Oracle: how to “group by” over a range?

Question

If I have a table like this:

pkey   age
----   ---
   1     8
   2     5
   3    12
   4    12
   5    22

I can "group by" to get a count of each age.

select age,count(*) n from tbl group by age;
age  n
---  -
  5  1
  8  1
 12  2
 22  1

What query can I use to group by age ranges?

  age  n
-----  -
 1-10  2
11-20  2
20+    1

I'm on 10gR2, but I'd be interested in any 11g-specific approaches as well.

Answer 1

add an age_range table and an age_range_id field to your table and group by that instead.

// excuse the DDL but you should get the idea

create table age_range(
age_range_id tinyint unsigned not null primary key,
name varchar(255) not null);

insert into age_range values 
(1, '18-24'),(2, '25-34'),(3, '35-44'),(4, '45-54'),(5, '55-64');

// again excuse the DML but you should get the idea

select
 count(*) as counter, p.age_range_id, ar.name
from
  person p
inner join age_range ar on p.age_range_id = ar.age_range_id
group by
  p.age_range_id, ar.name order by counter desc;

You can refine this idea if you like - add from_age to_age columns in the age_range table etc - but i'll leave that to you.

hope this helps :)

Answer 2

My approach:

select range, count(1) from (
select case 
  when age < 5 then '0-4' 
  when age < 10 then '5-9' 
  when age < 15 then '10-14' 
  when age < 20 then '15-20' 
  when age < 30 then '21-30' 
  when age < 40 then '31-40' 
  when age < 50 then '41-50' 
  else                '51+' 
end 
as range from
(select round(extract(day from feedback_update_time - feedback_time), 1) as age
from txn_history
) ) group by range  

• I have flexibility in defining the ranges
• I do not repeat the ranges in select and group clauses
• but some one please tell me, how to order them by magnitude!

Answer 3

Can you try the below solution:

SELECT count (1), '1-10'  where age between 1 and 10
union all 
SELECT count (1), '11-20'  where age between 11 and 20
union all
select count (1), '21+' where age >20
from age 

Answer 4

Try:

select to_char(floor(age/10) * 10) || '-' 
|| to_char(ceil(age/10) * 10 - 1)) as age, 
count(*) as n from tbl group by floor(age/10);

Answer 5

I had to get a count of samples by day. Inspired by @Clarkey I used TO_CHAR to extract the date of sample from the timestamp to an ISO-8601 date format and used that in the GROUP BY and ORDER BY clauses. (Further inspired, I also post it here in case it is useful to others.)

SELECT 
  TO_CHAR(X.TS_TIMESTAMP, 'YYYY-MM-DD') AS TS_DAY, 
  COUNT(*) 
FROM   
  TABLE X
GROUP BY
  TO_CHAR(X.TS_TIMESTAMP, 'YYYY-MM-DD')
ORDER BY
  TO_CHAR(X.TS_TIMESTAMP, 'YYYY-MM-DD') ASC
/

Answer 6

Here is a solution which creates a "range" table in a sub-query and then uses this to partition the data from the main table:

SELECT DISTINCT descr
  , COUNT(*) OVER (PARTITION BY descr) n
FROM age_table INNER JOIN (
  select '1-10' descr, 1 rng_start, 10 rng_stop from dual
  union (
  select '11-20', 11, 20 from dual
  ) union (
  select '20+', 21, null from dual
)) ON age BETWEEN nvl(rng_start, age) AND nvl(rng_stop, age)
ORDER BY descr;