Get detail days between two date (mysql query) - sql

I have data like this:
id | start_date | end_date
----------------------------
1 | 16-09-2019 | 22-12-2019
I want to get the following results:
id | month | year | days
------------------------
1 | 09 | 2019 | 15
1 | 10 | 2019 | 31
1 | 11 | 2019 | 30
1 | 12 | 2019 | 22
Is there a way to get that result ?

This is what you want to do:
SELECT id, EXTRACT(MONTH FROM start_date ) as month , EXTRACT(YEAR FROM start_date ) as year , DATEDIFF(end_date, start_date ) as days
From tbl

You can use MONTH() , YEAR() and DATEDIFF() functions
SELECT id, MONTH(start_date) as month, YEAR(start_date ) as year, DATEDIFF(end_date, start_date ) as days from table-name

One way is to create a Calendar table and use that.
select month,year, count(*)
from Calendar
where db_date between '2019-09-16'
and '2019-12-22'
group by month,year
CHECK DEMO HERE
Also you can use recursive CTE to achieve the same.

You can use a recursive CTE and aggregation:
with recursive cte as (
select id, start_date, end_date
from t
union all
select id, start_date + interval 1 day, end_date
from cte
where start_date < end_date
)
select id, year(start_date), month(start_date), count(*) as days
from cte
group by id, year(start_date), month(start_date);
Here is a db<>fiddle.

Related

Converting PostgreSQL recursive CTE to SQL Server

I'm having trouble adapting some recursive CTE code from PostgreSQL to SQL Server, from the book "Fighting Churn with Data"
This is the working PostgreSQL code:
with recursive
active_period_params as (
select interval '30 days' as allowed_gap,
'2021-09-30'::date as calc_date
),
active as (
-- anchor
select distinct account_id, min(start_date) as start_date
from subscription inner join active_period_params
on start_date <= calc_date
and (end_date > calc_date or end_date is null)
group by account_id
UNION
-- recursive
select s.account_id, s.start_date
from subscription s
cross join active_period_params
inner join active e on s.account_id=e.account_id
and s.start_date < e.start_date
and s.end_date >= (e.start_date-allowed_gap)::date
)
select account_id, min(start_date) as start_date
from active
group by account_id
This is my attempt at converting to SQL Server. It gets stuck in a loop. I believe the issue has to do with the UNION ALL required by SQL Server.
with
active_period_params as (
select 30 as allowed_gap,
cast('2021-09-30' as date) as calc_date
),
active as (
-- anchor
select distinct account_id, min(start_date) as start_date
from subscription inner join active_period_params
on start_date <= calc_date
and (end_date > calc_date or end_date is null)
group by account_id
UNION ALL
-- recursive
select s.account_id, s.start_date
from subscription s
cross join active_period_params
inner join active e on s.account_id=e.account_id
and s.start_date < e.start_date
and s.end_date >= dateadd(day, -allowed_gap, e.start_date)
)
select account_id, min(start_date) as start_date
from active
group by account_id
The subscription table is a list of subscriptions belonging to customers. A customer can have multiple subscriptions with overlapping dates or gaps between dates. null end_date means the subscription is currently active and has no defined end_date. Example data for a single customer (account_id = 15) below:
subscription
---------------------------------------------------
| id | account_id | start_date | end_date |
---------------------------------------------------
| 6 | 15 | 01/06/2021 | null |
| 5 | 15 | 01/01/2021 | null |
| 4 | 15 | 01/06/2020 | 01/02/2021 |
| 3 | 15 | 01/04/2020 | 15/05/2020 |
| 2 | 15 | 01/03/2020 | 15/05/2020 |
| 1 | 15 | 01/06/2019 | 01/01/2020 |
Expected query result (as produced by PostgreSQL code):
------------------------------
| account_id | start_date |
------------------------------
| 15 | 01/03/2020 |
Issue:
The SQL Server code above gets stuck in a loop and doesn't produce a result.
Description of the PostgreSQL code:
anchor block finds subs that are active as at the calc_date (30/09/2021) (id 5 & 6), and returns the min start_date (01/01/2021)
the recursion block then looks for any earlier subs that existed within the allowed_gap, which is 30 days prior to the min_start date found in 1). id 4 meets this criteria, so the new min start_date is 01/06/2020
recursion repeats and finds two subs within the allowed_gap (01/06/2020 - 30 days). Of these subs (id 2 & 3), the new min start_date is 01/03/2020
recursion fails to find an earlier sub within the allowed_gap (01/03/2020 - 30 days)
query returns a start date of 01/03/2020 for account_id 15
Any help appreciated!
It seems the issue is related to the way SQL Server deals with recursive CTEs.
This is a type of gaps-and-islands problem, and does not actually require recursion.
There are a number of solutions, here is one. Given your requirement, there may be more efficient methods, but this should get you started.
Using LAG we identify rows which are within the specified gap of the next row
We use a running COUNT to give each consecutive set of rows an ID
We group by that ID, and take the minimum start_date, filtering out non-qualifying groups
Group again to get the minimum per account
DECLARE #allowed_gap int = 30,
#calc_date datetime = cast('2021-09-30' as date);
WITH PrevValues AS (
SELECT *,
IsStart = CASE WHEN ISNULL(LAG(end_date) OVER (PARTITION BY account_id
ORDER BY start_date), '2099-01-01') < DATEADD(day, -#allowed_gap, start_date)
THEN 1 END
FROM subscription
),
Groups AS (
SELECT *,
GroupId = COUNT(IsStart) OVER (PARTITION BY account_id
ORDER BY start_date ROWS UNBOUNDED PRECEDING)
FROM PrevValues
),
ByGroup AS (
SELECT
account_id,
GroupId,
start_date = MIN(start_date)
FROM Groups
GROUP BY account_id, GroupId
HAVING COUNT(CASE WHEN start_date <= #calc_date
and (end_date > #calc_date or end_date is null) THEN 1 END) > 0
)
SELECT
account_id,
start_date = MIN(start_date)
FROM ByGroup
GROUP BY account_id;
db<>fiddle

How to SQL Query Last Day of the Month that has transaction?

My set of data are as follows:
+------+-------+-----+--------+
| Year | Month | Day | Amount |
+------+-------+-----+--------+
| 2019 | 01 | 01 | 10 |
| 2019 | 01 | 15 | 30 |
| 2019 | 01 | 29 | 40 |
| 2019 | 02 | 02 | 50 |
| 2019 | 02 | 22 | 60 |
| 2019 | 03 | 11 | 70 |
| 2019 | 03 | 31 | 80 |
+------+-------+-----+--------+
I just want to see the last record day of each month that has transaction.
My preferred result shown should look like this:
+------+-------+--------+
| Year | Month | Amount |
+------+-------+--------+
| 2019 | 01 | 40 |
| 2019 | 02 | 60 |
| 2019 | 03 | 80 |
+------+-------+--------+
For each combination of Year and Month, you want to get the maximum Day and Amount values:
SELECT Year, Month, max(Day) as Day, max(Amount) as Amount
FROM t
GROUP BY Year, Month;
Note that every column appearing in the SELECT clause but not in the GROUP BY clause must be aggregated (with max here).
That is, assuming Amount corresponds to the total per day, which is what your example suggests.
If your table contains more than one Amount per day, then you also need to sum up the amounts per day. I'd use something like:
SELECT Year, Month, max(Day) as Day, max(Amount) as Amount
FROM (
SELECT Year, Month, Day, sum(Amount) as Amount
FROM t
GROUP BY Year, Month, Day
) as tmp
GROUP BY Year, Month;
test me (with one more amount added to your example)
Or:
SELECT Year, Month, Day, sum(Amount) as Amount
FROM (
SELECT *, rank() over(partition by Year, Month order by Day desc) as r
FROM t
) as tmp
WHERE r = 1
GROUP BY Year, Month, Day;
Mind that you want to use rank() and not row_number() in here, as you need to give the same rank identifier to ties (same day).
Of course if you want you can wrap any of the queries above with:
SELECT Year, Month, Amount
FROM (<query>) as q;
to get rid of the day column.
You can use row number to achieve this:
SELECT [Year], [Month], [Amount]
FROM (
SELECT ROW_NUMBER() OVER(PARTITION BY CONCAT(YEAR,MONTH) ORDER BY DAY DESC) rn, *
FROM table) t
WHERE rn = 1
One method is a correlated subquery:
select t.*
from t
where t.day = (select max(t2.day)
from t t2
where t2.year = t.year and t2.month = t.month
);
Another common approach uses row_number():
select t.*
from (select t.*,
row_number() over (partition by year, month order by day desc) as seqnum
from t
) t
where seqnum = 1;
you can QUALIFY use below:
SELECT ROW_NUMBER() OVER(PARTITION BY YEAR, MONTH ORDER BY DAY DESC) as rn, *
FROM yourtable
QUALIFY rn = 1
Using ROW_NUMBER() function:
with cte_order as (select year, month, amount,row_number() over(partition by month order by day desc) as identity_row from test1)
select year, month, amount from cte_order where identity_row = 1;
Using QUALIFY() function:
select year, month, amount, day, row_number() over(partition by month order by day desc) as identity_row from test1 qualify identity_row = 1;
select year,month,amount from(SELECT year,month,max(day),amount FROM t GROUP BY year,month);

