Question

我在bigquery中存储了以下事务。

CUSID   PID TID YYYYMMDD
A01     P01 001 2017-01-01
A02     P01 002 2017-02-25
A02     P02 002 2017-02-25
A03     P02 003 2017-03-01
A03     P02 004 2017-03-05
A03     P02 004 2017-03-05
A04     P01 005 2017-03-10
A04     P03 005 2017-03-10
A04     P03 006 2017-03-11
A04     P03 007 2017-03-15

我想找到给定产品的两件事情如下：

1）X：购买客户的数量除以客户总数

2）Y：购买所有客户的平均日差距

因此，预期输出如下表

CUSID  PID  TID YYYYMMDD        X               Y
A01    P01  001 2017-01-01   3/4 = 0.75  AVG(0,0,0) = N/A (P01 does not have re-purchasing by A01, A02, and A04)
A02    P01  002 2017-02-25   3/4 = 0.50  AVG(0,0,0) = N/A (P01 is not re-purchased by A01, A02, and A04)
A02    P02  002 2017-02-25   2/4 = 0.50  AVG(0,4) = 4 (P02 is not re-purchased by A02 but it is re-purchased by A03 for 4 days.  Note: duplicated product in the same TID is excluded, e.g. TID = 004)
A03    P02  003 2017-03-01   2/4 = 0.50  AVG(0,4) = 4 (P02 is not re-purchased by A02 but it is re-purchased by A03 for 4 days.  Note: duplicated product in the same TID is excluded, e.g. TID = 004)
A03    P02  004 2017-03-05   2/4 = 0.50  AVG(0,4) = 4 (P02 is not re-purchased by A02 but it is re-purchased by A03 for 4 days.  Note: duplicated product in the same TID is excluded, e.g. TID = 004)
A03    P02  004 2017-03-05   2/4 = 0.50  AVG(0,4) = 4 (P02 is not re-purchased by A02 but it is re-purchased by A03 for 4 days.  Note: duplicated product in the same TID is excluded, e.g. TID = 004)
A04    P01  005 2017-03-10   3/4 = 0.75  AVG(0,0,0) = N/A (P01 is not re-purchased by A01, A02, and A04)
A04    P03  005 2017-03-10   1/4 = 0.25  AVG(1,4) = 2.5 (P03 is repurchased by A04 for 1 and 4 day gaps)
A04    P03  006 2017-03-11   1/4 = 0.25  AVG(1,4) = 2.5 (P03 is repurchased by A04 for 1 and 4 day gaps)
A04    P03  007 2017-03-15   1/4 = 0.25  AVG(1,4) = 2.5 (P03 is repurchased by A04 for 1 and 4 day gaps)

我可以提出你的建议吗？

Answer 1

以下是您所描述的内容它适用于BigQuery Standard SQL

#standardSQL
WITH data AS (
  SELECT 'A01' AS CUSID, 'P01' AS PID, '001' AS TID, DATE '2017-01-01' AS YYYYMMDD UNION ALL
  SELECT 'A02', 'P01', '002', DATE '2017-02-25' UNION ALL SELECT 'A02', 'P02', '002', DATE '2017-02-25' UNION ALL SELECT 'A03', 'P02', '003', DATE '2017-03-01' UNION ALL SELECT 'A03', 'P02', '004', DATE '2017-03-05' UNION ALL
  SELECT 'A03', 'P02', '004', DATE '2017-03-05' UNION ALL SELECT 'A04', 'P01', '005', DATE '2017-03-10' UNION ALL SELECT 'A04', 'P03', '005', DATE '2017-03-10' UNION ALL SELECT 'A04', 'P03', '006', DATE '2017-03-11' UNION ALL
  SELECT 'A04', 'P03', '007', DATE '2017-03-15' 
),
popularity AS (
  SELECT DISTINCT PID, 
    COUNT(DISTINCT CUSID) OVER(PARTITION BY PID) / COUNT(DISTINCT CUSID) OVER() AS X
  FROM data
),
gaps AS (
  SELECT CUSID, PID, TID, YYYYMMDD,
    DATE_DIFF(YYYYMMDD, LAG(YYYYMMDD) OVER(PARTITION BY CUSID, PID ORDER BY YYYYMMDD), DAY) AS gap
  FROM data
),
gaps_without_dups AS (
  SELECT CUSID, PID, YYYYMMDD,
    MAX(IFNULL(gap, 0)) AS gap
  FROM gaps
  GROUP BY CUSID, PID, YYYYMMDD
  HAVING gap > 0
),
average_gaps AS (
  SELECT PID, AVG(gap) AS Y
  FROM gaps_without_dups
  GROUP BY PID
)
SELECT CUSID, PID, TID, YYYYMMDD, X, Y
FROM data
LEFT JOIN popularity USING (PID)
LEFT JOIN average_gaps USING(PID)
-- ORDER BY TID, PID

输出符合预期

CUSID   PID TID YYYYMMDD    X       Y    
A01     P01 001 2017-01-01  0.75    null     
A02     P01 002 2017-02-25  0.75    null     
A02     P02 002 2017-02-25  0.5     4.0  
A03     P02 003 2017-03-01  0.5     4.0  
A03     P02 004 2017-03-05  0.5     4.0  
A03     P02 004 2017-03-05  0.5     4.0  
A04     P01 005 2017-03-10  0.75    null     
A04     P03 005 2017-03-10  0.25    2.5  
A04     P03 006 2017-03-11  0.25    2.5  
A04     P03 007 2017-03-15  0.25    2.5

Answer 2

这个查询也可以做到这一点（我假设如果一个事务id比另一个大，那么它的日期也更大）：

SELECT
  * EXCEPT(lead_date),
  AVG(CASE WHEN lead_date != date THEN DATE_DIFF(parse_DATE("%Y-%m-%d", lead_date), parse_DATE("%Y-%m-%d", date), DAY) END) OVER(PARTITION BY PID) Y
FROM(
  SELECT
    *,
    COUNT(DISTINCT cusid) OVER(PARTITION BY PID) / COUNT(DISTINCT cusid) OVER() X,
    LEAD(date) OVER(PARTITION BY PID, cusid ORDER BY TID) lead_date
  FROM
    data )

其中data是您的输入数据：

with data as(
select 'A01' as cusid, 'P01' as PID, 1 as TID, '2017-01-01' as date union all
select 'A02', 'P01', 2, '2017-02-25' UNION ALL
SELECT 'A02', 'P02', 2, '2017-02-25' UNION ALL
SELECT 'A03', 'P02', 3, '2017-03-01' UNION ALL
SELECT 'A03', 'P02', 4, '2017-03-05' UNION ALL
SELECT 'A03', 'P02', 4, '2017-03-05' UNION ALL
SELECT 'A04', 'P01', 5, '2017-03-10' UNION ALL
SELECT 'A04', 'P03', 5, '2017-03-10' UNION ALL
SELECT 'A04', 'P03', 6, '2017-03-11' UNION ALL
SELECT 'A04', 'P03', 7, '2017-03-15'
)

运行此类分析时，请务必使用analytical functions的概念。

BigQuery文档非常好，您可以安全地使用它来了解它的所有内容。这将使您能够使用更简单，更快速的查询来运行非常复杂的查询。

在bigquery或sql中查找购买客户的比率和来自多个客户的交易日差距的平均值

2 个答案: