3

试图获得一个显示从一个月到下一个月的保留率的基本表格。因此，如果有人上个月买了东西，而他们在下个月又买了，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都在下个月再次购买，那么您将获得 100%。

到目前为止，我只能手动计算东西。这给了我在这两个月中看到的行：

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

这是不对的，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。您主要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
4

2 回答 2

6

给定以下测试表（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。用作列名，因为我不使用基本类型作为标识符
。tstamp

我将使用窗口功能lag()来识别重复购买者。为了简短起见，我将聚合函数和窗口函数组合在一个查询级别中。请记住，窗口函数是在聚合函数之后应用的。

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

我扩展了您的问题，以提供交易数量和买家数量之间的差异。

OR NULLforrepeat_transaction用于转换FALSE为，因此在下一步中NULL不会计算这些值。count()

-> SQLfiddle。

于 2013-05-11T10:44:38.887 回答
0

这使用CASEandEXISTS来获得重复的交易：

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：再次阅读问题后从负 1 个月变为正 1 个月。我现在的理解是，如果有人在 2012-02 年购买了东西，然后在 2012-03 年再次购买了东西，那么他或她在 2012-02 年的交易计入当月留存。

于 2013-05-11T08:14:34.040 回答