SQL 30 day active user query - sql

I have a table of users and how many events they fired on a given date:
DATE
USERID
EVENTS
2021-08-27
1
5
2021-07-25
1
7
2021-07-23
2
3
2021-07-20
3
9
2021-06-22
1
9
2021-05-05
1
4
2021-05-05
2
2
2021-05-05
3
6
2021-05-05
4
8
2021-05-05
5
1
I want to create a table showing number of active users for each date with active user being defined as someone who has fired an event on the given date or in any of the preceding 30 days.
DATE
ACTIVE_USERS
2021-08-27
1
2021-07-25
3
2021-07-23
2
2021-07-20
2
2021-06-22
1
2021-05-05
5
I tried the following query which returned only the users who were active on the specified date:
SELECT COUNT(DISTINCT USERID), DATE
FROM table
WHERE DATE >= (CURRENT_DATE() - interval '30 days')
GROUP BY 2 ORDER BY 2 DESC;
I also tried using a window function with rows between but seems to end up getting the same result:
SELECT
DATE,
SUM(ACTIVE_USERS) AS ACTIVE_USERS
FROM
(
SELECT
DATE,
CASE
WHEN SUM(EVENTS) OVER (PARTITION BY USERID ORDER BY DATE ROWS BETWEEN 30 PRECEDING AND CURRENT ROW) >= 1 THEN 1
ELSE 0
END AS ACTIVE_USERS
FROM table
)
GROUP BY 1
ORDER BY 1
I'm using SQL:ANSI on Snowflake. Any suggestions would be much appreciated.

This is tricky to do as window functions -- because count(distinct) is not permitted. You can use a self-join:
select t1.date, count(distinct t2.userid)
from table t join
table t2
on t2.date <= t.date and
t2.date > t.date - interval '30 day'
group by t1.date;
However, that can be expensive. One solution is to "unpivot" the data. That is, do an incremental count per user of going "in" and "out" of active states and then do a cumulative sum:
with d as ( -- calculate the dates with "ins" and "outs"
select user, date, +1 as inc
from table
union all
select user, date + interval '30 day', -1 as inc
from table
),
d2 as ( -- accumulate to get the net actives per day
select date, user, sum(inc) as change_on_day,
sum(sum(inc)) over (partition by user order by date) as running_inc
from d
group by date, user
),
d3 as ( -- summarize into active periods
select user, min(date) as start_date, max(date) as end_date
from (select d2.*,
sum(case when running_inc = 0 then 1 else 0 end) over (partition by user order by date) as active_period
from d2
) d2
where running_inc > 0
group by user
)
select d.date, count(d3.user)
from (select distinct date from table) d left join
d3
on d.date >= start_date and d.date < end_date
group by d.date;

Related

Create months between two dates Snowflake SQL

I just want to generate the months between data range using SQL Query.
example
You can use a table generator:
select '2022-07-04'::date +
row_number() over(partition by 1 order by null) - 1 GENERATED_DATE
from table(generator(rowcount => 365))
;
Just change the start date and the number of days into the series. You can use the datediff function to calculate the number of days between the start end end dates.
Edit: I just realized the generator table function requires a constant for the number of rows. That's easily solvable. Just set a higher number of rows than you'll need and specify the end of the series in a qualify clause:
set startdate = (select '2022-04-15'::date);
set enddate = (select '2022-07-04'::date);
select $startdate::date +
row_number() over(partition by 1 order by null) - 1 GENERATED_DATE
from table(generator(rowcount => 100000))
qualify GENERATED_DATE <= $enddate
;
You can use a table generator in the CTE, and then select from the CTE and cartesian join to your table with data and use a case statement to see if the date in the generator is between your start and to dates.
Then select from it:
select user_id, x_date
from (
with dates as (
select '2019-01-01'::date + row_number() over(order by 0) x_date
from table(generator(rowcount => 1500))
)
select d.x_date, t.*,
case
when d.x_date between t.from_date and t.to_date then 'Y' else 'N' end target_date
from dates d, my_table t --deliberate cartesian join
)
where target_date = 'Y'
order by 1,2
Output:
USER_ID X_DATE
1 2/20/2019
1 2/21/2019
1 2/22/2019
1 2/23/2019
2 2/22/2019
2 2/23/2019
2 2/24/2019
2 2/25/2019
2 2/26/2019
2 2/27/2019
2 2/28/2019
3 3/1/2019
3 3/2/2019
3 3/3/2019
3 3/4/2019
3 3/5/2019
=======EDIT========
Based on your comments below, you are actually looking for something different than your original screenshots. Ok, so here we are still using the table generator, and then we're truncating the month to the first day of the month where the x-date is YES.
select distinct t.user_id, t.from_date, t.to_date, date_trunc('MONTH', z.x_date) as trunc_month
from (
with dates as (
select '2019-01-01'::date + row_number() over(order by 0) x_date
from table(generator(rowcount => 1500))
)
select d.x_date, t.*,
case
when d.x_date between t.from_date and t.to_date then 'Y' else 'N' end target_date
from dates d, my_table t
)z
join my_table t
on z.user_id = t.user_id
where z.target_date = 'Y'
order by 1,2
Output (modified User ID 3 to span 2 months):
USER_ID FROM_DATE TO_DATE TRUNC_MONTH
1 2/20/2019 2/23/2019 2/1/2019
2 2/22/2019 2/28/2019 2/1/2019
3 2/25/2019 3/5/2019 2/1/2019
3 2/25/2019 3/5/2019 3/1/2019

Count new entries day by day

I would like to count new id's in each day. Saying new, I mean new relative to the day before.
Assume we have a table:
Date
Id
2021-01-01
1
2021-01-02
4
2021-01-02
5
2021-01-02
6
2021-01-03
1
2021-01-03
5
2021-01-03
7
My desired output, would look like this:
Date
Count(NewId)
2021-01-01
1
2021-01-02
3
2021-01-03
2
You can use two levels of aggregation:
select date, count(*)
from (select id, min(date) as date
from t
group by id
) i
group by date
order by date;
If by "relative to the day before" you mean that you want to count someone as new whenever they have no record on the previous day, then use lag() . . . carefully:
select date,
sum(case when prev_date = date - interval '1' day then 0 else 1 end)
from (select t.*,
lag(date) over (partition by id order by date) as prev_date
from t
) t
group by date
order by date;
here is another way, probably the simplest :
select t1.Date, count(*) from table t1
where id not in (select id from table t2 where t2.date = t1.date- interval '1 day')
group by t1.Date
Maybe this other option could also do the job, but being honest I would prefer the #GordonLinoff answer:
select date, count(*)
from your_table t
where not exists (
select 1
from your_table tt
where tt.Id=t.id
and tt.date = date_sub(t.date,1)
)
group by date

sum values based on 7-day cycle in SQL Oracle

I have dates and some value, I would like to sum values within 7-day cycle starting from the first date.
date value
01-01-2021 1
02-01-2021 1
05-01-2021 1
07-01-2021 1
10-01-2021 1
12-01-2021 1
13-01-2021 1
16-01-2021 1
18-01-2021 1
22-01-2021 1
23-01-2021 1
30-01-2021 1
this is my input data with 4 groups to see what groups will create the 7-day cycle.
It should start with first date and sum all values within 7 days after first date included.
then start a new group with next day plus anothe 7 days, 10-01 till 17-01 and then again new group from 18-01 till 25-01 and so on.
so the output will be
group1 4
group2 4
group3 3
group4 1
with match_recognize would be easy current_day < first_day + 7 as a condition for the pattern but please don't use match_recognize clause as solution !!!
One approach is a recursive CTE:
with tt as (
select dte, value, row_number() over (order by dte) as seqnum
from t
),
cte (dte, value, seqnum, firstdte) as (
select tt.dte, tt.value, tt.seqnum, tt.dte
from tt
where seqnum = 1
union all
select tt.dte, tt.value, tt.seqnum,
(case when tt.dte < cte.firstdte + interval '7' day then cte.firstdte else tt.dte end)
from cte join
tt
on tt.seqnum = cte.seqnum + 1
)
select firstdte, sum(value)
from cte
group by firstdte
order by firstdte;
This identifies the groups by the first date. You can use row_number() over (order by firstdte) if you want a number.
Here is a db<>fiddle.

