Total Number of Records per Week - sql

I have a Postgres 9.1 database. I am trying to generate the number of records per week (for a given date range) and compare it to the previous year.
I have the following code used to generate the series:
select generate_series('2013-01-01', '2013-01-31', '7 day'::interval) as series
However, I am not sure how to join the counted records to the dates generated.
So, using the following records as an example:
Pt_ID exam_date
====== =========
1 2012-01-02
2 2012-01-02
3 2012-01-08
4 2012-01-08
1 2013-01-02
2 2013-01-02
3 2013-01-03
4 2013-01-04
1 2013-01-08
2 2013-01-10
3 2013-01-15
4 2013-01-24
I wanted to have the records return as:
series thisyr lastyr
=========== ===== =====
2013-01-01 4 2
2013-01-08 3 2
2013-01-15 1 0
2013-01-22 1 0
2013-01-29 0 0
Not sure how to reference the date range in the subsearch. Thanks for any assistance.

The simple approach would be to solve this with a CROSS JOIN like demonstrated by #jpw. However, there are some hidden problems:
The performance of an unconditional CROSS JOIN deteriorates quickly with growing number of rows. The total number of rows is multiplied by the number of weeks you are testing for, before this huge derived table can be processed in the aggregation. Indexes can't help.
Starting weeks with January 1st leads to inconsistencies. ISO weeks might be an alternative. See below.
All of the following queries make heavy use of an index on exam_date. Be sure to have one.
Only join to relevant rows
Should be much faster:
SELECT d.day, d.thisyr
, count(t.exam_date) AS lastyr
FROM (
SELECT d.day::date, (d.day - '1 year'::interval)::date AS day0 -- for 2nd join
, count(t.exam_date) AS thisyr
FROM generate_series('2013-01-01'::date
, '2013-01-31'::date -- last week overlaps with Feb.
, '7 days'::interval) d(day) -- returns timestamp
LEFT JOIN tbl t ON t.exam_date >= d.day::date
AND t.exam_date < d.day::date + 7
GROUP BY d.day
) d
LEFT JOIN tbl t ON t.exam_date >= d.day0 -- repeat with last year
AND t.exam_date < d.day0 + 7
GROUP BY d.day, d.thisyr
ORDER BY d.day;
This is with weeks starting from Jan. 1st like in your original. As commented, this produces a couple of inconsistencies: Weeks start on a different day each year and since we cut off at the end of the year, the last week of the year consists of just 1 or 2 days (leap year).
The same with ISO weeks
Depending on requirements, consider ISO weeks instead, which start on Mondays and always span 7 days. But they cross the border between years. Per documentation on EXTRACT():
week
The number of the week of the year that the day is in. By definition (ISO 8601), weeks start on Mondays and the first week of a
year contains January 4 of that year. In other words, the first
Thursday of a year is in week 1 of that year.
In the ISO definition, it is possible for early-January dates to be part of the 52nd or 53rd week of the previous year, and for
late-December dates to be part of the first week of the next year. For
example, 2005-01-01 is part of the 53rd week of year 2004, and
2006-01-01 is part of the 52nd week of year 2005, while 2012-12-31 is
part of the first week of 2013. It's recommended to use the isoyear
field together with week to get consistent results.
Above query rewritten with ISO weeks:
SELECT w AS isoweek
, day::text AS thisyr_monday, thisyr_ct
, day0::text AS lastyr_monday, count(t.exam_date) AS lastyr_ct
FROM (
SELECT w, day
, date_trunc('week', '2012-01-04'::date)::date + 7 * w AS day0
, count(t.exam_date) AS thisyr_ct
FROM (
SELECT w
, date_trunc('week', '2013-01-04'::date)::date + 7 * w AS day
FROM generate_series(0, 4) w
) d
LEFT JOIN tbl t ON t.exam_date >= d.day
AND t.exam_date < d.day + 7
GROUP BY d.w, d.day
) d
LEFT JOIN tbl t ON t.exam_date >= d.day0 -- repeat with last year
AND t.exam_date < d.day0 + 7
GROUP BY d.w, d.day, d.day0, d.thisyr_ct
ORDER BY d.w, d.day;
January 4th is always in the first ISO week of the year. So this expression gets the date of Monday of the first ISO week of the given year:
date_trunc('week', '2012-01-04'::date)::date
Simplify with EXTRACT()
Since ISO weeks coincide with the week numbers returned by EXTRACT(), we can simplify the query. First, a short and simple form:
SELECT w AS isoweek
, COALESCE(thisyr_ct, 0) AS thisyr_ct
, COALESCE(lastyr_ct, 0) AS lastyr_ct
FROM generate_series(1, 5) w
LEFT JOIN (
SELECT EXTRACT(week FROM exam_date)::int AS w, count(*) AS thisyr_ct
FROM tbl
WHERE EXTRACT(isoyear FROM exam_date)::int = 2013
GROUP BY 1
) t13 USING (w)
LEFT JOIN (
SELECT EXTRACT(week FROM exam_date)::int AS w, count(*) AS lastyr_ct
FROM tbl
WHERE EXTRACT(isoyear FROM exam_date)::int = 2012
GROUP BY 1
) t12 USING (w);
Optimized query
The same with more details and optimized for performance
WITH params AS ( -- enter parameters here, once
SELECT date_trunc('week', '2012-01-04'::date)::date AS last_start
, date_trunc('week', '2013-01-04'::date)::date AS this_start
, date_trunc('week', '2014-01-04'::date)::date AS next_start
, 1 AS week_1
, 5 AS week_n -- show weeks 1 - 5
)
SELECT w.w AS isoweek
, p.this_start + 7 * (w - 1) AS thisyr_monday
, COALESCE(t13.ct, 0) AS thisyr_ct
, p.last_start + 7 * (w - 1) AS lastyr_monday
, COALESCE(t12.ct, 0) AS lastyr_ct
FROM params p
, generate_series(p.week_1, p.week_n) w(w)
LEFT JOIN (
SELECT EXTRACT(week FROM t.exam_date)::int AS w, count(*) AS ct
FROM tbl t, params p
WHERE t.exam_date >= p.this_start -- only relevant dates
AND t.exam_date < p.this_start + 7 * (p.week_n - p.week_1 + 1)::int
-- AND t.exam_date < p.next_start -- don't cross over into next year
GROUP BY 1
) t13 USING (w)
LEFT JOIN ( -- same for last year
SELECT EXTRACT(week FROM t.exam_date)::int AS w, count(*) AS ct
FROM tbl t, params p
WHERE t.exam_date >= p.last_start
AND t.exam_date < p.last_start + 7 * (p.week_n - p.week_1 + 1)::int
-- AND t.exam_date < p.this_start
GROUP BY 1
) t12 USING (w);
This should be very fast with index support and can easily be adapted to intervals of choice.
The implicit JOIN LATERAL for generate_series() in the last query requires Postgres 9.3.
SQL Fiddle.

Using across joinshould work, I'm just going to paste the markdown output from SQL Fiddle below. It would seem that your sample output is incorrect for series 2013-01-08: the thisyr should be 2, not 3. This might not be the best way to do this though, my Postgresql knowledge leaves a lot to be desired.
SQL Fiddle
PostgreSQL 9.2.4 Schema Setup:
CREATE TABLE Table1
("Pt_ID" varchar(6), "exam_date" date);
INSERT INTO Table1
("Pt_ID", "exam_date")
VALUES
('1', '2012-01-02'),('2', '2012-01-02'),
('3', '2012-01-08'),('4', '2012-01-08'),
('1', '2013-01-02'),('2', '2013-01-02'),
('3', '2013-01-03'),('4', '2013-01-04'),
('1', '2013-01-08'),('2', '2013-01-10'),
('3', '2013-01-15'),('4', '2013-01-24');
Query 1:
select
series,
sum (
case
when exam_date
between series and series + '6 day'::interval
then 1
else 0
end
) as thisyr,
sum (
case
when exam_date + '1 year'::interval
between series and series + '6 day'::interval
then 1 else 0
end
) as lastyr
from table1
cross join generate_series('2013-01-01', '2013-01-31', '7 day'::interval) as series
group by series
order by series
Results:
| SERIES | THISYR | LASTYR |
|--------------------------------|--------|--------|
| January, 01 2013 00:00:00+0000 | 4 | 2 |
| January, 08 2013 00:00:00+0000 | 2 | 2 |
| January, 15 2013 00:00:00+0000 | 1 | 0 |
| January, 22 2013 00:00:00+0000 | 1 | 0 |
| January, 29 2013 00:00:00+0000 | 0 | 0 |

Related

Get days of the week from a date range in Postgres

So I have the following table :
id end_date name number_of_days start_date
1 "2022-01-01" holiday1 1 "2022-01-01"
2 "2022-03-20" holiday2 1 "2022-03-20"
3 "2022-04-09" holiday3 1 "2022-04-09"
4 "2022-05-01" holiday4 1 "2022-05-01"
5 "2022-05-04" holiday5 3 "2022-05-02"
6 "2022-07-12" holiday6 9 "2022-07-20"
I want to check if a week falls in a holiday range.
So far I can select the holidays that overlap with my choosen week( week_start_date, week_end_date) , but i cant get the exact days in which the overlap happens.
this is the query i'm using, i want to add a mechanism to detect the DAYS OF THE WEEK IN WHICH THE OVERLAP HAPPENS
SELECT * FROM holidays
where daterange(CAST(start_date AS date), CAST(end_date as date), '[]') && daterange('2022-07-18', '2022-07-26','[]')
THE CURRENT QUERY RETURNS THE OVERLLAPPING HOLIDA, (id = 6), however i'm trying to get the exact DAYS OF THE WEEK in which the overlap happens ( in this case, it should be monday,tuesday , wednesday)
You can use the * operator with tsranges, generate a series of dates with the lower and upper dates and finally with to_char print the days of the week, e.g.
SELECT
id, name, start_date, end_date, array_agg(dow) AS days
FROM (
SELECT *,
trim(
to_char(
generate_series(lower(overlap), upper(overlap),'1 day'),
'Day')) AS dow
FROM holidays
CROSS JOIN LATERAL (SELECT tsrange(start_date,end_date) *
tsrange('2022-07-18', '2022-07-26')) t (overlap)
WHERE tsrange(start_date,end_date) && tsrange('2022-07-18', '2022-07-26')) j
GROUP BY id,name,start_date,end_date,number_of_days;
id | name | start_date | end_date | days
----+----------+------------+------------+----------------------------
6 | holiday6 | 2022-07-12 | 2022-07-20 | {Monday,Tuesday,Wednesday}
(1 row)
Demo: db<>fiddle

Get count of susbcribers for each month in current year even if count is 0

I need to get the count of new subscribers each month of the current year.
DB Structure: Subscriber(subscriber_id, create_timestamp, ...)
Expected result:
date | count
-----------+------
2021-01-01 | 3
2021-02-01 | 12
2021-03-01 | 0
2021-04-01 | 8
2021-05-01 | 0
I wrote the following query:
SELECT
DATE_TRUNC('month',create_timestamp)
AS create_timestamp,
COUNT(subscriber_id) AS count
FROM subscriber
GROUP BY DATE_TRUNC('month',create_timestamp);
Which works but does not include months where the count is 0. It's only returning the ones that are existing in the table. Like:
"2021-09-01 00:00:00" 3
"2021-08-01 00:00:00" 9
First subquery is used for retrieving year wise each month row then LEFT JOIN with another subquery which is used to retrieve month wise total_count. COALESCE() is used for replacing NULL value to 0.
-- PostgreSQL (v11)
SELECT t.cdate
, COALESCE(p.total_count, 0) total_count
FROM (select generate_series('2021-01-01'::timestamp, '2021-12-15', '1 month') as cdate) t
LEFT JOIN (SELECT DATE_TRUNC('month',create_timestamp) create_timestamp
, SUM(subscriber_id) total_count
FROM subscriber
GROUP BY DATE_TRUNC('month',create_timestamp)) p
ON t.cdate = p.create_timestamp
Please check from url https://dbfiddle.uk/?rdbms=postgres_11&fiddle=20dcf6c1784ed0d9c5772f2487bcc221
get the count of new subscribers each month of the current year
SELECT month::date, COALESCE(s.count, 0) AS count
FROM generate_series(date_trunc('year', LOCALTIMESTAMP)
, date_trunc('year', LOCALTIMESTAMP) + interval '11 month'
, interval '1 month') m(month)
LEFT JOIN (
SELECT date_trunc('month', create_timestamp) AS month
, count(*) AS count
FROM subscriber
GROUP BY 1
) s USING (month);
db<>fiddle here
That's assuming every row is a "new subscriber". So count(*) is simplest and fastest.
See:
Join a count query on generate_series() and retrieve Null values as '0'
Generating time series between two dates in PostgreSQL

SQL - How to group/count items by age and status on every date of a year?

I am trying to build a query from multi-year data set (tickets table) of support tickets, with relevant columns of ticked_id, status, created_on date and closed_on date for each ticket. There is also a generic dates table I can join/query to a list of dates.
I'd like to create a "burn down" chart for this year that displays the number of open tickets that were at least a year old on any given date this year. I have been able to create tables that use a sum(case... statement to group by a date - for example to show how many tickets were created on a given week - but I can't figure out how to group by every day or week this year the number of tickets that were open on that day and at least a year old.
Any help is appreciated.
Example Data:
ticket_id | status | created_on | closed_on
--------------------------------------------
1 open 1/5/2019
2 open 1/26/2019
3 closed 1/28/2019 2/1/2020
4 open 6/1/2019
5 closed 6/5/2019 1/1/2020
Example Results I Seek:
Date (2020) | Count of Year+ Aged Tickets
------------------------------------------------
1/1/2020 0
1/2/2020 0
1/3/2020 0
1/4/2020 0
1/5/2020 1
1/6/2020 1
... (skipping dates here but want all dates in results)...
1/25/2020 1
1/26/2020 2
1/27/2020 2
1/28/2020 3
1/29/2020 3
1/30/2020 3
1/31/2020 3
2/1/2020 2
... (skipping dates here but want all dates up to current date in results)...
ticket_id 1 reached one year of age on 1/5/2020 and is still open
(remains in count)
ticket_id 2 reached one year of age on 1/26/2020 and is still open (remains in count)
ticket_id 3 reached one year of age on 1/28/2020 and was still open, adding to the count, but was closed on 2/1/2020, reducing the count
ticket_id 4 will only add to the count if it is still open on 6/1/2020, but not if it is closed before then
ticket_id 5 will never appear in the count because it never reached one year of age and is closed
One option is to build a sequential list of dates, then bring the table with a ‘left join` and conditional logic, and finally aggregate.
This would give the results you want for year 2020.
select d.dt, count(t.ticket_id) no_tickets
from (
select date '2020-01-01' + I * interval '1 day' dt
from generate_series(0, 365) i
) d
left join mytable t
on t.created_on + interval '1 year' <= d.dt
and (
t.closed_on is null
or t.closed_on > d.dt
)
group by d.dt
If your version of Redshift does not support generate_series(), you can emulate it a custom number table, or with row_number() against a large table (say mylargetable):
select d.dt, count(t.ticket_id) no_tickets
from (
select date '2020-01-01' + row_number() over(order by 1) * interval '1 day' dt
from mylargetable
) d
left join mytable t
on t.created_on + interval '1 year' <= d.dt
and (
t.closed_on is null
or t.closed_on > d.dt
)
where d.dt < date '2021-01-01'
group by d.dt
If ticket_id is unique then you can do this to get all ticket at least 1 year old
select ticket_id, created_on , status where status = 'open' and created_on <= dateadd(year,-1,getdate())
if you want to count number of ticket per month then
select count(ticket_id), month(created_on) , status where status = 'open' and created_on <= dateadd(year,-1,getdate())
group by month(created_on)

For LOOP in PostgreSQL

I have a table with the following columns:
(client_id, start_contract_date, end_contract_date)
Every client has a start_contract_date but some clients have a NULL for end_contract_date since they may still be active today.
If we check for a certain date D, a client is active if D is between start_contract_date and end_contract_date (or start_contract_date <= D, if end_contract_date is NULL.)
I want to count, for each month of each year, over 2016 until today, how many customers are active. My problem is that I do not know how to LOOP on the months and years.
I have a partial solution. I can count how many active clients for a specific month of a specific year.
SELECT 2016 as year , 7 as month, count(id_client)
FROM table
WHERE
EXTRACT(year from start_contract_date)<=2016
AND EXTRACT(month from start_contract_date)<=7
AND (EXTRACT(year from end_contract_date)>=2016 OR end_contract_date IS NULL)
AND (EXTRACT(month from end_contract_date)>=7 OR end_contract_date IS NULL)
;
So, how can I run a nested for loop that would be something like
FOR y IN 2016..2017
FOR m IN 1..12
I want the output to be
Year , Month , Count
2016 , 1 , 234
2016 , 2 , 54
…
2017 , 12 , 543
Use the function generate_series() to generate arbitrary series of months, e.g.:
select extract(year from d) as year, extract(month from d) as month
from generate_series('2017-11-01'::date, '2018-02-01', '1 month') d
year | month
------+-------
2017 | 11
2017 | 12
2018 | 1
2018 | 2
(4 rows)
Use the above and the function date_trunc() to extract year-month value from dates:
select extract(year from d) as year, extract(month from d) as month, count(id_client)
from generate_series('2016-01-01'::date, '2019-03-01', '1 month') d
left join my_table
on date_trunc('month', start_contract_date) <= date_trunc('month', d)
and (end_contract_date is null or date_trunc('month', end_contract_date) >= date_trunc('month', d))
group by d
order by d
Note also that the conditions in your query contain logical error.

Add Missing monthly dates in a timeseries data in Postgresql

I have monthly time series data in table where dates are as a last day of month. Some of the dates are missing in the data. I want to insert those dates and put zero value for other attributes.
Table is as follows:
id report_date price
1 2015-01-31 40
1 2015-02-28 56
1 2015-04-30 34
2 2014-05-31 45
2 2014-08-31 47
I want to convert this table to
id report_date price
1 2015-01-31 40
1 2015-02-28 56
1 2015-03-31 0
1 2015-04-30 34
2 2014-05-31 45
2 2014-06-30 0
2 2014-07-31 0
2 2014-08-31 47
Is there any way we can do this in Postgresql?
Currently we are doing this in Python. As our data is growing day by day and its not efficient to handle I/O just for one task.
Thank you
You can do this using generate_series() to generate the dates and then left join to bring in the values:
with m as (
select id, min(report_date) as minrd, max(report_date) as maxrd
from t
group by id
)
select m.id, m.report_date, coalesce(t.price, 0) as price
from (select m.*, generate_series(minrd, maxrd, interval '1' month) as report_date
from m
) m left join
t
on m.report_date = t.report_date;
EDIT:
Turns out that the above doesn't quite work, because adding months to the end of month doesn't keep the last day of the month.
This is easily fixed:
with t as (
select 1 as id, date '2012-01-31' as report_date, 10 as price union all
select 1 as id, date '2012-04-30', 20
), m as (
select id, min(report_date) - interval '1 day' as minrd, max(report_date) - interval '1 day' as maxrd
from t
group by id
)
select m.id, m.report_date, coalesce(t.price, 0) as price
from (select m.*, generate_series(minrd, maxrd, interval '1' month) + interval '1 day' as report_date
from m
) m left join
t
on m.report_date = t.report_date;
The first CTE is just to generate sample data.
This is a slight improvement over Gordon's query which fails to get the last date of a month in some cases.
Essentially you generate all the month end dates between the min and max date for each id (using generate_series) and left join on this generated table to show the missing dates with 0 price.
with minmax as (
select id, min(report_date) as mindt, max(report_date) as maxdt
from t
group by id
)
select m.id, m.report_date, coalesce(t.price, 0) as price
from (select *,
generate_series(date_trunc('MONTH',mindt+interval '1' day),
date_trunc('MONTH',maxdt+interval '1' day),
interval '1' month) - interval '1 day' as report_date
from minmax
) m
left join t on m.report_date = t.report_date
Sample Demo