Вычислить Average True Range с помощью SQL Server 2014

Question

Я хотел бы вычислить Average True Range с использованием MS SQL Server 2014.

Основная таблица выглядит следующим образом:

TIME     BID-OPEN BID-HIGH BID-LOW BID-CLOSE 
1993-05-09 21:00:00.000 1.5786 1.5786 1.5311 1.5346 
1993-05-10 21:00:00.000 1.5346 1.5551 1.5326 1.5391 
1993-05-11 21:00:00.000 1.5391 1.5521 1.5299 1.5306 
1993-05-12 21:00:00.000 1.5306 1.5451 1.5106 1.5256 
1993-05-13 21:00:00.000 1.5256 1.5416 1.5211 1.5361 
1993-05-16 21:00:00.000 1.5361 1.5534 1.5318 1.5361 
1993-05-17 21:00:00.000 1.5361 1.5451 1.5243 1.5361 
1993-05-18 21:00:00.000 1.5361 1.5506 1.5316 1.5431 
1993-05-19 21:00:00.000 1.5431 1.5686 1.5391 1.5566 
1993-05-20 21:00:00.000 1.5566 1.5665 1.5401 1.5481 
1993-05-23 21:00:00.000 1.5481 1.5481 1.5221 1.5356 
1993-05-24 21:00:00.000 1.5356 1.5491 1.5341 1.5436 
1993-05-25 21:00:00.000 1.5436 1.5567 1.5421 1.5476 
1993-05-26 21:00:00.000 1.5476 1.5656 1.5421 1.5621 
1993-05-27 21:00:00.000 1.5621 1.5714 1.5546 1.5621 
1993-05-30 21:00:00.000 1.5621 1.5631 1.5571 1.5621 
1993-05-31 21:00:00.000 1.5621 1.5706 1.5421 1.5611 
1993-06-01 21:00:00.000 1.5611 1.5611 1.5341 1.5341 
1993-06-02 21:00:00.000 1.5341 1.5526 1.5341 1.5411 
1993-06-03 21:00:00.000 1.5411 1.5566 1.5071 1.5081 
1993-06-06 21:00:00.000 1.5081 1.5306 1.5071 1.5246 
1993-06-07 21:00:00.000 1.5246 1.5306 1.5142 1.5221 
1993-06-08 21:00:00.000 1.5221 1.5271 1.5076 1.5148 
1993-06-09 21:00:00.000 1.5148 1.5326 1.5016 1.5301 
1993-06-10 21:00:00.000 1.5301 1.5401 1.5056 1.5231 
1993-06-13 21:00:00.000 1.5231 1.5331 1.5226 1.5281 
1993-06-14 21:00:00.000 1.5281 1.5391 1.5151 1.5171 
1993-06-15 21:00:00.000 1.5171 1.5283 1.4991 1.5063 
1993-06-16 21:00:00.000 1.5063 1.5186 1.4994 1.5153 
1993-06-17 21:00:00.000 1.5153 1.5171 1.4901 1.4941 
1993-06-20 21:00:00.000 1.4941 1.4988 1.4808 1.4888 
1993-06-21 21:00:00.000 1.4888 1.4911 1.4751 1.4788 
1993-06-22 21:00:00.000 1.4788 1.4846 1.4681 1.4715 
1993-06-23 21:00:00.000 1.4715 1.4776 1.4645 1.4704 
1993-06-24 21:00:00.000 1.4704 1.4886 1.4661 1.4806 

Токовый выход запроса, который вычисляет True Range заключается в следующем:

