I suspect I require some sort of windowing function to do this. I have the following item data as an example:
count | date
------+-----------
3 | 2017-09-15
9 | 2017-09-18
2 | 2017-09-19
6 | 2017-09-20
3 | 2017-09-21
So there are gaps in my data first off, and I have another query here:
select until_date, until_date - (lag(until_date) over ()) as delta_days from ranges
Which I have generated the following data:
until_date | delta_days
-----------+-----------
2017-09-08 |
2017-09-11 | 3
2017-09-13 | 2
2017-09-18 | 5
2017-09-21 | 3
2017-09-22 | 1
So I'd like my final query to produce this result:
start_date | ending_date | total_items
-----------+-------------+--------------
2017-09-08 | 2017-09-10 | 0
2017-09-11 | 2017-09-12 | 0
2017-09-13 | 2017-09-17 | 3
2017-09-18 | 2017-09-20 | 15
2017-09-21 | 2017-09-22 | 3
Which tells me the total count of items from the first table, per day, based on the custom ranges from the second table.
In this particular example, I would be summing up total_items BETWEEN start AND end (since there would be overlap on the dates, I'd subtract 1 from the end date to not count duplicates)
Anyone know how to do this?
Thanks!
Use the daterange type. Note that you do not have to calculate delta_days, just convert ranges to dataranges and use the operator <# - element is contained by.
with counts(count, date) as (
values
(3, '2017-09-15'::date),
(9, '2017-09-18'),
(2, '2017-09-19'),
(6, '2017-09-20'),
(3, '2017-09-21')
),
ranges (until_date) as (
values
('2017-09-08'::date),
('2017-09-11'),
('2017-09-13'),
('2017-09-18'),
('2017-09-21'),
('2017-09-22')
)
select daterange, coalesce(sum(count), 0) as total_items
from (
select daterange(lag(until_date) over (order by until_date), until_date)
from ranges
) s
left join counts on date <# daterange
where not lower_inf(daterange)
group by 1
order by 1;
daterange | total_items
-------------------------+-------------
[2017-09-08,2017-09-11) | 0
[2017-09-11,2017-09-13) | 0
[2017-09-13,2017-09-18) | 3
[2017-09-18,2017-09-21) | 17
[2017-09-21,2017-09-22) | 3
(5 rows)
Note, that in the dateranges above lower bounds are inclusive while upper bound are exclusive.
If you want to calculate items per day in the dateranges:
select
daterange, total_items,
round(total_items::dec/(upper(daterange)- lower(daterange)), 2) as items_per_day
from (
select daterange, coalesce(sum(count), 0) as total_items
from (
select daterange(lag(until_date) over (order by until_date), until_date)
from ranges
) s
left join counts on date <# daterange
where not lower_inf(daterange)
group by 1
) s
order by 1
daterange | total_items | items_per_day
-------------------------+-------------+---------------
[2017-09-08,2017-09-11) | 0 | 0.00
[2017-09-11,2017-09-13) | 0 | 0.00
[2017-09-13,2017-09-18) | 3 | 0.60
[2017-09-18,2017-09-21) | 17 | 5.67
[2017-09-21,2017-09-22) | 3 | 3.00
(5 rows)
I'm having issues trying to wrap my head around how to extract some time series stats from my Postgres DB.
For example, I have several stores. I record how many sales each store made each day in a table that looks like:
+------------+----------+-------+
| Date | Store ID | Count |
+------------+----------+-------+
| 2017-02-01 | 1 | 10 |
| 2017-02-01 | 2 | 20 |
| 2017-02-03 | 1 | 11 |
| 2017-02-03 | 2 | 21 |
| 2017-02-04 | 3 | 30 |
+------------+----------+-------+
I'm trying to display this data on a bar/line graph with different lines per Store and the blank dates filled in with 0.
