自联接，交叉联接和分组
https://stackoverflow.com/questions/1717766
-
19-09-2019 - |
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我已经从多个渠道获得的温度样本的表随着时间的推移，我想找到的最小，最大和平均温度在所有数据源在设定的时间间隔。乍一看，这是很容易，像这样完成的：
SELECT MIN(temp), MAX(temp), AVG(temp) FROM samples GROUP BY time;
然而，事情变得更加复杂（给我难倒点在哪里！）如果源下降进出和，而不是问题的间隔期间忽略丢失的来源我想用源最后一个已知的温度失踪的样本。使用日期时间以及构建跨越不均匀分布随时间进一步样本间隔（每分钟说）复杂的事情。
我认为它应该有可能通过执行对样品表，其中从所述第一表中的时间大于或等于所述第二个表时的自联接，以产生I想要的结果，然后计算用于聚合值行由源分组。不过，我很为难如何真正做到这一点。
下面是我的测试表：
+------+------+------+
| time | source | temp |
+------+------+------+
| 1 | a | 20 |
| 1 | b | 18 |
| 1 | c | 23 |
| 2 | b | 21 |
| 2 | c | 20 |
| 2 | a | 18 |
| 3 | a | 16 |
| 3 | c | 13 |
| 4 | c | 15 |
| 4 | a | 4 |
| 4 | b | 31 |
| 5 | b | 10 |
| 5 | c | 16 |
| 5 | a | 22 |
| 6 | a | 18 |
| 6 | b | 17 |
| 7 | a | 20 |
| 7 | b | 19 |
+------+------+------+
INSERT INTO samples (time, source, temp) VALUES (1, 'a', 20), (1, 'b', 18), (1, 'c', 23), (2, 'b', 21), (2, 'c', 20), (2, 'a', 18), (3, 'a', 16), (3, 'c', 13), (4, 'c', 15), (4, 'a', 4), (4, 'b', 31), (5, 'b', 10), (5, 'c', 16), (5, 'a', 22), (6, 'a', 18), (6, 'b', 17), (7, 'a', 20), (7, 'b', 19);
要尽我的最大，最小和平均计算，我想在中间表，看起来像这样：
+------+------+------+
| time | source | temp |
+------+------+------+
| 1 | a | 20 |
| 1 | b | 18 |
| 1 | c | 23 |
| 2 | b | 21 |
| 2 | c | 20 |
| 2 | a | 18 |
| 3 | a | 16 |
| 3 | b | 21 |
| 3 | c | 13 |
| 4 | c | 15 |
| 4 | a | 4 |
| 4 | b | 31 |
| 5 | b | 10 |
| 5 | c | 16 |
| 5 | a | 22 |
| 6 | a | 18 |
| 6 | b | 17 |
| 6 | c | 16 |
| 7 | a | 20 |
| 7 | b | 19 |
| 7 | c | 16 |
+------+------+------+
下面的查询让我接近我想要什么，但它需要源的第一个结果的温度值，而不是最近的一次在给定的时间间隔：
SELECT s.dt as sdt, s.mac, ss.temp, MAX(ss.dt) as maxdt FROM (SELECT DISTINCT dt FROM samples) AS s CROSS JOIN samples AS ss WHERE s.dt >= ss.dt GROUP BY sdt, mac HAVING maxdt <= s.dt ORDER BY sdt ASC, maxdt ASC;
+------+------+------+-------+
| sdt | mac | temp | maxdt |
+------+------+------+-------+
| 1 | a | 20 | 1 |
| 1 | c | 23 | 1 |
| 1 | b | 18 | 1 |
| 2 | a | 20 | 2 |
| 2 | c | 23 | 2 |
| 2 | b | 18 | 2 |
| 3 | b | 18 | 2 |
| 3 | a | 20 | 3 |
| 3 | c | 23 | 3 |
| 4 | a | 20 | 4 |
| 4 | c | 23 | 4 |
| 4 | b | 18 | 4 |
| 5 | a | 20 | 5 |
| 5 | c | 23 | 5 |
| 5 | b | 18 | 5 |
| 6 | c | 23 | 5 |
| 6 | a | 20 | 6 |
| 6 | b | 18 | 6 |
| 7 | c | 23 | 5 |
| 7 | b | 18 | 7 |
| 7 | a | 20 | 7 |
+------+------+------+-------+
更新：（！伟大的名字，顺便说一下） chadhoc给出了一个很好的解决方案，遗憾的是没有在MySQL的工作，因为它不支持他所使用的FULL JOIN。幸运的是，我相信一个简单UNION是一种有效的替代：
-- Unify the original samples with the missing values that we've calculated
(
SELECT time, source, temp
FROM samples
)
UNION
( -- Pull all the time/source combinations that we are missing from the sample set, along with the temp
-- from the last sampled interval for the same time/source combination if we do not have one
SELECT a.time, a.source, (SELECT t2.temp FROM samples AS t2 WHERE t2.time < a.time AND t2.source = a.source ORDER BY t2.time DESC LIMIT 1) AS temp
FROM
( -- All values we want to get should be a cross of time/temp
SELECT t1.time, s1.source
FROM
(SELECT DISTINCT time FROM samples) AS t1
CROSS JOIN
(SELECT DISTINCT source FROM samples) AS s1
) AS a
LEFT JOIN samples s
ON a.time = s.time
AND a.source = s.source
WHERE s.source IS NULL
)
ORDER BY time, source;
<强>更新2：的MySQL给出以下EXPLAIN输出chadhoc的代码：
+----+--------------------+------------+------+---------------+------+---------+------+------+-----------------------------+
| id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra |
+----+--------------------+------------+------+---------------+------+---------+------+------+-----------------------------+
| 1 | PRIMARY | temp | ALL | NULL | NULL | NULL | NULL | 18 | |
| 2 | UNION | <derived4> | ALL | NULL | NULL | NULL | NULL | 21 | |