Get last known record per month in BigQuery

Account balance collection, that shows the account balance of a customer at a given day:
+---------------+---------+------------+
| customer_id | value | timestamp |
+---------------+---------+------------+
| 1 | -500 | 2019-10-12 |
| 1 | -300 | 2019-10-11 |
| 1 | -200 | 2019-10-10 |
| 1 | 0 | 2019-10-09 |
| 2 | 200 | 2019-09-10 |
| 1 | 600 | 2019-09-02 |
+---------------+---------+------------+
Notice, that customer #2 had no updates to his account balance in October.
I want to get the last account balance per customer per month. If there has been no account balance update for a customer in a given month, the last known account balance should be transferred to the current month. The result should look like that:
+---------------+---------+------------+
| customer_id | value | timestamp |
+---------------+---------+------------+
| 1 | -500 | 2019-10-12 |
| 2 | 200 | 2019-10-10 |
| 2 | 200 | 2019-09-10 |
| 1 | 600 | 2019-09-02 |
+---------------+---------+------------+
Since the account balance of customer #2 was not updated in October but in September, we create a copy of the row from September changing the date to October. Any ideas how to achieve this in BigQuery?
Below is for BigQuery Standard SQL
#standardSQL
WITH customers AS (
SELECT DISTINCT customer_id FROM `project.dataset.table`
), months AS (
SELECT month FROM (
SELECT DATE_TRUNC(MIN(timestamp), MONTH) min_month, DATE_TRUNC(MAX(timestamp), MONTH) max_month
FROM `project.dataset.table`
), UNNEST(GENERATE_DATE_ARRAY(min_month, max_month, INTERVAL 1 MONTH)) month
)
SELECT customer_id,
IFNULL(value, LEAD(value) OVER(win)) value,
IFNULL(timestamp, DATE_ADD(LEAD(timestamp) OVER(win), INTERVAL DATE_DIFF(month, LEAD(month) OVER(win), MONTH) MONTH)) timestamp
FROM months, customers
LEFT JOIN (
SELECT DATE_TRUNC(timestamp, MONTH) month, customer_id,
ARRAY_AGG(STRUCT(value, timestamp) ORDER BY timestamp DESC LIMIT 1)[OFFSET(0)].*
FROM `project.dataset.table`
GROUP BY month, customer_id
) USING(month, customer_id)
WINDOW win AS (PARTITION BY customer_id ORDER BY month DESC)
if to apply to sample data from your question - as it is in below example
#standardSQL
WITH `project.dataset.table` AS (
SELECT 1 customer_id, -500 value, DATE '2019-10-12' timestamp UNION ALL
SELECT 1, -300, '2019-10-11' UNION ALL
SELECT 1, -200, '2019-10-10' UNION ALL
SELECT 2, 200, '2019-09-10' UNION ALL
SELECT 2, 100, '2019-08-11' UNION ALL
SELECT 2, 50, '2019-07-12' UNION ALL
SELECT 1, 600, '2019-09-02'
), customers AS (
SELECT DISTINCT customer_id FROM `project.dataset.table`
), months AS (
SELECT month FROM (
SELECT DATE_TRUNC(MIN(timestamp), MONTH) min_month, DATE_TRUNC(MAX(timestamp), MONTH) max_month
FROM `project.dataset.table`
), UNNEST(GENERATE_DATE_ARRAY(min_month, max_month, INTERVAL 1 MONTH)) month
)
SELECT customer_id,
IFNULL(value, LEAD(value) OVER(win)) value,
IFNULL(timestamp, DATE_ADD(LEAD(timestamp) OVER(win), INTERVAL DATE_DIFF(month, LEAD(month) OVER(win), MONTH) MONTH)) timestamp
FROM months, customers
LEFT JOIN (
SELECT DATE_TRUNC(timestamp, MONTH) month, customer_id,
ARRAY_AGG(STRUCT(value, timestamp) ORDER BY timestamp DESC LIMIT 1)[OFFSET(0)].*
FROM `project.dataset.table`
GROUP BY month, customer_id
) USING(month, customer_id)
WINDOW win AS (PARTITION BY customer_id ORDER BY month DESC)
-- ORDER BY month DESC, customer_id
result is
Row customer_id value timestamp
1 1 -500 2019-10-12
2 2 200 2019-10-10
3 1 600 2019-09-02
4 2 200 2019-09-10
5 1 null null
6 2 100 2019-08-11
7 1 null null
8 2 50 2019-07-12
The following query should mostly answer your question by creating a 'month-end' record for each customer for every month and getting the most recent balance:
with
-- Generate a set of months
month_begins as (
select dt from unnest(generate_date_array('2019-01-01','2019-12-01', interval 1 month)) dt
),
-- Get the month ends
month_ends as (
select date_sub(date_add(dt, interval 1 month), interval 1 day) as month_end_date from month_begins
),
-- Cross Join and group so we get 1 customer record for every month to account for
-- situations where customer doesn't change balance in a month
user_month_ends as (
select
customer_id,
month_end_date
from `project.dataset.table`
cross join month_ends
group by 1,2
),
-- Fan out so for each month end, you get all balances prior to month end for each customer
values_prior_to_month_end as (
select
customer_id,
value,
timestamp,
month_end_date
from `project.dataset.table`
inner join user_month_ends using(customer_id)
where timestamp <= month_end_date
),
-- Order by most recent balance before month end, even if it was more than 1+ months ago
ordered as (
select
*,
row_number() over (partition by customer_id, month_end_date order by timestamp desc) as my_row
from values_prior_to_month_end
),
-- Finally, select only the most recent record for each customer per month
final as (
select
* except(my_row)
from ordered
where my_row = 1
)
select * from final
order by customer_id, month_end_date desc
A few caveats:
I did not order results to match your desired result set, and I also kept a month-end date to illustrate the concept. You can easily change the ordering and exclude unneeded fields.
In the month_begins CTE, I set a range of months into the future, so your result set will contain the most recent balance of 'future months'. To make this a bit prettier, consider changing '2019-12-01' to 'current_date()' and your query will always return to the end of the current month.
Your timestamp field looks to be dates, so I used date logic, but you should be able to apply the same principles to use timestamp logic if your underlying fields are actual timestamps.
In your result set, I'm not sure why your 2nd row (customer 2) would have a timestamp of '2019-10-10', that seems arbitrary as customer 2 has no 2nd balance record.
I purposefully split the logic into several CTEs so I could comment on each step easier, you could definitely perform several steps in the same code block for a more condensed query.

