Question

尝试获取一个基本表，显示从一个月到下一个月的保留时间。因此，如果有人在上个月购买了某些东西，并且他们会在下个月购买，那么它就会被计算在内。

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。这给了我两个月都见过的行：

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tstamp作为列名，因为我不使用基类型作为标识符。

我将使用窗口函数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了窗口函数。

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

我延长了你的问题，以提供交易数量和买家数量之间的差异。

OR NULL repeat_transaction用于将FALSE转换为NULL，因此下一步count()不会计算这些值。

-> SQLfiddle.

Answer 2

这使用CASE和EXISTS来获取重复的交易：

SELECT
    *,
    CASE
        WHEN num_transactions = 0
        THEN 0
        ELSE round(100.0 * repeat_transactions / num_transactions, 2)
    END AS retention
FROM
    (
        SELECT
            to_char(timestamp, 'YYYY-MM') AS month,
            count(*) AS num_transactions,
            sum(CASE
                WHEN EXISTS (
                    SELECT 1
                    FROM transaction AS t
                    JOIN auth_user AS u
                    ON t.buyer_id = u.id
                    WHERE
                        date_trunc('month', transaction.timestamp)
                            + interval '1 month'
                            = date_trunc('month', t.timestamp)
                        AND auth_user.email = u.email
                )
                THEN 1
                ELSE 0
            END) AS repeat_transactions
        FROM
            transaction
            JOIN auth_user
            ON transaction.buyer_id = auth_user.id
        GROUP BY 1
    ) AS summary
ORDER BY 1;

编辑：再次阅读问题后，从减1个月更改为加1个月。我现在的理解是，如果有人在2012-02买东西，然后在2012-03再买东西，那么他或她在2012-02的交易就算作本月的保留。

如何使用SQL计算保留月份

2 个答案: