状态的平均持续时间 - 差距和孤岛

Question

我正在尝试计算处于 REPAIR 状态的设备的平均周转时间。

我能够创建一个查询，其中包含设备列表及其每天的快照状态。

+-----------------+--------------+--------+----------------------+------------+------------------+
| equipmentNumber | snapshotDate | status | previousSnapshotDate | prevStatus | statusChangeFlag |
+-----------------+--------------+--------+----------------------+------------+------------------+
|          123456 | 2018-04-29   | ONHIRE | 2018-04-28           | AVAILABLE  |                1 |
|          123456 | 2018-04-30   | ONHIRE | 2018-04-29           | ONHIRE     |                0 |
|          123456 | 2018-05-01   | ONHIRE | 2018-04-30           | ONHIRE     |                0 |
|          123456 | 2018-05-02   | REPAIR | 2018-05-01           | ONHIRE     |                1 |
|          123456 | 2018-05-03   | REPAIR | 2018-05-02           | REPAIR     |                0 |
|          123456 | 2018-05-04   | ONHIRE | 2018-05-03           | REPAIR     |                1 |
|          654321 | 2018-04-30   | REPAIR | 2018-04-29           | AVAILABLE  |                1 |
|          654321 | 2018-05-01   | REPAIR | 2018-04-30           | REPAIR     |                0 |
|          654321 | 2018-05-02   | REPAIR | 2018-05-01           | REPAIR     |                0 |
+-----------------+--------------+--------+----------------------+------------+------------------+

所以，在这个例子中，我们有 2 台设备，“123456”在 5 月 2 日和 5 月 3 日处于维修状态 2 天，“654321”在 4 月 30 日和 5 日处于维修状态 3 天/1 和 5/2。那将是 (2+3) / 2 = 2.5 天的平均维修周转时间。

我试过这个算法 (Detect consecutive dates ranges using SQL)，但它似乎不能完全满足我的需要。

Answer 1

该方法应该可以识别修复周期：

select equipmentNumber, min(snapshotDate), max(snapshotDate)
from (select t.*,
             row_number() over (partition by equipmentNumber order by snapshotDate) as seqnum
      from t
     ) t
where status = 'REPAIR'
group by equipmentNumber, dateadd(day, - seqnum, snapshotDate);

您可以使用子查询获取平均值：

select avg(datediff(day, minsd, maxsd) * 1.0)
from (select equipmentNumber, min(snapshotDate) as minsd, max(snapshotDate) as maxsd
      from (select t.*,
                   row_number() over (partition by equipmentNumber order by snapshotDate) as seqnum
            from t
           ) t
      where status = 'REPAIR'
      group by equipmentNumber, dateadd(day, - seqnum, snapshotDate)
     ) e;

Answer 2

我尝试使用递增 ID 列来回答差距和孤岛，如果不存在则创建一个，并且 ROW_NUMBER window 函数

CREATE TABLE T1
    ([equipmentNumber] int, [snapshotDate] datetime, [status] varchar(6), [previousSnapshotDate] datetime, [prevStatus] varchar(9), [statusChangeFlag] int)
;

INSERT INTO T1
    ([equipmentNumber], [snapshotDate], [status], [previousSnapshotDate], [prevStatus], [statusChangeFlag])
VALUES
    (123456, '2018-04-29 00:00:00', 'ONHIRE', '2018-04-28 00:00:00', 'AVAILABLE', 1),
    (123456, '2018-04-30 00:00:00', 'ONHIRE', '2018-04-29 00:00:00', 'ONHIRE', 0),
    (123456, '2018-05-01 00:00:00', 'ONHIRE', '2018-04-30 00:00:00', 'ONHIRE', 0),
    (123456, '2018-05-02 00:00:00', 'REPAIR', '2018-05-01 00:00:00', 'ONHIRE', 1),
    (123456, '2018-05-03 00:00:00', 'REPAIR', '2018-05-02 00:00:00', 'REPAIR', 0),
    (123456, '2018-05-04 00:00:00', 'ONHIRE', '2018-05-03 00:00:00', 'REPAIR', 1),
    (654321, '2018-04-30 00:00:00', 'REPAIR', '2018-04-29 00:00:00', 'AVAILABLE', 1),
    (654321, '2018-05-01 00:00:00', 'REPAIR', '2018-04-30 00:00:00', 'REPAIR', 0),
    (654321, '2018-05-02 00:00:00', 'REPAIR', '2018-05-01 00:00:00', 'REPAIR', 0)
;
;WITH cteX
AS(
    SELECT
         Id = ROW_NUMBER()OVER(ORDER BY T.equipmentNumber, T.snapshotDate)
        ,T.equipmentNumber
        ,T.snapshotDate
        ,T.[status]
        ,T.previousSnapshotDate
        ,T.prevStatus
        ,T.statusChangeFlag
    FROM dbo.T1 T
),cteIsland
AS(
    SELECT 
         Island = X.Id - ROW_NUMBER()OVER(ORDER BY X.Id)
        ,*
    FROM cteX X
    WHERE X.[status] = 'REPAIR'
)
SELECT * FROM cteIsland

注意 Island 列

Island  Id  equipmentNumber status
3       4   123456          REPAIR
3       5   123456          REPAIR
4       7   654321          REPAIR
4       8   654321          REPAIR
4       9   654321          REPAIR

使用 Island 列，您可以通过此 TSQL

获得所需的答案

;WITH cteX
AS(
    SELECT
         Id = ROW_NUMBER()OVER(ORDER BY T.equipmentNumber, T.snapshotDate)
        ,T.equipmentNumber
        ,T.snapshotDate
        ,T.[status]
        ,T.previousSnapshotDate
        ,T.prevStatus
        ,T.statusChangeFlag
    FROM dbo.T1 T
),cteIsland
AS(
    SELECT 
         Island = X.Id - ROW_NUMBER()OVER(ORDER BY X.Id)
        ,*
    FROM cteX X
    WHERE X.[status] = 'REPAIR'
)
SELECT 
     AvgDuration =SUM(Totals.IslandCounts) / (COUNT(Totals.IslandCounts) * 1.0)
FROM
(
    SELECT 
          IslandCounts = COUNT(I.Island)
         ,I.equipmentNumber
    FROM cteIsland I
    GROUP BY I.equipmentNumber
) Totals

回答

AvgDuration
2.50000000000000

这是SQLFiddle

状态的平均持续时间 - 差距和孤岛

Average Duration in Status - Gaps and Islands

sql

sql-server

gaps-and-islands