在 Snowflake 中根据日期和 Window function 过滤查询

Question

我被要求提取有关去年三种不同类型客户的信息（访问过一次、访问过 <10 次和访问过 10 次以上），看看他们返回的可能性是否与几个不同的因素相比。

出于这个原因，我创建了一个非常广泛的查询。 目前我有三张表的联合查询：客户信息、访问信息和员工信息。 我在 select 语句中创建了一个计算列：

COUNT(DISTINCT visitno) OVER(PARTITION BY clientid) as totalvisits

现在我只需要按 totalvisits 分组并按他们访问的日期进行过滤。

我试过了：

where visitdate> 01/01/2021
group by totalvisits
having total visits<10

但是我收到一个错误，指出 visitno 不是一个有效的 group by expression。

我可能做错了什么？

Answer 1

在雪花中，您可以使用QUALIFY子句过滤 window 函数后 window 聚合。

因此，查询将如下所示：

SELECT
  clientid,
  COUNT(DISTINCT visitno) OVER(PARTITION BY clientid) as totalvisits
FROM <your_table>
WHERE visitdate >= 2021-01-01::date
  AND visitdate < 2022-01-01::date
QUALIFY totalvisits < 10;

*不过，请确保visitdate事先有一个日期类型！

[参考下面的评论] ：如果您想查看历史总访问量，加上给定年份的总访问量，您可以执行以下操作：

SELECT
  clientid,
  YEAR(visitdate) as visit_date_year,
  COUNT(DISTINCT visitno) OVER (PARTITION BY clientid) as totalvisits,
  COUNT(DISTINCT visitno) OVER (PARTITION BY clientid, YEAR(visitdate) as total_visits_by_year
FROM <your_table>
QUALIFY total_visits_by_year < 10;

Answer 2

好的，让我们做一些假数据，然后做一些事情：

WITH fake_data(client_id, visit_date) as (
    SELECT * FROM VALUES
    -- this person has visted once
    (1, '2022-04-14'::date),
    -- this person has visited 3 timw in the year
    (3, '2022-04-13'::date),
    (3, '2022-03-13'::date),
    (3, '2022-02-13'::date),
    -- this person is a huge vistor, but 1 is outside the with in last year.
    (5, '2022-04-12'::date),
    (5, '2022-03-12'::date),
    (5, '2022-02-12'::date),
    (5, '2022-01-12'::date),
    (5, '2020-02-12'::date)
)
SELECT *,
    count(distinct visit_date) over (partition by client_id) as total_visits
FROM fake_data
WHERE visit_date >= dateadd('year', -1, '2022-04-14' /* CURRENT_DATE */)

繁荣：

客户编号	VISIT_DATE	TOTAL_VISITS 次
1个	2022-04-14	1个
3个	2022-04-13	3个
3个	2022-03-13	3个
3个	2022-02-13	3个
5个	2022-04-12	4个
5个	2022-03-12	4个
5个	2022-02-12	4个

现在将它们放入那些组/类别中。

SELECT *,
    count(distinct visit_date) over (partition by client_id) as total_visits,
    case 
        when total_visits = 1 then 1
        when total_visits <= 3 then 2
        when total_visits > 3 then 3
    end as group_id
FROM fake_data
WHERE visit_date >= dateadd('year', -1, '2022-04-14' /* CURRENT_DATE */)

现在一些数学，我将把它包装到一个子选择中（但也将一些东西压入其中）

WITH fake_data(client_id, visit_date) as (
    SELECT * FROM VALUES
    -- this person has visted once
    (1, '2022-04-14'::date),
    -- this person has visited 3 timw in the year
    (3, '2022-04-13'::date),
    (3, '2022-04-11'::date),
    (3, '2022-04-09'::date),
    -- this person is a huge vistor, but 1 is outside the with in last year.
    (5, '2022-04-12'::date),
    (5, '2022-03-12'::date),
    (5, '2022-02-12'::date),
    (5, '2022-01-12'::date),
    (5, '2020-02-12'::date)
)
SELECT group_id
    ,count(distinct client_id) as count_of_group_members
    ,sum(total_visits) as sum_of_group_visit
    ,avg(visit_gap_in_days) as avg_group_day_diff
    ,stddev(visit_gap_in_days) as stddev_group_day_diff
FROM (
SELECT *,
    count(distinct visit_date) over (partition by client_id) as total_visits,
    case 
        when total_visits = 1 then 1
        when total_visits <= 3 then 2
        when total_visits > 3 then 3
    end as group_id,
    lag(visit_date) over (partition by client_id order by visit_date) as prior_visit_date,
    datediff('day', prior_visit_date, visit_date) as visit_gap_in_days
FROM fake_data
WHERE visit_date >= dateadd('year', -1, '2022-04-14' /* CURRENT_DATE */)
)
GROUP BY 1
ORDER BY 1

群组编号	COUNT_OF_GROUP_MEMBERS 个成员	SUM_OF_GROUP_VISIT	AVG_GROUP_DAY_DIFF	STDDEV_GROUP_DAY_DIFF
1个	1个	1个
2个	1个	9	2个	0
3个	1个	16	30	1.732050808

Wozers，访问总和是错误的，我已经总结了我的总和..

所以这里给定count(distinct visitno)我不能求和，因为它变成了总和，而且我不能做 count(*) 因为我们刚刚注意到有重复项（否则不需要 distinct ）。 而且我假设您没有删除行，因为有一些“您需要的其他详细信息”

但无论如何。 这是关于 SQL 的伟大之处，你可以回答任何问题，但你必须知道问题，并了解数据，这样你才能知道哪些假设可以适用于你的数据。