如何计算每个阶段之间的平均时间差。
与实际数据集所面临的挑战是不是每个ID将经历的各个阶段..有些人会跳过一些阶段和日期并不适用于所有标识的像下面连续。
id date status
1 1/1/18 requirement
1 1/8/18 analysis
1 ? design
1 1/30/18 closed
2 2/1/18 requirement
2 2/18/18 closed
3 1/2/18 requirement
3 1/29/18 analysis
3 ? accepted
3 2/5/18 closed
？ - 我们缺少的日期，以及
Expected output
id date status time_spent
1 1/1/18 requirement 0
1 1/8/18 analysis 7
1 ? design
1 1/30/18 closed 22
2 2/1/18 requirement 0
2 2/18/18 closed 17
3 1/2/18 requirement 0
3 1/29/18 analysis 27
3 ? accepted
3 2/5/18 closed 24
status avg(timespent)
requirement 0
analysis 17
design
closed 21
您可以使用窗口函数LAG（或LEAD）获得的每个ID的一个（或下）状态的数据。这将让你计算在每个阶段所经过的时间。然后，经过计算每个阶段的平均时间。
这里是如何做到这一点的例子：
with input_data (id, dte, status) as (
SELECT 1, TO_DATE('1/1/18','MM/DD/YY'), 'requirement' FROM DUAL UNION ALL
SELECT 1, TO_DATE('1/8/18','MM/DD/YY'), 'analysis' FROM DUAL UNION ALL
SELECT 1, NULL, 'design' FROM DUAL UNION ALL
SELECT 1, TO_DATE('1/30/18','MM/DD/YY'), 'closed' FROM DUAL UNION ALL
SELECT 2, TO_DATE('2/1/18','MM/DD/YY'), 'requirement' FROM DUAL UNION ALL
SELECT 2, TO_DATE('2/18/18','MM/DD/YY'), 'closed' FROM DUAL UNION ALL
SELECT 3, TO_DATE('1/2/18','MM/DD/YY'), 'requirement' FROM DUAL UNION ALL
SELECT 3, TO_DATE('1/29/18','MM/DD/YY'), 'analysis' FROM DUAL UNION ALL
SELECT 3, NULL, 'accepted' FROM DUAL UNION ALL
SELECT 3, TO_DATE('2/5/18','MM/DD/YY'), 'closed' FROM DUAL ),
----- Solution begins here
data_with_elapsed_days as (
SELECT id.*, dte-nvl(lag(dte ignore nulls) over ( partition by id order by dte ), dte) elapsed
from input_data id)
SELECT status, avg(elapsed)
FROM data_with_elapsed_days d
group by status
order by decode(status,'requirement',1,'analysis',2,'design',3,'accepted',4,'closed',5,99);
+-------------+-------------------------------------------+
| STATUS | AVG(ELAPSED) |
+-------------+-------------------------------------------+
| requirement | 0 |
| analysis | 17 |
| design | |
| accepted | |
| closed | 15.33333333333333333333333333333333333333 |
+-------------+-------------------------------------------+
正如我在我的评论说，这个逻辑计算经过几天的时间从之前的状态给定的状态。因为，“规定”没有之前的状态，这种逻辑将始终显示在需求度过了0天。它可能会更好地从给定的状态到下一个状态计算时间。对于“关闭”，就没有下一个状态。你可以只留下空白或使用SYSDATE作为下一状态的数据。下面是一个例子：
with input_data (id, dte, status) as (
SELECT 1, TO_DATE('1/1/18','MM/DD/YY'), 'requirement' FROM DUAL UNION ALL
SELECT 1, TO_DATE('1/8/18','MM/DD/YY'), 'analysis' FROM DUAL UNION ALL
SELECT 1, NULL, 'design' FROM DUAL UNION ALL
SELECT 1, TO_DATE('1/30/18','MM/DD/YY'), 'closed' FROM DUAL UNION ALL
SELECT 2, TO_DATE('2/1/18','MM/DD/YY'), 'requirement' FROM DUAL UNION ALL
