Reputation: 85
How to write a concise sql to get subscription rate by month.
formula: subscription rate = subscription count/ trial count
NOTE: The tricky part is the subscription event should be attributed to the month that company started the trail.
| id | date | type |
|-------|------------|-------|
| 10001 | 2019-01-01 | Trial |
| 10001 | 2019-01-15 | Sub |
| 10002 | 2019-01-20 | Trial |
| 10002 | 2019-02-10 | Sub |
| 10003 | 2019-01-01 | Trial |
| 10004 | 2019-02-10 | Trial |
Based on the above table, the out output should be:
2019-01-01 2/3
2019-02-01 0/1
Upvotes: 1
Views: 404
Reputation: 1270613
You can do this with window functions. Assuming that there are not duplicate trial/subs:
select date_trunc('month', date) as yyyymm,
count(*) where (num_subs > 0) * 1.0 / count(*)
from (select t.*,
count(*) filter (where type = 'Sub') over (partition by id) as num_subs
from t
) t
where type = 'Trial'
group by yyyymm;
If an id
can have duplicate trials or subs, then I suggest that you ask a new question with more detail about the duplicates.
You an also do this with two levels of aggregation:
select trial_date,
count(sub_date) * 1.0 / count(*)
from (select id, min(date) filter (where type = 'trial') as trial_date,
min(date) filter (where type = 'sub') as sub_date
from t
group by id
) id
group by trial_date;
Upvotes: 0
Reputation: 222602
One option is a self-join to identify whether each trial eventually subscribed, then aggregation and arithmetics:
select
date_trunc('month', t.date) date_month
1.0 * count(s.id) / count(t.id) rate
from mytable t
left join mytable s on s.id = t.id and s.type = 'Sub'
where t.type = 'Trial'
group by date_trunc('month', t.date)
The syntax to truncate a date to the beginning of the month widely varies across databases. The above would work in Postgres. Alternatives are available in other databases, such as:
date_format(t.date, '%Y-%m-01') -- MySQL
trunc(t.date, 'mm') -- Oracle
datefromparts(year(t.date), month(t.date), 1) -- SQL Server
Upvotes: 2