I have been successful getting it to show the sum per day (combining all the stores into one sum) using generate_series, but I can't figure out how to separate it out so each store has a value for each day... the result being something like:
["Store ID 1", 10, 0, 11, 0]
["Store ID 2", 20, 0, 21, 0]
["Store ID 3", 0, 0, 0, 30]
It is necessary to build a cross join dates X stores:
select store_id, array_agg(total order by date) as total
from (
select store_id, date, coalesce(sum(total), 0) as total
from
t
right join (
generate_series(
(select min(date) from t),
(select max(date) from t),
'1 day'
) gs (date)
cross join
(select distinct store_id from t) s
) using (date, store_id)
group by 1,2
) s
group by 1
order by 1
;
store_id | total
----------+-------------
1 | {10,0,11,0}
2 | {20,0,21,0}
3 | {0,0,0,30}
Sample data:
create table t (date date, store_id int, total int);
insert into t (date, store_id, total) values
('2017-02-01',1,10),
('2017-02-01',2,20),
('2017-02-03',1,11),
('2017-02-03',2,21),
('2017-02-04',3,30);
Is there a unpivot equivalent function in PostgreSQL?
Create an example table:
CREATE TEMP TABLE foo (id int, a text, b text, c text);
INSERT INTO foo VALUES (1, 'ant', 'cat', 'chimp'), (2, 'grape', 'mint', 'basil');
You can 'unpivot' or 'uncrosstab' using UNION ALL:
SELECT id,
'a' AS colname,
a AS thing
FROM foo
UNION ALL
SELECT id,
'b' AS colname,
b AS thing
FROM foo
UNION ALL
SELECT id,
'c' AS colname,
c AS thing
FROM foo
ORDER BY id;
This runs 3 different subqueries on foo, one for each column we want to unpivot, and returns, in one table, every record from each of the subqueries.
But that will scan the table N times, where N is the number of columns you want to unpivot. This is inefficient, and a big problem when, for example, you're working with a very large table that takes a long time to scan.
Instead, use:
SELECT id,
unnest(array['a', 'b', 'c']) AS colname,
unnest(array[a, b, c]) AS thing
FROM foo
ORDER BY id;
This is easier to write, and it will only scan the table once.
array[a, b, c] returns an array object, with the values of a, b, and c as it's elements.
unnest(array[a, b, c]) breaks the results into one row for each of the array's elements.
You could use VALUES() and JOIN LATERAL to unpivot the columns.
Sample data:
CREATE TABLE test(id int, a INT, b INT, c INT);
INSERT INTO test(id,a,b,c) VALUES (1,11,12,13),(2,21,22,23),(3,31,32,33);
Query:
SELECT t.id, s.col_name, s.col_value
FROM test t
JOIN LATERAL(VALUES('a',t.a),('b',t.b),('c',t.c)) s(col_name, col_value) ON TRUE;
DBFiddle Demo
Using this approach it is possible to unpivot multiple groups of columns at once.
EDIT
Using Zack's suggestion:
SELECT t.id, col_name, col_value
FROM test t
CROSS JOIN LATERAL (VALUES('a', t.a),('b', t.b),('c',t.c)) s(col_name, col_value);
<=>
SELECT t.id, col_name, col_value
FROM test t
,LATERAL (VALUES('a', t.a),('b', t.b),('c',t.c)) s(col_name, col_value);
db<>fiddle demo
Great article by Thomas Kellerer found here
Unpivot with Postgres
Sometimes it’s necessary to normalize de-normalized tables - the opposite of a “crosstab” or “pivot” operation. Postgres does not support an UNPIVOT operator like Oracle or SQL Server, but simulating it, is very simple.
Take the following table that stores aggregated values per quarter:
create table customer_turnover
(
customer_id integer,
q1 integer,
q2 integer,
q3 integer,
q4 integer
);
And the following sample data:
customer_id | q1 | q2 | q3 | q4
------------+-----+-----+-----+----
1 | 100 | 210 | 203 | 304
2 | 150 | 118 | 422 | 257
3 | 220 | 311 | 271 | 269
But we want the quarters to be rows (as they should be in a normalized data model).
In Oracle or SQL Server this could be achieved with the UNPIVOT operator, but that is not available in Postgres. However Postgres’ ability to use the VALUES clause like a table makes this actually quite easy:
select c.customer_id, t.*
from customer_turnover c
cross join lateral (
values
(c.q1, 'Q1'),
(c.q2, 'Q2'),
(c.q3, 'Q3'),
(c.q4, 'Q4')
) as t(turnover, quarter)
order by customer_id, quarter;
will return the following result:
customer_id | turnover | quarter
------------+----------+--------
1 | 100 | Q1
1 | 210 | Q2
1 | 203 | Q3
1 | 304 | Q4
2 | 150 | Q1
2 | 118 | Q2
2 | 422 | Q3
2 | 257 | Q4
3 | 220 | Q1
3 | 311 | Q2
3 | 271 | Q3
3 | 269 | Q4
The equivalent query with the standard UNPIVOT operator would be:
select customer_id, turnover, quarter
from customer_turnover c
UNPIVOT (turnover for quarter in (q1 as 'Q1',
q2 as 'Q2',
q3 as 'Q3',
q4 as 'Q4'))
order by customer_id, quarter;
FYI for those of us looking for how to unpivot in RedShift.
The long form solution given by Stew appears to be the only way to accomplish this.
For those who cannot see it there, here is the text pasted below:
We do not have built-in functions that will do pivot or unpivot. However,
you can always write SQL to do that.
create table sales (regionid integer, q1 integer, q2 integer, q3 integer, q4 integer);
insert into sales values (1,10,12,14,16), (2,20,22,24,26);
select * from sales order by regionid;
regionid | q1 | q2 | q3 | q4
----------+----+----+----+----
1 | 10 | 12 | 14 | 16
2 | 20 | 22 | 24 | 26
(2 rows)
pivot query
create table sales_pivoted (regionid, quarter, sales)
as
select regionid, 'Q1', q1 from sales
UNION ALL
select regionid, 'Q2', q2 from sales
UNION ALL
select regionid, 'Q3', q3 from sales
UNION ALL
select regionid, 'Q4', q4 from sales
;
select * from sales_pivoted order by regionid, quarter;
regionid | quarter | sales
----------+---------+-------
1 | Q1 | 10
1 | Q2 | 12
1 | Q3 | 14
1 | Q4 | 16
2 | Q1 | 20
2 | Q2 | 22
2 | Q3 | 24
2 | Q4 | 26
(8 rows)
unpivot query
select regionid, sum(Q1) as Q1, sum(Q2) as Q2, sum(Q3) as Q3, sum(Q4) as Q4
from
(select regionid,
case quarter when 'Q1' then sales else 0 end as Q1,
case quarter when 'Q2' then sales else 0 end as Q2,
case quarter when 'Q3' then sales else 0 end as Q3,
case quarter when 'Q4' then sales else 0 end as Q4
from sales_pivoted)
group by regionid
order by regionid;
regionid | q1 | q2 | q3 | q4
----------+----+----+----+----
1 | 10 | 12 | 14 | 16
2 | 20 | 22 | 24 | 26
(2 rows)
Hope this helps, Neil
Pulling slightly modified content from the link in the comment from #a_horse_with_no_name into an answer because it works:
Installing Hstore
If you don't have hstore installed and are running PostgreSQL 9.1+, you can use the handy
CREATE EXTENSION hstore;
For lower versions, look for the hstore.sql file in share/contrib and run in your database.
Assuming that your source (e.g., wide data) table has one 'id' column, named id_field, and any number of 'value' columns, all of the same type, the following will create an unpivoted view of that table.
CREATE VIEW vw_unpivot AS
SELECT id_field, (h).key AS column_name, (h).value AS column_value
FROM (
SELECT id_field, each(hstore(foo) - 'id_field'::text) AS h
FROM zcta5 as foo
) AS unpiv ;
This works with any number of 'value' columns. All of the resulting values will be text, unless you cast, e.g., (h).value::numeric.
Just use JSON:
with data (id, name) as (
values (1, 'a'), (2, 'b')
)
select t.*
from data, lateral jsonb_each_text(to_jsonb(data)) with ordinality as t
order by data.id, t.ordinality;
This yields
|key |value|ordinality|
|----|-----|----------|
|id |1 |1 |
|name|a |2 |
|id |2 |1 |
|name|b |2 |
dbfiddle
I wrote a horrible unpivot function for PostgreSQL. It's rather slow but it at least returns results like you'd expect an unpivot operation to.
https://cgsrv1.arrc.csiro.au/blog/2010/05/14/unpivotuncrosstab-in-postgresql/
Hopefully you can find it useful..
Depending on what you want to do... something like this can be helpful.
with wide_table as (
select 1 a, 2 b, 3 c
union all
select 4 a, 5 b, 6 c
)
select unnest(array[a,b,c]) from wide_table
You can use FROM UNNEST() array handling to UnPivot a dataset, tandem with a correlated subquery (works w/ PG 9.4).
