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[英]How to calculate customer retention in SQL based on events (not dates)?
[英]How to calculate customer retention in SQL based on events?
我正在尝试创建一个 SQL 语句来找出哪个客户没有连续参加三个活动
表 1 - 客户:客户 ID、客户名称
+-------------+---------------+
| Customer ID | Customer Name |
+-------------+---------------+
| 01 | Customer 01 |
| 02 | Customer 02 |
| 03 | Customer 03 |
+-------------+---------------+
表 2 - 事件事件 ID、事件日期、事件名称
+----------------------------------+
| Event ID Event Date Event Name |
+----------------------------------+
| 01 01/01/2020 Event 01 |
| 02 01/15/2020 Event 02 |
| 03 02/15/2020 Event 03 |
| 04 03/13/2020 Event 04 |
| 05 05/17/2020 Event 05 |
| 06 06/20/2020 Event 06 |
+----------------------------------+
表 3 - 事件活动事件 ID、客户 ID
+----------+-------------+----+
| Event ID | Customer ID | |
+----------+-------------+----+
| 01 | | 01 |
| 01 | | 02 |
| 01 | | 03 |
| 02 | | 01 |
| 03 | | 01 |
| 03 | | 02 |
| 04 | | 01 |
| 05 | | 01 |
| 06 | | 01 |
| 06 | | 03 |
+----------+-------------+----+
现在我试图找到那些没有连续参加 3 次活动的客户。
因此,在给定的示例中,这将是客户 2 和客户 3。
我使用了史蒂夫的建议。 这里更新的 SQL 语句:
drop table if exists dbo.customer;
create table dbo.customer(
CustID int not null,
CustName varchar(20) not null);
insert dbo.customer(CustID, CustName) values
(1,'Cust 1'),
(2,'Cust 2'),
(3,'Cust 3'),
(4,'Cust 4'),
(5,'Cust 5')
;
drop table if exists dbo.events;
create table dbo.events(
EventID int not null,
EventDate date not null,
EventName varchar(20) not null);
insert dbo.events(EventId, EventDate, EventName) values
(1,'2020-01-01','Event 1'),
(2,'2020-01-15','Event 2'),
(3,'2020-02-15','Event 3'),
(4,'2020-03-13','Event 4'),
(5,'2020-05-17','Event 5'),
(6,'2020-06-20','Event 6');
drop table if exists dbo.eventactivity;
create table dbo.eventactivity(
EventID int not null,
CustID int not null);
insert dbo.eventactivity(EventID, CustID) values
(1,1),
(1,2),
(1,3),
(1,4),
(1,5),
(2,1),
(2,2),
(2,4),
(2,5),
(3,1),
(3,5),
(4,1),
(4,5),
(5,1),
(5,2),
(5,3),
(5,5),
(6,1),
(6,2),
(6,3);
(6,5);
和这里:
;with
events_sorted as (
select e.*, row_number() over (order by EventDate) seq from dbo.events e),
activity_lag as
(
select
a.*, e.seq,
lag(e.seq, 1, 0) over (partition by CustId order by e.seq) lag_seq,
iif(lag(e.seq, 1, 0) over (partition by CustId order by e.seq)=0, 1,
iif((e.seq-lag(e.seq, 1, 0) over (partition by CustId order by e.seq))>1, 1, 0)) seq_break
from dbo.eventactivity a
join events_sorted e on a.EventID=e.EventID
),
activity_lag_sum as (
select
alag.*, sum(seq_break) over (partition by CustId order by alag.seq) seq_grp
from
activity_lag alag
),
three_in_a_row_cte as (
select distinct CustId
from activity_lag_sum
group by CustID, seq_grp
having count(*)>=3
)
select *
from customer c
where not exists(select 1
from three_in_a_row_cte r
where c.CustID=r.CustID);
问题是,这会返回客户 2、客户 3、客户 4 - 客户 2 确实参加了 2 个活动,跳过了 2,参加了 2,因此客户 2 不应该在列表中。
有什么建议 ?
以下查询返回 CustId,它们具有:1) 跳过 3 个或更多事件,或 2) 总共参加了少于 3 个事件。
;with
events_sorted as (
select e.*, row_number() over (order by EventDate) seq from #events e),
activity_lag as
(
select
a.*, e.seq,
lag(e.seq, 1, 0) over (partition by CustId order by e.seq) lag_seq,
iif(lag(e.seq, 1, 0) over (partition by CustId order by e.seq)=0, 1,
iif((e.seq-lag(e.seq, 1, 0) over (partition by CustId order by e.seq))>1, 1, 0)) seq_break
from #eventactivity a
join events_sorted e on a.EventID=e.EventID
)
select distinct CustId
from activity_lag
where seq-lag_seq>3
union all
select CustId
from activity_lag
group by CustId
having count(*)<3;
结果
CustId
3
4
您只需要连续跳过 3 个或更多事件的客户,您可以通过从事件活动表本身查询来获取。 请在下面找到查询和查询结果:-
创建表
create table event_activity ("event_id" varchar(2),"customer_id" varchar(2))
insert into event_activity
values ('01','01'),('01','02'),('01','03'),('02','01'),('02','03'),('03','01'),
('03','02'),('04','01'),('04','02'),('05','02'),('06','01'),('06','03'),
('07','03'),('08','04'),('12','04'),('13','05')
以上查询将产生下表:-
event_id | customerid
---------------------
01 | 01
01 | 02
01 | 03
02 | 01
02 | 03
03 | 01
03 | 02
04 | 01
04 | 02
05 | 02
06 | 01
06 | 03
07 | 03
08 | 04
12 | 04
13 | 05
从上表中我们可以观察到除了客户 4 和 5 之外的所有客户都连续跳过了少于 3 个的事件。 根据您的问题,我们只需要 4 和 5,因为 4 连续跳过了 3 个事件,而 5 个仅参加了 1 个事件。
PS : - 在这里您可以发现客户 3 也跳过了 3 个活动,但在此之前他没有跳过任何活动,因此必须将其消除。
最终查询
select c.customer_id
from
(
select customer_id,
skipped_count,
lag(skipped_count,1) over (partition by customer_id order by event_id)
as ref
from
(
select customer_id,
event_id,
LAG(event_id,1) over (partition by customer_id order by event_id)
as previous_event,(event_id - LAG(event_id,1) over (partition by
customer_id order by event_id)-1) as skipped_count
from
(
select CONVERT(int,event_id) as event_id,
CONVERT(int, customer_id) as customer_id
from event_activity
)a
)b
)c
join
(
select convert(int,customer_id) as customer_id,
count(event_id) as count_event
from event_activity
group by customer_id
)d
on c.customer_id=d.customer_id
where (skipped_count >=3 and ref is null)
or count_event = 1
or (skipped_count >=3 and ref > 2)
输出
4
5
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