[英]How to count number of non-consecutive values in a column using SQL?
在這里跟進我的問題。 假設我在Oracle數據庫中有一個表,如下面的表(table_1),它跟蹤特定個人的服務參與:
name day srvc_ inv
bill 1 1
bill 2 1
bill 3 0
bill 4 0
bill 5 1
bill 6 0
susy 1 1
susy 2 0
susy 3 1
susy 4 0
susy 5 1
我的目標是獲得一個匯總表,列出所有獨特的個人,是否有服務參與和不同服務事件的數量(在這種情況下2為票據,3為susy),其中一個不同的服務事件由一個識別在幾天內打破活動。
為了獲得任何服務,我將使用以下查詢
SELECT table_1."Name", MAX(table_1."Name") AS "any_invl"
FROM table_1
GROUP BY table_1."Name"
但是,我不知道如何獲得服務涉及的數量(2)。 在R中使用靜態數據幀,您將使用運行長度編碼(請參閱我原來的問題),但我不知道如何在SQL中完成此操作。 此操作將在大量記錄上運行,因此將整個數據幀存儲為對象並在R中運行它是不切實際的。
編輯:我的期望輸出如下:
name any_invl n_srvc_inv
bill 1 2
susy 1 3
謝謝你的幫助!
像這樣的東西?
SQL> with test (name, day, srvc_inv) as
2 (select 'bill', 1, 1 from dual union all
3 select 'bill', 2, 1 from dual union all
4 select 'bill', 3, 0 from dual union all
5 select 'bill', 4, 0 from dual union all
6 select 'bill', 5, 1 from dual union all
7 select 'bill', 6, 0 from dual union all
8 select 'susy', 1, 1 from dual union all
9 select 'susy', 2, 0 from dual union all
10 select 'susy', 3, 1 from dual union all
11 select 'susy', 4, 0 from dual union all
12 select 'susy', 5, 1 from dual
13 ),
14 inter as
15 (select name, day, srvc_inv,
16 nvl(lead(srvc_inv) over (partition by name order by day), 0) lsrvc
17 from test
18 )
19 select name,
20 sum(case when srvc_inv <> lsrvc and lsrvc = 0 then 1
21 else 0
22 end) grp
23 from inter
24 group by name;
NAME GRP
---- ----------
bill 2
susy 3
SQL>
我建議使用lag()
。 這個想法是計算一個“1”,但只有當前面的值為零或為null
:
select name, count(*)
from (select t.*,
lag(srvc_inv) over (partition by name order by day) as prev_srvc_inv
from t
) t
where (prev_srvc_inv is null or prev_srvc_inv = 0) and
srvc_inv = 1
group by name;
您可以使用lag()
的默認值來簡化此操作:
select name, count(*)
from (select t.*,
lag(srvc_inv, 1, 0) over (partition by name order by day) as prev_srvc_inv
from t
) t
where prev_srvc_inv = 0 and srvc_inv = 1
group by name;
您可以嘗試以下查詢,具有LAG功能來處理srvc_invl中的更改
select name, 1 any_invl, count(case when diff = 1 then 1 end) n_srvc_inv
from (select name, day, srvc_inv - LAG(srvc_inv, 1, 0) OVER(ORDER BY name, day) diff
from tab
order by name, day) temp
group by name
這是小提琴,供您參考。
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