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# select random record with added constraints

Posted on 2013-11-12
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Greetings,
I have a table of emails and a number.

ie.
email1@email.com   5
email2@email.com   3
email3@email.com   1

By this count, I have 9 emails.
using some deranged mathematical calcuations, I read this as 5 x email1's, 3xemail2's, and 1xemail3 => 9 total.

I want to pick a random email from the list using the number as a weight against the total count.

in this case, email1 has a 5:9 chance of being selected; email2 has a 3:9 chance of being selected, and so on for email3.

How do I select the random email from the list using those numbers as weights?
Thanks.
0
Question by:Evan Cutler
• 2

LVL 11

Assisted Solution

Louis01 earned 150 total points
ID: 39643759
``````DECLARE @t1 TABLE (id int, email varchar(255), weight int);

INSERT INTO @t1(id, email, weight) VALUES(1, 'email1@email.com', 5);
INSERT INTO @t1(id, email, weight) VALUES(2, 'email2@email.com', 3);
INSERT INTO @t1(id, email, weight) VALUES(3, 'email3@email.com', 1);

--get data
DECLARE @id int, @wsum int, @wpoint int;

SELECT @wsum = SUM(weight) FROM @t1;
SELECT @wpoint = ROUND(((@wsum - 1) * RAND() + 1), 0);

SELECT @id = CASE WHEN @wpoint < 0 THEN @id ELSE t1.id END
, @wpoint = @wpoint - t1.weight
FROM @t1 t1;

SELECT email
FROM @t1
WHERE id = @id;
``````
0

LVL 33

Expert Comment

ID: 39643913
Nope, not really, Use the same weight 1 for all three address shows that it is not correct. First of all, it seems to be off-by-one at some point. And much worse: The statement where you calculate @id and @wpoint is basically undetermined. Cause the set is run thru in an arbitrary order which is not guaranteed. (At least when using SQL Server or ISO SQL). This can be demonstrated when using SQL Server:  Make the id column of @t1 the primary key and set the weights to 1. Also set the initial @wpoint value  to 1. Run it. Then change the id value of the first row id=1 to 10. Under normal circumstances this will change the result. Cause the rows are now run through in a different order.
0

LVL 33

Accepted Solution

ste5an earned 350 total points
ID: 39643943
Here are two solutions.

``````DECLARE @t1 TABLE
(
id INT ,
email VARCHAR(255) ,
[weight] INT
);

INSERT  INTO @t1
VALUES  ( 1, 'email1@email.com', 5 ),
( 2, 'email2@email.com', 3 ),
( 3, 'email3@email.com', 1 );

DECLARE @tally TABLE ( n INT );
INSERT  INTO @tally
VALUES  ( 1 ),
( 2 ),
( 3 ),
( 4 ),
( 5 ),
( 6 ),
( 7 ),
( 8 ),
( 9 ),
( 10 );

SELECT TOP 1
T1.*
FROM    @t1 T1
INNER JOIN @tally T2 ON T1.[weight] >= T2.n
ORDER BY NEWID();

DECLARE @wsum INT = 0;
SELECT  @wsum = SUM([weight])
FROM    @t1;
DECLARE @row INT = ROUND(( ( @wsum - 1 ) * RAND() + 1 ), 0);

WITH    Ordered
AS ( SELECT   T1.* ,
ROW_NUMBER() OVER ( ORDER BY [weight] ) AS RN
FROM     @t1 T1
INNER JOIN @tally T2 ON T1.[weight] >= T2.n
)
SELECT   id ,
email ,
[weight]
FROM    Ordered
WHERE   rn = @row;
``````

Both use a tally table to get the amount of rows correctly. Then the first uses the NEWID() function (SQL Server specific). This function is evaluated for each row, an gives us random GUID's. The second one shows the more obvious. You need an row number of the entire set and pick a random one from it.
0

LVL 9

Author Closing Comment

ID: 39668115
Thank you so much you guys.
This is awesome.

Evan
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