Postgres: select query with group by clause on a range of dates - sql

Question

My table contains answers from repeatable questionnaires that can be filled in a range of 30 days and are scheduled every 60 days.
Therefore, the answers from a single instance of a questionnaire are spread in a range of date that is always smaller tha 30 days and the first answer to the following repeatable questionnaire is at least 31 days after the last answer of the previous one.
How do I create a view that calculate a score (which is basically the sum of the answers of a single questionnaire) among the values whose dates are within 30 days from the start date (min date)?
Table raw_data
------------------------------------------------
user_name | question_id | answer | answer_date |
------------------------------------------------
user001 | 1 | 2 | 2019-02-04 |
user001 | 2 | 1 | 2019-02-04 |
user001 | 3 | 2 | 2019-02-05 |
user001 | 4 | 2 | 2019-02-05 |
user001 | 5 | 2 | 2019-02-09 |
user002 | 1 | 2 | 2019-01-09 |
user002 | 2 | 2 | 2019-01-10 |
user002 | 3 | 1 | 2019-02-01 |
user002 | 4 | 2 | 2019-02-01 |
user002 | 5 | 1 | 2019-02-01 |
user002 | 1 | 2 | 2019-03-11 |
user002 | 2 | 2 | 2019-03-11 |
user002 | 3 | 1 | 2019-03-12 |
user002 | 4 | 1 | 2019-03-13 |
user002 | 5 | 1 | 2019-03-14 |
Expected result
------------------------------
user_name | sum | start_date |
------------------------------
user001 | 9 | 2019-02-04 |
user002 | 8 | 2019-01-09 |
user002 | 7 | 2019-03-11 |
The solution I tried works for the first group only:
SELECT user_name, SUM(answer::int),
CASE
WHEN answer_date - MIN(answer_date) OVER (PARTITION BY user_name ORDER BY user_name ASC, answer_date ASC) < 30
THEN MIN(answer_date) OVER (PARTITION BY user_name ORDER BY user_name ASC, answer_date ASC)
ELSE answer_date END AS start_date,
FROM public.raw_data
GROUP BY user_name, answer_date

Answer 1

It's a classical gaps-and-islands problem. You'll find a lot under the tag I added.
An optimized query for your case could look like:
SELECT user_name
, sum(answer)
, min(answer_date) AS start_date
FROM (
SELECT user_name, answer, answer_date
, count(*) FILTER (WHERE step) OVER (PARTITION BY user_name ORDER BY answer_date) AS grp
FROM (
SELECT user_name, answer, answer_date
, lag(answer_date) OVER (PARTITION BY user_name ORDER BY answer_date) < answer_date - 30 AS step
FROM raw_data
) sub1
) sub2
GROUP BY user_name, grp
ORDER BY user_name, start_date; -- ORDER BY optional
db<>fiddle here
Closely related, with more explanation:
How to group timestamps into islands (based on arbitrary gap)?

Answer 2

Use lag() to find the gaps. Then a cumulative sum to assign a "question period" and then summarize:
select userid, min(answer_date) as start_date, sum(answer)
from (select rd.*,
count(*) filter (where prev_ad is null or prev_ad < answer_date - interval '30 day') over (partition by user_id) as period
from (select rd.*,
lag(answer_date) over (partition by user_id order by answer_date) as prev_ad
from raw_data rd
) rd
)
group by userid, period;

Answer 3

Thanks to #Gordon and to this
answer
I eventually found the missing step to determine my groups on a date range basis.
I will use the following query to create a view and SUM answers grouping by grp2
WITH query AS (
SELECT r.*,
SUM(CASE WHEN answer_date < prev_date + 30 THEN 0 ELSE 1 END) OVER (PARTITION BY user_name ORDER BY user_name ASC, answer_date ASC) AS grp
FROM (SELECT r.*,
LAG(answer_date) OVER (PARTITION BY user_name ORDER BY user_name ASC, answer_date ASC) AS prev_date
FROM raw_data r
) r
)
SELECT user_name, question_id, answer_date, answer, DENSE_RANK() OVER (ORDER BY user_name, grp) AS grp2
FROM query

Answer 4

You can use the query with row_number() window analytic function as below
with raw_data( user_name, question_id, answer, answer_date ) as
(
select 'user001',1,2, '2019-02-04' union all
select 'user001',2,1, '2019-02-04' union all
select 'user001',3,2, '2019-02-05' union all
select 'user001',4,2, '2019-02-05' union all
select 'user001',5,2, '2019-02-09' union all
select 'user002',1,2, '2019-01-09' union all
select 'user002',2,2, '2019-01-10' union all
select 'user002',3,1, '2019-02-01' union all
select 'user002',4,2, '2019-02-01' union all
select 'user002',5,1, '2019-02-01' union all
select 'user002',1,2, '2019-03-11' union all
select 'user002',2,2, '2019-03-11' union all
select 'user002',3,1, '2019-03-12' union all
select 'user002',4,1, '2019-03-13' union all
select 'user002',5,1, '2019-03-14'
)
select user_name, sum(answer) as sum, min(answer_date) as start_date
from
(
select row_number() over (partition by question_id order by user_name, answer_date) as rn,
t.*
from raw_data t
) t
group by user_name, rn
order by rn;
user_name sum start_date
--------- --- ----------
user001 9 2019-02-04
user002 8 2019-01-09
user002 7 2019-03-11
Demo