SELECT 2, TO_DATE('2/18/18','MM/DD/YY'), 'closed' FROM DUAL UNION ALL
SELECT 3, TO_DATE('1/2/18','MM/DD/YY'), 'requirement' FROM DUAL UNION ALL
SELECT 3, TO_DATE('1/29/18','MM/DD/YY'), 'analysis' FROM DUAL UNION ALL
SELECT 3, NULL, 'accepted' FROM DUAL UNION ALL
SELECT 3, TO_DATE('2/5/18','MM/DD/YY'), 'closed' FROM DUAL ),
----- Solution begins here
data_with_elapsed_days as (
SELECT id.*, nvl(lead(dte ignore nulls) over ( partition by id order by dte ), trunc(sysdate))-dte elapsed
from input_data id)
SELECT status, avg(elapsed)
FROM data_with_elapsed_days d
group by status
order by decode(status,'requirement',1,'analysis',2,'design',3,'accepted',4,'closed',5,99);
+-------------+------------------------------------------+
| STATUS | AVG(ELAPSED) |
+-------------+------------------------------------------+
| requirement | 17 |
| analysis | 14.5 |
| design | |
| accepted | |
| closed | 361.666666666666666666666666666666666667 |
+-------------+------------------------------------------+
我同意@MatthewMcPeak。您的要求似乎有点冤枉：你花requirement阶段的零天，但花费21天的平均上closed？ Fnord。
该解决方案对待呈现日期作为阶段的开始日期，并计算其与下一阶段的起始日期之间的差。
with cte as (
select status
, lead(dd ignore nulls) over (partition by id order by dd) - dd as dt_diff
from your_table)
select status, avg(dt_diff) as avg_ela
from cte
group by status
/
如果您希望包括每个d所有阶段，估计每个花（使用线性插值）的时间，那么你可以创建一个子查询的所有状态，并使用PARTITION OUTER JOIN加入他们，然后用LAG和LEAD找到日期范围内的地位是在和之间进行插值：
甲骨文设置：
CREATE TABLE data ( d, dt, status ) AS
SELECT 1, TO_DATE( '1/1/18', 'MM/DD/YY' ), 'requirement' FROM DUAL UNION ALL
SELECT 1, TO_DATE( '1/8/18', 'MM/DD/YY' ), 'analysis' FROM DUAL UNION ALL
SELECT 1, NULL, 'design' FROM DUAL UNION ALL
SELECT 1, TO_DATE( '1/30/18', 'MM/DD/YY' ), 'closed' FROM DUAL UNION ALL
SELECT 2, TO_DATE( '2/1/18', 'MM/DD/YY' ), 'requirement' FROM DUAL UNION ALL
SELECT 2, TO_DATE( '2/18/18', 'MM/DD/YY' ), 'closed' FROM DUAL UNION ALL
SELECT 3, TO_DATE( '1/2/18', 'MM/DD/YY' ), 'requirement' FROM DUAL UNION ALL
SELECT 3, TO_DATE( '1/29/18', 'MM/DD/YY' ), 'analysis' FROM DUAL UNION ALL
SELECT 3, NULL, 'accepted' FROM DUAL UNION ALL
SELECT 3, TO_DATE( '2/5/18', 'MM/DD/YY' ), 'closed' FROM DUAL;
查询：
WITH statuses ( status, id ) AS (
SELECT 'requirement', 1 FROM DUAL UNION ALL
SELECT 'analysis', 2 FROM DUAL UNION ALL
SELECT 'design', 3 FROM DUAL UNION ALL
SELECT 'accepted', 4 FROM DUAL UNION ALL
SELECT 'closed', 5 FROM DUAL
),
ranges ( d, dt, status, id, recent_dt, recent_id, next_dt, next_id ) AS (
SELECT d.d,
d.dt,
s.status,
s.id,
NVL(
d.dt,
LAG( d.dt, 1 )
IGNORE NULLS OVER ( PARTITION BY d.d ORDER BY s.id )
