SQL Query to select bottom 2 from each category

Question

In Mysql, I want to select the bottom 2 items from each category

Category Value
1        1.3
1        4.8
1        3.7
1        1.6
2        9.5
2        9.9
2        9.2
2        10.3
3        4
3        8
3        16

Giving me:

Category Value
1        1.3
1        1.6
2        9.5
2        9.2
3        4
3        8

Before I migrated from sqlite3 I had to first select a lowest from each category, then excluding anything that joined to that, I had to again select the lowest from each category. Then anything equal to that new lowest or less in a category won. This would also pick more than 2 in case of a tie, which was annoying... It also had a really long runtime.

My ultimate goal is to count the number of times an individual is in one of the lowest 2 of a category (there is also a name field) and this is the one part I don't know how to do. Thanks

Answer 1

SELECT c1.category, c1.value
FROM catvals c1
LEFT OUTER JOIN catvals c2
  ON (c1.category = c2.category AND c1.value > c2.value)
GROUP BY c1.category, c1.value
HAVING COUNT(*) < 2;

Tested on MySQL 5.1.41 with your test data. Output:

+----------+-------+
| category | value |
+----------+-------+
|        1 |  1.30 |
|        1 |  1.60 |
|        2 |  9.20 |
|        2 |  9.50 |
|        3 |  4.00 |
|        3 |  8.00 |
+----------+-------+

(The extra decimal places are because I declared the value column as NUMERIC(9,2).)

Like other solutions, this produces more than 2 rows per category if there are ties. There are ways to construct the join condition to resolve that, but we'd need to use a primary key or unique key in your table, and we'd also have to know how you intend ties to be resolved.

Answer 2

You could try this:

SELECT * FROM (
  SELECT c.*,
        (SELECT COUNT(*)
         FROM user_category c2
         WHERE c2.category = c.category
         AND c2.value < c.value) cnt
  FROM user_category c ) uc
WHERE cnt < 2

It should give you the desired results, but check if performance is ok.