Question

我正在使用来自汽车跟踪的数据的PostgreSQL数据库，该数据看起来与此类似。

+--------+---------------------+-------+---------+------------+------------+
| car_id |       datetime      | trip  |  speed  |  hub1_dist |  hub2_dist |
+--------+---------------------+-------+---------+------------+------------+
|    1   | 2014-12-20 12:12:12 |   1   |    82   |     6.7    |     2.1    |
|    1   | 2014-12-20 12:12:13 |   1   |    81   |     3.5    |     4.4    |
|    1   | 2014-12-20 12:12:14 |   1   |    85   |     7.8    |     3.8    |
|    1   | 2014-12-20 17:42:51 |   2   |    82   |     4.2    |     5.4    |
|    1   | 2014-12-20 17:42:52 |   2   |    76   |     2.5    |     2.1    |
|    2   | 2014-12-21 09:24:24 |   3   |    80   |     4.5    |     6.7    |
|    2   | 2014-12-21 09:24:25 |   3   |    84   |     5.3    |     3.5    |
|    2   | 2014-12-21 09:24:26 |   3   |    83   |     8.2    |     5.1    |
+--------+---------------------+-------+---------+------------+------------+

每辆车都有特定的car_id，并且trip会基于该car_id进行更改，或者datetime的更改大于5秒。为每个数据点注册speed，并计算到多个hubs的距离

我想根据到每个speed的最短距离，最后得到一个显示hub的表格。它应该看起来像这样：

+-------+------------------+------------------+
| trip  |    hub1_speed    |  speed_hubdist_2 |
+-------+------------------+------------------+
|   1   |        81        |        82        |
|   2   |        76        |        76        |
|   3   |        80        |        84        |
+-------+------------------+------------------+

我当前正在使用：

SELECT trip, hub1_dist, speed
FROM (
  SELECT trip, hub1_dist, speed,
         MIN(hub1_dist) OVER (PARTITION BY trip) AS MIN_hub1_dist
  FROM data_base
) t
WHERE hub1_dist = MIN_hub1_dist

那行得通。但是在某些情况下，我有50 hubs，所以我必须做50次。

Answer 1

我认为您只想进行汇总：

select trip, min(hub1_dist), min(hub2_dist), min(hub3_dist), . . . 
from data_base
group by trip;

您需要为每一列键入表达式。电子表格可以帮助您生成查询。或者，您可以执行以下操作：

select string_agg(replace('min(hub@n_dist) as hub@n_dist', '@n', gs.n::text), ', ')
from generate_series(1, 50, 1) gs(n)

编辑：

您可以使用数组函数获得最小距离的速度：

select trip, min(hub1_dist),
       (array_agg(speed order by hub1_dist))[1] as speed_at_min1,
       min(hub2_dist),
       (array_agg(speed order by hub2_dist))[1] as speed_at_min2,
       min(hub3_dist),
       (array_agg(speed order by hub3_dist))[1] as speed_at_min3,
       . . . 
from data_base
group by trip;

根据最小值选择数据

1 个答案: