如何计算每个阶段之间的平均时间差。
实际数据集面临的挑战是，并非每个ID都会经历所有阶段。有些会跳过阶段，并且日期对于所有Id来说都不是连续的，如下所示。
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
？-我们也缺少日期
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
答案 0 :(得分：0)
您可以使用窗口函数LAG（或LEAD）来获取每个ID的前一个（或下一个）状态的数据。这样一来，您就可以计算出每个阶段所花费的时间。然后，计算每个阶段花费的平均时间。
以下是如何执行此操作的示例：
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 |
+-------------+-------------------------------------------+
正如我在评论中所说，该逻辑将经过的天数计算为从先前状态到给定状态的时间。由于“需求”没有优先状态，因此此逻辑将始终显示零天的需求。计算从给定状态到 next 状态的时间可能会更好。对于“关闭”，将没有下一个状态。您可以将其保留为空白，也可以使用SYSDATE作为下一个状态的数据。这是一个示例：
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 |
+-------------+------------------------------------------+
答案 1 :(得分：0)
我同意@MatthewMcPeak。您的要求似乎有些奇怪：您在requirement阶段花费了零天，但在closed上平均花费了21天？弗洛德。
此解决方案将显示的日期视为该阶段的开始日期，并计算该日期与下一阶段的开始日期之差。
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
/
答案 2 :(得分：0)
如果您希望包括每个d的所有阶段并估计每个阶段所花费的时间（使用线性插值），则可以创建一个具有所有状态的子查询，并使用PARTITION OUTER JOIN来进行加入他们，然后使用LAG和LEAD查找状态所在的日期范围并在其之间进行插值：
Oracle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
查询2 ：
然后您可以轻松地将其更改为每种状态的平均值：
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
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