),
NVL2(
d.dt,
s.id,
LAG( CASE WHEN d.dt IS NOT NULL THEN s.id END, 1 )
IGNORE NULLS OVER ( PARTITION BY d.d ORDER BY s.id )
),
LEAD( d.dt, 1, d.dt )
IGNORE NULLS OVER ( PARTITION BY d.d ORDER BY s.id ),
LEAD( CASE WHEN d.dt IS NOT NULL THEN s.id END, 1, s.id + 1 )
IGNORE NULLS OVER ( PARTITION BY d.d ORDER BY s.id )
FROM data d
PARTITION BY ( d )
RIGHT OUTER JOIN statuses s
ON ( d.status = s.status )
)
SELECT d,
dt,
status,
( next_dt - recent_dt ) / (next_id - recent_id ) AS estimated_duration
FROM ranges;
输出：
D | DT | STATUS | ESTIMATED_DURATION -: | :-------- | :---------- | ---------------------------------------: 1 | 01-JAN-18 | requirement | 7 1 | 08-JAN-18 | analysis | 7.33333333333333333333333333333333333333 1 | null | design | 7.33333333333333333333333333333333333333 1 | null | accepted | 7.33333333333333333333333333333333333333 1 | 30-JAN-18 | closed | 0 2 | 01-FEB-18 | requirement | 4.25 2 | null | analysis | 4.25 2 | null | design | 4.25 2 | null | accepted | 4.25 2 | 18-FEB-18 | closed | 0 3 | 02-JAN-18 | requirement | 27 3 | 29-JAN-18 | analysis | 2.33333333333333333333333333333333333333 3 | null | design | 2.33333333333333333333333333333333333333 3 | null | accepted | 2.33333333333333333333333333333333333333 3 | 05-FEB-18 | closed | 0
问题2：
然后，你可以很容易地改变，要取平均值对各状态：
WITH statuses ( status, id ) AS (
SELECT 'requirement', 1 FROM DUAL UNION ALL
SELECT 'analysis', 2 FROM DUAL UNION ALL
SELECT 'design', 3 FROM DUAL UNION ALL
SELECT 'accepted', 4 FROM DUAL UNION ALL
SELECT 'closed', 5 FROM DUAL
),
ranges ( d, dt, status, id, recent_dt, recent_id, next_dt, next_id ) AS (
SELECT d.d,
d.dt,
s.status,
s.id,
NVL(
d.dt,
LAG( d.dt, 1 )
IGNORE NULLS OVER ( PARTITION BY d.d ORDER BY s.id )
),
NVL2(
d.dt,
s.id,
LAG( CASE WHEN d.dt IS NOT NULL THEN s.id END, 1 )
IGNORE NULLS OVER ( PARTITION BY d.d ORDER BY s.id )
),
LEAD( d.dt, 1, d.dt )
IGNORE NULLS OVER ( PARTITION BY d.d ORDER BY s.id ),
LEAD( CASE WHEN d.dt IS NOT NULL THEN s.id END, 1, s.id + 1 )
IGNORE NULLS OVER ( PARTITION BY d.d ORDER BY s.id )
FROM data d
PARTITION BY ( d )
RIGHT OUTER JOIN statuses s
ON ( d.status = s.status )
)
SELECT status,
AVG( ( next_dt - recent_dt ) / (next_id - recent_id ) ) AS estimated_duration
FROM ranges
GROUP BY status, id
ORDER BY id;
结果：
STATUS | ESTIMATED_DURATION :---------- | ---------------------------------------: requirement | 12.75 analysis | 4.63888888888888888888888888888888888889 design | 4.63888888888888888888888888888888888889 accepted | 4.63888888888888888888888888888888888889 closed | 0
分贝<>小提琴here