SQL - Identify consecutive numbers in a table - sql

Question

Is there a way to flag consecutive numbers in an SQL table?
Based on the values in 'value_group_4' column, is it possible to tag continous values? This needs to be done within groups of each 'date_group_1'
I tried using row_numbers, rank, dense_rank but unable to come up with a foolproof way.

Answer 1

This has nothing to do with consecutiveness. You simply want to mark all rows where date_group_1 and value_group_4 are not unique.
One way:
select
mytable.*,
case when exists
(
select null
from mytable agg
where agg.date_group_1 = mytable.date_group_1
and agg.value_group_4 = mytable.value_group_4
group by agg.date_group_1, agg.value_group_4
having count(*) > 1
) then 1 else 0 end as flag
from mytable
order by date_group_1, value_group_4;
In a later version of SQL Server you'd use COUNT OVER instead.

Answer 2

SQL tables represent unordered sets. There is no such thing as consecutive values, unless a column specifies the ordering. Your data does not have such an obvious column, but I'll assume one exists and just call it id for convenience.
With such a column, lag()/lead() does what you want:
select t.*,
(case when lag(value_group_4) over (partition by data_group1 order by id) = value_group_4
then 1
when lead(value_group_4) over (partition by data_group1 order by id) = value_group_4
then 1
else 0
end) as flag
from t;
On close inspection, value_group_3 may do what you want. So you can use that for the id.

Answer 3

If your version of SQL Server doesn't have a full suite of windowing functions it should be still possible. This problem looks like a last-non-null problem which Itzik Ben-Gan has good example here... http://www.itprotoday.com/software-development/last-non-null-puzzle
Also, look at Mikael Eriksson's answer here which uses no windowing functions.

Answer 4

If the order of your data is determined by the date_group_1, value_group_3 column values, then why not make it as simple as the following query:
select
*,
rank() over(partition by date_group_1 order by value_group_3) - 1 value_group_3,
case
when count(*) over(partition by date_group_1, value_group_3) > 1 then 1
else 0
end expected_result
from data;
Output:
| date_group_1 | category_group_2 | value_group_3 | value_group_3 | expected_result |
+--------------+------------------+---------------+---------------+-----------------+
| 2018-01-11 | A | 15.3 | 0 | 0 |
| 2018-01-11 | B | 17.3 | 1 | 1 |
| 2018-01-11 | A | 17.3 | 1 | 1 |
| 2018-01-11 | B | 21 | 3 | 0 |
| 2018-01-22 | A | 15.3 | 0 | 0 |
| 2018-01-22 | B | 17.3 | 1 | 0 |
| 2018-01-22 | A | 21 | 2 | 0 |
| 2018-01-22 | B | 23 | 3 | 0 |
| 2018-03-13 | A | 15.3 | 0 | 0 |
| 2018-03-13 | B | 17.3 | 1 | 1 |
| 2018-03-13 | A | 17.3 | 1 | 1 |
| 2018-03-13 | B | 23 | 3 | 0 |
| 2018-05-15 | A | 6 | 0 | 0 |
| 2018-05-15 | B | 6.3 | 1 | 0 |
| 2018-05-15 | A | 15 | 2 | 0 |
| 2018-05-15 | B | 16.3 | 3 | 1 |
| 2018-05-15 | A | 16.3 | 3 | 1 |
| 2018-05-15 | B | 22 | 5 | 0 |
| 2019-05-04 | A | 0 | 0 | 0 |
| 2019-05-04 | B | 7 | 1 | 0 |
| 2019-05-04 | A | 15.3 | 2 | 0 |
| 2019-05-04 | B | 17.3 | 3 | 0 |
Test it online with SQL Fiddle.