Find the n highest consecutive values in a set of rows - sql-server

Question

I have some data in a table as follows:
FileDate SumAmount
20150401 90.99
20150401 313
20150403 481.2
20150404 321.27
20150405 103
20150406 25
20150407 180.5
20150408 319.91
20150409 688
20150411 69
20150412 65
20150413 322
20150414 100
20150415 111.97
20150416 979.15
20150417 655.4
20150418 124
20150419 30
20150420 457
20150421 192.6
20150422 191.96
20150423 220
20150424 252.5
20150425 109.1
20150426 135.25
20150427 648.08
20150428 692
20150429 410.99
20150430 170
20150501 166.19
20150502 92
20150503 100
20150504 59
20150505 124.01
20150506 44.5
20150507 331.64
20150508 299.8
I am trying to devise a query that will find the highest 4 consecutive days values in the data.
Essentially, I think I need to partition by date and perform a row numbering over it but I can't seem to get the syntax right to evaluate the values.

Answer 1

So I use -3 in the join conditions since the day itself counts as one. Let me know what you think. Also I use day of year(DY) to ensure that it's only consecutive days and so I don't have to rank the dates manually. Hope this helps!
DECLARE #yourTable TABLE(FileDate DATE ,SumAmount FLOAT);
INSERT INTO #yourTable
VALUES ('20150401',90.99),
('20150402',313),
('20150403',481.2),
('20150404',321.27),
('20150405',103),
('20150406',25),
('20150407',180.5),
('20150408',319.91),
('20150409',688),
('20150411',69),
('20150412',65),
('20150413',322),
('20150414',100),
('20150415',111.97),
('20150416',979.15),
('20150417',655.4),
('20150418',124),
('20150419',30),
('20150420',457),
('20150421',192.6),
('20150422',191.96),
('20150423',220),
('20150424',252.5),
('20150425',109.1),
('20150426',135.25),
('20150427',648.08),
('20150428',692),
('20150429',410.99),
('20150430',170),
('20150501',166.19),
('20150502',92),
('20150503',100),
('20150504',59),
('20150505',124.01),
('20150506',44.5),
('20150507',331.64),
('20150508',299.8);
WITH CTE
AS
(
SELECT YEAR(FileDate) yr,DATEPART(DY,FileDate) dy,fileDate,SumAmount
FROM #yourTable
),
CTE_Max_Sum
AS
(
SELECT TOP 1 A.yr,A.dy,A.FileDate,SUM(B.SumAmount) consec4DaySum
FROM CTE A
INNER JOIN CTE B
ON B.dy BETWEEN A.dy - 3 AND A.dy
AND A.yr = B.yr
GROUP BY A.yr,A.dy,A.FileDate
ORDER BY SUM(B.SumAmount) DESC
)
SELECT A.*,B.consec4DaySum
FROM CTE A
INNER JOIN CTE_Max_Sum B
ON A.dy BETWEEN B.dy - 3 AND B.dy
AND A.yr = B.yr
Results:
yr dy fileDate SumAmount consec4DaySum
----------- ----------- ---------- ---------------------- ----------------------
2015 117 2015-04-27 648.08 1921.07
2015 118 2015-04-28 692 1921.07
2015 119 2015-04-29 410.99 1921.07
2015 120 2015-04-30 170 1921.07

Answer 2

You can use a CTE for that, joining every row with its three following rows (day-wise) and summing up. This Fiddle sadly does not work for me, it runs on my sql server and work for you. Watch out for recursion depth, without WHERE cte.Consecutive < 4 you quickly run into an error.
WITH cte (StartDate, EndDate, Consecutive, SumAmount)
AS (
SELECT t.FileDate, t.FileDate, 1, t.SumAmount FROM dbo.table30194903 t
UNION ALL
SELECT cte.StartDate, t.FileDate, cte.Consecutive + 1, cte.SumAmount + t.SumAmount
FROM dbo.table30194903 t INNER JOIN cte ON DATEADD(DAY, 1, cte.EndDate) = t.FileDate
WHERE cte.Consecutive < 5
)
SELECT *
FROM cte
WHERE cte.Consecutive = 4
ORDER BY cte.SumAmount DESC
EDIT: Had two errors in my query, it summed up wrong rows and showd the last day in the series.

Answer 3

I would like to add an answer using a subquery, however it does take more time compared to my cte...
SELECT t.FileDate, SUM(s.SumAmount)
FROM dbo.table30194903 t
LEFT JOIN dbo.table30194903 s ON t.FileDate <= s.FileDate AND DATEDIFF(DAY, t.FileDate, s.FileDate) < 4
GROUP BY t.FileDate
HAVING COUNT(s.SumAmount) = 4
ORDER BY SUM(s.SumAmount) DESC

Answer 4

I think the simplest way to get this is to use an APPLY to get the number of records in the n days following each row, and then limit this to where there are n dates, this ensures you have consecutive days. You can then just order by the sum and select the top 1:
DECLARE #n INT = 4;
SELECT TOP 1
FirstDate = t.FileDate,
FourDaySum = t2.Amount
FROM dbo.T
CROSS APPLY
( SELECT Amount = SUM(t2.SumAmount),
Dates = COUNT(DISTINCT t2.FileDate)
FROM dbo.T AS t2
WHERE t2.FileDate >= t.FileDate
AND t2.FileDate < DATEADD(DAY, #n, t.FileDate)
) AS t2
WHERE t2.Dates = #n
ORDER BY t2.Amount DESC;
Example on SQL Fiddle

Answer 5

How about a simply while block and sum the values of a range of dates?
DECLARE #startingDate DATETIME, #searchDate DATETIME;
DECLARE #maxSoFar INT, #sum INT, #daysRange INT;
SET #startingDate = convert(datetime, '20150401', 110)
SET #searchDate = #startingDate;
SET #daysRange = 3;
SET #maxSoFar = 0;
WHILE GETDATE()> #searchDate
BEGIN
--PRINT #searchDate
--PRINT DATEADD(DAY,#daysRange,#searchDate)
SELECT #sum = SUM(SumAmount) FROM MyTable WHERE FileDate >= #searchDate AND FileDate <= DATEADD(DAY,#daysRange,#searchDate)
IF #sum > #maxSoFar
BEGIN
SET #maxSoFar = #sum;
END
SET #searchDate = DATEADD(DAY,1,#searchDate)
END