Situation
Using Python 3, Django 1.9, Cubes 1.1, and Postgres 9.5.
These are my datatables in pictorial form:
The same in text format:
Store table
------------------------------
| id | code | address |
|-----|------|---------------|
| 1 | S1 | Kings Row |
| 2 | S2 | Queens Street |
| 3 | S3 | Jacks Place |
| 4 | S4 | Diamonds Alley|
| 5 | S5 | Hearts Road |
------------------------------
Product table
------------------------------
| id | code | name |
|-----|------|---------------|
| 1 | P1 | Saucer 12 |
| 2 | P2 | Plate 15 |
| 3 | P3 | Saucer 13 |
| 4 | P4 | Saucer 14 |
| 5 | P5 | Plate 16 |
| and many more .... |
|1000 |P1000 | Bowl 25 |
|----------------------------|
Sales table
----------------------------------------
| id | product_id | store_id | amount |
|-----|------------|----------|--------|
| 1 | 1 | 1 |7.05 |
| 2 | 1 | 2 |9.00 |
| 3 | 2 | 3 |1.00 |
| 4 | 2 | 3 |1.00 |
| 5 | 2 | 5 |1.00 |
| and many more .... |
| 1000| 20 | 4 |1.00 |
|--------------------------------------|
The relationships are:
Sales belongs to Store
Sales belongs to Product
Store has many Sales
Product has many Sales
What I want to achieve
I want to use cubes to be able to do a display by pagination in the following manner:
Given the stores S1-S3:
-------------------------
| product | S1 | S2 | S3 |
|---------|----|----|----|
|Saucer 12|7.05|9 | 0 |
|Plate 15 |0 |0 | 2 |
| and many more .... |
|------------------------|
Note the following:
Even though there were no records in sales for Saucer 12 under Store S3, I displayed 0 instead of null or none.
I want to be able to do sort by store, say descending order for, S3.
The cells indicate the SUM total of that particular product spent in that particular store.
I also want to have pagination.
What I tried
This is the configuration I used:
"cubes": [
{
"name": "sales",
"dimensions": ["product", "store"],
"joins": [
{"master":"product_id", "detail":"product.id"},
{"master":"store_id", "detail":"store.id"}
]
}
],
"dimensions": [
{ "name": "product", "attributes": ["code", "name"] },
{ "name": "store", "attributes": ["code", "address"] }
]
This is the code I used:
result = browser.aggregate(drilldown=['Store','Product'],
order=[("Product.name","asc"), ("Store.name","desc"), ("total_products_sale", "desc")])
I didn't get what I want.
I got it like this:
----------------------------------------------
| product_id | store_id | total_products_sale |
|------------|----------|---------------------|
| 1 | 1 | 7.05 |
| 1 | 2 | 9 |
| 2 | 3 | 2.00 |
| and many more .... |
|---------------------------------------------|
which is the whole table with no pagination and if the products not sold in that store it won't show up as zero.
My question
How do I get what I want?
Do I need to create another data table that aggregates everything by store and product before I use cubes to run the query?
Update
I have read more. I realised that what I want is called dicing as I needed to go across 2 dimensions. See: https://en.wikipedia.org/wiki/OLAP_cube#Operations
Cross-posted at Cubes GitHub issues to get more attention.
This is a pure SQL solution using crosstab() from the additional tablefunc module to pivot the aggregated data. It typically performs better than any client-side alternative. If you are not familiar with crosstab(), read this first:
PostgreSQL Crosstab Query
And this about the "extra" column in the crosstab() output:
Pivot on Multiple Columns using Tablefunc
SELECT product_id, product
, COALESCE(s1, 0) AS s1 -- 1. ... displayed 0 instead of null
, COALESCE(s2, 0) AS s2
, COALESCE(s3, 0) AS s3
, COALESCE(s4, 0) AS s4
, COALESCE(s5, 0) AS s5
FROM crosstab(
'SELECT s.product_id, p.name, s.store_id, s.sum_amount
FROM product p
JOIN (
SELECT product_id, store_id
, sum(amount) AS sum_amount -- 3. SUM total of product spent in store
FROM sales
GROUP BY product_id, store_id
) s ON p.id = s.product_id
ORDER BY s.product_id, s.store_id;'
, 'VALUES (1),(2),(3),(4),(5)' -- desired store_id's
) AS ct (product_id int, product text -- "extra" column
, s1 numeric, s2 numeric, s3 numeric, s4 numeric, s5 numeric)
ORDER BY s3 DESC; -- 2. ... descending order for S3
Produces your desired result exactly (plus product_id).
