Thread: Group by a range of values

On 8/1/20 6:34 AM, Torsten Grust wrote:
> Hi,
> 
> maybe this does the job already (/ is integer division):
> 
> SELECT i, 1 + (i-1) / 3
> FROM   generate_series(1,10) AS i;
> 
> An expression like (i-1) / 3 could, of course, also be used as 
> partitioning criterion in GROUP BY and/or window functions.
> 
> Cheers,
>    —T
> 
> On Sat, Aug 1, 2020 at 2:15 PM Mike Martin <redtux1@gmail.com 
> <mailto:redtux1@gmail.com>> wrote:
> 
>     Say I have a field of ints, as for example created by with
>     ordinality or generate_series, is it possible to group by a range? eq
> 
>     1,2,3,4,5,6,7,8,9,10
>     step 3
>     so output is
> 
>     1 1
>     2 1
>     3 1
>     4 2
>     5 2
>     6 2
>     7 3
>     8 3
>     9 3
>     10 4
> 
>     thanks
> 
> 
> 
> -- 
> | Torsten Grust
> | Torsten.Grust@gmail.com <mailto:Torsten.Grust@gmail.com>
> 
My version is as follows, the point being that the "grouping" requested 
is simply an ordering of the step mechanism.  Naturally this series is 
generated in order shown but real data for val likely won't be.

test=# with ts as (select generate_series(1,10) as val) select s.val, 
(s.val /3)+1 as ord from ts as s order by ord;
  val | ord
-----+-----
    1 |   1
    2 |   1
    3 |   2
    4 |   2
    5 |   2
    6 |   3
    7 |   3
    8 |   3
    9 |   4
   10 |   4
(10 rows)