具有索引日期条件的每个主题的 SQL 日期减法

Question

我有一个主题记录表，其中每个主题作为 1 个或多个具有开始/结束日期和索引日期的行：

SUBJECT	BEGIN_DATE	END_DATE	INQUIRY_DATE
1	1988-01-01	2010-04-05	2022-05-06
1	2010-04-06	2022-10-02	2022-05-06
2	1996-09-24	2005-08-08	2022-10-01
2	2016-11-21	2022-04-04	2022-10-01
3	2005-01-01	2021-02-12	2022-03-21
4	1999-12-31	2015-07-16	2022-08-15
4	2015-07-20	2020-04-01	2022-08-15
4	2020-12-31	2022-10-01	2022-08-15

我想知道的是，对于每个主题，开始/结束范围是否连续涵盖查询日期之前的 3 个月（结束日期和下一个开始日期之间的间隔不超过 14 天）？例如受试者 1：他们在 1988-01-01 和 2022-10-02 之间进行了连续观察，所以是的，这个受试者的询问日期已经涵盖了 3 个月之前；而主题 2 则没有。

使用 SQL (snowflake) 并从类似这样的东西开始（显然不完整，只是不确定如何到达我想要的地方）：

with lookback as (
select *, INQUIRY_DATE - interval '3 months' as look_back_3m from tbl)

select *, case when diff >= 14 then 1 else 0 end as flag from (
select SUBJECT, BEGIN_DATE - lag(END_DATE) over(partition by subject order by BEGIN_DATE) as diff from tbl) z

Answer 1

第一步是您要过滤掉任何在您的 INQUIRY_DATE 之前 4 个月之前结束或在您的 INQUIRY_DATE 之后 1 个月开始的行，因为它们在您定义的 3 个月 + 14 天规则之外。

所以一些增强的数据：

with your_table(subject, begin_date, end_date, inquiry_date) as (
    select * from values
    (1, '1988-01-01'::date, '2010-04-05'::date, '2022-05-06'::date),
    (1, '2010-04-06'::date, '2022-10-02'::date, '2022-05-06'::date),
    (2, '1996-09-24'::date, '2005-08-08'::date, '2022-10-01'::date),
    (2, '2016-11-21'::date, '2022-04-04'::date, '2022-10-01'::date),
    (3, '2005-01-01'::date, '2021-02-12'::date, '2022-03-21'::date),
    (4, '1999-12-31'::date, '2015-07-16'::date, '2022-08-15'::date),
    (4, '2015-07-20'::date, '2020-04-01'::date, '2022-08-15'::date),
    (4, '2020-12-31'::date, '2022-10-01'::date, '2022-08-15'::date),
    (5, '2022-05-01'::date, '2022-06-01'::date, '2022-08-15'::date),
    (5, '2022-06-01'::date, '2022-07-01'::date, '2022-08-15'::date),
    (5, '2022-07-01'::date, '2022-08-01'::date, '2022-08-15'::date),
    (5, '2022-08-01'::date, '2022-10-01'::date, '2022-08-15'::date),
    (6, '2022-05-01'::date, '2022-05-10'::date, '2022-08-15'::date),
    (6, '2022-06-01'::date, '2022-06-10'::date, '2022-08-15'::date),
    (6, '2022-07-01'::date, '2022-07-10'::date, '2022-08-15'::date),
    (6, '2022-08-01'::date, '2022-10-01'::date, '2022-08-15'::date)
)

select * 
from your_table
where begin_date <= inquiry_date
   and end_date >= dateadd(month, -4, inquiry_date);

给出：

SUBJECT	BEGIN_DATE	END_DATE	INQUIRY_DATE
1	2010-04-06	2022-10-02	2022-05-06
4	2020-12-31	2022-10-01	2022-08-15
5	2022-05-01	2022-06-01	2022-08-15
5	2022-06-01	2022-07-01	2022-08-15
5	2022-07-01	2022-08-01	2022-08-15
5	2022-08-01	2022-10-01	2022-08-15
6	2022-05-01	2022-05-10	2022-08-15
6	2022-06-01	2022-06-10	2022-08-15
6	2022-07-01	2022-07-10	2022-08-15
6	2022-08-01	2022-10-01	2022-08-15

