可能的最短查询，递归检查的大查询

Question

假设我在大查询中有一个数据，一个月有数百万行。 例如：

|---------------------|------------------|
|      date           |     user         |
|---------------------|------------------|
|          01-12-2019 |   xyz            |
|---------------------|------------------|
|          02-12-2019 |   xyz            |
|---------------------|------------------|
|          03-12-2019 |   abc            |
|---------------------|------------------|

现在我想检索每日数据，接下来 14 天的重复用户计数，即 2019 年 1 月 12 日首次访问的用户，然后是未来 14 天内再次访问的重复用户计数（02-12-2019 - 15-12-2019）。我想出了检索相同的方法，但是对于某些特定日期，使用下面的查询。

 SELECT '2019-12-01' AS visit_date, COUNT(DISTINCT user) AS visitors_count
 FROM `user_data`
 WHERE
 date = '2019-12-01' AND user IN (SELECT user FROM `user_data`
 WHERE date between DATE_ADD('2019-12-01', INTERVAL 1 DAY) AND DATE_ADD('2019-12-01', INTERVAL 
 14 DAY) )
 GROUP BY 1

我可以使用的一种方法是 UNION ALL，这可能不是最好的解决方案，这就是为什么我愿意知道一些最佳实践，我必须养成这种情况的习惯。

Answer 1

内联查询在这里应该非常有效：

select 
    date,
    (
        select count(distinct u1.user) 
        from user_data u1 
        where u1.date 
            between date_add(u.date, interval 1 day) 
            and date_add(u.date, interval 14 day)
    ) visitors_count
from (select distinct date from user_data) u

为了提高性能，您需要在(date, user) 上建立索引。

如果您有一个包含表中可用日期列表的参考表，则可以使用 if 而不是 select distinct: 这也可以加快查询速度。

Answer 2

以下是 BigQuery 标准 SQL

#standardSQL
SELECT visit_day, COUNT(DISTINCT IF(visits_next_14_days, NULL, user)) AS repeating_visitors_count 
FROM (
  SELECT visit_day, user, 0 = COUNT(1) OVER(
    PARTITION BY user 
    ORDER BY UNIX_DATE(visit_day) 
    RANGE BETWEEN 1 FOLLOWING AND 14 FOLLOWING
  ) visits_next_14_days
  FROM `project.dataset.user_data`
)
GROUP BY visit_day

Answer 3

您可以通过使用union all 和聚合来解决此问题。关键是保持日期进出。所以：

with ud as (
      select user, date, 1 as inc
      from user_data
      union all
      select user, date_add(date, interval 15 day), -1 as inc
      from user_data
     )
select date,
       sum(inc) as change_on_day,
       sum(sum(inc)) over (order by date) as total_on_day
from ud
group by date
order by date;

编辑：

您可以修改上述内容以获得客户的第一个积极公司和最后一个：

with ud as (
      select user, date, 1 as inc
      from (select ud.*,
                   lag(date) over (partition by user order by date) as prev_date
            from user_data ud
           ) ud
      where prev_date is null or prev_date < date_add(date, interval -14 day)
      union all
      select user, date_add(date, interval 15 day), -1 as inc
      from (select ud.*,
                   lead(date) over (partition by user order by date) as lead_date
            from user_data ud
           ) ud
      where next_date is null or next_date < date_add(date, interval 14 day)
     )
select date,
       sum(inc) as change_on_day,
       sum(sum(inc)) over (order by date) as total_on_day
from ud
group by date
order by date;