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

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

What you are looking for, is basically the data for a histogram.

You would have the age (or age-range) on the x-axis and the count n (or frequency) on the y-axis.

In the simplest form, one could simply count the number of each distinct age value like you already described:

SELECT age, count(*)
FROM tbl
GROUP BY age

When there are too many different values for the x-axis however, one may want to create groups (or clusters or buckets). In your case, you group by a constant range of 10.

We can avoid writing a WHEN ... THEN line for each range - there could be hundreds if it were not about age. Instead, the approach by @MatthewFlaschen is preferable for the reasons mentioned by @NitinMidha.

Now let's build the SQL...

First, we need to split the ages into range-groups of 10 like so:

• 0-9
• 10-19
• 20 - 29
• etc.

This can be achieved by dividing the age column by 10 and then calculating the result's FLOOR:

FLOOR(age/10)

"FLOOR returns the largest integer equal to or less than n" http://docs.oracle.com/cd/E11882_01/server.112/e26088/functions067.htm#SQLRF00643

Then we take the original SQL and replace age with that expression:

SELECT FLOOR(age/10), count(*)
FROM tbl
GROUP BY FLOOR(age/10)

This is OK, but we cannot see the range, yet. Instead we only see the calculated floor values which are 0, 1, 2 ... n.

To get the actual lower bound, we need to multiply it with 10 again so we get 0, 10, 20 ... n:

FLOOR(age/10) * 10

We also need the upper bound of each range which is lower bound + 10 - 1 or

FLOOR(age/10) * 10 + 10 - 1

Finally, we concatenate both into a string like this:

TO_CHAR(FLOOR(age/10) * 10) || '-' || TO_CHAR(FLOOR(age/10) * 10 + 10 - 1)

This creates '0-9', '10-19', '20-29' etc.

Now our SQL looks like this:

SELECT 
TO_CHAR(FLOOR(age/10) * 10) || ' - ' || TO_CHAR(FLOOR(age/10) * 10 + 10 - 1),
COUNT(*)
FROM tbl
GROUP BY FLOOR(age/10)

Finally, apply an order and nice column aliases:

SELECT 
TO_CHAR(FLOOR(age/10) * 10) || ' - ' || TO_CHAR(FLOOR(age/10) * 10 + 10 - 1) AS range,
COUNT(*) AS frequency
FROM tbl
GROUP BY FLOOR(age/10)
ORDER BY FLOOR(age/10)

However, in more complex scenarios, these ranges might not be grouped into constant chunks of size 10, but need dynamical clustering. Oracle has more advanced histogram functions included, see http://docs.oracle.com/cd/E16655_01/server.121/e15858/tgsql_histo.htm#TGSQL366

Credits to @MatthewFlaschen for his approach; I only explained the details.

Answer 4

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 5

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;

Answer 6

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);