Select distinct users group by time range - sql

Question

I have a table with the following info
|date | user_id | week_beg | month_beg|
SQL to create table with test values:
CREATE TABLE uniques
(
date DATE,
user_id INT,
week_beg DATE,
month_beg DATE
)
INSERT INTO uniques VALUES ('2013-01-01', 1, '2012-12-30', '2013-01-01')
INSERT INTO uniques VALUES ('2013-01-03', 3, '2012-12-30', '2013-01-01')
INSERT INTO uniques VALUES ('2013-01-06', 4, '2013-01-06', '2013-01-01')
INSERT INTO uniques VALUES ('2013-01-07', 4, '2013-01-06', '2013-01-01')
INPUT TABLE:
| date | user_id | week_beg | month_beg |
| 2013-01-01 | 1 | 2012-12-30 | 2013-01-01 |
| 2013-01-03 | 3 | 2012-12-30 | 2013-01-01 |
| 2013-01-06 | 4 | 2013-01-06 | 2013-01-01 |
| 2013-01-07 | 4 | 2013-01-06 | 2013-01-01 |
OUTPUT TABLE:
| date | time_series | cnt |
| 2013-01-01 | D | 1 |
| 2013-01-01 | W | 1 |
| 2013-01-01 | M | 1 |
| 2013-01-03 | D | 1 |
| 2013-01-03 | W | 2 |
| 2013-01-03 | M | 2 |
| 2013-01-06 | D | 1 |
| 2013-01-06 | W | 1 |
| 2013-01-06 | M | 3 |
| 2013-01-07 | D | 1 |
| 2013-01-07 | W | 1 |
| 2013-01-07 | M | 3 |
I want to calculate the number of distinct user_id's for a date:
For that date
For that week up to that date (Week to date)
For the month up to that date (Month to date)
1 is easy to calculate.
For 2 and 3 I am trying to use such queries:
SELECT
date,
'W' AS "time_series",
(COUNT DISTINCT user_id) COUNT (user_id) OVER (PARTITION BY week_beg) AS "cnt"
FROM user_subtitles
SELECT
date,
'M' AS "time_series",
(COUNT DISTINCT user_id) COUNT (user_id) OVER (PARTITION BY month_beg) AS "cnt"
FROM user_subtitles
Postgres does not allow window functions for DISTINCT calculation, so this approach does not work.
I have also tried out a GROUP BY approach, but it does not work as it gives me numbers for whole week/months.
Whats the best way to approach this problem?