Querying for an ID that has the most number of reads

Suppose I have a table like the one below:
+----+-----------+
| ID | TIME |
+----+-----------+
| 1 | 12-MAR-15 |
| 2 | 23-APR-14 |
| 2 | 01-DEC-14 |
| 1 | 01-DEC-15 |
| 3 | 05-NOV-15 |
+----+-----------+
What I want to do is for each year ( the year is defined as DATE), list the ID that has the highest count in that year. So for example, ID 1 occurs the most in 2015, ID 2 occurs the most in 2014, etc.
What I have for a query is:
SELECT EXTRACT(year from time) "YEAR", COUNT(ID) "ID"
FROM table
GROUP BY EXTRACT(year from time)
ORDER BY COUNT(ID) DESC;
But this query just counts how many times a year occurs, how do I fix it to highest count of an ID in that year?
Output:
+------+----+
| YEAR | ID |
+------+----+
| 2015 | 2 |
| 2012 | 2 |
+------+----+
Expected Output:
+------+----+
| YEAR | ID |
+------+----+
| 2015 | 1 |
| 2014 | 2 |
+------+----+
Starting with your sample query, the first change is simply to group by the ID as well as by the year.
SELECT EXTRACT(year from time) "YEAR" , id, COUNT(*) "TOTAL"
FROM table
GROUP BY EXTRACT(year from time), id
ORDER BY EXTRACT(year from time) DESC, COUNT(*) DESC
With that, you could find the rows you want by visual inspection (the first row for each year is the ID with the most rows).
To have the query just return the rows with the highest totals, there are several different ways to do it. You need to consider what you want to do if there are ties - do you want to see all IDs tied for highest in a year, or just an arbitrary one?
Here is one approach - if there is a tie, this should return just the lowest of the tied IDs:
WITH groups AS (
SELECT EXTRACT(year from time) "YEAR" , id, COUNT(*) "TOTAL"
FROM table
GROUP BY EXTRACT(year from time), id
)
SELECT year, MIN(id) KEEP (DENSE_RANK FIRST ORDER BY total DESC)
FROM groups
GROUP BY year
ORDER BY year DESC
You need to count per id and then apply a RANK on that count:
SELECT *
FROM
(
SELECT EXTRACT(year from time) "YEAR" , ID, COUNT(*) AS cnt
, RANK() OVER (PARTITION BY "YEAR" ORDER BY COUNT(*) DESC) AS rnk
FROM table
GROUP BY EXTRACT(year from time), ID
) dt
WHERE rnk = 1
If this return multiple rows with the same high count per year and you want just one of them randomly, you can switch to a ROW_NUMBER.
This should do what you're after, I think:
with sample_data as (select 1 id, to_date('12/03/2015', 'dd/mm/yyyy') time from dual union all
select 2 id, to_date('23/04/2014', 'dd/mm/yyyy') time from dual union all
select 2 id, to_date('01/12/2014', 'dd/mm/yyyy') time from dual union all
select 1 id, to_date('01/12/2015', 'dd/mm/yyyy') time from dual union all
select 3 id, to_date('05/11/2015', 'dd/mm/yyyy') time from dual)
-- End of creating a subquery to mimick a table called "sample_data" containing your input data.
-- See SQL below:
select yr,
id most_frequent_id,
cnt_id_yr cnt_of_most_freq_id
from (select to_char(time, 'yyyy') yr,
id,
count(*) cnt_id_yr,
dense_rank() over (partition by to_char(time, 'yyyy') order by count(*) desc) dr
from sample_data
group by to_char(time, 'yyyy'),
id)
where dr = 1;
YR MOST_FREQUENT_ID CNT_OF_MOST_FREQ_ID
---- ---------------- -------------------
2014 2 2
2015 1 2

Select distinct users group by time range

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