Count daily fidelity - sql

Question

I have the below table and I would like to count, day by day, the number of distinct people who logged in everyday. For example, for day 1, everyone logged in, so it's 4. For day 4, there's just one person ID who logged in everyday since day 1, so the count would be 1.
DAY
PERSON_ID
1
01
1
02
1
03
1
04
2
01
2
02
2
03
3
01
4
02
4
01
Expected output.
DAY
PEOPLE_LOGGED_EVERYDAY
PEOPLE
1
4
01, 02, 03, 04
2
3
01, 02, 03
3
1
01
4
1
01
EDIT: the query should also work on the below data.
with t ( DAY, PERSON_ID ) AS(
SELECT 10, '01' FROM DUAL UNION ALL
SELECT 10, '02' FROM DUAL UNION ALL
SELECT 10, '04' FROM DUAL UNION ALL
SELECT 10, '04' FROM DUAL UNION ALL
SELECT 12, '01' FROM DUAL UNION ALL
SELECT 12, '02' FROM DUAL UNION ALL
SELECT 12, '03' FROM DUAL UNION ALL
SELECT 13, '04' FROM DUAL UNION ALL
SELECT 13, '01' FROM DUAL UNION ALL
SELECT 14, '02' FROM DUAL UNION ALL
SELECT 14, '01' FROM DUAL)
Expected output:
DAY
PEOPLE_LOGGED_EVERYDAY
PEOPLE
EXPLANATION
10
3
01, 02, 04
Three unique people in day 10
12
2
01, 02
Day 11 does not have values, so it's not included. From those in day 10, only 2 appear in day 12
13
1
01
From those in day 10 and 12, only 01 appears in day 13
14
1
01
From those in day 10, 12 and 13, only 01 appears in day 14

Answer 1

You can use listagg() with group by clause. If day is always start from the 1 and increases by 1 then you can use below query. He with the help of exits I have selected only those person_id which are available in all the previous days.
create table yourtable(DAY int, PERSON_ID varchar(10));
insert into yourtable values(1, '01');
insert into yourtable values(1, '02');
insert into yourtable values(1, '03');
insert into yourtable values(1, '04');
insert into yourtable values(2, '01');
insert into yourtable values(2, '02');
insert into yourtable values(2, '03');
insert into yourtable values(3, '01');
insert into yourtable values(4, '02');
insert into yourtable values(4, '01');
Query:
select day, count(person_id) as PEOPLE_LOGGED_EVERYDAY, LISTAGG(person_id,',') WITHIN GROUP(ORDER BY person_id) AS PEOPLE
from yourtable a
where exists (select 1 from yourtable b where b.day<=a.day and a.person_id=b.person_id
group by person_id having count(day)=a.day)
group by day;
Output:
DAY
PEOPLE_LOGGED_EVERYDAY
PEOPLE
1
4
01,02,03,04
2
3
01,02,03
3
1
01
4
1
01
db<fiddle here
Instead of day sequence if you had increasing dates in day column:
create table yourtable(DAY date, PERSON_ID varchar(10));
insert into yourtable values(date '2021-01-01', '01');
insert into yourtable values(date '2021-01-01', '02');
insert into yourtable values(date '2021-01-01', '03');
insert into yourtable values(date '2021-01-01', '04');
insert into yourtable values(date '2021-01-02', '01');
insert into yourtable values(date '2021-01-02', '02');
insert into yourtable values(date '2021-01-02', '03');
insert into yourtable values(date '2021-01-03', '01');
insert into yourtable values(date '2021-01-04', '02');
insert into yourtable values(date '2021-01-04', '01');
Query:
select day, count(person_id) as PEOPLE_LOGGED_EVERYDAY, LISTAGG(person_id,',') WITHIN GROUP(ORDER BY person_id) AS PEOPLE
from yourtable a
where exists (select 1 from yourtable b where b.day<=a.day and a.person_id=b.person_id
group by person_id having count(day)=( max(day)- min(day))+1)
group by day;
Output:
DAY
PEOPLE_LOGGED_EVERYDAY
PEOPLE
01-JAN-21
4
01,02,03,04
02-JAN-21
3
01,02,03
03-JAN-21
1
01
04-JAN-21
1
01
db<fiddle here
Revised answer
create table yourtable(DAY int, PERSON_ID varchar(10));
insert into yourtable(day,person_id)
with cte ( DAY, PERSON_ID ) AS(
SELECT 10, '01' FROM DUAL UNION ALL
SELECT 10, '02' FROM DUAL UNION ALL
SELECT 10, '04' FROM DUAL UNION ALL
SELECT 10, '04' FROM DUAL UNION ALL
SELECT 12, '01' FROM DUAL UNION ALL
SELECT 12, '02' FROM DUAL UNION ALL
SELECT 12, '03' FROM DUAL UNION ALL
SELECT 13, '04' FROM DUAL UNION ALL
SELECT 13, '01' FROM DUAL UNION ALL
SELECT 14, '02' FROM DUAL UNION ALL
SELECT 14, '01' FROM DUAL)
select * from cte ;
Query#1 (for Oracle 19c and later)
select day, count(person_id) as PEOPLE_LOGGED_EVERYDAY, LISTAGG(distinct person_id,',') WITHIN GROUP(ORDER BY person_id) AS PEOPLE
from yourtable a
where exists (select 1 from yourtable b where b.day<=a.day and a.person_id=b.person_id
group by person_id having count(DISTINCT day)=(select COUNT( distinct DAY) from yourtable where day<=a.day))
group by day;
Query#1 (for Oracle 18c and earlier)
select day, count(person_id) as PEOPLE_LOGGED_EVERYDAY, LISTAGG( person_id,',') WITHIN GROUP(ORDER BY person_id) AS PEOPLE
from
(
select distinct day, person_id
from yourtable a
where exists (select 1 from yourtable b where b.day<=a.day and a.person_id=b.person_id
group by person_id having count(DISTINCT day)=(select COUNT( distinct DAY) from yourtable where day<=a.day))
)t group by day
Output:
DAY
PEOPLE_LOGGED_EVERYDAY
PEOPLE
10
3
01,02,04
12
2
01,02
13
1
01
14
1
01
db<fiddle here

