Irregular grouping of timestamp variable - sql

Question

I have a table organized as follows:
id lateAt
1231235 2019/09/14
1242123 2019/09/13
3465345 NULL
5676548 2019/09/28
8986475 2019/09/23
Where lateAt is a timestamp of when a certain loan's payment became late. So, for each current date - I need to look at these numbers daily - there's a certain amount of entries which are late for 0-15, 15-30, 30-45, 45-60, 60-90 and 90+ days.
This is my desired output:
lateGroup Count
0-15 20
15-30 22
30-45 25
45-60 32
60-90 47
90+ 57
This is something I can easily calculate in R, but to get the results back to my BI dashboard I'd have to create a new table in my database, which I don't think is a good practice. What is the SQL-native approach to this problem?

Answer 1

I would define the "late groups" using a range, the join against the number of days:
with groups (grp) as (
values
(int4range(0,15, '[)')),
(int4range(15,30, '[)')),
(int4range(30,45, '[)')),
(int4range(45,60, '[)')),
(int4range(60,90, '[)')),
(int4range(90,null, '[)'))
)
select grp, count(t.user_id)
from groups g
left join the_table t on g.grp #> current_date - t.late_at
group by grp
order by grp;
int4range(0,15, '[)') creates a range from 0 (inclusive) and 15 (exclusive)
Online example: https://rextester.com/QJSN89445

Answer 2

The quick and dirty way to do this in SQL is:
SELECT '0-15' AS lateGroup,
COUNT(*) AS lateGroupCount
FROM my_table t
WHERE (CURRENT_DATE - t.lateAt) >= 0
AND (CURRENT_DATE - t.lateAt) < 15
UNION
SELECT '15-30' AS lateGroup,
COUNT(*) AS lateGroupCount
FROM my_table t
WHERE (CURRENT_DATE - t.lateAt) >= 15
AND (CURRENT_DATE - t.lateAt) < 30
UNION
SELECT '30-45' AS lateGroup,
COUNT(*) AS lateGroupCount
FROM my_table t
WHERE (CURRENT_DATE - t.lateAt) >= 30
AND (CURRENT_DATE - t.lateAt) < 45
-- Etc...
For production code, you would want to do something more like Ross' answer.

Answer 3

You didn't mention which DBMS you're using, but nearly all of them will have a construct known as a "value constructor" like this:
select bins.lateGroup, bins.minVal, bins.maxVal FROM
(VALUES
('0-15',0,15),
('15-30',15.0001,30), -- increase by a small fraction so bins don't overlap
('30-45',30.0001,45),
('45-60',45.0001,60),
('60-90',60.0001,90),
('90-99999',90.0001,99999)
) AS bins(lateGroup,minVal,maxVal)
If your DBMS doesn't have it, then you can probably use UNION ALL:
SELECT '0-15' as lateGroup, 0 as minVal, 15 as maxVal
union all SELECT '15-30',15,30
union all SELECT '30-45',30,45
Then your complete query, with the sample data you provided, would look like this:
--- example from SQL Server 2012 SP1
--- first let's set up some sample data
create table #temp (id int, lateAt datetime);
INSERT #temp (id, lateAt) values
(1231235,'2019-09-14'),
(1242123,'2019-09-13'),
(3465345,NULL),
(5676548,'2019-09-28'),
(8986475,'2019-09-23');
--- here's the actual query
select lateGroup, count(*) as Count
from #temp as T,
(VALUES
('0-15',0,15),
('15-30',15.0001,30), -- increase by a small fraction so bins don't overlap
('30-45',30.0001,45),
('45-60',45.0001,60),
('60-90',60.0001,90),
('90-99999',90.0001,99999)
) AS bins(lateGroup,minVal,maxVal)
) AS bins(lateGroup,minVal,maxVal)
where datediff(day,lateAt,getdate()) between minVal and maxVal
group by lateGroup
order by lateGroup
--- remove our sample data
drop table #temp;
Here's the output:
lateGroup Count
15-30 2
30-45 2
Note: rows with null lateAt are not counted.

Answer 4

I think you can do it all in one clear query :
with cte_lategroup as
(
select *
from (values(0,15,'0-15'),(15,30,'15-30'),(30,45,'30-45')) as t (mini, maxi, designation)
)
select
t2.designation
, count(*)
from test t
left outer join cte_lategroup t2
on current_date - t.lateat >= t2.mini
and current_date - lateat < t2.maxi
group by t2.designation;
With a preset like yours :
create table test
(
id int
, lateAt date
);
insert into test
values (1231235, to_date('2019/09/14', 'yyyy/mm/dd'))
,(1242123, to_date('2019/09/13', 'yyyy/mm/dd'))
,(3465345, null)
,(5676548, to_date('2019/09/28', 'yyyy/mm/dd'))
,(8986475, to_date('2019/09/23', 'yyyy/mm/dd'));