Question

我有一个由患者及其合并症（或通常被称为诊断代码）组成的假设数据库：

PATIENT_ID | HOSPITAL | DIAG_01 | DIAG_02 | DIAG_03 | DIAG_04 ...up to Diag_14
 Patient 1 |  Hosp1   |  C01    |  C02    |  NULL   |  NULL
 Patient 2 |  Hosp1   |  A01    |  A02    |  A03    |  P37
 Patient 3 |  Hosp2   |  A01    |  D10    |  X01    |  NULL

患者合并症（其中最多可记录14例）定义患者可能患有的任何疾病/疾病。特别是C00-C97和D00-D48涉及癌症ICD-10代码。

我想为每家医院制作一份摘要，其中列出了有多少患者在诊断栏中有癌症ICD-10代码的情况。

基于上面的例子，Hosp1和Hosp2都有1名患者，每名患有癌症。患者1的C01和C02诊断代码仅计为一名患有癌症的患者。然而，如果患者1再次入院，那将被视为癌症患者的另一例。

我尝试创建一个包含所有癌症合并症的表格，并在所有14个诊断列上使用JOIN查询，但无法计算。

任何想法或帮助都会受到大量的重视。

Answer 1

您的数据布局不正确。你应该有一个联结表。但是，你不是。您可以使用union all：

执行所需操作

select hospital, count(distinct patient_id)
from ((select patient_id, hospital, diag_01 as diag
       from hypothetical
      ) union all
      (select patient_id, hospital, diag_02 as diag
       from hypothetical
      )
      union all
      . . .
      (select patient_id, hospital, diag_14 as diag
       from hypothetical
      )
     ) phd
where diag in ('C01', 'C02');

Answer 2

使用union all

SELECT * FROM /**customize your select here could be count() etc **/
 (SELECT PATIENT_ID,HOSPITAL,DIAGNOSIS
FROM (
    SELECT PATIENT_ID, HOSPITAL, LEFT(DIAG_01,1) AS DIAGNOSIS FROM table_name WHERE DIAG_01 IS NOT NULL UNION ALL   
    SELECT PATIENT_ID, HOSPITAL, LEFT(DIAG_02,1) AS DIAGNOSIS FROM table_name WHERE DIAG_02 IS NOT NULL UNION ALL
    SELECT PATIENT_ID, HOSPITAL, LEFT(DIAG_03,1) AS DIAGNOSIS FROM table_name WHERE DIAG_03 IS NOT NULL UNION ALL
    SELECT PATIENT_ID, HOSPITAL, LEFT(DIAG_04,1) AS DIAGNOSIS FROM table_name WHERE DIAG_04 IS NOT NULL
      ) AS T1
GROUP BY PATIENT_ID,HOSPITAL,DIAGNOSIS
) AS T2
WHERE DIAGNOSIS IN ('C','D') /**where **/

SQL - 跨多个列计算出现次数

2 个答案: