I have a problem with a SQL query. I have a list of dates in one column, I would like to create pairs of dates. The dates are sequenced, so I have to match the first date with the second and create a record, then the third date with the fourth date and create a record etc .. as in the following example:
ID DATA
50 10/04/2019
50 12/04/2019
50 13/04/2019
50 17/04/2019
50 18/04/2019
50 19/04/2019
ID DATA_START DATA_END
50 10/04/2019 12/04/2019
50 13/04/2019 17/04/2019
50 18/04/2019 19/04/2019
Thanks very much everyone for the help
You should mark your rows that should be grouped together (into single row) and which date will have which role (start or end).
Here's the code:
with a as (
/*Source data*/
select 50 as id, convert(date, '2019-04-10', 23) as dt union all
select 50 as id, convert(date, '2019-04-12', 23) as dt union all
select 50 as id, convert(date, '2019-04-13', 23) as dt union all
select 50 as id, convert(date, '2019-04-17', 23) as dt union all
select 50 as id, convert(date, '2019-04-18', 23) as dt union all
select 50 as id, convert(date, '2019-04-19', 23) as dt
)
select
id,
[1] as dt_start,
[0] as dt_end
from (
select
id,
dt,
/*
the first row (with modulo = 1) is date from
and the second row (with modulo = 0) is date to
*/
(row_number() over(partition by id order by dt)) % 2 as dt_role,
/*Integer division by 2 will group rows together*/
(row_number() over(partition by id order by dt) + 1) / 2 as dt_group
from a
) as s
pivot (
max(dt) for dt_role in ([0], [1])
) as p
GO
id | dt_start | dt_end
-: | :--------- | :---------
50 | 2019-04-10 | 2019-04-12
50 | 2019-04-13 | 2019-04-17
50 | 2019-04-18 | 2019-04-19
db<>fiddle here
Related
I have records of No. of calls coming to a call center. When a call comes into a call center a ticket is open.
So, let's say ticket 1 (T1) is open on 8/1/19 and it stays open till 8/5/19. So, if a person ran a query everyday then on 8/1 it will show 1 ticket open...same think on day 2 till day 5....I want to get records by day to see how many tickets were open for each day.....
In short, Frequency Distribution by Day.
Ticket Open_date Close_date
T1 8/1/2019 8/5/2019
T2 8/1/2019 8/6/2019
Result:
Result
Date # Tickets_Open
8/1/2019 2
8/2/2019 2
8/3/2019 2
8/4/2019 2
8/5/2019 2
8/6/2019 1
8/7/2019 0
8/8/2019 0
8/9/2019 0
8/10/2019 0
We can handle your requirement via the use of a calendar table, which stores all dates covering the full range in your data set.
WITH dates AS (
SELECT '2019-08-01' AS dt UNION ALL
SELECT '2019-08-02' UNION ALL
SELECT '2019-08-03' UNION ALL
SELECT '2019-08-04' UNION ALL
SELECT '2019-08-05' UNION ALL
SELECT '2019-08-06' UNION ALL
SELECT '2019-08-07' UNION ALL
SELECT '2019-08-08' UNION ALL
SELECT '2019-08-09' UNION ALL
SELECT '2019-08-10'
)
SELECT
d.dt,
COUNT(t.Open_date) AS num_tickets_open
FROM dates d
LEFT JOIN tickets t
ON d.dt BETWEEN t.Open_date AND t.Close_date
GROUP BY
d.dt;
Note that in practice if you expect to have this reporting requirement in the long term, you might want to replace the dates CTE above with a bona-fide table of dates.
This solution generates the list of dates from the tickets table using CTE recursion and calculates the count:
WITH Tickets(Ticket, Open_date, Close_date) AS
(
SELECT "T1", "8/1/2019", "8/5/2019"
UNION ALL
SELECT "T2", "8/1/2019", "8/6/2019"
),
Ticket_dates(Ticket, Dates) as
(
SELECT t1.Ticket, CONVERT(DATETIME, t1.Open_date)
FROM Tickets t1
UNION ALL
SELECT t1.Ticket, DATEADD(dd, 1, CONVERT(DATETIME, t1.Dates))
FROM Ticket_dates t1
inner join Tickets t2 on t1.Ticket = t2.Ticket
where DATEADD(dd, 1, CONVERT(DATETIME, t1.Dates)) <= CONVERT(DATETIME, t2.Close_date)
)
SELECT CONVERT(varchar, Dates, 1), count(*)
FROM Ticket_dates
GROUP by Dates
ORDER by Dates
A "general purpose" trick is to generate a series of numbers, which can be done using CTE's but there are many alternatives, and from that create the needed range of dates. Once that exists then you can left join your ticket data to this and then count by date.
CREATE TABLE mytable(
Ticket VARCHAR(8) NOT NULL PRIMARY KEY
,Open_date DATE NOT NULL
,Close_date DATE NOT NULL
);
INSERT INTO mytable(Ticket,Open_date,Close_date) VALUES ('T1','8/1/2019','8/5/2019');
INSERT INTO mytable(Ticket,Open_date,Close_date) VALUES ('T2','8/1/2019','8/6/2019');
Also note I am using a cross apply in this example to "attach" the min and max dates of your tickets to each numbered row. You would need to include your own logic on what data to select here.
;WITH
cteDigits AS (
SELECT 0 AS digit UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL
SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9
)
, cteTally AS (
SELECT
[1s].digit
+ [10s].digit * 10
+ [100s].digit * 100 /* add more like this as needed */
AS num
FROM cteDigits [1s]
CROSS JOIN cteDigits [10s]
CROSS JOIN cteDigits [100s] /* add more like this as needed */
)
select
n.num + 1 rownum
, dateadd(day,n.num,ca.min_date) as on_date
, count(t.Ticket) as tickets_open
from cteTally n
cross apply (select min(Open_date), max(Close_date) from mytable) ca (min_date, max_date)
left join mytable t on dateadd(day,n.num,ca.min_date) between t.Open_date and t.Close_date
where dateadd(day,n.num,ca.min_date) <= ca.max_date
group by
n.num + 1
, dateadd(day,n.num,ca.min_date)
order by
rownum
;
result:
+--------+---------------------+--------------+
| rownum | on_date | tickets_open |
+--------+---------------------+--------------+
| 1 | 01.08.2019 00:00:00 | 2 |
| 2 | 02.08.2019 00:00:00 | 2 |
| 3 | 03.08.2019 00:00:00 | 2 |
| 4 | 04.08.2019 00:00:00 | 2 |
| 5 | 05.08.2019 00:00:00 | 2 |
| 6 | 06.08.2019 00:00:00 | 1 |
+--------+---------------------+--------------+
I have the following data set:
I want to create a new column that sums the last 7 days of sales. So the query result should look be the following:
Pls help
Thanks!
In standard SQL, you would use a window function -- assuming you have data for each day:
select t.*,
sum(sales) over (partition by itemid order by date rows between 6 preceding and current row) as sales_7
from t;
use sum() aggregate function and group by
select country,itemid,year,monthnumber,week sum(sales) as sales_last_7days from your_table
where date>=DATEADD(day, -7, getdate()) and date< getdate()
group by country,itemid,year,monthnumber,week
with window:
select (list other columns here), sum(sum(sales)) over
(partition by week
order by day
rows between 6 preceding and current row)
from table
group by date, week;
note that week doesen't change group by beacause a date is reffered to one week only, but it is needed in window.
Seems you are working with SQL Server if so, then you can use apply :
select t.*, t1.[last7day]
from table t outer apply
(select sum(t1.sales) as [last7day]
from table t1
where t.itemid = t1.itemid and
t1.date <= dateadd(day, -6, t.dt)
) t1;
If you don't have exactly one day for each row, for example if you have a list of transactions...
The below example completely confused me the first time I saw it, so I've tried to comment as much as I can to explain what's happening.
Suppose we have a table tbl with date column dt and amount column amt, and for each date in tbl we want to return a rolling sum of the amount from the current day and the past 6 days.
select distinct -- see note after code on what this distinct is doing.
dt
, ( -- Has to be in brackets to denote we're returning 1 value per row.
-- for each row of T1:
select sum(b.amt) -- the sum of amounts in T2. The where clause will restrict which rows in T2 will be summed.
from tbl T2
where T2.dt between T1.dt - 6 and T1.dt -- for each row in T1, give me all rows in T2 where the date is between 6 days before this T1 row's date and T1 row's date, giving us our rolling sum
-- WARNING: CHECK YOUR VERSION OF SQL FOR HOW TO SUBTRACT DAYS FROM A DATE, I'VE MADE IT (T1.dt - 6) FOR SIMPLICITY
-- we don't need a group by, because we're returning one value for each row in T1
)
from tbl T1
We have a main version of tbl, aliased T1. We then have a secondary table, aliased T2. For each row in T1, we're going to ask for a set of rows in T2 that we're going to sum before giving it to our main query.
To understand what's happening, run the code without the distinct. You'll notice that we have the same number of rows as in tbl, because the T2 statement is happening for every row in T1.
Notes:
If you have any days for which no rows exist in your table you will not get a calculation for this day. To be certain this doesn't happen, join your table to a table containing a distinct list of consecutive dates, and use this as your date column.
If you have nulls in your amount column the calculation will still work, but if the rolling average contains only nulls you will have null instead of 0 as your result. If that troubles you convert all your nulls to zero's before (or after) you use the query.
The beginning of the period will have a 'ramp up'. But this would be the same whatever method you use to do a rolling sum. If it bothers you, don't return the first 6 days.
Finally a worked example if you're playing along at home using SQL Server:
with tbl as (
-- a list of transactions from 1.10.2019 to 14.10.2019
select cast('2019-10-01' as date) dt, 1 amt
union select cast('2019-10-02' as date), 4
union select cast('2019-10-01' as date), 10
union select cast('2019-10-03' as date), 3
union select cast('2019-10-04' as date), 20
union select cast('2019-10-04' as date), 2
union select cast('2019-10-04' as date), 12
union select cast('2019-10-04' as date), 17
union select cast('2019-10-05' as date), null -- a whole week of null values because we all had the week off... I hope this data wasn't important
union select cast('2019-10-06' as date), null
union select cast('2019-10-07' as date), null
union select cast('2019-10-08' as date), null
union select cast('2019-10-09' as date), null
union select cast('2019-10-10' as date), null
union select cast('2019-10-10' as date), null
union select cast('2019-10-10' as date), null
union select cast('2019-10-11' as date), null
union select cast('2019-10-12' as date), 1
union select cast('2019-10-12' as date), 1
union select cast('2019-10-12' as date), 1
union select cast('2019-10-12' as date), 1
union select cast('2019-10-12' as date), 1
union select cast('2019-10-12' as date), 1
union select cast('2019-10-13' as date), 2
union select cast('2019-10-14' as date), 1000
)
select distinct
a.dt
, (
select sum(b.amt)
from tbl b
where b.dt between dateadd(dd, -6, a.dt) and a.dt
) past_7_days_amt
from tbl a
Returns:
+------------+-----------------+
| dt | past_7_days_amt |
+------------+-----------------+
| 2019-10-01 | 11 |
| 2019-10-02 | 15 |
| 2019-10-03 | 18 |
| 2019-10-04 | 69 |
| 2019-10-05 | 69 |
| 2019-10-06 | 69 |
| 2019-10-07 | 69 |
| 2019-10-08 | 58 |
| 2019-10-09 | 54 |
| 2019-10-10 | 51 |
| 2019-10-11 | NULL |
| 2019-10-12 | 1 |
| 2019-10-13 | 3 |
| 2019-10-14 | 1003 |
+------------+-----------------+
I have a problem with a query in Oracle.
My table contains all of the loan applications from last year. Some of the customers have more than one application. I want to aggregate those applications as follows:
For each customer, I want to find his first application (let's call it A) in the last year and then I want to find out what was the last application in 30 days interval, counting from the first application (say B is the last one). Next, I need to find the application following B and again find for it the last one in 30 days interval, as in the previous step. What I want as the result is the table with the latest and earliest applications on each customer's interval. It is also possible that the first one is the same as the last one.
How could I do this in Oracle without plsql? Is this possible? Should I use cumulative sums of time intervals for it? (but then the starting point for each sum depends on the counted sum..)
Let's say the table has a following form:
application_id (unique) | customer_id (not unique) | create_date
1 1 2017-01-02 <- first
2 1 2017-01-10 <- middle
3 1 2017-01-30 <- last
4 1 2017-05-02 <- first and last
5 1 2017-06-02 <- first
6 1 2017-06-30 <- middle
7 1 2017-06-30 <- middle
8 1 2017-07-01 <- last
What I expect is:
application_id (unique) | customer_id (not unique) | create_date
1 1 2017-01-02 <- first
3 1 2017-01-30 <- last
4 1 2017-05-02 <- first and last
5 1 2017-06-02 <- first
8 1 2017-07-01 <- last
Thanks in advance for help.
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE table_name ( application_id, customer_id, create_date ) AS
SELECT 1, 1, DATE '2017-01-02' FROM DUAL UNION ALL -- <- first
SELECT 2, 1, DATE '2017-01-10' FROM DUAL UNION ALL -- <- middle
SELECT 3, 1, DATE '2017-01-30' FROM DUAL UNION ALL -- <- last
SELECT 4, 1, DATE '2017-05-02' FROM DUAL UNION ALL -- <- first and last
SELECT 5, 1, DATE '2017-06-02' FROM DUAL UNION ALL -- <- first
SELECT 6, 1, DATE '2017-06-30' FROM DUAL UNION ALL -- <- middle
SELECT 7, 1, DATE '2017-06-30' FROM DUAL UNION ALL -- <- middle
SELECT 8, 1, DATE '2017-07-01' FROM DUAL -- <- last
Query 1:
WITH data ( application_id, customer_id, create_date, first_date, grp ) AS (
SELECT t.application_id,
t.customer_id,
t.create_date,
t.create_date,
1
FROM table_name t
WHERE application_id = 1
UNION ALL
SELECT t.application_id,
t.customer_id,
t.create_date,
CASE WHEN t.create_date <= d.first_date + INTERVAL '30' DAY
THEN d.first_date
ELSE t.create_date
END,
CASE WHEN t.create_date <= d.first_date + INTERVAL '30' DAY
THEN grp
ELSE grp + 1
END
FROM data d
INNER JOIN table_name t
ON ( d.customer_id = t.customer_id
AND d.application_id + 1 = t.application_id )
)
SELECT application_id,
customer_id,
create_date,
grp
FROM (
SELECT d.*,
ROW_NUMBER() OVER ( PARTITION BY customer_id, grp ORDER BY create_date ASC ) AS rn_a,
ROW_NUMBER() OVER ( PARTITION BY customer_id, grp ORDER BY create_date DESC ) AS rn_d
FROM data d
)
WHERE rn_a = 1
OR rn_d = 1
Results:
| APPLICATION_ID | CUSTOMER_ID | CREATE_DATE | GRP |
|----------------|-------------|----------------------|-----|
| 1 | 1 | 2017-01-02T00:00:00Z | 1 |
| 3 | 1 | 2017-01-30T00:00:00Z | 1 |
| 4 | 1 | 2017-05-02T00:00:00Z | 2 |
| 5 | 1 | 2017-06-02T00:00:00Z | 3 |
| 8 | 1 | 2017-07-01T00:00:00Z | 3 |
Currently I have data in a table as shown below:
date id value
1-Jan-13 1 100
2-Jan-13 1 100
3-Jan-13 1 100
4-Jan-13 1 200
5-Jan-13 1 200
6-Jan-13 1 100
7-Jan-13 1 100
I am trying to group the records based on the id and val and version records with startdate and end date .
Desired output:
start date end date id value
1-Jan-13 3-Jan-13 1 100
4-Jan-13 5-Jan-13 1 200
6-Jan-13 7-Jan-13 1 100
I'm not an expert in Teradata but you most likely, since windowing functions are supported (specifically ROW_NUMBER), be able to do something like this
SELECT MIN(date) start_date, MAX(date) end_date, id, value
FROM
(
SELECT date, id, value,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY date) -
ROW_NUMBER() OVER (PARTITION BY id, value ORDER BY date) island
FROM table1
) q
GROUP BY id, value, island
ORDER BY start_date, end_date
Sample output:
| START_DATE | END_DATE | ID | VALUE |
|------------|------------|----|-------|
| 2013-01-01 | 2013-01-03 | 1 | 100 |
| 2013-01-04 | 2013-01-05 | 1 | 200 |
| 2013-01-06 | 2013-01-07 | 1 | 100 |
Here is SQLFiddle demo (It's a SQL Server demo, but should work as expected in Teradata)
The ROW_NUMBER version can be further simplified: modified SQL Fiddle
For Teradata:
SELECT
id,val,MIN(dt),MAX(dt)
FROM
(
SELECT
id,val,dt,
dt - ROW_NUMBER() OVER (PARTITION BY id ORDER BY val, dt) AS dummy
FROM table1
) AS dt
GROUP BY 1,2,dummy
And there are some hardly known functions in TD13.10 for processing time series data:
WITH cte(id,val,pd) AS
(
SELECT id, val, PERIOD(dt, dt+1) AS pd
FROM table1
)
SELECT
id, val,
BEGIN(pd) AS start_dt,
LAST(pd) AS end_dt
FROM
TABLE (TD_NORMALIZE_MEET
(NEW VARIANT_TYPE(cte.id,cte.val)
,cte.pd)
RETURNS (id INTEGER
,val INTEGER
,pd PERIOD(DATE)
,Nrm_Count INTEGER)
HASH BY id
LOCAL ORDER BY id, val, pd
) A
ORDER BY start_dt, end_dt
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