How to get back aggregate values across 2 dimensions using Python Cubes? - postgresql

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

Related

Efficient way to retrieve all values from a column that start with other values from the same column in PostgreSQL

For the sake of simplicity, suppose you have a table with numbers like:
| number |
----------
|123 |
|1234 |
|12345 |
|123456 |
|111 |
|1111 |
|2 |
|700 |
What would be an efficient way of retrieving the shortest numbers (call them roots or whatever) and all values derived from them, eg:
| root | derivatives |
--------------------------------
| 123 | 1234, 12345, 123456 |
| 111 | 1111 |
Numbers 2 & 700 are excluded from the list because they're unique, and thus have no derivatives.
An output as the above would be ideal, but since it's probably difficult to achieve, the next best thing would be something like below, which I can then post-process:
| root | derivative |
-----------------------
| 123 | 1234 |
| 123 | 12345 |
| 123 | 123456 |
| 111 | 1111 |
My naive initial attempt to at least identify roots (see below) has been running for 4h now with a dataset of ~500k items, but the real one I'd have to inspect consists of millions.
select number
from numbers n1
where exists(
select number
from numbers n2
where n2.number <> n1.number
and n2.number like n1.number || '_%'
);
This works if number is an integer or bigint:
select min(a.number) as root, b.number as derivative
from nums a
cross join lateral generate_series(1, 18) as gs(power)
join nums b
on b.number / (10^gs.power)::bigint = a.number
group by b.number
order by root, derivative;
EDIT: I moved a non-working query to the bottom. It fails for reasons outlined by #Morfic in the comments.
We can do a similar and simpler join using like for character types:
select min(a.number) as root, b.number as derivative
from numchar a
join numchar b on b.number like a.number||'%'
and b.number != a.number
group by b.number
order by root, derivative;
Updated fiddle.
Faulty Solution Follows
If number is a character type, then try this:
with groupings as (
select number,
case
when number like (lag(number) over (order by number))||'%' then 0
else 1
end as newgroup
from numchar
), groupnums as (
select number, sum(newgroup) over (order by number) as groupnum
from groupings
), matches as (
select min(number) over (partition by groupnum) as root,
number as derivative
from groupnums
)
select *
from matches
where root != derivative;
There should be only a single sort on groupnum in this execution since the column is your table's primary key.
db<>fiddle here

PostgreSQL - Setting null values to missing rows in a join statement

SQL newbie here. I'm trying to write a query that generates a scoring table, setting null to a student's grades in a module for which they haven't yet taken their exams (on PostgreSQL).
So I start with tables that look something like this:
student_evaluation:
|student_id| module_id | course_id |grade |
|----------|-----------|-----------|-------|
| 1 | 1 | 1 |3 |
| 1 | 1 | 1 |7 |
| 1 | 2 | 1 |8 |
| 2 | 4 | 2 |9 |
course_module:
| module_id | course_id |
| ---------- | --------- |
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 2 |
In our use case, a course is made up of several modules. Each module has a single exam, but a student who failed his exam may have a couple of retries. The same module may also be present in different courses, but an exam attempt only counts for one instance of the module (ie. student A passed module 1's exam on course 1. If course 2 also has module 1, student A has to retake the same exam for course 2 if he also has access to that course).
So the output should look like this:
student_id
module_id
course_id
grade
1
1
1
3
1
1
1
7
1
2
1
8
1
3
1
null
2
4
2
9
I feel like this should have been a simple task, but I think I have a very flawed understanding of how outer and cross joins work. I have tried stuff like:
SELECT se.student_id, se.module_id, se.course_id, se.grade FROM student_evaluation se
RIGHT OUTER JOIN course_module ON course_module.course_id = se.course_id
AND course_module.module_id = se.module_id
or
SELECT se.student_id, se.module_id, se.course_id, se.grade FROM student_evaluation se
CROSS JOIN course_module WHERE course_module.course_id = se.course_id
Neither worked. These all feel wrong, but I'm lost as to what would be the proper way to go about this.
Thank you in advance.
I think you need both join types: first use a cross join to build a list of all combinations of students and courses, then use an outer join to add the grades.
SELECT sc.student_id,
sc.module_id,
sc.course_id,
se.grade
FROM student_evaluation se
RIGHT JOIN (SELECT s.student_id,
c.module_id,
c.course_id
FROM (SELECT DISTINCT student_id
FROM student_evaluation) AS s
CROSS JOIN course_module AS c) AS sc
USING (course_id));

