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I am using count and group by to get the number of subscribers registered each day:

  SELECT created_at, COUNT(email)  
    FROM subscriptions 
GROUP BY created at;

Result:

created_at  count
-----------------
04-04-2011  100
05-04-2011   50
06-04-2011   50
07-04-2011  300

I want to get the cumulative total of subscribers every day instead. How do I get this?

created_at  count
-----------------
04-04-2011  100
05-04-2011  150
06-04-2011  200
07-04-2011  500
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With larger datasets, window functions are the most efficient way to perform these kinds of queries -- the table will be scanned only once, instead of once for each date, like a self-join would do. It also looks a lot simpler. :) PostgreSQL 8.4 and up have support for window functions.

This is what it looks like:

SELECT created_at, sum(count(email)) OVER (ORDER BY created_at)
FROM subscriptions
GROUP BY created_at;

Here OVER creates the window; ORDER BY created_at means that it has to sum up the counts in created_at order.


Edit: If you want to remove duplicate emails within a single day, you can use sum(count(distinct email)). Unfortunately this won't remove duplicates that cross different dates.

If you want to remove all duplicates, I think the easiest is to use a subquery and DISTINCT ON. This will attribute emails to their earliest date (because I'm sorting by created_at in ascending order, it'll choose the earliest one):

SELECT created_at, sum(count(email)) OVER (ORDER BY created_at)
FROM (
    SELECT DISTINCT ON (email) created_at, email
    FROM subscriptions ORDER BY email, created_at
) AS subq
GROUP BY created_at;

If you create an index on (email, created_at), this query shouldn't be too slow either.


(If you want to test, this is how I created the sample dataset)

create table subscriptions as
   select date '2000-04-04' + (i/10000)::int as created_at,
          'foofoobar@foobar.com' || (i%700000)::text as email
   from generate_series(1,1000000) i;
create index on subscriptions (email, created_at);

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