Selecting unique pairs of equal and opposite rows

All we need is an easy explanation of the problem, so here it is.

I am trying to select unique record "pairs", where a pair is 2 rows that have equal and opposite numeric values (eg 10 and -10). Once a record has been paired, it can’t be used in another set of pairs. Consider this table:

CREATE TABLE vals (
    id text,
    scalar numeric
);

INSERT INTO vals (id, scalar)
VALUES
    ('A', 10),
    ('B', -10),
    ('C', 10),
    ('D', -10),
    ('E', 10);

I have tried multiple variations of DISTINCT, DISTINCT ON (), and UNIQUE, but the closest I’ve come so far is the following:

WITH all_matching AS (
    SELECT
        v.id id1,
        v2.id id2
    FROM vals v
    JOIN vals v2 ON
        v.scalar = (v2.scalar * -1)
    WHERE v.scalar > 0
), unique_left_id AS (
    SELECT DISTINCT ON (id1) *
    FROM all_matching
)

SELECT * FROM unique_left_id;

Which outputs:

id1 | id2
----+----
 A  | B
 C  | B
 E  | B

The obvious problem is that the id B is being used to pair off against all of the other transactions, when I only want to use it once. If I select distinct on id2 after the final transaction, I would be left with a single pair instead of 2 pairs.

How to solve :

I know you bored from this bug, So we are here to help you! Take a deep breath and look at the explanation of your problem. We have many solutions to this problem, But we recommend you to use the first method because it is tested & true method that will 100% work for you.

Method 1

WITH
cte AS (
    SELECT id, 
           scalar, 
           ROW_NUMBER() OVER (PARTITION BY scalar ORDER BY id) rn
    FROM vals
)
SELECT t1.id, t2.id
FROM cte t1
JOIN cte t2 ON t1.scalar = -t2.scalar
           AND t1.scalar > t2.scalar
           AND t1.rn = t2.rn

https://dbfiddle.uk/?rdbms=postgres_12&fiddle=2561e29b69b587e8c827169b361bde07

PS. This code does not process scalar IN (0, NULL).

Note: Use and implement method 1 because this method fully tested our system.
Thank you 🙂

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

Leave a Reply