过滤掉具有不间断时间范围的条目
[英]filter out entries with an uninterrupted Timeframe
问题描述
我有如下原始数据
INSP_EQP SENDTIME SENDTIME_NOW
Item A 00h 10
Item A 01h 10
Item A 02h 10
Item A 03h 10
Item A 04h 10
Item A 05h 10
Item A 06h 10
Item A 07h 10
Item A 09h 10
Item A 10h 10
Item B 00h 10
Item B 01h 10
Item B 02h 10
Item B 03h 10
Item B 04h 10
Item B 05h 10
Item B 06h 10
Item B 07h 10
Item B 08h 10
Item B 09h 10
Item B 10h 10
我想选择哪个条目与当前时间（SENDTIME_NOW = 10）只有连续的时间范围（至少 5 次）而不会中断。 在这种情况下，只需保留项目 B 并删除项目 A。我想要的结果是：
INSP_EQP SENDTIME SENDTIME_NOW
Item B 00h 10
Item B 01h 10
Item B 02h 10
Item B 03h 10
Item B 04h 10
Item B 05h 10
Item B 06h 10
Item B 07h 10
Item B 08h 10
Item B 09h 10
Item B 10h 10
我试图找到很多功能，但似乎没有 SQL oracle 功能支持这一点。 谁能告诉我是否应该使用 SQL 或使用其他语言来执行此操作。 谢谢
2 个解决方案
解决方案1
0 2022-09-12 03:15:03
您可以为此使用match_recognize ：
- 您可以检查哪些行适合您的条件，例如：
DBFiddle： https://dbfiddle.uk/RsuH2bG2
with t(INSP_EQP, SENDTIME, SENDTIME_NOW) as (
select 'Item A', '00h', 10 from dual union all
select 'Item A', '01h', 10 from dual union all
select 'Item A', '02h', 10 from dual union all
select 'Item A', '03h', 10 from dual union all
select 'Item A', '04h', 10 from dual union all
select 'Item A', '05h', 10 from dual union all
select 'Item A', '06h', 10 from dual union all
select 'Item A', '07h', 10 from dual union all
select 'Item A', '09h', 10 from dual union all
select 'Item A', '10h', 10 from dual union all
select 'Item B', '00h', 10 from dual union all
select 'Item B', '01h', 10 from dual union all
select 'Item B', '02h', 10 from dual union all
select 'Item B', '03h', 10 from dual union all
select 'Item B', '04h', 10 from dual union all
select 'Item B', '05h', 10 from dual union all
select 'Item B', '06h', 10 from dual union all
select 'Item B', '07h', 10 from dual union all
select 'Item B', '08h', 10 from dual union all
select 'Item B', '09h', 10 from dual union all
select 'Item B', '10h', 10 from dual
)
select *
from (-- inline view to make columns more readable
select
INSP_EQP
,to_number(rtrim(sendtime,'h')) SENDTIME_num
,SENDTIME
,SENDTIME_NOW
from t
)
match_recognize (
partition by INSP_EQP
order by SENDTIME_num
measures
classifier() as cls,
count(*) as cnt,
first(SENDTIME_num) as n_first,
last (SENDTIME_num) as n_last
ALL ROWS PER MATCH
PATTERN (ok* err*)
DEFINE
ok as (ok.SENDTIME_num = 0 or ok.SENDTIME_num = prev(ok.SENDTIME_num)+1)
)
返回：
INSP_EQP SENDTIME_NUM CLS CNT N_FIRST N_LAST SENDTIME SENDTIME_NOW
---------- ------------ ----- ---------- ---------- ---------- -------- ------------
Item A 0 OK 1 0 0 00h 10