Answer 1

Count all rows
SELECT date, '1_D' AS time_series, count(DISTINCT user_id) AS cnt
FROM uniques
GROUP BY 1
UNION ALL
SELECT DISTINCT ON (1)
date, '2_W', count(*) OVER (PARTITION BY week_beg ORDER BY date)
FROM uniques
UNION ALL
SELECT DISTINCT ON (1)
date, '3_M', count(*) OVER (PARTITION BY month_beg ORDER BY date)
FROM uniques
ORDER BY 1, time_series
Your columns week_beg and month_beg are 100 % redundant and can easily be replaced by
date_trunc('week', date + 1) - 1 and date_trunc('month', date) respectively.
Your week seems to start on Sunday (off by one), therefore the + 1 .. - 1.
The default frame of a window function with ORDER BY in the OVER clause uses is RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW. That's exactly what you need.
Use UNION ALL, not UNION.
Your unfortunate choice for time_series (D, W, M) does not sort well, I renamed to make the final ORDER BY easier.
This query can deal with multiple rows per day. Counts include all peers for a day.
More about DISTINCT ON:
Select first row in each GROUP BY group?
DISTINCT users per day
To count every user only once per day, use a CTE with DISTINCT ON:
WITH x AS (SELECT DISTINCT ON (1,2) date, user_id FROM uniques)
SELECT date, '1_D' AS time_series, count(user_id) AS cnt
FROM x
GROUP BY 1
UNION ALL
SELECT DISTINCT ON (1)
date, '2_W'
,count(*) OVER (PARTITION BY (date_trunc('week', date + 1)::date - 1)
ORDER BY date)
FROM x
UNION ALL
SELECT DISTINCT ON (1)
date, '3_M'
,count(*) OVER (PARTITION BY date_trunc('month', date) ORDER BY date)
FROM x
ORDER BY 1, 2
DISTINCT users over dynamic period of time
You can always resort to correlated subqueries. Tend to be slow with big tables!
Building on the previous queries:
WITH du AS (SELECT date, user_id FROM uniques GROUP BY 1,2)
,d AS (
SELECT date
,(date_trunc('week', date + 1)::date - 1) AS week_beg
,date_trunc('month', date)::date AS month_beg
FROM uniques
GROUP BY 1
)
SELECT date, '1_D' AS time_series, count(user_id) AS cnt
FROM du
GROUP BY 1
UNION ALL
SELECT date, '2_W', (SELECT count(DISTINCT user_id) FROM du
WHERE du.date BETWEEN d.week_beg AND d.date )
FROM d
GROUP BY date, week_beg
UNION ALL
SELECT date, '3_M', (SELECT count(DISTINCT user_id) FROM du
WHERE du.date BETWEEN d.month_beg AND d.date)
FROM d
GROUP BY date, month_beg
ORDER BY 1,2;
SQL Fiddle for all three solutions.
Faster with dense_rank()
#Clodoaldo came up with a major improvement: use the window function dense_rank(). Here is another idea for an optimized version. It should be even faster to exclude daily duplicates right away. The performance gain grows with the number of rows per day.
Building on a simplified and sanitized data model
- without the redundant columns
- day as column name instead of date
date is a reserved word in standard SQL and a basic type name in PostgreSQL and shouldn't be used as identifier.
CREATE TABLE uniques(
day date -- instead of "date"
,user_id int
);
Improved query:
WITH du AS (
SELECT DISTINCT ON (1, 2)
day, user_id
,date_trunc('week', day + 1)::date - 1 AS week_beg
,date_trunc('month', day)::date AS month_beg
FROM uniques
)
SELECT day, count(user_id) AS d, max(w) AS w, max(m) AS m
FROM (
SELECT user_id, day
,dense_rank() OVER(PARTITION BY week_beg ORDER BY user_id) AS w
,dense_rank() OVER(PARTITION BY month_beg ORDER BY user_id) AS m
FROM du
) s
GROUP BY day
ORDER BY day;
SQL Fiddle demonstrating the performance of 4 faster variants. It depends on your data distribution which is fastest for you.
All of them are about 10x as fast as the correlated subqueries version (which isn't bad for correlated subqueries).

Answer 2

Without correlated subqueries. SQL Fiddle
with u as (
select
"date", user_id,
date_trunc('week', "date" + 1)::date - 1 week_beg,
date_trunc('month', "date")::date month_beg
from uniques
)
select
"date", count(distinct user_id) D,
max(week_dr) W, max(month_dr) M
from (
select
user_id, "date",
dense_rank() over(partition by week_beg order by user_id) week_dr,
dense_rank() over(partition by month_beg order by user_id) month_dr
from u
) s
group by "date"
order by "date"

Answer 3

Try
SELECT
*
FROM
(
SELECT dates, count(user_id), 'D' as timesereis FROM users_data GROUP BY dates
UNION
SELECT max(dates), count(user_id), 'W' FROM users_data GROUP BY date_part('year',dates)+date_part('week',dates)
UNION
SELECT max(dates), count(user_id), 'M' FROM users_data GROUP BY date_part('year',dates)+date_part('week',dates)
) tEMP order by dates, timesereis
SQLFIDDLE

Answer 4

Try queries like this
SELECT count(distinct user_id), date_format(date, '%Y-%m-%d') as date_period
FROM uniques
GROUP By date_period