| 2 | UNION | s | ALL | NULL | NULL | NULL | NULL | 18 | Using where |
| 4 | DERIVED | <derived6> | ALL | NULL | NULL | NULL | NULL | 3 | |
| 4 | DERIVED | <derived5> | ALL | NULL | NULL | NULL | NULL | 7 | |
| 6 | DERIVED | temp | ALL | NULL | NULL | NULL | NULL | 18 | Using temporary |
| 5 | DERIVED | temp | ALL | NULL | NULL | NULL | NULL | 18 | Using temporary |
| 3 | DEPENDENT SUBQUERY | t2 | ALL | NULL | NULL | NULL | NULL | 18 | Using where; Using filesort |
| NULL | UNION RESULT | <union1,2> | ALL | NULL | NULL | NULL | NULL | NULL | Using filesort |
+----+--------------------+------------+------+---------------+------+---------+------+------+-----------------------------+
我能得到查尔斯的代码，像这样的工作：
SELECT T.time, S.source,
COALESCE(
D.temp,
(
SELECT temp FROM samples
WHERE source = S.source AND time = (
SELECT MAX(time)
FROM samples
WHERE
source = S.source
AND time < T.time
)
)
) AS temp
FROM (SELECT DISTINCT time FROM samples) AS T
CROSS JOIN (SELECT DISTINCT source FROM samples) AS S
LEFT JOIN samples AS D
ON D.source = S.source AND D.time = T.time
它的解释是：
+----+--------------------+------------+------+---------------+------+---------+------+------+-----------------+
| id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra |
+----+--------------------+------------+------+---------------+------+---------+------+------+-----------------+
| 1 | PRIMARY | <derived5> | ALL | NULL | NULL | NULL | NULL | 3 | |
| 1 | PRIMARY | <derived4> | ALL | NULL | NULL | NULL | NULL | 7 | |
| 1 | PRIMARY | D | ALL | NULL | NULL | NULL | NULL | 18 | |
| 5 | DERIVED | temp | ALL | NULL | NULL | NULL | NULL | 18 | Using temporary |
| 4 | DERIVED | temp | ALL | NULL | NULL | NULL | NULL | 18 | Using temporary |
| 2 | DEPENDENT SUBQUERY | temp | ALL | NULL | NULL | NULL | NULL | 18 | Using where |
| 3 | DEPENDENT SUBQUERY | temp | ALL | NULL | NULL | NULL | NULL | 18 | Using where |
+----+--------------------+------------+------+---------------+------+---------+------+------+-----------------+
解决方案
我想你会得到更好的性能利用的排序/窗口化功能在MySQL，但不幸的是我不知道的还有TSQL执行。下面是一个符合ANSI标准解决方案，将工作，虽然：
-- Full join across the sample set and anything missing from the sample set, pulling the missing temp first if we do not have one
select coalesce(c1.[time], c2.[time]) as dt, coalesce(c1.source, c2.source) as source, coalesce(c2.temp, c1.temp) as temp
from samples c1
full join ( -- Pull all the time/source combinations that we are missing from the sample set, along with the temp
-- from the last sampled interval for the same time/source combination if we do not have one
select a.time, a.source,
(select top 1 t2.temp from samples t2 where t2.time < a.time and t2.source = a.source order by t2.time desc) as temp
from
( -- All values we want to get should be a cross of time/samples
select t1.[time], s1.source
from
(select distinct [time] from samples) as t1
cross join
(select distinct source from samples) as s1
) a
left join samples s
on a.[time] = s.time
and a.source = s.source
where s.source is null
) c2
on c1.time = c2.time
and c1.source = c2.source
order by dt, source
其他提示
我知道这看起来很复杂，但它的格式来解释自己... 它应该工作...希望你只有三个来源...如果你有渠道比这将不工作的任意数量......在这种情况下，看到第二个查询... 编辑：删除第一次尝试
编辑：如果你不知道提前的来源，你将不得不做，你创建一个中间结果集“填补”缺失值东西.. 是这样的：
第二编辑：移除的需要聚结由移动逻辑来检索最近温度读数用于从Select子句每个源到加入条件
Select T.Time, Max(Temp) MaxTemp,
Min(Temp) MinTemp, Avg(Temp) AvgTemp
From
(Select T.TIme, S.Source, D.Temp
From (Select Distinct Time From Samples) T
Cross Join
(Select Distinct Source From Samples) S
Left Join Samples D
On D.Source = S.Source
And D.Time =
(Select Max(Time)
From Samples
Where Source = S.Source
And Time <= T.Time)) Z
Group By T.Time
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