views:

389

answers:

2

Given a set of random numeric values in a database, how do I generate a limited list of ranges where each range contains at least one value? The ranges should not overlap and ideally have a similar amount of values in them. Ideally their boundaries should also be multiples of 10, 100, 1000 etc...

For example:

Values: 100,150,180,300,400,500,600,650,700
results in
4 ranges: 100-180(2), 180-300(1), 300-600(3), 600-800(3)

How could this be done in C# or T-SQL?

Thanks

+1  A: 

In MS SQL 2005+:

SELECT range, (MIN(getprev) + MIN(value)) / 2 as range_start, (MAX(getnext) + MAX(value)) / 2 AS range_end, COUNT(*) as range_values
FROM (
  SELECT value,
         NTILE(4) OVER (ORDER BY value ) AS range,
         (
         SELECT TOP 1 value
         FROM values li
         WHERE li.value < lo.value
         ORDER BY
           li.value DESC
         ) AS getprev,
         (
         SELECT TOP 1 value
         FROM values li
         WHERE li.value > lo.value
         ORDER BY
           li.value
         ) AS getnext
  FROM values lo
) vo
GROUP BY range
ORDER BY range

In your case:

1   100   240   3
2   240   450   2
3   450   625   2
4   625   700   2
Quassnoi
Can you do this in Linq? :)
AndyD
You can do this in a stored procedure and call it from LINQ :)
Quassnoi
A: 

Assume we want ranges with one value in them each: Values: 100 150 180 300 Just produce these random numbers:

- a: 0 <= 100, eg: 50
- b: 100 <= 150, eg: 125
- c: 150 <= 180, eg: 165
- d: 180 <= 300, eg: 200
- e: 300 <=    , eg: 350

...

Ranges: 50-125 (1), 125-180(1), 180-200(1), 200 - 350 (1) ...: each one contains exactly one number.

Now to have more then one number in each range, just start skippin numbers, for example skip c (can pick a random chance to skip one): skip c:

50-125(1), 125-200(2), 200-350(1)
Carra
I'm sure if I get your answer... could you write some pseudo code for your algorithm?
AndyD