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Home/ Questions/Q 8238331
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Editorial Team
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Editorial Team
Asked: June 7, 20262026-06-07T19:46:39+00:00 2026-06-07T19:46:39+00:00

Given the following table grp | ind | val ———————- a | 1 |

  • 0

Given the following table

  grp |   ind |   val
----------------------
    a |     1 |     1
    a |     2 |     1
    a |     3 |     1
    a |     4 |     2
    a |     5 |     2
    a |     6 |     4
    a |     7 |     2
    b |     1 |     1
    b |     2 |     1
    b |     3 |     1
    b |     4 |     3
    b |     5 |     3
    b |     6 |     4

I need to select the following:

  grp |   ind |   val
----------------------
    a |     1 |     1
    a |     4 |     2
    a |     6 |     4
    a |     7 |     2
    b |     1 |     1
    b |     4 |     3
    b |     6 |     4

That is for each ‘grp’, each record where the ‘val’ is different to the proceeding ‘val’ (ordered by ‘index’) So each record where the ‘value’ “steps”.

what would be the most efficient way to achieve this?

thanks.

Here is a script to create the test case:

create temp table test_table
(
    grp character varying,
    ind numeric,
    val numeric
);
insert into test_table values
    ('a', 1 , 1),
    ('a', 2 , 1),
    ('a', 3 , 1),
    ('a', 4 , 2),
    ('a', 5 , 2),
    ('a', 6 , 4),
    ('a', 7 , 2),
    ('b', 1 , 1),
    ('b', 2 , 1),
    ('b', 3 , 1),
    ('b', 4 , 3),
    ('b', 5 , 3),
    ('b', 6 , 4);
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-07T19:46:41+00:00Added an answer on June 7, 2026 at 7:46 pm
    select grp,
           ind,
           val
    from (
       select grp, 
              ind, 
              val,
              lag(val,1,0::numeric) over (partition by grp order by ind) - val as diff
       from test_table
    ) t
    where diff <> 0;
    
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