Question

我计算患者再入院率，需要找出患者在一定间隔内重新入院的情况，以及频率。我承认数据如下：

Subscriber_id   New_Admission_date
01              2016-06-02 
02              2016-06-01 
03              2016-06-10 
04              2016-06-08 
02              2016-06-04 
02              2016-06-30 
03              2016-06-28

为了找到患者在14天内重新入院的内容以及承认之间的间隔时间，我有以下代码：

select ra.Subscriber_id, DATEDIFF(d,ra.first_ad,ra.last_ad) as interval
from
(
    select j.Subscriber_ID, 
    min(j.New_admission_date) as first_ad, 
    max (j.New_Admission_Date) as last_ad
    from June_inpatients as j
    inner join
        (select j.Subscriber_ID, count(Subscriber_ID) as total
        from June_inpatients as j
        group by Subscriber_ID
        having count(Subscriber_ID) >1 ) as r
    on j.Subscriber_ID = r.Subscriber_ID
    group by j.Subscriber_ID
) as ra
where DATEDIFF(d,ra.first_ad,ra.last_ad) < 15

问题是某些患者，例如示例数据中的患者ID 02，有2个以上的患者。我的代码错过了任何中介，因为它使用min()和max()。当有三次录取时，我如何找到患者的第一次录取和第二次录取之间的间隔，然后找到第二次录取和第三次录取之间的间隔？

Answer 1

假设您至少使用SQL 2012，则可以使用Lag功能。

LAG / LEAD的想法是我们可以查询返回的上一行/下一行的数据。

在下面的完整示例中，我使用LAG两次，一次是订阅者，一次是约会。按订户和日期排序可确保前一行/下一行的顺序正确。然后我限制我的where子句以确保：

上一行是针对同一订户的
日期在15天内

DECLARE @tbl TABLE (
pkey INT NOT NULL PRIMARY KEY IDENTITY,
subscriber INT NOT NULL,
dt DATETIME NOT NULL
);

INSERT INTO @tbl
        ( subscriber, dt )
VALUES  
( 1, '2016-06-02'),
( 2, '2016-06-01'),
(3, '2016-06-10'),
(4, '2016-06-08'),
(2, '2016-06-04'),
(2, '2016-06-30'),
(3, '2016-06-28');

SELECT *
FROM @tbl
ORDER BY subscriber, dt

; WITH tmp AS (
SELECT subscriber, dt, 
LAG(subscriber) OVER (ORDER BY subscriber, dt) previousSubscriber,
LAG(dt) OVER (ORDER BY subscriber, dt) previousDt
FROM @tbl
--ORDER BY subscriber, dt
)
SELECT tmp.*, DATEDIFF(DAY, previousDt, dt)
FROM tmp
WHERE tmp.subscriber = previousSubscriber
    AND DATEDIFF(DAY, previousDt, dt) < 15

Answer 2

如果您使用的是SQL Server 2012或更高版本，请尝试以下操作：

;WITH
    cte As
    (
        SELECT      Subscriber_id,
                    LAG(New_Admission_Date, 1) OVER (PARTITION BY Subscriber_id ORDER BY New_Admission_Date)    AS PreviousAdmissionDate,
                    New_Admission_Date
        FROM        AdmissionTable
    )

SELECT      *
FROM        cte
WHERE       DATEDIFF(DAY, PreviousAdmissionDate, New_Admission_Date) <= 14

Answer 3

这将在没有滞后功能的情况下工作。

;WITH J
AS
(
    SELECT ROW_NUMBER() OVER(PARTITION BY J1.Subscriber_id ORDER BY J1.New_Admission_date ) ROW_ID ,*
    FROM June_inpatients  J1
)
SELECT J1.Subscriber_id, J1.New_Admission_date Previous_Admission_date ,
      J2.Subscriber_id, J2.New_Admission_date , DATEDIFF(DD,J1.New_Admission_date,J2.New_Admission_date) Interval
FROM J J1
INNER JOIN J J2 ON J1.Subscriber_id = J2.Subscriber_id AND J1.ROW_ID = J2.ROW_ID -1 
WHERE DATEDIFF(DD,J1.New_Admission_date,J2.New_Admission_date)<15

当行数超过2行时，如何在行间找到日期间隔？

3 个答案: