在另一个 table 中查找与值最接近的匹配项

Question

我有一个 table_a，在 Postgre 中每个时间戳都有很多行和列SQL 13. 我试图找到 X 列中的值最接近基准值的行从另一个 table 获得。第二个 table 每个时间戳只有一个基准值。对于每个时间戳，我需要 return table_a 的大部分列。当直接提供基准值时，下面的查询工作正常。

如何从 table_b 中获取基准值以用于此查询？

简单地将 table_b.benchmark 替换为 (SELECT benchmark FROM table_b WHERE table_a.timestamp = table_b.timestamp) 导致 'relation "t1" does not exist' 错误。

也找不到有效的连接。

table_a:

+-----------------+-----+---------------+
|    timestamp    |  x  | other_columns |
+-----------------+-----+---------------+
| 2020-01-01 8:00 | 100 |               |
| 2020-01-01 8:00 | 200 |               |
| 2020-01-01 8:00 | 300 |               |
| 2020-01-01 8:00 | 400 |               |
| 2020-01-01 8:00 | 500 |               |
| ...             |     |               |
| 2020-01-01 9:00 | 100 |               |
| 2020-01-01 9:00 | 200 |               |
| 2020-01-01 9:00 | 300 |               |
| 2020-01-01 9:00 | 400 |               |
| 2020-01-01 9:00 | 500 |               |
| ...             |     |               |
+-----------------+-----+---------------+

table_b:

+-----------------+-----------+
|    timestamp    | benchmark |
+-----------------+-----------+
| 2020-01-01 8:00 |       340 |
| 2020-01-01 9:00 |       380 |
| ...             |           |
+-----------------+-----------+

预期结果：

+-----------------+-----+
|    timestamp    |  x  |
+-----------------+-----+
| 2020-01-01 8:00 | 300 |
| 2020-01-01 9:00 | 400 |
| ...             |     |
+-----------------+-----+

SQL查询：

WITH date_filter AS (
    SELECT *
    FROM table_a
    WHERE timestamp >= {start_date} and timestamp < {end_date}
    )

SELECT DISTINCT t1.timestamp, t1.x, t1.etc
FROM date_filter AS t1

INNER JOIN (
    SELECT timestamp, MIN(ABS(x - (table_b.benchmark))) AS target_value
    FROM t1
    GROUP BY timestamp
    ) AS t2

ON t2.timestamp = t1.timestamp AND t2.target_value = ABS(x - (table_b.benchmark))
ORDER BY timestamp ASC;```

Answer 1

我建议横向连接：

select b.*, a.x
from table_b b left join lateral
     (select a.*
      from table_a a
      where a.timestamp = b.timestamp
      order by abs(a.x - b.benchmark)
      limit 1
     ) b
     on 1=1;

Answer 2

一个选项使用横向连接：

select b.timestamp, a.x
from table_b b
cross join lateral (
    select a.*
    from table_a a
    where a.timestamp = b.timestamp
    order by abs(a.x - b.benchmark)
    limit 1
) a

您也可以使用 distinct on:

select distinct on (b.timestamp) b.timestamp, a.x
from table_b b
inner join table_a a on a.timestamp = b.timestamp
order by b.timestamp, abs(a.x - b.benchmark)

在另一个 table 中查找与值最接近的匹配项

Find closest match to value in another table

sql

postgresql

datetime

lateral-join

postgresql-13