查询付费客户和流失客户的数量？

Posted 2023-03-31

【中文标题】查询付费客户和流失客户的数量？

【英文标题】：Query to find the number of paying customers and churned customers?

【发布时间】：2015-11-23 17:24:35

【问题描述】：

我有一个如下所示的表paid_users：

http://sqlfiddle.com/#!15/d25ba

我正在尝试确定按月-年分组的付费客户和按月-年分组的流失客户。本质上，有付款人和用户。付款人是为该特定用户付款的人。如果没有payment_stop_date，则表示付款人仍在为用户付款。 payment_stop_date 指示付款人是否/何时停止为用户付款。

我想找出查询结果应该是多少付费客户：

Month-Year | New Paying Customers | Churned Paying Customers
------------------------------------------------------------
11-2014    | 1                    |
12-2014    |                      | 1
01-2015    | 1                    |
04-2015    |                      |
06-2015    | 2                    |
07-2015    | 1                    |
10-2015    |                      | 1

查看 payor_id 3453，她在 11-2014 年开始为 user_id 3182 付款，因此她将被纳入 11-2014 组。然而，她在 12-2014 年停止为两个用户付费，因此被纳入了 12-2014 年流失的群体。如果付款人完全停止向我们付款（即，他们本可以为一个人付款然后取消。或者在这种情况下，payor_id 3453 为 2 个用户付款然后取消），则付款人被视为流失的付款客户。 Payor_3453 然后在 01-2015 开始为 user_id 4716 付款，因此她随后被包含在 01-2015 组中。

我很难为此编写查询，因为它不一定是不同的 payor_id，因为 payor_id 3453 两次被视为新的付费客户

【参考方案1】：

不确定我是否理解正确：您想知道每个月有多少客户开始为第一个用户付费，有多少客户停止为最后一个用户付费？

解决方案看起来相当复杂，但也许并不那么容易。

with months as 
(
    select * from 
    generate_series('2014-06-01', now() at time zone 'utc', interval '1 month') as month
    cross join paid_users
)
, sums as
(
    select month, payor_id, joiners, leavers, sum(net) over (partition by payor_id order by month) 
    from
    (
        select month, payor_id, joiners, leavers, coalesce(joiners,0) - coalesce(leavers, 0) as net
        from
        (
            select payor_id, month, count(*) as joiners
            from months
            where payment_start_date >= month
            and payment_start_date < month + interval '1 month'
            group by month, payor_id
        ) as t
        full join
        (
            select payor_id, month, count(*) as leavers
            from months
            where payment_stop_date >= month
            and payment_stop_date < month + interval '1 month'
            group by month, payor_id
        ) as u
        using (month, payor_id)
    ) as v
)

select * from sums
order by payor_id, sum

以上内容应为您提供每位客户的付费用户总数

        month        | payor_id | joiners | leavers | sum 
---------------------+----------+---------+---------+-----
 2014-06-01 00:00:00 |     1725 |       1 |         |   1
 2014-06-01 00:00:00 |     1929 |       1 |         |   1
 2015-10-01 00:00:00 |     1929 |         |       1 |   0
 2014-06-01 00:00:00 |     1986 |       1 |         |   1
 2014-11-01 00:00:00 |     3453 |       2 |         |   2
 2014-12-01 00:00:00 |     3453 |         |       2 |   0
 2015-01-01 00:00:00 |     3453 |       1 |         |   1
 2015-03-01 00:00:00 |     3453 |       1 |         |   2
 2015-04-01 00:00:00 |     3453 |       2 |       1 |   3
 2015-05-01 00:00:00 |     3453 |         |       1 |   2
 2015-06-01 00:00:00 |     3453 |         |       1 |   1
 2015-10-01 00:00:00 |     3453 |       1 |         |   2
 2015-07-01 00:00:00 |     6499 |       1 |         |   1
 2015-08-01 00:00:00 |     6499 |       3 |         |   4
 2015-10-01 00:00:00 |     6499 |         |       1 |   3
 2015-11-01 00:00:00 |     6499 |         |       1 |   2

所以新客户是总和为 0 到非零总和的客户，流失客户是总和为 0 的客户？

select month, new, churned from
(
    (
        select month, count(*) as churned
        from sums
        where sum = 0
        group by month
    ) as l
    full join
    (
        select month, count(*) as new
        from (
            select month, payor_id, sum, coalesce(lag(sum) over (partition by payor_id order by month), 0) as prev_sum
            from sums
            order by payor_id, month
        ) as t
        where prev_sum = 0 and sum > 0
        group by month
    ) as r
    using (month)
)
order by month

输出

        month        | new | churned 
---------------------+-----+---------
 2014-06-01 00:00:00 |   3 |        
 2014-11-01 00:00:00 |   1 |        
 2014-12-01 00:00:00 |     |       1
 2015-01-01 00:00:00 |   1 |        
 2015-07-01 00:00:00 |   1 |        
 2015-10-01 00:00:00 |     |       1

希望这会有所帮助。如果有人知道更简单的方法，我会很高兴听到它。

【讨论】：

这似乎很合理，你肯定比我更好地解释了这个问题的想法 - 非常感谢你的帮助！数字看起来正确，逻辑很有意义:)