FROM UNNEST() is more powerful & flexible than the typical method of using FROM (VALUES .... ) to unpivot datasets. This is b/c FROM UNNEST() is variadic (with n-ary arity). By using a correlated subquery the need for the lateral ORDINAL clause is eliminated, & Postgres keeps the resulting parallel columnar sets in the proper ordinal sequence.
This is, BTW, FAST -- in practical use spawning 8 million rows in < 15 seconds on a 24-core system.
WITH _students AS ( /** CTE **/
SELECT * FROM
( SELECT 'jane'::TEXT ,'doe'::TEXT , 1::INT
UNION
SELECT 'john'::TEXT ,'doe'::TEXT , 2::INT
UNION
SELECT 'jerry'::TEXT ,'roe'::TEXT , 3::INT
UNION
SELECT 'jodi'::TEXT ,'roe'::TEXT , 4::INT
) s ( fn, ln, id )
) /** end WITH **/
SELECT s.id
, ax.fanm -- field labels, now expanded to two rows
, ax.anm -- field data, now expanded to two rows
, ax.someval -- manually incl. data
, ax.rankednum -- manually assigned ranks
,ax.genser -- auto-generate ranks
FROM _students s
,UNNEST /** MULTI-UNNEST() BLOCK **/
(
( SELECT ARRAY[ fn, ln ]::text[] AS anm -- expanded into two rows by outer UNNEST()
/** CORRELATED SUBQUERY **/
FROM _students s2 WHERE s2.id = s.id -- outer relation
)
,( /** ordinal relationship preserved in variadic UNNEST() **/
SELECT ARRAY[ 'first name', 'last name' ]::text[] -- exp. into 2 rows
AS fanm
)
,( SELECT ARRAY[ 'z','x','y'] -- only 3 rows gen'd, but ordinal rela. kept
AS someval
)
,( SELECT ARRAY[ 1,2,3,4,5 ] -- 5 rows gen'd, ordinal rela. kept.
AS rankednum
)
,( SELECT ARRAY( /** you may go wild ... **/
SELECT generate_series(1, 15, 3 )
AS genser
)
)
) ax ( anm, fanm, someval, rankednum , genser )
;
RESULT SET:
+--------+----------------+-----------+----------+---------+-------
| id | fanm | anm | someval |rankednum| [ etc. ]
+--------+----------------+-----------+----------+---------+-------
| 2 | first name | john | z | 1 | .
| 2 | last name | doe | y | 2 | .
| 2 | [null] | [null] | x | 3 | .
| 2 | [null] | [null] | [null] | 4 | .
| 2 | [null] | [null] | [null] | 5 | .
| 1 | first name | jane | z | 1 | .
| 1 | last name | doe | y | 2 | .
| 1 | | | x | 3 | .
| 1 | | | | 4 | .
| 1 | | | | 5 | .
| 4 | first name | jodi | z | 1 | .
| 4 | last name | roe | y | 2 | .
| 4 | | | x | 3 | .
| 4 | | | | 4 | .
| 4 | | | | 5 | .
| 3 | first name | jerry | z | 1 | .
| 3 | last name | roe | y | 2 | .
| 3 | | | x | 3 | .
| 3 | | | | 4 | .
| 3 | | | | 5 | .
+--------+----------------+-----------+----------+---------+ ----
Here's a way that combines the hstore and CROSS JOIN approaches from other answers.
It's a modified version of my answer to a similar question, which is itself based on the method at https://blog.sql-workbench.eu/post/dynamic-unpivot/ and another answer to that question.
-- Example wide data with a column for each year...
WITH example_wide_data("id", "2001", "2002", "2003", "2004") AS (
VALUES
(1, 4, 5, 6, 7),
(2, 8, 9, 10, 11)
)
-- that is tided to have "year" and "value" columns
SELECT
id,
r.key AS year,
r.value AS value
FROM
example_wide_data w
CROSS JOIN
each(hstore(w.*)) AS r(key, value)
WHERE
-- This chooses columns that look like years
-- In other cases you might need a different condition
r.key ~ '^[0-9]{4}$';
It has a few benefits over other solutions:
By using hstore and not jsonb, it hopefully minimises issues with type conversions (although hstore does convert everything to text)
The columns don't need to be hard coded or known in advance. Here, columns are chosen by a regex on the name, but you could use any SQL logic based on the name, or even the value.
It doesn't require PL/pgSQL - it's all SQL