求group by 后满足求差条件后的最小值对应的SQL

2014-06-26 11:47:22 1.3k 阅读 AI 写同款 · GEO 优化AI 选题数据监测一键分发 ›

python+SQL group by后排序取最大值/最小值取根据时间最小值的行 # 这样得到的不是按照ts最小排序的第一行数据 # grouped=data.groupby(['t1.date','t1.imei','address']).head(1) #43626行 grouped=data.groupby(['t1.date','t1.imei']).apply(lambda t: t[t.ts==t.ts.min()]) grouped.hea... 阅读详情

表名T1
数据格式如下.we
C1 C2 C3
A1 ATIME1 112
A1 ATIME2 321
A1 ATIME3 4324
B1 BTIME1 213
B2 BTIME2 434
D1 DTIME1 55

想拉出数据要求如下
以C1 列分组。
在A1 中
如果ATIME1，ATIME2 ， ATIME3 时间差小于 5分钟只取最小的那条。如果大于5分钟则几条都要

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测试230

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以学习为目的 2014-07-01

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引用 50 楼 blandwolf 的回复:

[quote=引用 48 楼 tnt230 的回复:] [quote=引用 47 楼 blandwolf 的回复:] 早上过来，又写了个，用CTE作了个递归，应该可以了吧

with t1 as(
select 'A1' c1,'2014-6-26 18:00:00' c2
union all select 'A1','2014-6-26 18:02:00'
union all select 'A1','2014-6-26 18:05:00'
union all select 'A1','2014-6-26 18:07:00'
union all select 'A1','2014-6-26 18:11:00'
union all select 'A1','2014-6-26 18:12:00'
union all select 'A1','2014-6-26 18:14:00'
union all select 'B1','2014-6-26 18:02:00'
union all select 'B1','2014-6-26 18:06:00'
union all select 'B1','2014-6-26 18:08:00'
union all select 'B1','2014-6-26 18:12:00'
union all select 'B1','2014-6-26 18:14:00'
)
,t2 as(
select (row_number() over (partition by c1 order by c2)) id,*,min(c2) over(partition by c1) c3
from t1
)
,t3 as(
select *
from t2
where id=1
union all
select t2.id,t2.c1,t2.c2,
case when datediff(mi,t3.c3,t2.c2)>=5 then t2.c2 else t3.c3 end
from t2 inner join t3
on t2.c1=t3.c1 and t2.id=t3.id+1
)
select * from t3 where c2=c3 order by c1,c2

都是和最小时间比较的,其实最小时间再变得如分钟时 1,3,7,8,9, 11,12 希望结果1,7,9,12 其实在8被过滤后就是9开始了,应该算的. 逻辑伪码大概是这样 DO(string C1 ,string C2,string C3) //记录的一行,C1类型,C2 开始时间,C3 结束时间 { //一条条读记录X,Y,Z 没下条记录跳出 if (X==C1) , if (Y- C2)< 5, { C3 = Z 覆盖原来结束时间 DO(C1.C2.C3) } else { SAVE()保存记录 DO(C1.Y.Z) } } [/quote] 9开始了，后面为什么又是12呢，不是要差5啊看不懂了～～能不能上20条实际数据和预期结果[/quote] 我46楼就说了，楼主估计给出的数据不全面，所以有些规律肯定没体现出来，所以写出来的不符合要求

测试230 2014-07-01

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引用 50 楼 blandwolf 的回复:

[quote=引用 48 楼 tnt230 的回复:] [quote=引用 47 楼 blandwolf 的回复:] 早上过来，又写了个，用CTE作了个递归，应该可以了吧

with t1 as(
select 'A1' c1,'2014-6-26 18:00:00' c2
union all select 'A1','2014-6-26 18:02:00'
union all select 'A1','2014-6-26 18:05:00'
union all select 'A1','2014-6-26 18:07:00'
union all select 'A1','2014-6-26 18:11:00'
union all select 'A1','2014-6-26 18:12:00'
union all select 'A1','2014-6-26 18:14:00'
union all select 'B1','2014-6-26 18:02:00'
union all select 'B1','2014-6-26 18:06:00'
union all select 'B1','2014-6-26 18:08:00'
union all select 'B1','2014-6-26 18:12:00'
union all select 'B1','2014-6-26 18:14:00'
)
,t2 as(
select (row_number() over (partition by c1 order by c2)) id,*,min(c2) over(partition by c1) c3
from t1
)
,t3 as(
select *
from t2
where id=1
union all
select t2.id,t2.c1,t2.c2,
case when datediff(mi,t3.c3,t2.c2)>=5 then t2.c2 else t3.c3 end
from t2 inner join t3
on t2.c1=t3.c1 and t2.id=t3.id+1
)
select * from t3 where c2=c3 order by c1,c2