Item A 1 OK 2 0 1 01h 10
Item A 2 OK 3 0 2 02h 10
Item A 3 OK 4 0 3 03h 10
Item A 4 OK 5 0 4 04h 10
Item A 5 OK 6 0 5 05h 10
Item A 6 OK 7 0 6 06h 10
Item A 7 OK 8 0 7 07h 10
Item A 9 ERR 9 0 9 09h 10
Item A 10 ERR 10 0 10 10h 10
Item B 0 OK 1 0 0 00h 10
Item B 1 OK 2 0 1 01h 10
Item B 2 OK 3 0 2 02h 10
Item B 3 OK 4 0 3 03h 10
Item B 4 OK 5 0 4 04h 10
Item B 5 OK 6 0 5 05h 10
Item B 6 OK 7 0 6 06h 10
Item B 7 OK 8 0 7 07h 10
Item B 8 OK 9 0 8 08h 10
Item B 9 OK 10 0 9 09h 10
Item B 10 OK 11 0 10 10h 10
如您所见，“项目 A”有 2 行带有“错误”，因此您现在可以将它们过滤掉。 例如使用分析函数：
DBFiddle 2： https://dbfiddle.uk/W5b-cE7L
select *
from (
select
v.*
,count(case when cls='ERR' then 1 end)over(partition by insp_eqp) cnt_errs
from (-- inline view to make columns more readable
select
INSP_EQP
,to_number(rtrim(sendtime,'h')) SENDTIME_num
,SENDTIME
,SENDTIME_NOW
from t
)
match_recognize (
partition by INSP_EQP
order by SENDTIME_num
measures
classifier() as cls,
count(*) as cnt,
first(SENDTIME_num) as n_first,
last (SENDTIME_num) as n_last
ALL ROWS PER MATCH
PATTERN (ok* err*)
DEFINE
ok as (ok.SENDTIME_num = 0 or ok.SENDTIME_num = prev(ok.SENDTIME_num)+1)
) v
)
where cnt_errs=0
它返回：
INSP_EQP SENDTIME_NUM CLS CNT N_FIRST N_LAST SENDTIME SENDTIME_NOW CNT_ERRS
---------- ------------ ----- ---------- ---------- ---------- -------- ------------ ----------
Item B 0 OK 1 0 0 00h 10 0
Item B 1 OK 2 0 1 01h 10 0
Item B 2 OK 3 0 2 02h 10 0
Item B 3 OK 4 0 3 03h 10 0
Item B 4 OK 5 0 4 04h 10 0
Item B 5 OK 6 0 5 05h 10 0
Item B 6 OK 7 0 6 06h 10 0
Item B 7 OK 8 0 7 07h 10 0
Item B 8 OK 9 0 8 08h 10 0
Item B 9 OK 10 0 9 09h 10 0
Item B 10 OK 11 0 10 10h 10 0
11 rows selected.
解决方案2
0 2022-09-12 09:29:24
最后基于这篇文章中的 1 个答案，我能够得到我想要的结果。 但似乎有什么不对劲，让我再也看不到那个人的回复了。 无论如何，非常感谢
select INSP_EQP
,SENDTIME
,CHAMBER
,AREA
,SENDTIME_NOW
from (
select INSP_EQP
,CHAMBER
,AREA
,SENDTIME
,SENDTIME_NOW
,SENDTIME_INT
,rn
,pre_2
,SENDTIME_INT-2
,case when RN = 1 and SENDTIME_INT-2 = pre_2 then 1 ELSE null END match
,count(case when RN = 1 and SENDTIME_INT-2 = pre_2 then 1 ELSE null end) over(partition by INSP_EQP) as chk
from (
select INSP_EQP
,SENDTIME
, SENDTIME_INT
,CHAMBER
,AREA
,SENDTIME_NOW
,row_number() over (partition by INSP_EQP order by SENDTIME_INT desc) as rn
,lag(SENDTIME_INT, 2) over(partition by INSP_EQP order by SENDTIME_INT) as pre_2
from t
) t
) t
where chk = 1
order by SENDTIME_INT
)
GROUP BY INSP_EQP
,SENDTIME_NOW
,CHAMBER
,AREA
ORDER BY COUNT(SENDTIME)
如何通过不间断的序列进行GROUP条目？
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