替换每个分区的 NULL 值答案

【问题标题】：Replace NULL values per partition替换每个分区的 NULL 值
【发布时间】：2021-07-04 12:08:26
【问题描述】：

我想为每个session_id 的device 列中的NULL 值填充一个关联的非NULL 值。我怎样才能做到这一点？

这里是示例数据：

+------------+-------+---------+
| session_id | step  | device  |
+------------+-------+---------+
| 351acc     | step1 |         |
| 351acc     | step2 |         |
| 351acc     | step3 | mobile  |
| 351acc     | step4 | mobile  |
| 350bca     | step1 | desktop |
| 350bca     | step2 |         |
| 350bca     | step3 |         |
| 350bca     | step4 | desktop |
+------------+-------+---------+

期望的输出：

+------------+-------+---------+
| session_id | step  | device  |
+------------+-------+---------+
| 351acc     | step1 | mobile  |
| 351acc     | step2 | mobile  |
| 351acc     | step3 | mobile  |
| 351acc     | step4 | mobile  |
| 350bca     | step1 | desktop |
| 350bca     | step2 | desktop |
| 350bca     | step3 | desktop |
| 350bca     | step4 | desktop |
+------------+-------+---------+

【问题讨论】：

标签： sql postgresql null window-functions coalesce

【解决方案1】：

正确排序的window function first_value() 可能最便宜：

SELECT session_id, step
     , COALESCE(device
              , first_value(device) OVER (PARTITION BY session_id ORDER BY device IS NULL, step)
               ) AS device
FROM   tbl
ORDER  BY session_id DESC, step;

db小提琴here

ORDER BY device IS NULL, step 将NULL 值排序在最后，因此选择最早的具有非空值的step。见：

Sorting null values after all others, except special

如果session_id 的 notnull 设备始终相同，您可以简化为仅ORDER BY device IS NULL。而且你不需要COALESCE。

【讨论】：

【解决方案2】：

根据您的数据示例，每个会话有一个设备，因此您只需添加一个子查询即可从其他行获取值

WITH j (session_id, step, device) AS (
  VALUES ('351acc','step1',NULL),
         ('351acc','step2',NULL),
         ('351acc','step3','mobile'),
         ('351acc','step4','mobile'),
         ('350bca','step1','desktop'),
         ('350bca','step2',NULL),
         ('350bca','step3',NULL),
         ('350bca','step4','desktop')
) 
SELECT session_id,step,
  (SELECT DISTINCT device 
   FROM j q2
   WHERE q2.session_id = q1.session_id AND q2.device IS NOT NULL) AS device
FROM j q1 ORDER BY session_id,step;

 session_id | step  | device  
------------+-------+---------
 350bca     | step1 | desktop
 350bca     | step2 | desktop
 350bca     | step3 | desktop
 350bca     | step4 | desktop
 351acc     | step1 | mobile
 351acc     | step2 | mobile
 351acc     | step3 | mobile
 351acc     | step4 | mobile
(8 Zeilen)

演示：db<>fiddle

【讨论】：

【解决方案3】：

select session_id, step,coalesce(device, max(device) over (partition by session_id order by step desc)) device
from table

【讨论】：