Weighted random value (1-10) e.g.3 will be 50 % of records

Question

Hello,
i need to generate random values into my table. For the simplification, lets say that i need to generate values in range from 1 to 10. For this there is a lot of ways.
e.g. UPDATE test_number SET random_value = round (dbms_random.value (1,10))
But what I need is something like weighted random values. I need random values to be inserted into column, but I need e.g. 3 to be inserted more times than other values e.g. if i have 100 rows i want to have 3 in 50 random rows
Do you have any idea how to achieve it?
Thanks a lot for any help or reply !

Answer 1

Hi,
You can do something like this:
WITH     got_p_num     AS
     SELECT     LEVEL                         AS id
     ,     FLOOR (dbms_random.value (1, 19))     AS p_num
     FROM     dual
     CONNECT BY     LEVEL     <= 10
SELECT       id
,       p_num
,       CASE
          WHEN  p_num     < 3     THEN p_num
          WHEN  p_num     < 12     THEN 3
                                 ELSE p_num - 8
       END     AS r_num
FROM       got_p_num
ORDER BY  r_num
;If you want the number 3 to appear on half of the rows; that means that, out of 18 random numbers, you expect 3 to occur 9 times, and the other numbers to appear once each. The query above generates random integers in the range 1 through 18 inclusive, and calls this number p_num. The CASE statement maps p_num to what you really want, a n integer between 1 and 10, with 3 occurring 9/18 (= 1/2) of the time.
Instead of using a CASE expression, you might want to create a table, that has one row for every number that can be in the output, and the probability of picking that number. You could then join to that table to convert the raw random number to the id that you want.
Edited by: Frank Kulash on Oct 6, 2011 3:36 PM