如何遍历 SQL 表和 select 数据中的峰值？

Question

Lets say I have the following table:

Id|name|spike|timestamp
1 |John|15   |111
2 |Jim |12   |112
3 |Jeff|13   |113
4 |Joe |4    |114
5 |Jess|0    |115
6 |Jill|0    |116
7 |Jey |13   |117
8 |Joy |15   |118
9 |Jess|14   |119
10|Joe |0    |120

I need to iterate through the table and select data where spike > 10 and separate rows into different sets of data. An acceptable query for the top table should result in:

Id|name|spike|timestamp
1 |John|15   |111
2 |Jim |12   |112
3 |Jeff|13   |113

and

Id|name|spike|timestamp
7 |Jey |13   |117
8 |Joy |15   |118
9 |Jess|14   |119

I need to process all the spikes in the table.

Edit: I don't know the number of islands in the table or how far apart they are.

Answer 1

If you do not have an idea for number of islands then use this my dynamic approach solution. here is the demo .

with cte as
(
    select
        *,
        id - row_number() over (order by id) as rnk
    from myTable
    where spike > 10
),
islands as
(
  select
      *,
      (select count(distinct rnk)::int from cte) as total_islands
  from cte
),
buckets as
(
  select
    Id,
    name,
    spike,
    timestamp,
    NTILE(total_islands) Over (Order by id) as island
  from islands
)  

select *
from buckets
where island = 2

for your case you can directly use window function NTILE which will bucket your data in two parts. Here is the demo .

first create a cte

with cte as
(
  select
    Id,
    name,
    spike,
    timestamp,
    NTILE(2) Over (Order by id) as nums
  from myTable
  where spike > 10
)  

Then run first query to get first part

select  
    Id,
    name,
    spike,
    timestamp
from cte
where nums = 1;

Output:

| id  | name | spike | timestamp |
| --- | ---- | ----- | --------- |
| 1   | John | 15    | 111       |
| 2   | Jim  | 12    | 112       |
| 3   | Jeff | 13    | 113       |

Now run the second query to get second part

select  
    Id,
    name,
    spike,
    timestamp
from cte
where nums = 2;

Output:

| id  | name | spike | timestamp |
| --- | ---- | ----- | --------- |
| 7   | Jey  | 13    | 117       |
| 8   | Joy  | 15    | 118       |
| 9   | Jess | 14    | 119       |

Answer 2

You can split your data into islands using a CTE to calculate rows according to timestamp and also partitioned by spike > 10 and then taking DENSE_RANK() over their difference to compute the island number. You can then select from that based on the island number:

WITH CTE AS (
  SELECT *,
         ROW_NUMBER() OVER (ORDER BY timestamp) AS rn,
         ROW_NUMBER() OVER (PARTITION BY spike > 10 ORDER BY timestamp) AS sn
  FROM data
),
islands AS (
  SELECT id, name, spike, timestamp,
         DENSE_RANK() OVER (ORDER BY rn - sn) AS island
  FROM CTE
  WHERE spike > 10
)
SELECT *
FROM islands 
WHERE island = 2

Output:

id  name    spike   timestamp   island
7   Jey     13      117         2
8   Joy     15      118         2
9   Jess    14      119         2

Demo on dbfiddle

Note that if you can have duplicate timestamp values but they do increase with id , you should change the ORDER BY timestamp clauses to ORDER BY id .

Answer 3

There are many ways to deal with gaps-and-islands problems. In this case, a cumulative sum uniquely identifies each group.

select t.*
from (select t.*,
             count(*) filter (where spike <= 10) over (order by timestamp) as island
      from t
     ) t
where spike > 10;

This counts the number of spike values less than or equal to 10 . This is constant for each group of consecutive "spikey" numbers.

If you want the islands enumerated, just use dense_rank() :

select t.*, dense_rank() over (order by island) as grouping_number
from (select t.*,
             count(*) filter (where spike <= 10) over (order by timestamp) as island
      from t
     ) t
where spike > 10;

Here is a db<>fiddle.