userid,year(now())-left(birth,6) as age,

case when year(now())-left(birth,6)<=12 then '0-12岁'

when year(now())-left(birth,6)<=25 then '13-25岁'

when year(now())-left(birth,6)<=30 then '26-30岁'

when year(now())-left(birth,6)<=50 then '30-50岁'

when year(now())-left(birth,6)<=70 then '51-70岁'

else 'over70' end as al

from test.user ;

先求出每个age对应的年龄段，再求出每个年龄段的次数。

select t.al,count(t.ct) as 次数 from

(select a.userid,count(a.userid) as ct,year(now())-left(b.birth,6) as age,

case when year(now())-left(b.birth,6)<=12 then '0-12岁'

when year(now())-left(b.birth,6)<=25 then '13-25岁'

when year(now())-left(b.birth,6)<=30 then '26-30岁'

when year(now())-left(b.birth,6)<=50 then '30-50岁'

when year(now())-left(b.birth,6)<=70 then '51-70岁'

else 'over70' end as al

from test.orderinfo a,test.user b

where a.userid= b.userid and a.ispaid='已支付'

and year(now())-left(b.birth,6) !='2020'

group by a.userid) t

group by t.al;

select t.al,avg(t.ct) as 频次 from

(select a.userid,count(a.userid) as ct,year(now())-left(b.birth,6) as age,

case when year(now())-left(b.birth,6)<=12 then '0-12岁'

when year(now())-left(b.birth,6)<=25 then '13-25岁'

when year(now())-left(b.birth,6)<=30 then '26-30岁'

when year(now())-left(b.birth,6)<=50 then '30-50岁'

when year(now())-left(b.birth,6)<=70 then '51-70岁'

else 'over70' end as al

from test.orderinfo a,test.user b

where a.userid= b.userid and a.ispaid='已支付'

and year(now())-left(b.birth,6) !='2020'

group by a.userid) t

group by t.al

order by avg(t.ct) desc;

4、消费top20%的用户贡献了多少消费额度

计算每个用户消费的总额，

select userid,sum(price) as sum from test.orderinfo

where ispaid ="已支付" group by userid order by sum desc;

top20用户的数量和消费额度

select floor(count(1)*0.2) from (select userid,sum(price) as sum

from test.orderinfo where ispaid ="已支付"

group by userid order by sum desc) t;

select sum(t.sum) as top20 from

(select userid,sum(price) as sum from test.orderinfo where ispaid ="已支付"

group by userid order by sum desc limit 17129 ) t;

计算全部的消费额

用top20的消费额/消费总额=2700/3683 约为73.9%。

select a.userid,a.sum,round(a.sum/(select sum(price) as total from test.orderinfo) ,4) as 贡献率 from

(select userid,sum(price) as sum

from test.orderinfo where ispaid ="已支付"

group by userid order by sum desc) a

order by a.sum desc;

5、统计多次消费用户，第一次和最后一次的消费间隔

select t.userid,timestampdiff(day,min(t.time),max(t.time)) as timediif from

(select userid,str_to_date(concat(left(paidtime,4),'-0',mid(paidtime,6,1),'-0',mid(paidtime,8,1)),'%Y-%m-%d') as time

from test.orderinfo) t

where t.time is not null

group by t.userid

having timestampdiff(day,min(t.time),max(t.time))!= 0;

这里用到了字符串拼接函数concat()，时间间隔函数timestampdiff()，以天为单位，消费多次的用户的时间间隔

6、统计三月份消费的复购率和回购率

复购率：本月消费一次以上用户的占比

思路：对用户进行分组，用总的用户数>1的数量/总数

select count(if (t.ct>1,1,null))/count(1) as 复购率

from (select userid,count(userid) as ct from test.orderinfo where ispaid="已支付" and mid(paidtime,6,1)="3"

group by userid) t

每个月购买用户的复购率

select s.time,s.复购人数, round(s.复购人数/(select count(distinct userid) as zs from test.orderinfo

where ispaid='已支付'),2) as 复购率

from (select a.userid,a.time,count(if(a.ct>1,1,0)) as 复购人数 from

(select userid,count(userid) as ct,left(paidtime,6) as time from

test.orderinfo where ispaid='已支付'

group by left(paidtime,6),userid) a

group by a.time) s

回购率：三月购买的人，4月依旧购买。

思路：先算出每月购买的用户，分成两个表，将两个表左联结。以时间为计数求出本月购买人数和下月回购人数。

select a.t1,count(a.t1) as 本月购买人数,count(b.t2) as 下月回购人数, round(count(b.t2)/count(a.t1),3) as 回购率

from

(select userid,left(paidtime,6) as t1 from test.orderinfo

where ispaid='已支付'

group by userid,left(paidtime,6)) a

left join (

select userid,left(paidtime,6) as t2 from test.orderinfo

where ispaid='已支付'

group by userid,left(paidtime,6)) b

on a.userid =b.userid and a.t1 = concat(left(b.t2,5),mid(b.t2,6,1)-1)

group by a.t1;

总结从每个月的订单数来看，3、4月份的订单数较多，5月份只有几个。

从每个月以及全年的男女消费频次来看，男女消费频次是差不多的，差别不大。

从各年龄段的消费频次来看，70岁以上的频次最高，其次是30-50岁用户，而30-50岁用户消费的总次数最高。

根据二八法则，分析得出top20%消费的用户消费总额占总额的73%。

用户第一次消费和最后一次消费的时间间隔多数为1天，说明用户生命周期较短，回头客较少。

3、4月份的用户复购率较大，，约为50-60%，，3月购买用户在4月的回购率较高，而4月回购率就非常低了。

利用sql和现有数据得出以上分析结论，可以从结论中推导为什么4月用户回购率变少了，用户流失较大，i由于数据有限，因此还需要作进一步分析。

本次用到的新韩淑：

concat()

str_to_date() 函数，(str1，str2....，'%Y%m%d')

case when then

…… end

窗口函数rank()和dense_rank() 两者的区别

更新表

update table tablename set