sql - 锯齿图案局部最大值的每日总计

Question

我有多个可以临时重置的单调计数器。这些计数器在绘制图表时表现出锯齿行为(但它们并不是严格递增的)。我想要一份显示每个计数器每日最大值总和的月度报告。

到目前为止，我的策略是在计数器小于计数器前一个采样(也小于或等于下一个)的行上放置“1”。然后计算该列的运行总计，以识别没有重置的系列。

然后，我对每日间隔进行分组，计算当天每个系列的最大最小值，然后将这些部分相加以获得当天的总计。

我所拥有的可以工作，但运行需要大约 10 秒。执行计划显示了两大类:一种在 cteData 中，我认为另一种在 cteSeries 中。我觉得我应该能够消除其中一个，但我不知道如何做到这一点。

这段代码的结果是(我现在可以看到实际上是跳过了间隔边界的样本):

interval  tagname total
2020-01-01    alpha   3
2020-01-01    bravo   4
2020-01-02    alpha   3
2020-01-02    bravo   4

IF OBJECT_ID('tempdb..#counter_data') IS NOT NULL
    DROP TABLE #counter_data;

CREATE TABLE #counter_data(
    t_stamp DATETIME NOT NULL
    ,tagname VARCHAR(32) NOT NULL
    ,val REAL NULL
    PRIMARY KEY(t_stamp, tagname)
);

INSERT INTO #counter_data(t_stamp, tagname, val)
VALUES
     ('2020-01-01 04:00', 'alpha', 0)
    ,('2020-01-01 04:00', 'bravo', 0)
    ,('2020-01-01 08:00', 'alpha', 1)
    ,('2020-01-01 08:00', 'bravo', 1)
    ,('2020-01-01 12:00', 'alpha', 2)
    ,('2020-01-01 12:00', 'bravo', 2)
    ,('2020-01-01 16:00', 'alpha', 0)
    ,('2020-01-01 16:00', 'bravo', 3)
    ,('2020-01-01 20:00', 'alpha', 1)
    ,('2020-01-01 20:00', 'bravo', 4)
    ,('2020-01-02 04:00', 'alpha', 2)
    ,('2020-01-02 04:00', 'bravo', 5)
    ,('2020-01-02 08:00', 'alpha', 3)
    ,('2020-01-02 08:00', 'bravo', 6)
    ,('2020-01-02 12:00', 'alpha', 0)
    ,('2020-01-02 12:00', 'bravo', 7)
    ,('2020-01-02 16:00', 'alpha', 1)
    ,('2020-01-02 16:00', 'bravo', 8)
    ,('2020-01-02 20:00', 'alpha', 2)
    ,('2020-01-02 20:00', 'bravo', 9)
;

DECLARE @dateStart AS DATETIME = '2020-01-01';
DECLARE @dateEnd AS DATETIME = DATEADD(month, 2, @dateStart);

WITH cteData AS(
    SELECT
        t_stamp
        ,tagname
        ,val
        ,CASE 
            WHEN val < LAG(val) OVER(PARTITION BY tagname ORDER BY t_stamp)
                AND val <= LEAD(val) OVER(PARTITION BY tagname ORDER BY t_stamp)
                THEN 1 
            ELSE 0
            END AS rn
    FROM #counter_data
    WHERE 
        t_stamp >= @dateStart AND t_stamp < @dateEnd
        AND tagname IN(
            'alpha'
            ,'bravo'
        )
)
,cteSeries AS(
    SELECT
        CAST(t_stamp AS DATE) AS interval
        ,tagname
        ,val
        ,SUM(rn) OVER(PARTITION BY tagname ORDER BY t_stamp) AS series
    FROM cteData
)
,cteSubtotal AS(
    SELECT 
        interval
        ,tagname
        ,MAX(val) - MIN(val) AS subtotal
    FROM cteSeries
    GROUP BY interval, tagname, series
)
,cteGrandTotal AS(
    SELECT
        interval
        ,tagname
        ,SUM(subtotal) AS total
    FROM cteSubtotal
    GROUP BY interval, tagname
)
SELECT *
FROM cteGrandTotal
ORDER BY interval, tagname

Answer 1

我只需通过与前一行进行比较来计算每行中计数器的增加:

with cte
as
(
    SELECT *,isnull(lag(val) over (partition by tagname order by t_stamp),0) as previousVal
    FROM counter_data
)
SELECT cast(t_stamp as date),tagname, sum(case when val>previousVal then val-previousval else val end )
FROM cte
GROUP BY cast(t_stamp as date),tagname;