如何使用SQL计算保留月份

Question

试图获得一个基本表,显示从一个月到下一个月的保留.因此,如果有人在上个月购买了某些东西,那么他们会在下个月购买它.

month, num_transactions, repeat_transactions, retention
2012-02, 5, 2, 40%
2012-03, 10, 3, 30%
2012-04, 15, 8, 53%

因此,如果上个月购买的每个人在下个月再次购买,那么您将获得100%.

到目前为止,我只能手动计算东西.这给了我两个月里看到的行:

select count(*) as num_repeat_buyers from 

(select distinct
  to_char(transaction.timestamp, 'YYYY-MM') as month,
  auth_user.email
from
  auth_user,
  transaction
where
  auth_user.id = transaction.buyer_id and
  to_char(transaction.timestamp, 'YYYY-MM') = '2012-03'
) as table1,


(select distinct
  to_char(transaction.timestamp, 'YYYY-MM') as month,
  auth_user.email
from
  auth_user,
  transaction
where
  auth_user.id = transaction.buyer_id and
  to_char(transaction.timestamp, 'YYYY-MM') = '2012-04'
) as table2
where table1.email = table2.email

这不对,但我觉得我可以使用一些Postgres的窗口函数.请记住,窗口函数不允许您指定WHERE子句.您几乎可以访问前面的行和前面的行:

select month, count(*) as num_transactions, count(*) over (PARTITION BY month ORDER BY month)
from 
    (select distinct
      to_char(transaction.timestamp, 'YYYY-MM') as month,
      auth_user.email
    from
      auth_user,
      transaction
    where
      auth_user.id = transaction.buyer_id
    order by
      month
    ) as transactions_by_month
group by
    month

Answer 1

给出以下测试表(您应该提供):

CREATE TEMP TABLE transaction (buyer_id int, tstamp timestamp);
INSERT INTO transaction VALUES 
 (1,'2012-01-03 20:00')
,(1,'2012-01-05 20:00')
,(1,'2012-01-07 20:00')  -- multiple transactions this month
,(1,'2012-02-03 20:00')  -- next month
,(1,'2012-03-05 20:00')  -- next month
,(2,'2012-01-07 20:00')
,(2,'2012-03-07 20:00')  -- not next month
,(3,'2012-01-07 20:00')  -- just once
,(4,'2012-02-07 20:00'); -- just once

表auth_user与问题无关.
使用tstampas name作为名称,因为我不使用基类型作为标识符.

我将使用窗口功能lag()来识别重复的买家.为了简短起见,我将聚合和窗口函数组合在一个查询级别中.请记住,在聚合函数之后应用窗口函数.

WITH t AS (
   SELECT buyer_id
         ,date_trunc('month', tstamp) AS month
         ,count(*) AS item_transactions
         ,lag(date_trunc('month', tstamp)) OVER (PARTITION BY  buyer_id
                                           ORDER BY date_trunc('month', tstamp)) 
          = date_trunc('month', tstamp) - interval '1 month'
            OR NULL AS repeat_transaction
   FROM   transaction
   WHERE  tstamp >= '2012-01-01'::date
   AND    tstamp <  '2012-05-01'::date -- time range of interest.
   GROUP  BY 1, 2
   )
SELECT month
      ,sum(item_transactions) AS num_trans
      ,count(*) AS num_buyers
      ,count(repeat_transaction) AS repeat_buyers
      ,round(
          CASE WHEN sum(item_transactions) > 0
             THEN count(repeat_transaction) / sum(item_transactions) * 100
             ELSE 0
          END, 2) AS buyer_retention
FROM   t
GROUP  BY 1
ORDER  BY 1;

结果:

  month  | num_trans | num_buyers | repeat_buyers | buyer_retention_pct
---------+-----------+------------+---------------+--------------------
 2012-01 |         5 |          3 |             0 |               0.00
 2012-02 |         2 |          2 |             1 |              50.00
 2012-03 |         2 |          2 |             1 |              50.00

我扩展了你的问题,以提供交易数量和买家数量之间的差异.

该OR NULL对repeat_transaction用于转换FALSE到NULL,因此这些值不要被计入count()下一步骤.

- > SQLfiddle.