然后你想找到一系列的差距并使它们引起争议。

就像是：

select *
    ,lag(end_date) over(partition by subject order by begin_date) as prior_end
    ,datediff(days, prior_end, begin_date) as days_of_gap
    ,(begin_date - 14) <= prior_end as prior_overlapped
    ,iff(prior_overlapped, null, begin_date) as group_start_date
from your_table
where begin_date <= inquiry_date
   and end_date >= dateadd(month, -4, inquiry_date);

然后您想问，任何链条是否在 INQUIRY_DATE 之前/之日开始 - 3 个月并在 INQUIRY_DATE 之日/之后结束。

select
    subject
    ,begin_date
    ,end_date
    ,group_date
    ,dateadd(month, -3, inquiry_date) as req_start_date
    ,datediff(days, group_date, end_date) as days_of_streak
    ,end_date >= inquiry_date as rule_1
    ,group_date <= dateadd(month, -3, inquiry_date) as rule_2
from (
    select *
        ,lag(group_start_date) ignore nulls over (partition by subject order by begin_date) as l_g_d
        ,nvl(group_start_date, l_g_d) as group_date
    from (
        select *
            ,lag(end_date) over(partition by subject order by begin_date) as prior_end
            ,iff((begin_date - 14) <= prior_end, null, begin_date) as group_start_date
        from your_table
        where begin_date <= inquiry_date
           and end_date >= dateadd(month, -4, inquiry_date)
    )
)
order by subject, begin_date
;

给出：

SUBJECT	BEGIN_DATE	END_DATE	GROUP_DATE	REQ_START_DATE	DAYS_OF_STREAK	RULE_1	RULE_2
1	2010-04-06	2022-10-02	2010-04-06	2022-02-06	4,562	TRUE	TRUE
4	2020-12-31	2022-10-01	2020-12-31	2022-05-15	639	TRUE	TRUE
5	2022-05-01	2022-06-01	2022-05-01	2022-05-15	31	FALSE	TRUE
5	2022-06-01	2022-07-01	2022-05-01	2022-05-15	61	FALSE	TRUE
5	2022-07-01	2022-08-01	2022-05-01	2022-05-15	92	FALSE	TRUE
5	2022-08-01	2022-10-01	2022-05-01	2022-05-15	153	TRUE	TRUE
6	2022-05-01	2022-05-10	2022-05-01	2022-05-15	9	FALSE	TRUE
6	2022-06-01	2022-06-10	2022-06-01	2022-05-15	9	FALSE	FALSE
6	2022-07-01	2022-07-10	2022-07-01	2022-05-15	9	FALSE	FALSE
6	2022-08-01	2022-10-01	2022-08-01	2022-05-15	61	TRUE	FALSE

我们想要两个规则都为真的行..

因此：

select
    subject
    ,group_date
    ,inquiry_date
from (
    select *
        ,lag(group_start_date) ignore nulls over (partition by subject order by begin_date) as l_g_d
        ,nvl(group_start_date, l_g_d) as group_date
    from (
        select *
            ,lag(end_date) over(partition by subject order by begin_date) as prior_end
            ,iff((begin_date - 14) <= prior_end, null, begin_date) as group_start_date
        from your_table
        where begin_date <= inquiry_date
           and end_date >= dateadd(month, -4, inquiry_date)
    )
)
where end_date >= inquiry_date
    and group_date <= dateadd(month, -3, inquiry_date)
order by subject, begin_date
;

给出：

SUBJECT	GROUP_DATE	INQUIRY_DATE
1	2010-04-06	2022-05-06
4	2020-12-31	2022-08-15
5	2022-05-01	2022-08-15