SQL-识别表格中的连续数字

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

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 3

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 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 .