如何计算一个月内不存在于前一个月的用户？

Question

我正在尝试按月计算上个月不存在的唯一用户。因此，如果用户有 1 月的记录，然后有 2 月的另一个记录，那么我只会计算该用户的 1 月。

user_id    time
a1         1/2/17
a1         2/10/17
a2         2/18/17
a4         2/5/17
a5         3/25/17

我的结果应该是这样的

Month   User Count
January     1
February    2
March       1

Answer 1

我对 BigQuery 不是很熟悉，但这里是我使用 TSQL 解决问题的方法。我想您可以在 BigQuery 中使用类似的逻辑。

1)。首先按 user_id 排序数据，然后按时间排序。在 TSQL 中，您可以通过以下方式完成此操作并将其存储在公用表表达式中，您将在此之后的步骤中查询该表达式。

;WITH cte AS
(
select  ROW_NUMBER() OVER (PARTITION BY [user_id] ORDER BY [time]) AS rn,*
from    dbo.employees
)

2)。接下来只查询 rn = 1 的行（特定用户的第一次出现）并按月分组。

select    DATENAME(month, [time]) AS [Month], count(*) AS user_count 
from      cte
where     rn = 1    
group by  DATENAME(month, [time])

这是假设 2017 年是您要处理的唯一年份。如果您处理的时间超过一年，您可能希望第 2 步看起来像这样：

select    year([time]) as [year],  DATENAME(month, [time]) AS [month], 
          count(*) AS user_count 
from      cte
where     rn = 1    
group by  year([time]), DATENAME(month, [time])

Answer 2

首先按用户 ID 和月份聚合。然后使用lag()查看用户上个月是否在场：

with du as (
      select date_trunc(time, month) as yyyymm, user_id
      from t
      group by date_trunc(time, month)
     )
select yyyymm, count(*)
from (select du.*,
             lag(yyyymm) over (partition by user_id order by yyyymm) as prev_yyyymm
      from du
     ) du
where prev_yyyymm is not null or
      prev_yyyymm < date_add(yyyymm, interval 1 month)
group by yyyymm;

注意：这使用了date 函数，但timestamp 也存在类似的函数。

Answer 3

我理解问题的方式是-仅当同一用户在上个月出现时才将用户排除在给定月份。但是，如果同一用户在给定前几个月出现，但以前没有出现 - 用户应该被计算在内。

如果这是正确的 - 请尝试以下 BigQuery 标准 SQL

#standardSQL
SELECT Year, Month, COUNT(DISTINCT user_id) AS User_Count
FROM (
  SELECT *,
    DATE_DIFF(time, LAG(time) OVER(PARTITION BY user_id ORDER BY time), MONTH) AS flag
  FROM (
    SELECT 
      user_id,
      DATE_TRUNC(PARSE_DATE('%x', time), MONTH) AS time, 
      EXTRACT(YEAR FROM PARSE_DATE('%x', time)) AS Year, 
      FORMAT_DATE('%B', PARSE_DATE('%x', time)) AS Month
    FROM yourTable
    GROUP BY 1, 2, 3, 4
  )
)
WHERE IFNULL(flag, 0) <> 1
GROUP BY Year, Month, time
ORDER BY time 

您可以使用下面的示例使用您问题中的虚拟数据来测试/玩上面的内容

#standardSQL
WITH yourTable AS (
  SELECT 'a1' AS user_id, '1/2/17' AS time UNION ALL
  SELECT 'a1', '2/10/17' UNION ALL
  SELECT 'a2', '2/18/17' UNION ALL
  SELECT 'a4', '2/5/17' UNION ALL
  SELECT 'a5', '3/25/17' 
)
SELECT Year, Month, COUNT(DISTINCT user_id) AS User_Count
FROM (
  SELECT *,
    DATE_DIFF(time, LAG(time) OVER(PARTITION BY user_id ORDER BY time), MONTH) AS flag
  FROM (
    SELECT 
      user_id,
      DATE_TRUNC(PARSE_DATE('%x', time), MONTH) AS time, 
      EXTRACT(YEAR FROM PARSE_DATE('%x', time)) AS Year, 
      FORMAT_DATE('%B', PARSE_DATE('%x', time)) AS Month
    FROM yourTable
    GROUP BY 1, 2, 3, 4
  )
)
WHERE IFNULL(flag, 0) <> 1
GROUP BY Year, Month, time
ORDER BY time  

输出是

Year    Month       User_Count   
2017    January     1    
2017    February    2    
2017    March       1    

Answer 4

试试这个查询：

SELECT 
    t1.d,
    count(DISTINCT t1.user_id)
FROM 
(
    SELECT 
        EXTRACT(MONTH FROM time) AS d,
        --EXTRACT(MONTH FROM time)-1 AS d2,
        user_id
    FROM nbitra.tmp
) t1
LEFT JOIN 
(
    SELECT 
        EXTRACT(MONTH FROM time) AS d,
        user_id
    FROM nbitra.tmp
) t2
    ON t1.d = t2.d+1
WHERE 
(
    t1.user_id <> t2.user_id --User is in previous month
    OR t2.user_id IS NULL --To handle january, since there is no previous month to compare to
)
GROUP BY t1.d;