我使用找到的答案here建立了一个查询,这非常有帮助。我已经添加了一些东西来满足我的需求。我添加的一件事是ROW_NUMBER()
,以便计算在任何时间长度内30天内有人重新接收了多少次。我已按照第一个答案中的建议和问题that was posted here将cte
结果插入到临时表中。这并没有解决思路,序列长度和序列计数问题。
这是查询:
-- CREATE TABLE TO STORE CTE RESULTS
DECLARE @PPR TABLE(
VISIT1 VARCHAR(20) -- THIS IS A UNIQUE VALUE
, READMIT VARCHAR(20) -- THIS IS A UNIQUE VALUE
, MRN VARCHAR(10) -- THIS IS UNIQUE TO A PERSON
, INIT_DISC DATETIME
, RA_ADM DATETIME
, R1 INT
, R2 INT
, INTERIM1 VARCHAR(20)
, RA_COUNT INT
, FLAG VARCHAR(2)
);
-- THE CTE THAT WILL GET USED TO POPULATE THE ABOVE TABLE
WITH cte AS (
SELECT PTNO_NUM
, Med_Rec_No
, Dsch_Date
, Adm_Date
, ROW_NUMBER() OVER (
PARTITION BY MED_REC_NO
ORDER BY PtNo_Num
) AS r
FROM smsdss.BMH_PLM_PtAcct_V
WHERE Plm_Pt_Acct_Type = 'I'
AND PtNo_Num < '20000000'
)
-- INSERT CTE RESULTS INTO PPR TABLE
INSERT INTO @PPR
SELECT
c1.PtNo_Num AS [INDEX]
, c2.PtNo_Num AS [READMIT]
, c1.Med_Rec_No AS [MRN]
, c1.Dsch_Date AS [INITIAL DISCHARGE]
, c2.Adm_Date AS [READMIT DATE]
, C1.r
, C2.r
, DATEDIFF(DAY, c1.Dsch_Date, c2.Adm_Date) AS INTERIM1
, ROW_NUMBER() OVER (
PARTITION BY C1.MED_REC_NO
ORDER BY C1.PTNO_NUM ASC
) AS [RA COUNT]
, CASE
WHEN DATEDIFF(DAY, c1.Dsch_Date, c2.Adm_Date) <= 30
THEN 1
ELSE 0
END [FLAG]
FROM cte C1
INNER JOIN cte C2
ON C1.Med_Rec_No = C2.Med_Rec_No
WHERE C1.Adm_Date <> C2.Adm_Date
AND C1.r + 1 = C2.r
ORDER BY C1.Med_Rec_No, C1.Dsch_Date
-- MANIPULATE PPR TABLE
SELECT PPR.VISIT1
, PPR.READMIT
, PPR.MRN
, PPR.INIT_DISC
, PPR.RA_ADM
--, PPR.R1
--, PPR.R2
, PPR.INTERIM1
--, PPR.RA_COUNT
, PPR.FLAG
-- THE BELOW DOES NOT WORK AT ALL
, CASE
WHILE (SELECT PPR.INTERIM1 FROM @PPR PPR) <= 30
BEGIN
ROW_NUMBER() OVER (PARTITION BY PPR.MRN, PPR.VISIT1
ORDER BY PPR.VISIT1
)
IF (SELECT PPR.INTERIM1 FROM @PPR PPR) > 30
BREAK
END
END
FROM @PPR PPR
WHERE PPR.MRN = 'A NUMBER'
当前输出示例:
INDEX | READMIT | MRN | INIT DISCHARGE | RA DATE | INTERIM | RACOUNT | FLAG | FLAG_2
12345 | 12349 | 123 | 2005-07-05 | 2005-07-09| 4 | 1 | 1 | 0
12349 | 12351 | 123 | 2005-07-11 | 2005-07-15| 4 | 2 | 1 | 0
因此,第三行显然不是30天内的重新接受,而只是患者回到医院的时间点,因此RA_Count回到1并且标志变为0,因为它不是30天重新接受。
我应该创建一个表而不是使用cte
吗?