SQL - Query to return active subscriptions on a given day

I have a table that shows when a user signs up for a subscription and when their membership will expire. A user can purchase a new subscription even if their current one is in force.
userid|purchasedate|expirydate
1 |2019-01-01 |2019-02-01
2 |2019-01-02 |2019-02-02
3 |2019-01-03 |2019-02-03
3 |2019-01-04 |2019-03-03
I need a SQL query that will GROUP BY the date and return the number of active subscriptions on that date. So it would return:
date |count
2019-01-01|1
2019-01-02|2
2019-01-03|3
2019-01-04|3
Below is for BigQuery Standard SQL
#standardSQL
SELECT day, COUNT(DISTINCT userid) active_subscriptions
FROM (SELECT AS STRUCT MIN(purchasedate) min_date, MAX(expirydate) max_date FROM `project.dataset.table`),
UNNEST(GENERATE_DATE_ARRAY(min_date, max_date)) day
JOIN `project.dataset.table`
ON day BETWEEN purchasedate AND expirydate
GROUP BY day
You can test, play with above using dummy data from your question as in below example
#standardSQL
WITH `project.dataset.table` AS (
SELECT 1 userid, DATE '2019-01-01' purchasedate, DATE '2019-02-01' expirydate UNION ALL
SELECT 2, '2019-01-02', '2019-02-02' UNION ALL
SELECT 3, '2019-01-03', '2019-02-03' UNION ALL
SELECT 3, '2019-01-04', '2019-03-03'
)
SELECT day, COUNT(DISTINCT userid) active_subscriptions
FROM (SELECT AS STRUCT MIN(purchasedate) min_date, MAX(expirydate) max_date FROM `project.dataset.table`),
UNNEST(GENERATE_DATE_ARRAY(min_date, max_date)) day
JOIN `project.dataset.table`
ON day BETWEEN purchasedate AND expirydate
GROUP BY day
with below output
Row day active_subscriptions
1 2019-01-01 1
2 2019-01-02 2
3 2019-01-03 3
4 2019-01-04 3
5 2019-01-05 3
6 2019-01-06 3
... ... ...
... ... ...
31 2019-01-31 3
32 2019-02-01 3
33 2019-02-02 2
34 2019-02-03 1
35 2019-02-04 1
... ... ...
... ... ...
61 2019-03-02 1
62 2019-03-03 1
You need a list of dates and count(distinct):
select d.dte, count(distinct t.userid) as num_users
from (select distinct purchase_date as dte from t) d left join
t
on d.dte >= t.dte and
d.dte <= t.expiry_date
group by d.dte
order by d.dte;
EDIT:
BigQuery can be fickle about inequalities in the on clause. Here is another approach:
select dte, count(distinct t.userid) as num_users
from t cross join
unnest(generate_date_array(t.purchase_date, t.expiry_date, interval 1 day)) dte
group by dte
order by dte;
You can use a where clause to filter down to particular dates.
I make the table name 'test_expirydate' and use your data
and this one work
select
tb1.expirydate,
count(*) as total
from test_expirydate as tb1
left join (
select
expirydate
from test_expirydate as tb2
group by userid
) as tb2
on tb1.expirydate >= tb2.expirydate
group by tb1.expirydate
I don't sure is it work in other case or not but it fine with current data
Oh, I interpret that the left column should be the expiration date.

Add Missing monthly dates in a timeseries data in Postgresql

I have monthly time series data in table where dates are as a last day of month. Some of the dates are missing in the data. I want to insert those dates and put zero value for other attributes.
Table is as follows:
id report_date price
1 2015-01-31 40
1 2015-02-28 56
1 2015-04-30 34
2 2014-05-31 45
2 2014-08-31 47
I want to convert this table to
id report_date price
1 2015-01-31 40
1 2015-02-28 56
1 2015-03-31 0
1 2015-04-30 34
2 2014-05-31 45
2 2014-06-30 0
2 2014-07-31 0
2 2014-08-31 47
Is there any way we can do this in Postgresql?
Currently we are doing this in Python. As our data is growing day by day and its not efficient to handle I/O just for one task.
Thank you
You can do this using generate_series() to generate the dates and then left join to bring in the values:
with m as (
select id, min(report_date) as minrd, max(report_date) as maxrd
from t
group by id
)
select m.id, m.report_date, coalesce(t.price, 0) as price
from (select m.*, generate_series(minrd, maxrd, interval '1' month) as report_date
from m
) m left join
t
on m.report_date = t.report_date;
EDIT:
Turns out that the above doesn't quite work, because adding months to the end of month doesn't keep the last day of the month.
This is easily fixed:
with t as (
select 1 as id, date '2012-01-31' as report_date, 10 as price union all
select 1 as id, date '2012-04-30', 20
), m as (
select id, min(report_date) - interval '1 day' as minrd, max(report_date) - interval '1 day' as maxrd
from t
group by id
)
select m.id, m.report_date, coalesce(t.price, 0) as price
from (select m.*, generate_series(minrd, maxrd, interval '1' month) + interval '1 day' as report_date
from m
) m left join
t
on m.report_date = t.report_date;
The first CTE is just to generate sample data.
This is a slight improvement over Gordon's query which fails to get the last date of a month in some cases.
Essentially you generate all the month end dates between the min and max date for each id (using generate_series) and left join on this generated table to show the missing dates with 0 price.
with minmax as (
select id, min(report_date) as mindt, max(report_date) as maxdt
from t
group by id
)
select m.id, m.report_date, coalesce(t.price, 0) as price
from (select *,
generate_series(date_trunc('MONTH',mindt+interval '1' day),
date_trunc('MONTH',maxdt+interval '1' day),
interval '1' month) - interval '1 day' as report_date
from minmax
) m
left join t on m.report_date = t.report_date
Sample Demo