SQL: calculate MAU by window function - sql

Question

I'm trying unsuccessfully to calculate a MAU- monthly distinct active users, by using window functions.
I need the calculation for each day during the month, for the preceding 30 days
This is what I have so far:
select
t.datee
, t.app,i.sourcee
, i.campaign
, t.mobile
, sum(count(distinct t.user_id)) over (
PARTITION BY
date_trunc('month',datee)
, t.app
, i.sourcee
, i.campaign
, t.mobile
ORDER BY datee asc
ROWS BETWEEN 30 PRECEDING AND CURRENT ROW
)
FROM dim_x i
JOIN agg_y t
ON i.app=t.app
AND i.mobile=t.mobile
WHERE t.datee>=CURRENT_DATE-30
AND t.datee<CURRENT_DATE
GROUP BY 1,2,3,4,5
order by 1 desc
But all I get is a sum of active users by all days instead of sum of distinct users. I'm using Vertica db.
Any suggestions?

Answer 1

I'm not getting, really, why you should need an OLAP expression for that.
Aren't you looking for the total number of distinct users per:
year-month combination out of datee
app
sourcee (whatever that might be)
campaign
mobile (probably mobile number)
?
A simple GROUP BY would do, as far as I'm concerned. If I disregard sourcee, campaign and mobile, selecting just from one table: input for argument's sake, with some sample data I just made up, this query:
SELECT
YEAR(datee) * 100 + MONTH(datee) AS yearmonth
, app
, COUNT(DISTINCT user_id) AS monthly_active_users
FROM input
GROUP BY 1,2
ORDER BY 1
;
... would return:
YEARMONTH|app |monthly_active_users
201,601|app-a| 2
201,601|app-b| 2
201,602|app-a| 2
201,602|app-b| 2
201,603|app-a| 2
201,603|app-b| 2
201,604|app-a| 2
201,604|app-b| 2
201,605|app-a| 2
201,605|app-b| 2
201,606|app-a| 1
201,606|app-b| 1
Just editing my previous answer. You seem to need the running COUNT DISTINCT of user id-s , partitioned by several expressions.
With the input from the WITH clause below, would you need a report like this (only showing the first 12 rows of 53, ordered by datee, app)?
datee |app |user_id |running_active_users
2016-01-01|app-a|arthur | 1
2016-01-04|app-b|ford | 1
2016-01-07|app-a|trillian| 2
2016-01-10|app-b|zaphod | 2
2016-01-13|app-a|arthur | 2
2016-01-16|app-b|ford | 2
2016-01-19|app-a|trillian| 2
2016-01-22|app-b|zaphod | 2
2016-01-25|app-a|arthur | 2
2016-01-28|app-b|ford | 2
2016-01-31|app-a|trillian| 2
2016-02-03|app-b|zaphod | 2
?
If that's the case, I don't see the reason for existence of your GROUP BY clause, though.
Below is the query with GROUP BY as above with test data returning the results above in a WITH clause. Regard that input as the join between your two tables.
WITH
input(datee,app,user_id) AS (
SELECT DATE '2016-01-01','app-a','arthur'
UNION ALL SELECT DATE '2016-01-04','app-b','ford'
UNION ALL SELECT DATE '2016-01-07','app-a','trillian'
UNION ALL SELECT DATE '2016-01-10','app-b','zaphod'
UNION ALL SELECT DATE '2016-01-25','app-a','arthur'
UNION ALL SELECT DATE '2016-01-28','app-b','ford'
UNION ALL SELECT DATE '2016-03-04','app-b','ford'
UNION ALL SELECT DATE '2016-03-25','app-a','arthur'
UNION ALL SELECT DATE '2016-04-09','app-b','ford'
UNION ALL SELECT DATE '2016-04-30','app-a','arthur'
UNION ALL SELECT DATE '2016-05-06','app-a','trillian'
UNION ALL SELECT DATE '2016-05-09','app-b','zaphod'
UNION ALL SELECT DATE '2016-05-15','app-b','ford'
UNION ALL SELECT DATE '2016-06-05','app-a','arthur'
UNION ALL SELECT DATE '2016-01-13','app-a','arthur'
UNION ALL SELECT DATE '2016-01-16','app-b','ford'
UNION ALL SELECT DATE '2016-01-31','app-a','trillian'
UNION ALL SELECT DATE '2016-02-03','app-b','zaphod'
UNION ALL SELECT DATE '2016-02-06','app-a','arthur'
UNION ALL SELECT DATE '2016-02-09','app-b','ford'
UNION ALL SELECT DATE '2016-02-12','app-a','trillian'
UNION ALL SELECT DATE '2016-02-15','app-b','zaphod'
UNION ALL SELECT DATE '2016-02-18','app-a','arthur'
UNION ALL SELECT DATE '2016-02-21','app-b','ford'
UNION ALL SELECT DATE '2016-02-24','app-a','trillian'
UNION ALL SELECT DATE '2016-02-27','app-b','zaphod'
UNION ALL SELECT DATE '2016-03-01','app-a','arthur'
UNION ALL SELECT DATE '2016-03-10','app-b','zaphod'
UNION ALL SELECT DATE '2016-03-13','app-a','arthur'
UNION ALL SELECT DATE '2016-03-16','app-b','ford'
UNION ALL SELECT DATE '2016-03-28','app-b','ford'
UNION ALL SELECT DATE '2016-03-31','app-a','trillian'
UNION ALL SELECT DATE '2016-04-06','app-a','arthur'
UNION ALL SELECT DATE '2016-04-12','app-a','trillian'
UNION ALL SELECT DATE '2016-04-15','app-b','zaphod'
UNION ALL SELECT DATE '2016-04-27','app-b','zaphod'
UNION ALL SELECT DATE '2016-05-03','app-b','ford'
UNION ALL SELECT DATE '2016-05-27','app-b','ford'
UNION ALL SELECT DATE '2016-05-30','app-a','trillian'
UNION ALL SELECT DATE '2016-01-19','app-a','trillian'
UNION ALL SELECT DATE '2016-01-22','app-b','zaphod'
UNION ALL SELECT DATE '2016-03-07','app-a','trillian'
UNION ALL SELECT DATE '2016-03-19','app-a','trillian'
UNION ALL SELECT DATE '2016-03-22','app-b','zaphod'
UNION ALL SELECT DATE '2016-04-03','app-b','zaphod'
UNION ALL SELECT DATE '2016-04-18','app-a','arthur'
UNION ALL SELECT DATE '2016-04-21','app-b','ford'
UNION ALL SELECT DATE '2016-04-24','app-a','trillian'
UNION ALL SELECT DATE '2016-05-12','app-a','arthur'
UNION ALL SELECT DATE '2016-05-18','app-a','trillian'
UNION ALL SELECT DATE '2016-05-21','app-b','zaphod'
UNION ALL SELECT DATE '2016-05-24','app-a','arthur'
UNION ALL SELECT DATE '2016-06-02','app-b','zaphod'
)
SELECT
YEAR(datee) * 100 + MONTH(datee) AS YEARMONTH
, app
, COUNT(DISTINCT user_id) AS monthly_active_users
FROM input
GROUP BY 1,2
ORDER BY 1
;