SQL：差距和岛屿，分组日期

Question

我正在尝试将日期分组在 3 天内，并根据 30 天内的再入院分配分数。 MRN 每次重新入院将获得 3 分。在下面修改我的查询的任何帮助都会很棒。

示例：

CREATE TABLE #z      (
    ID INT IDENTITY(1,1), 
    OrganizationMrn INT,
    VisitDate DATE, 
    CATEGORY VARCHAR(15) )

INSERT #z(OrganizationMrn, VisitDate, CATEGORY)
VALUES 
(1, '1/2/2016','Inpatient'),
(1, '1/5/2016','Inpatient'),  
(1, '1/7/2016','Inpatient'),  
(1, '1/8/2016','Inpatient'), 
(1, '1/9/2016','Inpatient'),  
(1, '2/4/2016','Inpatient'), 
(1, '6/2/2016','Inpatient'),
(1, '6/3/2016','Inpatient'),
(1, '6/5/2016','Inpatient'),  
(1, '6/6/2016','Inpatient'), 
(1, '6/8/2016','Inpatient'),  
(1, '7/1/2016','Inpatient'),  
(1, '8/1/2016','Inpatient'),  
(1, '8/4/2016','Inpatient'),  
(1, '8/15/2016','Inpatient'), 
(1, '8/18/2016','Inpatient'), 
(1, '8/28/2016','Inpatient'),
(1, '10/12/2016','Inpatient'),
(1, '10/15/2016','Inpatient'),
(1, '11/17/2016','Inpatient'),
(1, '12/20/2016','Inpatient')

期望输出：我真的只需要实际访问次数、OrganizationMrn 和积分。（日期分组时（实际访问），第一个日期应用于 30 天内的再入院）。

ACTUAL Visits   Grouped Dates               Re-admissions       Points
1/2/2016        (grouped 1/2, 1/5)
1/7/2016        (grouped 1/7, 1/8, 1/9)     Readmit from 1/2    (3 points)  
2/4/2016                                    Readmit from 1/7    (3 points)
6/2/2016        (grouped 6/2, 6/3, 6/5)
6/6/2016        (grouped 6/6, 6/8)          Readmit from 6/2    (3 points)  
7/1/2016                                    Readmit from 6/6    (3 points)
8/1/2016        (grouped 8/1, 8/4)
8/15/2016       (grouped 8/15, 8/18)        Readmit from 8/1    (3 points)  
8/28/2016                                   Readmit from 8/15   (3 points)
10/12/2016      (grouped 10/12, 10/15)
11/17/2016  
12/20/2016  
___________________________________________ 6 total readmits    (18 total points)

下面的查询使用 gaps 和 islands 对彼此相隔 3 天以内的日期进行分组。但是，如果日期是连续的，则开始/结束日期被分组。（示例：下面的查询将 [1/2, 1/5, 1/7/, 1/8, 1/9] 分组为一行；日期应分为两行 [1/2, 1/5] 和 [1/7/, 1/8, 1/9]）。

一旦分组的日期有单独的行，我需要在 30 天内为每次重新入院分配 3 分。（每个 OrganizationMrn 的实际访问彼此相隔 30 天）。上面所需的输出部分描述了我的示例中的日期应该如何分组。

;WITH StartingPoints AS (
    SELECT OrganizationMrn, VisitDate, ROW_NUMBER() OVER (ORDER BY VisitDate) AS Sequence FROM #z AS A 
    WHERE a.category = 'Inpatient' AND NOT EXISTS (
        SELECT * FROM #z AS B 
        WHERE B.OrganizationMrn = A.OrganizationMrn AND 
              B.VisitDate >= DATEADD(DAY, -4, A.VisitDate) AND
              B.VisitDate < A.VisitDate AND
              B.Category = 'Inpatient'  )   ),
EndingPoints AS (
    SELECT OrganizationMrn, VisitDate, ROW_NUMBER() OVER (ORDER BY VisitDate) AS Sequence FROM #z AS A 
    WHERE a.category = 'Inpatient' AND NOT EXISTS (
        SELECT * FROM #z AS B 
        WHERE B.OrganizationMrn = A.OrganizationMrn AND 
              B.VisitDate <= DATEADD(DAY, 4, A.VisitDate) AND 
              B.VisitDate > A.VisitDate AND
              B.Category = 'Inpatient'  )   )
SELECT S.OrganizationMrn, S.VisitDate AS StartDate, E.VisitDate AS EndDate, CEILING((DATEDIFF(DAY, S.VisitDate, E.VisitDate) + 1) / 4.0) AS Points
FROM StartingPoints AS S 
    JOIN EndingPoints AS E ON (E.Sequence = S.Sequence)
ORDER BY S.OrganizationMrn DESC

Answer 1

此答案适用于您提供的示例。如果您的桌子很小且入场人数有限，这可能会很有用。（它在日期上使用递归）。

WITH a AS (
    SELECT
        z1.VisitDate
        , z1.OrganizationMrn
        , (SELECT MIN(VisitDate) FROM #z WHERE VisitDate > DATEADD(day, 3, z1.VisitDate)) AS NextDay
    FROM
        #z z1
    WHERE
        CATEGORY = 'Inpatient'
), a1 AS ( 
    SELECT
        OrganizationMrn
        , MIN(VisitDate) AS VisitDate
        , MIN(NextDay) AS NextDay
    FROM
        a
    GROUP BY
        OrganizationMrn
), b AS (
    SELECT
        VisitDate
        , OrganizationMrn
        , NextDay
        , 1 AS OrderRow
    FROM
        a1


    UNION ALL

    SELECT
        a.VisitDate
        , a.OrganizationMrn
        , a.NextDay
        , b.OrderRow +1 AS OrderRow
    FROM
        a
        JOIN b
        ON a.VisitDate = b.NextDay
), c AS (
SELECT
    VisitDate
    , (SELECT MAX(VisitDate) FROM b WHERE b1.VisitDate > VisitDate) AS PreviousVisitDate
FROM
    b b1
)
SELECT
    c1.VisitDate
    , CASE 
        WHEN DATEDIFF(day,c1.PreviousVisitDate,c1.VisitDate) < 30 THEN PreviousVisitDate
        ELSE NULL
     END AS ReAdmissionFrom
    , CASE
        WHEN DATEDIFF(day,c1.PreviousVisitDate,c1.VisitDate) < 30 THEN 3
        ELSE 0
    END AS Points
FROM
    c c1

SQL：差距和岛屿，分组日期

SQL: Gaps and Islands, Grouped dates

sql

sql-server

tsql

gaps-and-islands