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Home/ Questions/Q 8735615
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T10:06:06+00:00 2026-06-13T10:06:06+00:00

Example input: SELECT * FROM test; id | percent —-+———- 1 | 50 2

  • 0

Example input:

SELECT * FROM test;
 id | percent   
----+----------
  1 | 50 
  2 | 35   
  3 | 15   
(3 rows)

How would you write such query, that on average 50% of time i could get the row with id=1, 35% of time row with id=2, and 15% of time row with id=3?

I tried something like SELECT id FROM test ORDER BY p * random() DESC LIMIT 1, but it gives wrong results. After 10,000 runs I get a distribution like: {1=6293, 2=3302, 3=405}, but I expected the distribution to be nearly: {1=5000, 2=3500, 3=1500}.

Any ideas?

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  1. Editorial Team
    Editorial Team
    2026-06-13T10:06:07+00:00Added an answer on June 13, 2026 at 10:06 am

    This should do the trick:

    WITH CTE AS (
        SELECT random() * (SELECT SUM(percent) FROM YOUR_TABLE) R
    )
    SELECT *
    FROM (
        SELECT id, SUM(percent) OVER (ORDER BY id) S, R
        FROM YOUR_TABLE CROSS JOIN CTE
    ) Q
    WHERE S >= R
    ORDER BY id
    LIMIT 1;
    

    The sub-query Q gives the following result:

    1  50
    2  85
    3  100
    

    We then simply generate a random number in range [0, 100) and pick the first row that is at or beyond that number (the WHERE clause). We use common table expression (WITH) to ensure the random number is calculated only once.

    BTW, the SELECT SUM(percent) FROM YOUR_TABLE allows you to have any weights in percent – they don’t strictly need to be percentages (i.e. add-up to 100).

    [SQL Fiddle]

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