如何计算回访用户

Question

我正在学习 SQL。现在我有一个包含列的表：user_id、event_timestamp 和 event_name。我需要统计每个月的新用户 (I) 和第二个月返回站点的用户 (II)（例如，如果用户第一次出现是在 2 月份，他们在 3 月份使用该站点，则应该统计他们。我想我计算了第一列（I），但我不知道如何计算第二列。 因此，结果应该有一个包含“月和年”、“每月新用户”和“returning_users”列的表。

select
    distinct date_trunc('month', u.date_timestamp) as month_and_year,
    count(*) as count_users
from (select distinct on (t.user_id) *
    from example_table.table as t
    order by t.user_id, t.date_timestamp
 ) as u
group by month_and_year
order by month_and_year

所以，答案中的解决方案有效，但我仍然有问题。我不确定，但我认为它并不像我想要的那样工作。我在这样的真实基础上尝试过：

select date_trunc('month', u.ship_date) as month_and_year,
       count(distinct case when date_trunc('month', u.ship_date) = date_trunc('month', u.min_date) then cust_id end) as num_starts,
       count(distinct case when date_trunc('month', u.ship_date) = date_trunc('month', u.min_date + interval '1 month') then cust_id end) as num_returning
from (select sh.*,
             min(ship_date) over (partition by cust_id) as min_date
      from shipping.shipment as sh
     ) u

group by month_and_year
order by month_and_year

我有一张这样的桌子：

+----------------------------+------------+---------------+
| month_and_year             | num_starts | num_returning |
+----------------------------+------------+---------------+
| January 1, 2016, 12:00 AM  | 6          | 0             |
+----------------------------+------------+---------------+
| February 1, 2016, 12:00 AM | 8          | 1             |
+----------------------------+------------+---------------+
| March 1, 2016, 12:00 AM    | 16         | 0             |
+----------------------------+------------+---------------+
| April 1, 2016, 12:00 AM    | 29         | 1             |
+----------------------------+------------+---------------+
| May 1, 2016, 12:00 AM      | 23         | 9             |
+----------------------------+------------+---------------+
| June 1, 2016, 12:00 AM     | 13         | 10            |
+----------------------------+------------+---------------+
| July 1, 2016, 12:00 AM     | 4          | 5             |
+----------------------------+------------+---------------+
| August 1, 2016, 12:00 AM   | 0          | 2             |
+----------------------------+------------+---------------+

如您所见，看起来 7 月和 8 月返回的用户比刚出现的用户多。那是因为这个查询显示谁在本月返回，但我想知道，例如，在 2 月出现的有多少人在 下 个月返回（所以， f.e. 三月）。我想现在第二个数字在 num_returning 的下面一行。你能帮我做对吗？

Answer 1

使用窗口函数引入第一个月。然后使用条件聚合：

select date_trunc('month', u.date_timestamp) as month_and_year,
       count(distinct case when date_trunc('month', u.date_timestamp) = date_trunc('month', u.dt) then user_id end) as num_starts,
       count(distinct case when date_trunc('month', u.date_timestamp) > date_trunc('month', u.dt) then user_id end) as num_returning
    count(*) as count_users
from (select t.*,
             min(u.date_timestamp) over (partition by user_id) as min_dt
      from example_table.table as t
     ) u
group by month_and_year
order by month_and_year;

其实我觉得在子查询中将数据减少到每个用户每月一行会更有效率：

select yyyymm,
       count(*) filter (where yyyymm = min_yyyymm) as num_starts,
       count(*) filter (where yyyymm > min_yyyymm) as num_returns
from (select distinct on (user_id, date_trunc('month', u.date_timestamp)),
             t.*,
             date_trunc('month', u.date_timestamp) as yyyymm,
             min(date_trunc('month', u.date_timestamp)) over (partition by user_id) as min_yyyymm
      from example_table.table as t
      order by user_id, date_trunc('month', u.dt), u.dt
     ) u
group by yyyymm
order by yyyymm;

编辑：

如果你想要他们开始后一个月返回的数字，你也可以添加它：

select yyyymm,
       count(*) filter (where yyyymm = min_yyyymm) as num_starts,
       count(*) filter (where yyyymm > min_yyyymm) as num_returns,
       count(*) filter (where yyyymm = min_yyyymm + interval '1 month') as num_returns_second_month