CODEWITHSUNDEEP
postgresqllagcumulative-sum
RohanDsouza
RohanDsouza

Reputation: 412

Lag calculation over in postgresql

I have a table with following data: computed column : current

| id | Date (dd/mm/yyyy) | Factor | Actual | Current  |
|----|-------------------|--------|--------|----------|
| 1  |     04/01/2017    | 0.5    | 100    | 100      |
| 2  |     04/02/2017    | 0.5    | 120    | 100      |
| 3  |     04/03/2017    | 0.5    | 120    | 110      |
| 4  |     04/04/2017    | 0.5    | 115    | 115      |
| 5  |     04/05/2017    | 0.5    | 125    | 115      |
| 6  |     04/06/2017    | 0.5    | 100    | 120      |
| 7  |     04/07/2017    | 0.5    | 100    | 110      |

Current row = current of previous row + factor * (actual of previous row - current of previous row)

    For id = 1, current = same as actual = 100
    For id = 2, current = 100 + 0.5 * (100 - 100) = 100
    For id = 3, current = 100 + 0.5 * (120 - 100) = 110
    For id = 4, current = 110 + 0.5 * (120 - 110) = 115 
and so on...

How to achieve in postgresql using query?

Upvotes: 1

Views: 933

Answers (1)

klin
klin

Reputation: 121774

You need a recursive query.

with recursive my_table_with_rn as 
(
    select *, row_number() over (order by id) as rn
    from my_table
),

rec_query(rn, id, date, factor, actual, current) as 
(
    select rn, id, date, factor, actual, actual
    from my_table_with_rn
    where rn = 1

union all

    select 
        t.rn, t.id, t.date, t.factor, t.actual, 
        p.current + t.factor * (p.actual - p.current)
    from rec_query p
    join my_table_with_rn t on t.rn = p.rn + 1
)

select id, date, factor, actual, current
from rec_query
order by id;

Note, that row_number() was added to work in cases when ids are not consecutive (it's not necessary for actual data, you can use id instead of rn).

Test it here.

Upvotes: 1

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