SQL frequency distribution query to count ranges with group-by and include 0 counts

Question

Given:

table 'thing':

age
---
3.4
3.4
10.1
40
45
49

I want to count the number of things for each 10-year range, e.g.,

age_range | count
----------+-------
        0 |     2
        10|     1
        20|     0
        30|     0
        40|     3

This query comes close:

SELECT FLOOR(age / 10) as age_range, COUNT(*)
FROM thing
GROUP BY FLOOR(age / 10) ORDER BY FLOOR(age / 10);

Output:

 age_range | count 
-----------+-------
         0 |     1
         1 |     2
         4 |     3

However, it doesn't show me the ranges which have 0 counts. How can I modify the query so that it also shows the ranges in between with 0 counts?

I found similar stackoverflow questions for counting ranges, some for 0 counts, but they involve having to specify each range (either hard-coding the ranges into the query, or putting the ranges in a table). I would prefer to use a generic query like that above where I do not have to explicitly specify each range (e.g., 0-10, 10-20, 20-30, ...). I'm using PostgreSQL 9.1.3.

Is there a way to modify the simple query above to include 0 counts?

Similar:
Oracle: how to "group by" over a range?
Get frequency distribution of a decimal range in MySQL

Answer 1

You need some way to invent the table of age ranges. Row number usually works nicely. Do a cartesian product against a big table to get lots of numbers.

WITH RANGES AS (
SELECT (rownum - 1) * 10 AS age_range
  FROM ( SELECT row_number() OVER() as rownum
           FROM pg_tables
       ) n
      ,( SELECT ceil( max(age) / 10 )  range_end
           FROM thing
       ) m
  WHERE  n. rownum <= range_end
)
SELECT r.age_range, COUNT(t.age) AS count
  FROM ranges r
  LEFT JOIN thing t ON r.age_range = FLOOR(t.age / 10) * 10
  GROUP BY r.age_range
  ORDER BY r.age_range;

EDIT: mu is too short has a much more elegant answer, but if you didn't have a generate_series function on the db, ... :)