To include products that have never been sold replace [INNER] JOIN with LEFT [OUTER] JOIN.
SQL Fiddle with base query.
The tablefunc module is not installed on sqlfiddle.
Major points
Read the basic explanation in the reference answer for crosstab().
I am including with product_id because product.name is hardly unique. This might otherwise lead to sneaky errors conflating two different products.
You don't need the store table in the query if referential integrity is guaranteed.
ORDER BY s3 DESC works, because s3 references the output column where NULL values have been replaced with COALESCE. Else we would need DESC NULLS LAST to sort NULL values last:
PostgreSQL sort by datetime asc, null first?
For building crosstab() queries dynamically consider:
Dynamic alternative to pivot with CASE and GROUP BY
I also want to have pagination.
That last item is fuzzy. Simple pagination can be had with LIMIT and OFFSET:
Displaying data in grid view page by page
I would consider a MATERIALIZED VIEW to materialize results before pagination. If you have a stable page size I would add page numbers to the MV for easy and fast results.
To optimize performance for big result sets, consider:
SQL syntax term for 'WHERE (col1, col2) < (val1, val2)'
Optimize query with OFFSET on large table
I have three tables in Postgres. They are all about a single event (an occurrence, not "sports event"). Each table is about a specific item during the event.
table_header columns
gid, start_timestamp, end_timestamp, location, positions
table_item1 columns
gid, side, visibility, item1_timestamp
table_item2 columns
gid, position_id, name, item2_timestamp
I've tried the following query:
SELECT h.gid, h.location, h.start_timestamp, h.end_timestamp, i1.side,
i1.visibility, i2.position_id, i2.name, i2.item2_timestamp AS timestamp
FROM tablet_header AS h
LEFT OUTER JOIN table_item1 i1 on (i1.gid = h.gid)
LEFT OUTER JOIN table_item2 i2 on (i2.gid = i1.gid AND
i1.item1_timestamp = i2.item2_timestamp)
WHERE h.start_timestamp BETWEEN '2016-03-24 12:00:00'::timestamp AND now()::timestamp
The problem is that I'm losing some data from rows when item1_timestamp and item2_timestamp do not match.
So if I have in table_item1 and table_item2:
gid | item1_timestamp | side gid | item2_timestamp | name
---------------------------- -----------------------------------
1 | 17:00:00 | left 1 | 17:00:00 | charlie
1 | 17:00:05 | right 1 | 17:00:03 | frank
1 | 17:00:10 | left 1 | 17:00:06 | dee
I would want the final output to be:
gid | timestamp | side | name
-----------------------------
1 | 17:00:00 | left | charlie
1 | 17:00:03 | | frank
1 | 17:00:05 | right |
1 | 17:00:06 | | dee
1 | 17:00:10 | left |
based purely on the timestamp (and gid). Naturally I would have the header info in there too, but that's trivial.
I tried playing around with the query I posted used different JOINs and UNIONs, but I cannot seem to get it right. The one I posted gives the best results I could manage, but it's incomplete.
Side note: every minute or so there will be a new "event". So the gid will be unique to each event and the query needs to ensure that each dataset is paired with data from the same gid. Which is the reason for my i1.gid = h.gid lines. Data between different events should not be compared.
select t1.gid, t1.timestamp, t1.side, t2.name
from t1
left join t2 on t2.timestamp=t1.timestamp and t2.gid=t1.gid
union
select t1.gid, t1.timestamp, t1.side, t2.name
from t2
left join t1 on t2.timestamp=t1.timestamp and t2.gid=t1.gid
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