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ID我想为我的数据中的每个单元标记至少 31 天的每个窗口中的第一个日期。

ROW  ID    INDEX_DATE
1    ABC   1/1/2019
2    ABC   1/7/2019
3    ABC   1/21/2019
4    ABC   2/2/2019
5    ABC   2/9/2019
6    ABC   3/6/2019
7    DEF   1/5/2019
8    DEF   2/1/2019
9    DEF   2/8/2019

所需的行是 1、4、6、7 和 9；这些要么是INDEX_DATE给定的第一个ID，要么发生在先前标记的至少 31 天之后INDEX_DATE。我发现的每个建议都使用LAG()或LEAD与窗口函数一起使用，但我只能得到这些来比较相邻的行。例如，第 4 行需要与第 1 行进行比较，以便在 31 天窗口完成后被识别为第一行。

我尝试了以下方法：
数据

DROP TABLE tTest IF EXISTS; 
CREATE TEMP TABLE tTest
(
ROWN    INT,
ID VARCHAR(3),
INDEX_DATE DATE
) ;
GO

INSERT INTO tTEST VALUES (1, 'ABC', '1/1/2019');
INSERT INTO tTEST VALUES (2, 'ABC', '1/7/2019');
INSERT INTO tTEST VALUES (3, 'ABC', '1/21/2019');
INSERT INTO tTEST VALUES (4, 'ABC', '2/2/2019');
INSERT INTO tTEST VALUES (5, 'ABC', '2/9/2019');
INSERT INTO tTEST VALUES (6, 'ABC', '3/6/2019');
INSERT INTO tTEST VALUES (7, 'DEF', '1/5/2019');
INSERT INTO tTEST VALUES (8, 'DEF', '2/1/2019');
INSERT INTO tTEST VALUES (9, 'DEF', '2/8/2019');
GO

询问：

DROP TABLE TTEST2 IF EXISTS;
CREATE TEMP TABLE TTEST2 AS (
WITH 
RN_CTE(ROWN, ID, INDEX_DATE, RN) AS (
        SELECT *, ROW_NUMBER() OVER (PARTITION BY ID ORDER BY INDEX_DATE)
        FROM tTEST),
MIN_CTE(ROWN, ID, INDEX_DATE, RN) AS (SELECT * FROM RN_CTE WHERE RN=1),
DIFF_CTE(ROWN,ID, INDEX_DATE, RN, DAY_DIFF) AS (
        SELECT RN.*, DATE(RN.INDEX_DATE + INTERVAL '30 DAYS') 
        FROM RN_CTE AS RN
                JOIN MIN_CTE AS MC ON RN.ID=MC.ID
        WHERE RN.RN=1
           OR RN.INDEX_DATE > MC.INDEX_DATE + INTERVAL '30 DAYS' ),
MIN_DIFF_CTE AS (
        SELECT ID, DAY_DIFF, MIN(ROWN) AS MIN_ROW
        FROM DIFF_CTE
        GROUP BY ID, DAY_DIFF)
SELECT T.*
FROM MIN_DIFF_CTE AS MDC
        JOIN tTEST AS T ON MDC.MIN_ROW = T.ROWN
ORDER BY ID, INDEX_DATE
);

结果：

SELECT * FROM TTEST2 ORDER BY ID, INDEX_DATE;    
ROWN ID INDEX_DATE
1   ABC 2019-01-01
4   ABC 2019-02-02
5   ABC 2019-02-09
6   ABC 2019-03-06
7   DEF 2019-01-05
9   DEF 2019-02-08

INDEX_DATE = 2019-02-09 的第 5 行不应出现在输出中，因为它在第 4 行的 INDEX_DATE 之后不到 31 天。
4

1 回答 1

1

像这样的东西。CTE 为每个 ID 转换和 31 天滚动定位最小 ROW 值的唯一窗口。

数据

drop table if exists #tTEST;
go
select * INTO #tTEST from (values 
(1, 'abc', '1/1/2019'),
(2, 'abc', '1/7/2019'),
(3, 'abc', '1/21/2019'),
(4, 'abc', '2/2/2019'),
(5, 'abc', '2/9/2019'),
(6, 'abc', '3/6/2019'),
(7, 'def', '1/5/2019'),
(8, 'def', '2/1/2019'),
(9, 'def', '2/8/2019')) V([ROW], ID, INDEX_DATE);

询问

;with
rn_cte([ROW], ID, INDEX_DATE, rn) as (
    select *, row_number() over (partition by ID order by INDEX_DATE)
    from #tTEST),
min_cte([ROW], ID, INDEX_DATE, rn) as (select * from rn_cte where rn=1),
diff_cte([ROW], ID, INDEX_DATE, rn, day_diff) as (
    select rn.*, datediff(d, mc.INDEX_DATE, rn.INDEX_DATE)/31
    from rn_cte rn
         join min_cte mc on rn.ID=mc.ID
    where rn.rn=1 
          or datediff(d, mc.INDEX_DATE, rn.INDEX_DATE)/31>0),
min_diff_cte as (
    select ID, day_diff, min([ROW]) min_row
    from diff_cte 
    group by ID, day_diff)
select t.*
from min_diff_cte mdc
     join #tTEST t on mdc.min_row=t.ROW
order by 1;

输出

ROW ID  INDEX_DATE
1   abc 1/1/2019
4   abc 2/2/2019
6   abc 3/6/2019
7   def 1/5/2019
9   def 2/8/2019
于 2020-09-08T01:49:52.793 回答