Best way to interpolate values in SQL

问题:

I have a table with rate at certain date :

          Rates

Id  |     Date      |  Rate
----+---------------+-------
 1  |   01/01/2011  |  4.5
 2  |   01/04/2011  |  3.2
 3  |   04/06/2011  |  2.4
 4  |   30/06/2011  |  5

I want to get the output rate base on a simple linear interpolation.

So if I enter 17/06/2011:

Date        Rate
----------  -----
01/01/2011  4.5
01/04/2011  3.2
04/06/2011  2.4
17/06/2011  
30/06/2011  5.0

the linear interpolation is (5 + 2,4) / 2 = 3,7

Is there a way to do a simple query (SQL Server 2005), or this kind of stuff need to be done in a programmatic way (C#...) ?

回答1:

Something like this (corrected):

SELECT CASE WHEN next.Date IS NULL  THEN prev.Rate
            WHEN prev.Date IS NULL  THEN next.Rate
            WHEN next.Date = prev.Date  THEN prev.Rate
              ELSE ( DATEDIFF(d, prev.Date, @InputDate) * next.Rate 
                   + DATEDIFF(d, @InputDate, next.Date) * prev.Rate
                   ) / DATEDIFF(d, prev.Date, next.Date)
       END AS interpolationRate 
FROM
  ( SELECT TOP 1 
        Date, Rate 
    FROM Rates
    WHERE Date <= @InputDate
    ORDER BY Date DESC
  ) AS prev
  CROSS JOIN
  ( SELECT TOP 1 
        Date, Rate 
    FROM Rates
    WHERE Date >= @InputDate
    ORDER BY Date ASC
  ) AS next

回答2:

The trick with CROSS JOIN here is it wont return any records if either of the table does not have rows (1 * 0 = 0) and the query may break. Better way to do is use FULL OUTER JOIN with inequality condition (to avoid getting more than one row)

( SELECT TOP 1 
        Date, Rate 
    FROM Rates
    WHERE Date <= @InputDate
    ORDER BY Date DESC
  ) AS prev
  FULL OUTER JOIN
  ( SELECT TOP 1 
        Date, Rate 
    FROM Rates
    WHERE Date >= @InputDate
    ORDER BY Date ASC
  ) AS next
ON (prev.Date <> next.Date) [or Rate depending on what is unique]

回答3:

As @Mark already pointed out, the CROSS JOIN has its limitations. As soon as the target value falls outside the range of defined values no records will be returned.

Also the above solution is limited to one result only. For my project I needed an interpolation for a whole list of x values and came up with the following solution. Maybe it is of interested to other readers too?

-- generate some grid data values in table #ddd:
CREATE TABLE #ddd (id int,x float,y float, PRIMARY KEY(id,x));
INSERT INTO  #ddd VALUES (1,3,4),(1,4,5),(1,6,3),(1,10,2),
                         (2,1,4),(2,5,6),(2,6,5),(2,8,2);
SELECT * FROM #ddd;                         

-- target x-values in table #vals (results are to go into column yy):
CREATE TABLE #vals (xx float PRIMARY KEY,yy float null, itype int);
INSERT INTO  #vals (xx) VALUES (1),(3),(4.3),(9),(12);

-- do the actual interpolation
WITH valstyp AS (
  SELECT id ii,xx,
         CASE WHEN min(x)<xx THEN CASE WHEN max(x)>xx THEN 1 ELSE 2 END ELSE 0 END flag,
         min(x) xmi,max(x) xma 
  FROM #vals INNER JOIN #ddd ON id=1 GROUP BY xx,id
), ipol AS (
  SELECT v.*,(b.x-xx)/(b.x-a.x) f,a.y ya,b.y yb 
  FROM valstyp v 
  INNER JOIN #ddd a ON a.id=ii AND a.x=(SELECT max(x) FROM #ddd WHERE id=ii 
             AND (flag=0 AND x=xmi OR flag=1 AND x<xx OR flag=2 AND x<xma))
  INNER JOIN #ddd b ON b.id=ii AND b.x=(SELECT min(x) FROM #ddd WHERE id=ii 
             AND (flag=0 AND x>xmi OR flag=1 AND x>xx OR flag=2 AND x=xma))
)
UPDATE v SET yy=ROUND(f*ya+(1-f)*yb,8),itype=flag FROM #vals v INNER JOIN ipol i ON i.xx=v.xx;

-- list the interpolated results table:
SELECT * FROM #vals

When running the above script you will get the following data grid points in table #ddd

id x  y 
-- -- - 
1  3  4 
1  4  5 
1  6  3 
1  10 2 
2  1  4 
2  5  6 
2  6  5 
2  8  2 

[[ The table contains grid points for two identities (id=1 and id=2). In my example I referenced only the 1-group by using where id=1 in the valstyp CTE. This can be changed to suit your requirements. ]]

and the results table #vals with the interpolated data in column yy:

xx  yy   itype 
--- ---- ----- 
1   2    0     
3   4    0     
4.3 4.7  1     
9   2.25 1     
12  1.5  2     

The last column itype indicates the type of interpolation/extapolation that was used to calculate the value:

0:  extrapolation to lower end
1:  interpolation within given data range
2:  extrapolation to higher end

This working example can be found here.