我想要添加的是链长和链数。以下是一些定义:
链长:在随后的访问后30天内连续多少次重新入院。
例如
INDEX | READMIT | MRN | INITIAL DISCHARGE | READMIT DATE | CHAIN LEN | Count
123 | 133 | 1236 | 2009-05-13 | 2009-06-12 | 1 | 1
133 | 145 | 1236 | 2009-06-16 | 2009-07-04 | 2 | 1
145 | 157 | 1236 | 2009-07-06 | 2009-07-15 | 3 | 1
165 | 189 | 1236 | 2011-01-01 | 2011-01-12 | 1 | 2
189 | 195 | 1236 | 2011-02-06 | 2011-03-01 | 2 | 2
链计数将是多少链:所以在上表中会有2.我试图使用case
语句来建立链长
这是一个SQL小提琴,其中包含一些示例数据,因为它将在执行CTE
之前显示SQL Fiddle
谢谢,
答案 0 :(得分:4)
更新#1:如果两个事件之间的最大差异为30天,则会链接两个事件。 [COUNT]值是按人数生成的。
您可以调整以下使用recursive common table expression的示例:
CREATE TABLE dbo.Events (
EventID INT IDENTITY(1,1) PRIMARY KEY,
EventDate DATE NOT NULL,
PersonID INT NOT NULL
);
GO
INSERT dbo.Events (EventDate, PersonID)
VALUES
('2014-01-01', 1), ('2014-01-05', 1), ('2014-02-02', 1), ('2014-03-30', 1), ('2014-04-04', 1),
('2014-01-11', 2), ('2014-02-02', 2),
('2014-01-03', 3), ('2014-03-03', 3);
GO
DECLARE @EventsWithNum TABLE (
EventID INT NOT NULL,
EventDate DATE NOT NULL,
PersonID INT NOT NULL,
EventNum INT NOT NULL,
PRIMARY KEY (EventNum, PersonID)
);
INSERT @EventsWithNum
SELECT crt.EventID, crt.EventDate, crt.PersonID,
ROW_NUMBER() OVER(PARTITION BY crt.PersonID ORDER BY crt.EventDate, crt.EventID) AS EventNum
FROM dbo.Events crt;
WITH CountingSequentiaEvents
AS (
SELECT crt.EventID, crt.EventDate, crt.PersonID, crt.EventNum,
1 AS GroupNum,
1 AS GroupEventNum
FROM @EventsWithNum crt
WHERE crt.EventNum = 1
UNION ALL
SELECT crt.EventID, crt.EventDate, crt.PersonID, crt.EventNum,
CASE
WHEN DATEDIFF(DAY, prev.EventDate, crt.EventDate) <= 30 THEN prev.GroupNum
ELSE prev.GroupNum + 1
END AS GroupNum,
CASE
WHEN DATEDIFF(DAY, prev.EventDate, crt.EventDate) <= 30 THEN prev.GroupEventNum + 1
ELSE 1
END AS GroupEventNum
FROM @EventsWithNum crt JOIN CountingSequentiaEvents prev ON crt.PersonID = prev.PersonID
AND crt.EventNum = prev.EventNum + 1
)
SELECT x.EventID, x.EventDate, x.PersonID,
x.GroupEventNum AS [CHAIN LEN],
x.GroupNum AS [Count]
FROM CountingSequentiaEvents x
ORDER BY x.PersonID, x.EventDate
-- 1000 means 1000 + 1 = maximum 1001 events / person
OPTION (MAXRECURSION 1000); -- Please read http://msdn.microsoft.com/en-us/library/ms175972.aspx (section Guidelines for Defining and Using Recursive Common Table Expressions)
输出:
EventID EventDate PersonID CHAIN LEN Count
------- ---------- -------- --------- -----
1 2014-01-01 1 1 1
2 2014-01-05 1 2 1
3 2014-02-02 1 3 1
------- ---------- -------- --------- -----
4 2014-03-30 1 1 2
5 2014-04-04 1 2 2
------- ---------- -------- --------- -----
6 2014-01-11 2 1 1
7 2014-02-02 2 2 1
------- ---------- -------- --------- -----
8 2014-01-03 3 1 1
------- ---------- -------- --------- -----
9 2014-03-03 3 1 2
------- ---------- -------- --------- -----
如你所见
对于最后一个语句,执行计划包含两个Index Seek
运算符,因为PRIMARY KEY (EventNum, PersonID)
上定义的此约束@EventsWithNum
强制SQL Server创建(在本例中)带有a的聚簇索引复合键EventNum, PersonID
。
此外,我们可以看到INSERT @EventsWithNum ...
的估算费用高于WITH CountingSequentiaEvents (...) SELECT ... FROM CountingSequentiaEvents ...
的估算费用。
答案 1 :(得分:2)
解决方案是从一个视图开始,该视图总结了每个VisitID的数据。我使用了一个视图而不是CTE,因为它似乎是你在一次以上的时间里找到我们的东西。
create view vReadmits
as
select t.VisitID,
t.UID,
min(r.AdmitDT) ReadmittedDT,
min(r.VisitID) NextVisitID,
sum(case when r.AdmitDT < dateadd(d, 30, isnull(t.DischargeDT, t.AdmitDT))
then 1 else 0 end) ReadmitNext30
from t
left join t as r
on t.UID = r.UID
and t.VisitID < r.VisitID
group by t.VisitID,
t.UID
这将获取每个VisitID并找到该UID的下一个VisitID。同时,它总结了不到30天的未来访问。它使用ISNULL()来解释丢失的DischargeDT。
然后,您可以在CTE中添加链的逻辑。然后,您可以加入视图和CTE以在视图中包含列。
with Chains as
(
select v.UID,
sum(case when r.ReadmittedDT < dateadd(d, 30, v.ReadmittedDT)
then 0 else 1 end) as ChainCount
from vReadmits v
left join vReadmits r
on r.NextVisitID = v.VisitID
group by v.UID
)
select t.UID,
t.VisitId,
t.AdmitDT,
t.DischargeDT,
v.NextVisitID,
v.ReadmitNext30,
v.ReadmittedDT,
c.ChainCount
from t
join vReadmits v
on t.VisitID = v.VisitID
inner join Chains c
on v.UID = c.UID
order by t.UID, t.VisitID
以下是SQLFiddle
我所做的假设是,如果VisitID大于另一个,那么它的AdmitDT也会更大。这应该是这种情况(特别是对于相同的UID),但如果不是,您将更改视图以使用AdmitDTs而不是VisitID。
答案 2 :(得分:1)
要记住关于CTE的一个缺点。
每次引用它们时,它都会重新执行查询。
在您的查询中,您将引用CTE两次(c1和c2)。这意味着它正在执行两次查询。最好将CTE结果存储在表变量或临时表中,然后执行连接。