都是和最小时间比较的,其实最小时间再变得如分钟时 1,3,7,8,9, 11,12 希望结果1,7,9,12 其实在8被过滤后就是9开始了,应该算的. 逻辑伪码大概是这样 DO(string C1 ,string C2,string C3) //记录的一行,C1类型,C2 开始时间,C3 结束时间 { //一条条读记录X,Y,Z 没下条记录跳出 if (X==C1) , if (Y- C2)< 5, { C3 = Z 覆盖原来结束时间 DO(C1.C2.C3) } else { SAVE()保存记录 DO(C1.Y.Z) } } [/quote] 9开始了，后面为什么又是12呢，不是要差5啊看不懂了～～能不能上20条实际数据和预期结果[/quote] 你的语句应该没问题，我看错了，谢谢你。

测试230 2014-07-01

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引用 50 楼 blandwolf 的回复:

[quote=引用 48 楼 tnt230 的回复:] [quote=引用 47 楼 blandwolf 的回复:] 早上过来，又写了个，用CTE作了个递归，应该可以了吧

with t1 as(
select 'A1' c1,'2014-6-26 18:00:00' c2
union all select 'A1','2014-6-26 18:02:00'
union all select 'A1','2014-6-26 18:05:00'
union all select 'A1','2014-6-26 18:07:00'
union all select 'A1','2014-6-26 18:11:00'
union all select 'A1','2014-6-26 18:12:00'
union all select 'A1','2014-6-26 18:14:00'
union all select 'B1','2014-6-26 18:02:00'
union all select 'B1','2014-6-26 18:06:00'
union all select 'B1','2014-6-26 18:08:00'
union all select 'B1','2014-6-26 18:12:00'
union all select 'B1','2014-6-26 18:14:00'
)
,t2 as(
select (row_number() over (partition by c1 order by c2)) id,*,min(c2) over(partition by c1) c3
from t1
)
,t3 as(
select *
from t2
where id=1
union all
select t2.id,t2.c1,t2.c2,
case when datediff(mi,t3.c3,t2.c2)>=5 then t2.c2 else t3.c3 end
from t2 inner join t3
on t2.c1=t3.c1 and t2.id=t3.id+1
)
select * from t3 where c2=c3 order by c1,c2

都是和最小时间比较的,其实最小时间再变得如分钟时 1,3,7,8,9, 11,12 希望结果1,7,9,12 其实在8被过滤后就是9开始了,应该算的. 逻辑伪码大概是这样 DO(string C1 ,string C2,string C3) //记录的一行,C1类型,C2 开始时间,C3 结束时间 { //一条条读记录X,Y,Z 没下条记录跳出 if (X==C1) , if (Y- C2)< 5, { C3 = Z 覆盖原来结束时间 DO(C1.C2.C3) } else { SAVE()保存记录 DO(C1.Y.Z) } } 不好意思,我举例错了. 是这种情况, [/quote] 9开始了，后面为什么又是12呢，不是要差5啊看不懂了～～能不能上20条实际数据和预期结果[/quote] 不好意思,我打错了期望结果1，7，12

向东流 2014-06-30

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引用 48 楼 tnt230 的回复:

[quote=引用 47 楼 blandwolf 的回复:] 早上过来，又写了个，用CTE作了个递归，应该可以了吧

with t1 as(
select 'A1' c1,'2014-6-26 18:00:00' c2
union all select 'A1','2014-6-26 18:02:00'
union all select 'A1','2014-6-26 18:05:00'
union all select 'A1','2014-6-26 18:07:00'
union all select 'A1','2014-6-26 18:11:00'
union all select 'A1','2014-6-26 18:12:00'
union all select 'A1','2014-6-26 18:14:00'
union all select 'B1','2014-6-26 18:02:00'
union all select 'B1','2014-6-26 18:06:00'
union all select 'B1','2014-6-26 18:08:00'
union all select 'B1','2014-6-26 18:12:00'
union all select 'B1','2014-6-26 18:14:00'
)
,t2 as(
select (row_number() over (partition by c1 order by c2)) id,*,min(c2) over(partition by c1) c3
from t1
)
,t3 as(
select *
from t2
where id=1
union all
select t2.id,t2.c1,t2.c2,
case when datediff(mi,t3.c3,t2.c2)>=5 then t2.c2 else t3.c3 end
from t2 inner join t3
on t2.c1=t3.c1 and t2.id=t3.id+1
)
select * from t3 where c2=c3 order by c1,c2

chen357313771 2014-06-30

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引用 45 楼 chen357313771 的回复:

也写了个，应该能满足吧


WITH T1(C1,C2) AS(
SELECT		'A1' ,'2014-6-26 18:00:00' 
UNION ALL SELECT 'A1','2014-6-26 18:02:00'
UNION ALL SELECT 'A1','2014-6-26 18:06:00'
UNION ALL SELECT 'A1','2014-6-26 18:08:00'
UNION ALL SELECT 'A1','2014-6-26 18:12:00'
UNION ALL SELECT 'A1','2014-6-26 18:14:00'
UNION ALL SELECT 'B1','2014-6-23 17:00:00'
UNION ALL SELECT 'B1','2014-6-23 17:01:00'
UNION ALL SELECT 'B1','2014-6-23 17:08:00'
UNION ALL SELECT 'B1','2014-6-23 17:18:00'
UNION ALL SELECT 'C1','2014-6-24 19:00:00'
UNION ALL SELECT 'C1','2014-6-24 19:07:00'
UNION ALL SELECT 'C1','2014-6-24 19:08:00'
UNION ALL SELECT 'C1','2014-6-24 19:18:00'
)
,t2 AS (
	SELECT *
		,ROW_NUMBER() OVER(ORDER BY c1,c2) AS ID
		,ROW_NUMBER() OVER(PARTITION BY c1 ORDER BY c2) AS ID1
		,MIN(c2) OVER(PARTITION BY c1) AS mintime
	FROM t1
)
,t3 AS (
	SELECT *,CAST(mintime AS DATETIME ) AS c3 FROM t2 WHERE ID=1 AND ID1=1
	UNION ALL
    SELECT A.*,CASE WHEN A.c1=B.c1 THEN (CASE WHEN  DATEADD(MINUTE,5,CAST(B.c3 AS DATETIME))>=CAST(A.c2 AS datetime) 
					THEN CAST(B.c3 AS datetime) ELSE CAST(A.c2 AS DATETIME) END ) ELSE CAST(A.c2 AS datetime) END
	FROM t2 AS A,t3 AS B
	WHERE A.ID=B.ID+1  
)
SELECT ID1, c1,c2 
FROM t3 
WHERE c2=c3

/*
ID1	c1	c2
1	A1	2014-6-26 18:00:00
3	A1	2014-6-26 18:06:00
5	A1	2014-6-26 18:12:00
1	B1	2014-6-23 17:00:00
3	B1	2014-6-23 17:08:00
4	B1	2014-6-23 17:18:00
1	C1	2014-6-24 19:00:00
2	C1	2014-6-24 19:07:00
4	C1	2014-6-24 19:18:00
*/

介个不行么？

测试230 2014-06-30

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引用 47 楼 blandwolf 的回复:

早上过来，又写了个，用CTE作了个递归，应该可以了吧

with t1 as(
select 'A1' c1,'2014-6-26 18:00:00' c2
union all select 'A1','2014-6-26 18:02:00'
union all select 'A1','2014-6-26 18:05:00'
union all select 'A1','2014-6-26 18:07:00'
union all select 'A1','2014-6-26 18:11:00'
union all select 'A1','2014-6-26 18:12:00'
union all select 'A1','2014-6-26 18:14:00'
union all select 'B1','2014-6-26 18:02:00'
union all select 'B1','2014-6-26 18:06:00'
union all select 'B1','2014-6-26 18:08:00'
union all select 'B1','2014-6-26 18:12:00'
union all select 'B1','2014-6-26 18:14:00'
)
,t2 as(
select (row_number() over (partition by c1 order by c2)) id,*,min(c2) over(partition by c1) c3
from t1
)
,t3 as(
select *
from t2
where id=1
union all
select t2.id,t2.c1,t2.c2,
case when datediff(mi,t3.c3,t2.c2)>=5 then t2.c2 else t3.c3 end
from t2 inner join t3
on t2.c1=t3.c1 and t2.id=t3.id+1
)
select * from t3 where c2=c3 order by c1,c2

向东流 2014-06-27

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早上过来，又写了个，用CTE作了个递归，应该可以了吧

with t1 as(
select 'A1' c1,'2014-6-26 18:00:00' c2
union all select 'A1','2014-6-26 18:02:00'
union all select 'A1','2014-6-26 18:05:00'
union all select 'A1','2014-6-26 18:07:00'
union all select 'A1','2014-6-26 18:11:00'
union all select 'A1','2014-6-26 18:12:00'
union all select 'A1','2014-6-26 18:14:00'
union all select 'B1','2014-6-26 18:02:00'
union all select 'B1','2014-6-26 18:06:00'
union all select 'B1','2014-6-26 18:08:00'
union all select 'B1','2014-6-26 18:12:00'
union all select 'B1','2014-6-26 18:14:00'
)
,t2 as(
select (row_number() over (partition by c1 order by c2)) id,*,min(c2) over(partition by c1) c3
from t1
)
,t3 as(
select *
from t2
where id=1
union all
select t2.id,t2.c1,t2.c2,
case when datediff(mi,t3.c3,t2.c2)>=5 then t2.c2 else t3.c3 end
from t2 inner join t3
on t2.c1=t3.c1 and t2.id=t3.id+1
)
select * from t3 where c2=c3 order by c1,c2

以学习为目的 2014-06-27

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引用 41 楼 tnt230 的回复:

[quote=引用 36 楼 galenkeny 的回复:] [quote=引用 31 楼 tnt230 的回复:] [quote=引用 27 楼 blandwolf 的回复:] 1 A1 2014-6-26 18:00:00 2 A1 2014-6-26 18:01:00 3 A1 2014-6-26 18:05:01 你预期的结果是什么呢？

用我这租数据 1 A1 2014-6-26 18:00:00 2 A1 2014-6-26 18:04:00 3 A1 2014-6-26 18:08:00 要1,3 [/quote]

declare @galenkeny table(C1 VARCHAR(10),C2 DATETIME)
insert into @galenkeny
SELECT 'A1', '2014-6-26 18:00:00' union all
select 'A1', '2014-6-26 18:04:00' union all
select 'A1', '2014-6-26 18:08:00' 

;WITH cte AS 
(SELECT *,rn=ROW_NUMBER()OVER(PARTITION BY C1 ORDER BY C2 DESC ) FROM @galenkeny)

