SQL AVG()为3列返回错误的结果

时间:2013-07-02 12:39:14

标签: sql sql-server-2008 tsql

我正在编写一个应该给我一个count()和三个avg()的查询。 count()工作正常,但avg()函数返回错误结果。我正在使用的数据如下所示:

MD Name |    PT | Med Staff | LOS | DRG Bench | LOS - Bench
MCP     | 12345 | Ortho SX  |  5  |    4      |       1
MCP     | 25879 | Ortho SX  |  3  |    5      |      -2
MCP     | 98556 | Ortho SX  |  4  |    5      |      -1
... 

我想要的输出是:

MD Name | # PT | Med Staff | Avg LOS | Avg DRG Bench | AVG LOS - Bench
MCP     |   3  | Ortho SX  |    4    |       4.66    |       0

我得到的平均值不正确。我有一个案例,特别是我有以下内容:

MD Name | LOS    | Bench  | LOS - Bench
MCP     | 2.0000 | 1.8000 |    0.2000
MCP     | 1.0000 | 1.7000 |   -0.7000
MCP     | 25.0000| 4.9000 |   20.1000
MCP     | 4.0000 | 2.2000 |    1.8000

对于和AVG LOS我得到9.000000而不是8.000000,对于AVG Bench我得到2.780000而不是2.65而对于LOS-Bench我得到6.220000而不是5.35,这些是显着的差异我必须是正确的两位小数。

这是我正在使用的SQL,SQL Server 2008

DECLARE @STARTDATE DATETIME
DECLARE @ENDATE DATETIME

SET @STARTDATE = '2013-05-01'
SET @ENDATE = '2013-05-31'

SELECT DISTINCT pv.pract_rpt_name AS 'PHYSICIAN'
, COUNT(DISTINCT vr.pt_id) AS '# PTS' 
--, pv.spclty_desc AS 'SPECIALTY'
, pv.med_staff_dept AS 'MED STAFF'
, AVG(vr.len_of_stay) AS 'LOS'
, AVG(vr.drg_std_days_stay) AS 'DRG LOS BENCH'
, AVG(vr.len_of_stay - vr.drg_std_days_stay) AS 'LOS - DRG BENCH'

FROM smsmir.vst_rpt vr
LEFT OUTER JOIN smsmir.pyr_plan pp <-- removed and fixed
ON vr.pt_id = pp.pt_id <-- removed and fixed
JOIN smsdss.pract_dim_v pv
ON vr.adm_pract_no = pv.src_pract_no

WHERE vr.adm_dtime BETWEEN @STARTDATE AND @ENDATE
AND vr.vst_type_cd = 'I'
AND pv.spclty_desc != 'NO DESCRIPTION'
--AND pv.spclty_desc NOT LIKE 'HOSPITALIST%'
AND vr.drg_std_days_stay IS NOT NULL
AND pv.pract_rpt_name != '?'
AND pv.orgz_cd = 's0x0'
AND pv.med_staff_dept IN (
'INTERNAL MEDICINE',
'FAMILY PRACTICE',
'SURGERY'
)
GROUP BY pv.pract_rpt_name, pv.med_staff_dept
ORDER BY pv.med_staff_dept, AVG(vr.len_of_stay - vr.drg_std_days_stay)DESC

感谢您的时间和精力。

2 个答案:

答案 0 :(得分:1)

查询中的OUTER JOIN可能会影响AVG功能正在运行的行数。如果您不需要它(我无法看到您的查询中其他地方引用该表的任何地方)尝试删除它。

答案 1 :(得分:1)

唯一的可能性是在表中的行中存在NULL选择... 如果有一个空列,AVG将忽略它,而不是计算它......

DECLARE @STARTDATE DATETIME
DECLARE @ENDATE DATETIME

SET @STARTDATE = '2013-05-01'
SET @ENDATE = '2013-05-31'

SELECT DISTINCT pv.pract_rpt_name AS 'PHYSICIAN'
, COUNT(DISTINCT vr.pt_id) AS '# PTS' 
--, pv.spclty_desc AS 'SPECIALTY'
, pv.med_staff_dept AS 'MED STAFF'
, AVG(ISNULL(vr.len_of_stay,0)) AS 'LOS'
, AVG(ISNULL(vr.drg_std_days_stay,0)) AS 'DRG LOS BENCH'
, AVG(ISNULL((vr.len_of_stay - vr.drg_std_days_stay),0)) AS 'LOS - DRG BENCH'

FROM smsmir.vst_rpt vr
LEFT OUTER JOIN smsmir.pyr_plan pp
ON vr.pt_id = pp.pt_id
JOIN smsdss.pract_dim_v pv
ON vr.adm_pract_no = pv.src_pract_no

WHERE vr.adm_dtime BETWEEN @STARTDATE AND @ENDATE
AND vr.vst_type_cd = 'I'
AND pv.spclty_desc != 'NO DESCRIPTION'
--AND pv.spclty_desc NOT LIKE 'HOSPITALIST%'
AND vr.drg_std_days_stay IS NOT NULL
AND pv.pract_rpt_name != '?'
AND pv.orgz_cd = 's0x0'
AND pv.med_staff_dept IN (
'INTERNAL MEDICINE',
'FAMILY PRACTICE',
'SURGERY'
)
GROUP BY pv.pract_rpt_name, pv.med_staff_dept
ORDER BY pv.med_staff_dept, AVG(vr.len_of_stay - vr.drg_std_days_stay)DESC