I am working on query to get cumulative distinct count of uids on daily basis.
Example : Say there are 2 uids (100,200) appeared on date 2016-11-01 and they also appea
Please try the following...
SELECT date AS date
COUNT( uid ) AS daily_cumulative_count
FROM ( SELECT leftTable.date AS date,
rightTable.uid AS uid
FROM sample_table AS leftTable
JOIN sample_table AS rightTable ON leftTable.date >= rightTable.date
GROUP BY leftTable.date,
rightTable.uid
) AS allUIDSForDateFinder
GROUP BY date;
This statement starts by joining one instance of sample_table
to another in such a way that each record in leftTable
has associated with it a copy of each record from rightTable
that has an earlier or equal date
value. This effectively attaches a list to each date
of all uid
values that have occurred up to and including that date
value.
The resulting dataset is refined to unique date
and uid
combinations through use of GROUP BY
.
The refined dataset from the subquery allUIDSForDateFinder
is then grouped by date
by the main body of the query, and a COUNT()
of uid
values associated with each group is performed.
If you have any questions or comments, then please feel free to post a Comment accordingly.
WITH firstseen AS (
SELECT uid, MIN(date) date
FROM sample_table
GROUP BY 1
)
SELECT DISTINCT date, COUNT(uid) OVER (ORDER BY date) daily_cumulative_count
FROM firstseen
ORDER BY 1
Using SELECT DISTINCT
because (date, COUNT(uid))
will be duplicated many times.
Explanation: for each date dt
, it counts uid from the earliest date up to dt
, because we are specifying ORDER BY date
and it defaults to BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
.
easiest way:
SELECT *, count(*) over (order by fst_date ) cum_uids
FROM (
SELECT uid, min(date) fst_date FROM t GROUP BY uid
) t
or something like this
You can use exists
to check if an id was present on any of the previous dates. Then get the running sum and find the max value for each group which would get you the daily distinct cumulative count.
select dt, max(col) as daily_cumulative_count
from (select t1.*,
sum(case when not exists (select 1 from t where t1.dt > dt and id = t1.uid) then 1 else 0 end) over(order by dt) col
from t t1) x
group by dt