SELECT C1,C2 FROM  cte
WHERE rn IN(1,3) ORDER BY C2

[/quote] 这个不满足需求[/quote] 你给出的例子这个是满足你期望结果的。或者你多列出数据，写出正确的期望结果

chen357313771 2014-06-27

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也写了个，应该能满足吧


WITH T1(C1,C2) AS(
SELECT		'A1' ,'2014-6-26 18:00:00' 
UNION ALL SELECT 'A1','2014-6-26 18:02:00'
UNION ALL SELECT 'A1','2014-6-26 18:06:00'
UNION ALL SELECT 'A1','2014-6-26 18:08:00'
UNION ALL SELECT 'A1','2014-6-26 18:12:00'
UNION ALL SELECT 'A1','2014-6-26 18:14:00'
UNION ALL SELECT 'B1','2014-6-23 17:00:00'
UNION ALL SELECT 'B1','2014-6-23 17:01:00'
UNION ALL SELECT 'B1','2014-6-23 17:08:00'
UNION ALL SELECT 'B1','2014-6-23 17:18:00'
UNION ALL SELECT 'C1','2014-6-24 19:00:00'
UNION ALL SELECT 'C1','2014-6-24 19:07:00'
UNION ALL SELECT 'C1','2014-6-24 19:08:00'
UNION ALL SELECT 'C1','2014-6-24 19:18:00'
)
,t2 AS (
	SELECT *
		,ROW_NUMBER() OVER(ORDER BY c1,c2) AS ID
		,ROW_NUMBER() OVER(PARTITION BY c1 ORDER BY c2) AS ID1
		,MIN(c2) OVER(PARTITION BY c1) AS mintime
	FROM t1
)
,t3 AS (
	SELECT *,CAST(mintime AS DATETIME ) AS c3 FROM t2 WHERE ID=1 AND ID1=1
	UNION ALL
    SELECT A.*,CASE WHEN A.c1=B.c1 THEN (CASE WHEN  DATEADD(MINUTE,5,CAST(B.c3 AS DATETIME))>=CAST(A.c2 AS datetime) 
					THEN CAST(B.c3 AS datetime) ELSE CAST(A.c2 AS DATETIME) END ) ELSE CAST(A.c2 AS datetime) END
	FROM t2 AS A,t3 AS B
	WHERE A.ID=B.ID+1  
)
SELECT ID1, c1,c2 
FROM t3 
WHERE c2=c3

/*
ID1	c1	c2
1	A1	2014-6-26 18:00:00
3	A1	2014-6-26 18:06:00
5	A1	2014-6-26 18:12:00
1	B1	2014-6-23 17:00:00
3	B1	2014-6-23 17:08:00
4	B1	2014-6-23 17:18:00
1	C1	2014-6-24 19:00:00
2	C1	2014-6-24 19:07:00
4	C1	2014-6-24 19:18:00
*/

东家来了 2014-06-27

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create table loginlog(UserID bigint, LoginTime datetime) insert into loginlog(UserID, LoginTime) values (201401, '2014-06-27 08:00:43'), (201401, '2014-06-27 08:03:43'), (201401, '2014-06-27 08:05:43'), (201401, '2014-06-27 08:08:43'), (201402, '2014-06-27 08:01:43'), (201402, '2014-06-27 08:02:43'), (201402, '2014-06-27 08:07:43'), (201402, '2014-06-27 08:09:43'), (201403, '2014-06-27 08:01:43'), (201403, '2014-06-27 08:04:43'), (201403, '2014-06-27 08:06:43'), (201403, '2014-06-27 08:09:43'), (201401, '2014-06-27 09:00:43'), (201401, '2014-06-27 09:03:43'), (201401, '2014-06-27 09:05:43'), (201401, '2014-06-27 09:08:43'), (201402, '2014-06-27 09:01:43'), (201402, '2014-06-27 09:02:43'), (201402, '2014-06-27 09:07:43'), (201402, '2014-06-27 09:09:43'), (201403, '2014-06-27 09:01:43'), (201403, '2014-06-27 09:04:43'), (201403, '2014-06-27 09:06:43'), (201403, '2014-06-27 09:09:43') with cte as ( select UserID, LoginTime, (DATEPART(MINUTE,LoginTime)/5)*5 'GroupID', ROW_NUMBER() over(partition by UserID, (DATEPART(MINUTE,LoginTime)/5)*5, CONVERT(varchar(14), LoginTime) order by LoginTime ) 'Num' from loginlog) select UserID, LoginTime from cte where Num = 1

twtiqfn 2014-06-27

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有点难啊，我写不了来，水平还不行，哈

测试230 2014-06-26

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引用 3 楼 xdashewan 的回复:

[quote=引用 2 楼 tnt230 的回复:] max(c2) - min(c2) > 5分钟,你只比了最大和最小值,还有中间的忽略了

你要逐条比较差5分钟？因为你需求没写要逐条比较[/quote] 就是要每次都比, 打个比方. 对于用户登录数的统计. 同一用户在5分钟内的重复登录只算时间最早的那次

xdashewan 2014-06-26

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引用 2 楼 tnt230 的回复:

max(c2) - min(c2) > 5分钟,你只比了最大和最小值,还有中间的忽略了

你要逐条比较差5分钟？因为你需求没写要逐条比较

测试230 2014-06-26

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引用 1 楼 xdashewan 的回复:

先做group by ，然后用T1和group by后的子集联，条件是 C1相等 and （max(c3) - min(c3) > 5分钟）or （C3 = min(c3)）

有点问题没明白, max(c2) - min(c2) > 5分钟,你只比了最大和最小值,还有中间的忽略了

xdashewan 2014-06-26

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先做group by ，然后用T1和group by后的子集联，条件是 C1相等 and （max(c3) - min(c3) > 5分钟）or （C3 = min(c3)）

xdashewan 2014-06-26

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WITH cte as
 (
    SELECT top 1 C1,C2,C3 FROM T1 WHERE C1 = 'A1'
    UNION ALL
    SELECT top 1 a.C1,a.C2,a.C3 FROM T1 a,cte b
    where a.C1 = b.C1 and a.C3 - b.C3 > 5分钟   //时间比较自己写
)
SELECT *  from cte

拿去自己改改，没环境测试，我只写了a1数据，如果要其他一起，你自己改 UNION上面代码，用分组排序取每个c1的第一条数据

测试230 2014-06-26

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引用 36 楼 galenkeny 的回复:

[quote=引用 31 楼 tnt230 的回复:] [quote=引用 27 楼 blandwolf 的回复:] 1 A1 2014-6-26 18:00:00 2 A1 2014-6-26 18:01:00 3 A1 2014-6-26 18:05:01 你预期的结果是什么呢？

用我这租数据 1 A1 2014-6-26 18:00:00 2 A1 2014-6-26 18:04:00 3 A1 2014-6-26 18:08:00 要1,3 [/quote]

declare @galenkeny table(C1 VARCHAR(10),C2 DATETIME)
insert into @galenkeny
SELECT 'A1', '2014-6-26 18:00:00' union all
select 'A1', '2014-6-26 18:04:00' union all
select 'A1', '2014-6-26 18:08:00' 

;WITH cte AS 
(SELECT *,rn=ROW_NUMBER()OVER(PARTITION BY C1 ORDER BY C2 DESC ) FROM @galenkeny)

SELECT C1,C2 FROM  cte
WHERE rn IN(1,3) ORDER BY C2

[/quote] 这个不满足需求

以学习为目的 2014-06-26

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引用 38 楼 tnt230 的回复:

[quote=引用 32 楼 xdashewan 的回复:] [quote=引用 26 楼 tnt230 的回复:] 恩,你说的有道理,刚才我正好也在考虑这情况 ,

思路我早给你了，可惜你只顾看现成sql，完全不理会[/quote] 我看了阿,问题真不知道要怎么写, 你这种做法是不是也要逐个比较? A 18:00:00 A 18:01:00 A 18:07:00 A 18:08:00 2,3比较大于5,是不是第2条也被记了呢? [/quote] 当然按照你的需求，是不记第二条的

测试230 2014-06-26

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引用 37 楼 blandwolf 的回复:

with t1 as( select 'A1' c1,'2014-6-26 18:00:00' c2 union all select 'A1','2014-6-26 18:02:00' union all select 'A1','2014-6-26 18:06:00' union all select 'A1','2014-6-26 18:08:00' union all select 'A1','2014-6-26 18:12:00' union all select 'A1','2014-6-26 18:14:00' ) 你要什么，这样就基本能判断你的需求了

需求就是5分钟内登录数只算一次,并且是时间最早的那次. 从你这个数据就三要1,3,5. 逻辑上应该是两个逐行比较,如果小于5记第1条,(算1次登录) 同时跳过第2条从第三条开始,(第三条算新登录) 如果大于5两条都记(算两次次登录),

测试230 2014-06-26

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引用 32 楼 xdashewan 的回复:

[quote=引用 26 楼 tnt230 的回复:] 恩,你说的有道理,刚才我正好也在考虑这情况 ,

思路我早给你了，可惜你只顾看现成sql，完全不理会[/quote] 我看了阿,问题真不知道要怎么写, 你这种做法是不是也要逐个比较? A 18:00:00 A 18:01:00 A 18:07:00 A 18:08:00 2,3比较大于5,是不是第2条也被记了呢?

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