Answer 2

In Standard SQL, I would approach this by doing the following:
Enumerate the days for each person.
Determine the earliest day for each person.
Filter where the earliest day is "1" and the enumeration equals the days.
Then aggregate:
select day, count(*),
listagg(person_id, ',') within group (order by person_id)
from (select t.*,
row_number() over (partition by person_id order by day) as seqnum,
min(day) over (partition by person_id) as min_day
from t
) t
where seqnum = day and min_day = 1
group by day
order by day;
Note only is this simpler than using match recognize, but I would guess that the performance would be much better too.

Answer 3

You can use either:
SELECT DAY,
COUNT(DISTINCT person_id) AS num_people
FROM (
SELECT t.*,
DENSE_RANK() OVER (ORDER BY day)
- DENSE_RANK() OVER (PARTITION BY person_id ORDER BY day) AS day_grp
FROM table_name t
)
WHERE day_grp = 0
GROUP BY day
ORDER BY day
or MATCH_RECOGNIZE to find the successive days:
SELECT day,
COUNT(
DISTINCT
CASE cls WHEN 'CONSECUTIVE_DAYS' THEN person_id END
) AS num_people
FROM (
SELECT t.*,
DENSE_RANK() OVER (ORDER BY day) AS day_rank
FROM table_name t
)
MATCH_RECOGNIZE(
PARTITION BY person_id
ORDER BY day
MEASURES
classifier() AS cls
ALL ROWS PER MATCH
PATTERN ( ^ consecutive_days* )
DEFINE
consecutive_days AS COALESCE( PREV(day_rank) + 1, 1 ) = day_rank
)
GROUP BY day
ORDER BY day
Which, for the sample data:
CREATE TABLE table_name ( DAY, PERSON_ID ) AS
SELECT 1, '01' FROM DUAL UNION ALL
SELECT 1, '02' FROM DUAL UNION ALL
SELECT 1, '03' FROM DUAL UNION ALL
SELECT 1, '04' FROM DUAL UNION ALL
SELECT 2, '01' FROM DUAL UNION ALL
SELECT 2, '02' FROM DUAL UNION ALL
SELECT 2, '03' FROM DUAL UNION ALL
SELECT 3, '01' FROM DUAL UNION ALL
SELECT 3, '02' FROM DUAL UNION ALL
SELECT 4, '01' FROM DUAL;
Outputs:
DAY
NUM_PEOPLE
1
4
2
3
3
2
4
1
and for the sample data:
CREATE TABLE table_name ( DAY, PERSON_ID ) AS
SELECT 10, '01' FROM DUAL UNION ALL
SELECT 10, '02' FROM DUAL UNION ALL
SELECT 10, '04' FROM DUAL UNION ALL
SELECT 10, '04' FROM DUAL UNION ALL
SELECT 12, '01' FROM DUAL UNION ALL
SELECT 12, '02' FROM DUAL UNION ALL
SELECT 12, '03' FROM DUAL UNION ALL
SELECT 13, '04' FROM DUAL UNION ALL
SELECT 13, '01' FROM DUAL UNION ALL
SELECT 14, '02' FROM DUAL UNION ALL
SELECT 14, '01' FROM DUAL
Outputs:
DAY
NUM_PEOPLE
10
3
12
2
13
1
14
1
db<>fiddle here