如何解决SQL查找具有下一个最佳时间戳匹配的行对
我的挑战是找到时间戳相邻的行对，并仅保留值字段距离最小（正值之差）的那对行
表measurement从带有时间戳和值的不同传感器收集数据。
id | sensor_id | timestamp | value
---+-----------+-----------+------
1 | 1 | 12:00:00 | 5
2 | 2 | 12:01:00 | 6
3 | 1 | 12:02:00 | 4
4 | 2 | 12:02:00 | 7
5 | 2 | 12:03:00 | 3
6 | 1 | 12:05:00 | 3
7 | 2 | 12:06:00 | 4
8 | 2 | 12:07:00 | 5
9 | 1 | 12:08:00 | 6
传感器的值从其时间戳记到下一个记录的时间戳记（相同的sensor_id）都是有效的。
图形表示
下方的绿线表示传感器1（蓝线）和传感器2（红线）的值随时间的距离。
我的目标是
- 仅合并两个与时间戳逻辑匹配的传感器的记录（以获取绿线）
- 在以下位置查找实例的局部最小值
- 12:01:00（在12:00:00时，传感器2没有记录）
- 12:05:00
- 12:08:00
真实表位于PostgreSQL数据库中，包含约500万条记录的15个传感器。
测试数据
create table measurement (
id serial,sensor_id integer,timestamp timestamp,value integer)
;
insert into measurement (sensor_id,timestamp,value)
values
(1,'2020-08-16 12:00:00',5),(2,'2020-08-16 12:01:00',6),(1,'2020-08-16 12:02:00',4),7),'2020-08-16 12:03:00',3),'2020-08-16 12:05:00','2020-08-16 12:06:00','2020-08-16 12:07:00','2020-08-16 12:08:00',6)
;
我的方法
选择2个任意传感器（通过特定的sensor_id），进行自我连接，并仅保留具有先前时间戳的传感器2的记录（对于具有传感器1的时间戳的传感器2的最大时间戳，
select
*
from (
select
*,row_number() over (partition by m1.timestamp order by m2.timestamp desc) rownum
from measurement m1
join measurement m2
on m1.sensor_id <> m2.sensor_id
and m1.timestamp >= m2.timestamp
--arbitrarily sensor_ids 1 and 2
where m1.sensor_id = 1
and m2.sensor_id = 2
) foo
where rownum = 1
union --vice versa
select
*
from (
select
*,row_number() over (partition by m2.timestamp order by m1.timestamp desc) rownum
from measurement m1
join measurement m2
on m1.sensor_id <> m2.sensor_id
and m1.timestamp <= m2.timestamp
--arbitrarily sensor_ids 1 and 2
where m1.sensor_id = 1
and m2.sensor_id = 2
) foo
where rownum = 1
;
但这会返回一对12:00:00，其中传感器2没有数据（不是大问题）
在实际表上，语句执行不会在几个小时后结束（大问题）。
我发现了某些类似的问题，但与我的问题不符
谢谢！
解决方法
您可以使用几个横向连接。例如：
with
t as (select distinct timestamp as ts from measurement)
select
t.ts,s1.value as v1,s2.value as v2,abs(s1.value - s2.value) as distance
from t,lateral (
select value
from measurement m
where m.sensor_id = 1 and m.timestamp <= t.ts
order by timestamp desc
limit 1
) s1,lateral (
select value
from measurement m
where m.sensor_id = 2 and m.timestamp <= t.ts
order by timestamp desc
limit 1
) s2
order by t.ts
结果：
ts v1 v2 distance
--------------------- -- -- --------
2020-08-16 12:01:00.0 5 6 1
2020-08-16 12:02:00.0 4 7 3
2020-08-16 12:03:00.0 4 3 1
2020-08-16 12:05:00.0 3 3 0
2020-08-16 12:06:00.0 3 4 1
2020-08-16 12:07:00.0 3 5 2
2020-08-16 12:08:00.0 6 5 1
请参见DB Fiddle上的运行示例。
此外，如果您要所有时间戳，甚至是12:00:00之类的不匹配时间戳，也可以执行以下操作：
with
t as (select distinct timestamp as ts from measurement)
select
t.ts,abs(s1.value - s2.value) as distance
from t
left join lateral (
select value
from measurement m
where m.sensor_id = 1 and m.timestamp <= t.ts
order by timestamp desc
limit 1
) s1 on true
left join lateral (
select value
from measurement m
where m.sensor_id = 2 and m.timestamp <= t.ts
order by timestamp desc
limit 1
) s2 on true
order by t.ts
在这种情况下，无法计算距离。
结果：
ts v1 v2 distance
--------------------- -- ------ --------
2020-08-16 12:00:00.0 5 <null> <null>
2020-08-16 12:01:00.0 5 6 1
2020-08-16 12:02:00.0 4 7 3
2020-08-16 12:03:00.0 4 3 1
2020-08-16 12:05:00.0 3 3 0
2020-08-16 12:06:00.0 3 4 1
2020-08-16 12:07:00.0 3 5 2
2020-08-16 12:08:00.0 6 5 1
第一步是计算每个时间戳记之间的差异。一种方法是使用横向联接和条件聚合：
select t.timestamp,max(m.value) filter (where s.sensor_id = 1) as value_1,max(m.value) filter (where s.sensor_id = 2) as value_2,abs(max(m.value) filter (where s.sensor_id = 2) -
max(m.value) filter (where s.sensor_id = 1)
) as diff
from (values (1),(2)) s(sensor_id) cross join
(select distinct timestamp
from measurement
where sensor_id in (1,2)
) t left join lateral
(select m.value
from measurement m
where m.sensor_id = s.sensor_id and
m.timestamp <= t.timestamp
order by m.timestamp desc
limit 1
) m
on 1=1
group by timestamp;
现在的问题是，差异何时输入局部最小值。对于您的样本数据，本地最小值全部为一个时间单位。这意味着您可以使用lag()和lead()来找到它们：
with t as (
select t.timestamp,abs(max(m.value) filter (where s.sensor_id = 2) -