Time     H-L  H-Cp L-Cp TR 
1993-05-09 21:00:00.000 0.0475 NULL NULL 0.0475 
1993-05-10 21:00:00.000 0.0225 0.0205 0.002 0.0225 
1993-05-11 21:00:00.000 0.0222 0.013 0.0092 0.0222 
1993-05-12 21:00:00.000 0.0345 0.0145 0.02 0.0345 
1993-05-13 21:00:00.000 0.0205 0.016 0.0045 0.0205 
1993-05-16 21:00:00.000 0.0216 0.0173 0.0043 0.0216 
1993-05-17 21:00:00.000 0.0208 0.009 0.0118 0.0208 
1993-05-18 21:00:00.000 0.019 0.0145 0.0045 0.019 
1993-05-19 21:00:00.000 0.0295 0.0255 0.004 0.0295 
1993-05-20 21:00:00.000 0.0264 0.0099 0.0165 0.0264 
1993-05-23 21:00:00.000 0.026 0  0.026 0.026 
1993-05-24 21:00:00.000 0.015 0.0135 0.0015 0.015 
1993-05-25 21:00:00.000 0.0146 0.0131 0.0015 0.0146 
1993-05-26 21:00:00.000 0.0235 0.018 0.0055 0.0235 
1993-05-27 21:00:00.000 0.0168 0.0093 0.0075 0.0168 
1993-05-30 21:00:00.000 0.006 0.001 0.005 0.006 
1993-05-31 21:00:00.000 0.0285 0.0085 0.02 0.0285 
1993-06-01 21:00:00.000 0.027 0  0.027 0.027 
1993-06-02 21:00:00.000 0.0185 0.0185 0  0.0185 
1993-06-03 21:00:00.000 0.0495 0.0155 0.034 0.0495 
1993-06-06 21:00:00.000 0.0235 0.0225 0.001 0.0235 
1993-06-07 21:00:00.000 0.0164 0.006 0.0104 0.0164 
1993-06-08 21:00:00.000 0.0195 0.005 0.0145 0.0195 
1993-06-09 21:00:00.000 0.031 0.0178 0.0132 0.031 
1993-06-10 21:00:00.000 0.0345 0.01 0.0245 0.0345 
1993-06-13 21:00:00.000 0.0105 0.01 0.0005 0.0105 
1993-06-14 21:00:00.000 0.024 0.011 0.013 0.024 
1993-06-15 21:00:00.000 0.0292 0.0112 0.018 0.0292 
1993-06-16 21:00:00.000 0.0192 0..0069 0.0192 
1993-06-17 21:00:00.000 0.027 0.0018 0.0252 0.027 
1993-06-20 21:00:00.000 0.018 0.0047 0.0133 0.018 
1993-06-21 21:00:00.000 0.016 0.0023 0.0137 0.016 
1993-06-22 21:00:00.000 0.0165 0.0058 0.0107 0.0165 
1993-06-23 21:00:00.000 0.0131 0.0061 0.007 0.0131 
1993-06-24 21:00:00.000 0.0225 0.0182 0.0043 0.0225 

Используя этот SQL-запрос: (спасибо за Lad2025)

WITH cte AS 
(
    SELECT 
    t.[Time] 
    ,[H-L] = Round([BID-HIGH]-[BID-LOW],5) 
    ,[H-Cp] = Abs(Round([BID-HIGH]-[prev_BID-CLOSE],5)) 
    ,[L-Cp] = Abs(Round([BID-LOW]-[prev_BID-CLOSE],5)) 
    FROM (SELECT *, 
       [prev_BID-CLOSE] = LAG([BID-CLOSE]) OVER(ORDER BY [Time]) 
     FROM [#tbl_GBP-USD_1-Day]) AS t 
) 
SELECT * 
FROM cte 
CROSS APPLY (SELECT MAX(v) AS v 
      FROM (VALUES ([H-L]),([H-Cp]),([L-Cp])) AS value(v) 
) AS sub([TR]); 

Пример запроса ATR ниже - 10-дневный ATR (средний истинный диапазон).

Как вы можете видеть, первые 9 строк пустые, так как эти данные необходимы для вычисления первого 10-дневного среднего значения [TR]. Затем каждая строка после первого вычисления ATR представляет собой среднее значение последнего [TR] 10 дней и так далее.

TIME     H-L  H-Cp L-Cp TR  ATR 
1993/05/09 21:00:00.000 0.0475 NULL NULL 0.0475 
1993/05/10 21:00:00.000 0.0225 0.0205 0.002 0.0225 
1993/05/11 21:00:00.000 0.0222 0.013 0.0092 0.0222 
1993/05/12 21:00:00.000 0.0345 0.0145 0.02 0.0345 
1993/05/13 21:00:00.000 0.0205 0.016 0.0045 0.0205 
1993/05/16 21:00:00.000 0.0216 0.0173 0.0043 0.0216 
1993/05/17 21:00:00.000 0.0208 0.009 0.0118 0.0208 
1993/05/18 21:00:00.000 0.019 0.0145 0.0045 0.019 
1993/05/19 21:00:00.000 0.0295 0.0255 0.004 0.0295 
1993/05/20 21:00:00.000 0.0264 0.0099 0.0165 0.0264 0.02645 
1993/05/23 21:00:00.000 0.026 0  0.026 0.026 0.0243 
1993/05/24 21:00:00.000 0.015 0.0135 0.0015 0.015 0.02355 
1993/05/25 21:00:00.000 0.0146 0.0131 0.0015 0.0146 0.02279 
1993/05/26 21:00:00.000 0.0235 0.018 0.0055 0.0235 0.02169 
1993/05/27 21:00:00.000 0.0168 0.0093 0.0075 0.0168 0.02132 
1993/05/30 21:00:00.000 0.006 0.001 0.005 0.006 0.01976 
1993/05/31 21:00:00.000 0.0285 0.0085 0.02 0.0285 0.02053 
1993/06/01 21:00:00.000 0.027 0  0.027 0.027 0.02133 
1993/06/02 21:00:00.000 0.0185 0.0185 0  0.0185 0.02023 
1993/06/03 21:00:00.000 0.0495 0.0155 0.034 0.0495 0.02254 
1993/06/06 21:00:00.000 0.0235 0.0225 0.001 0.0235 0.02229 
1993/06/07 21:00:00.000 0.0164 0.006 0.0104 0.0164 0.02243 
1993/06/08 21:00:00.000 0.0195 0.005 0.0145 0.0195 0.02292 
1993/06/09 21:00:00.000 0.031 0.0178 0.0132 0.031 0.02367 
1993/06/10 21:00:00.000 0.0345 0.01 0.0245 0.0345 0.02544 
1993/06/13 21:00:00.000 0.0105 0.01 0.0005 0.0105 0.02589 
1993/06/14 21:00:00.000 0.024 0.011 0.013 0.024 0.02544 
1993/06/15 21:00:00.000 0.0292 0.0112 0.018 0.0292 0.02566 
1993/06/16 21:00:00.000 0.0192 0..0069 0.0192 0.02573 
1993/06/17 21:00:00.000 0.027 0.0018 0.0252 0.027 0.02348 
1993/06/20 21:00:00.000 0.018 0.0047 0.0133 0.018 0.02293 
1993/06/21 21:00:00.000 0.016 0.0023 0.0137 0.016 0.02289 
1993/06/22 21:00:00.000 0.0165 0.0058 0.0107 0.0165 0.02259 
1993/06/23 21:00:00.000 0.0131 0.0061 0.007 0.0131 0.0208 
1993/06/24 21:00:00.000 0.0225 0.0182 0.0043 0.0225 0.0196 

Просто для дальнейшего расширения.

[H-L] = High (minus) Low, 
[H-Cp] = High (Minus) Previous Day Close, 
[L-Cp] = Low (Minus) Previous Day Close 
[TR] = MAX Value of [H-L],[H-Cp],[L-Cp] 
[ATR] = Average of TR over X amount of days 

X in this case is 10 days 

Мне требуется инструкция SQL для добавления к текущему коду TR запроса выше.

Благодарим вас за помощь!