PostgreSQL One ID multiple values

I have a Postgres table where one id may have multiple Channel values as follows
ID |Channel | Column 3 | Column 4
_____|________|__________|_________
1 | Sports | x | null
1 | Organic| x | z
2 | Organic| null | q
3 | Arts | b | w
3 | Organic| e | r
4 | Sports | sp | t
No ID will have a duplicate channel name, and no ID will be both Sports and Arts. That is, ID 1 could have a Sports and Organic channel, a Sports and Arts channel, but not two sports or two organic entries and not a Sports and Arts channel. I want all IDs to be in the query, but if there is a non-organic channel I prefer that. The result I would want would be
ID |Channel | Column 3 | Column 4
_____|________|__________|_________
1 | Sports | x | null
2 | Organic| null | q
3 | Arts | b | w
4 | Sports | sp | t
I feel like there is some CTE here, a rank and partition or something that could do the trick, but I'm just not getting it. I'm only including Columns 3 and 4 to show there are extra columns.
Does anyone have any ideas on the code to deploy here?
You could use DISTINCT ON with an appropriate ORDER BY clause:
SELECT DISTINCT ON (id)
id, channel, column3, column4
FROM atable
ORDER BY id, channel = 'Organic';
This relies on the fact that FALSE < TRUE.
I ended up using a rank over function
ROW_NUMBER () over (partition by salesforce_id order by case when channel is organic then 0 else 1 end desc, timestamp desc) as id_rank
I didn't include in the original question that I had a timestamp! This works now. Thanks

SELECT DISTINCT on a ordered subquery's table

I'm working on a problem, involving these two tables.
books
isbn | title | author
------------+-----------------------------------------+------------------
1840918626 | Hogwarts: A History | Bathilda Bagshot
3458400871 | Fantastic Beasts and Where to Find Them | Newt Scamander
9136884926 | Advanced Potion-Making | Libatius Borage
transactions
id | patron_id | isbn | checked_out_date | checked_in_date
----+-----------+------------+------------------+-----------------
1 | 1 | 1840918626 | 2012-05-05 | 2012-05-06
2 | 4 | 9136884926 | 2012-05-05 | 2012-05-06
3 | 2 | 3458400871 | 2012-05-05 | 2012-05-06
4 | 3 | 3458400871 | 2018-04-29 | 2018-05-02
5 | 2 | 9136884926 | 2018-05-03 | NULL
6 | 1 | 3458400871 | 2018-05-03 | 2018-05-05
7 | 5 | 3458400871 | 2018-05-05 | NULL
the query "Make a list of all book titles and denote whether or not a copy of that book is checked out." so pretty much just the first table with a checked out column.
im trying to SELECT DISTINCT on a sub query with the checkout books first, but that doesn't work. I've researched and others say to accomplish this use a GROUP BY clause instead of DISTINCT but the examples they provide are one column queries and when more columns are added it doesn't work.
this is my closest attempt
SELECT DISTINCT ON (title)
title, checked_out
FROM(
SELECT b.title, t.checked_in_date IS NULL AS checked_out
FROM transactions t
natural join books b
ORDER BY checked_out DESC
) t;
or you can join only transactions where books are not checked in:
SELECT b.title, t.isbn IS NOT NULL AS checked_out
, t.checked_out_date
FROM books b
LEFT JOIN transactions t ON t.isbn = b.isbn AND t.checked_in_date IS NULL
ORDER BY checked_out DESC
I adjusted your attempt a little bit. Basically I changed the way your data is joined
SELECT DISTINCT ON (title)
title, checked_out
FROM(
SELECT b.title, t.checked_in_date IS NULL AS checked_out
FROM books b
LEFT OUTER JOIN transactions t USING (isbn)
ORDER BY checked_out DESC
) t;

Equivalent to unpivot() in PostgreSQL

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