max(m.value) filter (where s.sensor_id = 1)
) as diff
from (values (1),(2)) s(sensor_id) cross join
(select distinct timestamp
from measurement
where sensor_id in (1,2)
) t left join lateral
(select m.value
from measurement m
where m.sensor_id = s.sensor_id and
m.timestamp <= t.timestamp
order by m.timestamp desc
limit 1
) m
on 1=1
group by timestamp
)
select *
from (select t.*,lag(diff) over (order by timestamp) as prev_diff,lead(diff) over (order by timestamp) as next_diff
from t
) t
where (diff < prev_diff or prev_diff is null) and
(diff < next_diff or next_diff is null);
这可能不是一个合理的假设。因此，在应用此逻辑之前，请滤除相邻的重复值：
select *
from (select t.*,lead(diff) over (order by timestamp) as next_diff
from (select t.*,lag(diff) over (order by timestamp) as test_for_dup
from t
) t
where test_for_dup is distinct from diff
) t
where (diff < prev_diff or prev_diff is null) and
(diff < next_diff or next_diff is null)
Here是db 小提琴。
,填充缺失值需要窗口函数和与两个传感器交叉的每分钟的笛卡尔积。
invars cte接受参数。
with invars as (
select '2020-08-16 12:00:00'::timestamp as start_ts,'2020-08-16 12:08:00'::timestamp as end_ts,array[1,2] as sensor_ids
),创建minute x sensor_id的矩阵
calendar as (
select g.minute,s.sensor_id,sensor_ids[1] as sid1,sensor_ids[2] as sid2
from invars i
cross join generate_series(
i.start_ts,i.end_ts,interval '1 minute'
) as g(minute)
cross join unnest(i.sensor_ids) as s(sensor_id)
),每次从mgrp获得新值时都查找sensor_id
gaps as (
select c.minute,c.sensor_id,m.value,sum(case when m.value is null then 0 else 1 end)
over (partition by c.sensor_id
order by c.minute) as mgrp,c.sid1,c.sid2
from calendar c
left join measurement m
on m.timestamp = c.minute
and m.sensor_id = c.sensor_id
),通过结转最新值来插补缺失值
interpolated as (
select minute,sensor_id,coalesce(
value,first_value(value) over
(partition by sensor_id,mgrp
order by minute)
) as value,sid1,sid2
from gaps
)
执行distance计算（{{1}可能是sum()或max()，这没什么区别。
min()结果：
select minute,sum(value) filter (where sensor_id = sid1) as value1,sum(value) filter (where sensor_id = sid2) as value2,abs(
sum(value) filter (where sensor_id = sid1)
- sum(value) filter (where sensor_id = sid2)
) as distance
from interpolated
group by minute
order by minute;
窗口功能和检查相邻区域。 （您将需要一个额外的反自加入来删除重复项，并为稳定婚姻问题发明决胜局）
SELECT id,ztimestamp,value
--,prev_ts,next_ts,(ztimestamp - prev_ts) AS prev_span,(next_ts - ztimestamp) AS next_span,(sensor_id <> prev_sensor) AS prev_valid,(sensor_id <> next_sensor) AS next_valid,CASE WHEN (sensor_id <> prev_sensor AND sensor_id <> next_sensor) THEN
CASE WHEN (ztimestamp - prev_ts) < (next_ts - ztimestamp) THEN prev_id ELSE next_id END
WHEN (sensor_id <> prev_sensor) THEN prev_id
WHEN (sensor_id <> next_sensor) THEN next_id
ELSE NULL END AS best_neigbor
FROM (
SELECT id,value,lag(id) OVER www AS prev_id,lead(id) OVER www AS next_id,lag(sensor_id) OVER www AS prev_sensor,lead(sensor_id) OVER www AS next_sensor,lag(ztimestamp) OVER www AS prev_ts,lead(ztimestamp) OVER www AS next_ts
FROM measurement
WINDOW www AS (order by ztimestamp)
) q
ORDER BY ztimestamp,sensor_id
;
结果：
DROP SCHEMA
CREATE SCHEMA
SET
CREATE TABLE
INSERT 0 9
id | sensor_id | ztimestamp | value | prev_span | next_span | prev_valid | next_valid | best_neigbor
----+-----------+---------------------+-------+-----------+-----------+------------+------------+--------------
1 | 1 | 2020-08-16 12:00:00 | 5 | | 00:01:00 | | t | 2
2 | 2 | 2020-08-16 12:01:00 | 6 | 00:01:00 | 00:01:00 | t | t | 3
3 | 1 | 2020-08-16 12:02:00 | 4 | 00:01:00 | 00:00:00 | t | t | 4
4 | 2 | 2020-08-16 12:02:00 | 7 | 00:00:00 | 00:01:00 | t | f | 3
5 | 2 | 2020-08-16 12:03:00 | 3 | 00:01:00 | 00:02:00 | f | t | 6
6 | 1 | 2020-08-16 12:05:00 | 3 | 00:02:00 | 00:01:00 | t | t | 7
7 | 2 | 2020-08-16 12:06:00 | 4 | 00:01:00 | 00:01:00 | t | f | 6
8 | 2 | 2020-08-16 12:07:00 | 5 | 00:01:00 | 00:01:00 | f | t | 9
9 | 1 | 2020-08-16 12:08:00 | 6 | 00:01:00 | | t | | 8
(9 rows)
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