Answer 1

Нечто подобное может сделать, используя тот же подход, как с TR:

WITH cteRange AS (
    SELECT 
     t.[Time], 
     Round(t.[BID-HIGH]-t.[BID-LOW],5) AS [H-L], 
     Abs(Round(t.[BID-HIGH]-t.[prev_BID-CLOSE],5)) AS [H-Cp], 
     Abs(Round(t.[BID-LOW]-t.[prev_BID-CLOSE],5)) AS [L-Cp] 
    FROM (
     SELECT *, [prev_BID-CLOSE] = LAG([BID-CLOSE]) OVER (ORDER BY [Time]) 
     FROM [#tbl_GBP-USD_1-Day] 
    ) AS t 
), cteTrueRange AS (
    SELECT * 
    FROM cteRange 
    CROSS APPLY (
     SELECT MAX(v) AS v 
     FROM (VALUES ([H-L]), ([H-Cp]), ([L-Cp])) AS value(v) 
    ) AS sub([TR]) 
), cteTrueRange10 AS (
    SELECT 
     *, 
     LAG([TR], 1) OVER (ORDER BY [Time]) AS [TR1], 
     LAG([TR], 2) OVER (ORDER BY [Time]) AS [TR2], 
     LAG([TR], 3) OVER (ORDER BY [Time]) AS [TR3], 
     LAG([TR], 4) OVER (ORDER BY [Time]) AS [TR4], 
     LAG([TR], 5) OVER (ORDER BY [Time]) AS [TR5], 
     LAG([TR], 6) OVER (ORDER BY [Time]) AS [TR6], 
     LAG([TR], 7) OVER (ORDER BY [Time]) AS [TR7], 
     LAG([TR], 8) OVER (ORDER BY [Time]) AS [TR8], 
     LAG([TR], 9) OVER (ORDER BY [Time]) AS [TR9] 
    FROM cteTrueRange 
) 
SELECT [Time], [H-L], [H-Cp], [L-Cp], [TR], [TRA] 
    FROM cteTrueRange10 
    CROSS APPLY (
     SELECT CASE WHEN [TR9] IS NOT NULL THEN AVG(v) END AS v 
     FROM (VALUES ([TR]), ([TR1]), ([TR2]), ([TR3]), ([TR4]), ([TR5]), ([TR6]), ([TR7]), ([TR8]), ([TR9])) AS value(v) 
    ) AS sub([TRA]); 

А вот другое решение объяснил в комментарии, где вы можете иметь его в среднем за любое количество дней (100 в этом случай):

WITH cteRange AS (
    SELECT 
     t.[Time], 
     ROUND(t.[BID-HIGH]-t.[BID-LOW],5) AS [H-L], 
     ABS(ROUND(t.[BID-HIGH]-t.[prev_BID-CLOSE],5)) AS [H-Cp], 
     ABS(ROUND(t.[BID-LOW]-t.[prev_BID-CLOSE],5)) AS [L-Cp] 
    FROM (
     SELECT *, [prev_BID-CLOSE] = LAG(u.[BID-CLOSE]) OVER (ORDER BY u.[Time]) 
     FROM [#tbl_GBP-USD_1-Day] u 
    ) AS t 
), cteTrueRange AS (
    SELECT *, ROW_NUMBER() OVER (ORDER BY [time]) AS ix 
    FROM cteRange 
    CROSS APPLY (
     SELECT MAX(v) AS v 
     FROM (VALUES ([H-L]), ([H-Cp]), ([L-Cp])) AS value(v) 
    ) AS sub([TR]) 
) 
SELECT t.[Time], t.[H-L], t.[H-Cp], t.[L-Cp], t.[TR], CASE WHEN LAG(t.[TR], 100) OVER (ORDER BY t.[time]) IS NOT NULL THEN (
     SELECT ROUND(AVG([TR]), 5) FROM cteTrueRange t2 WHERE (t2.ix > t.ix-100) AND (t2.ix <= t.ix) 
    ) END AS [TRA] 
    FROM cteTrueRange t; 

LAG([TR], 100) вызова здесь используется для блокирования средних от игнорирования NULL - нормальное поведение AVG() является просто игнорировать аннулирует, которые не удовлетворили бы ваше требование вернуть нуль для первого п-1 д AYS.