How do I generate a random number for each row in a T-SQL select?
Asked Answered
C

21

413

I need a different random number for each row in my table. The following seemingly obvious code uses the same random value for each row.

SELECT table_name, RAND() magic_number 
FROM information_schema.tables 

I'd like to get an INT or a FLOAT out of this. The rest of the story is I'm going to use this random number to create a random date offset from a known date, e.g. 1-14 days offset from a start date.

This is for Microsoft SQL Server 2000.

Condolence answered 25/6, 2009 at 17:17 Comment(5)
Is there a solution to this that does not use NEWID()? I want to be able to generate the same sequence of random numbers for a given seed.Perpetrate
@Rory Ask that as new question, it will get more attention. (My answer would be to use fixed tables of random numbers, eg. For example this famous standard set of random number: rand.org/pubs/monograph_reports/MR1418/index.html )Condolence
Look @ RAND (Transact-SQL)Enforcement
RAND was introduced in 2005, this question was asked in 2009, which organizations still used SQL 2000 because that was the 1st version good enough to use forever.Condolence
Rory MacLeod asked, "Is there a solution to this that does not use NEWID()? I want to be able to generate the same sequence of random numbers for a given seed." The answer is yes, but its a bit convoluted. 1. Create a view that returns select rand() 2. Create a UDF that selects the value from the view. 3. Before selecting your data, seed the rand() function. 4. Use the UDF in your select statement. I'll post a full example belowUsually
E
619

Take a look at SQL Server - Set based random numbers which has a very detailed explanation.

To summarize, the following code generates a random number between 0 and 13 inclusive with a uniform distribution:

ABS(CHECKSUM(NewId())) % 14

To change your range, just change the number at the end of the expression. Be extra careful if you need a range that includes both positive and negative numbers. If you do it wrong, it's possible to double-count the number 0.

A small warning for the math nuts in the room: there is a very slight bias in this code. CHECKSUM() results in numbers that are uniform across the entire range of the sql Int datatype, or at least as near so as my (the editor) testing can show. However, there will be some bias when CHECKSUM() produces a number at the very top end of that range. Any time you get a number between the maximum possible integer and the last exact multiple of the size of your desired range (14 in this case) before that maximum integer, those results are favored over the remaining portion of your range that cannot be produced from that last multiple of 14.

As an example, imagine the entire range of the Int type is only 19. 19 is the largest possible integer you can hold. When CHECKSUM() results in 14-19, these correspond to results 0-5. Those numbers would be heavily favored over 6-13, because CHECKSUM() is twice as likely to generate them. It's easier to demonstrate this visually. Below is the entire possible set of results for our imaginary integer range:

Checksum Integer: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Range Result:     0 1 2 3 4 5 6 7 8 9 10 11 12 13  0  1  2  3  4  5

You can see here that there are more chances to produce some numbers than others: bias. Thankfully, the actual range of the Int type is much larger... so much so that in most cases the bias is nearly undetectable. However, it is something to be aware of if you ever find yourself doing this for serious security code.

Edwardedwardian answered 25/6, 2009 at 17:22 Comment(11)
This linked page had the solution: ABS(CHECKSUM(NewId())) % 14Condolence
% 14 would return numbers between 0 and 13Overshoe
@Dennis Palmer, just add 1Antarctica
Just to expand on the bias bit, if you're generating numbers in a very large range, the bias gets bigger. For example if you're using 2/3 of the Integer number space, you're twice as likely to get numbers in the lower half of the produced number space than in the upper half, so even for even some ad-hoc problems this could have a big impact. Also, I believe that if you're % number is a power of 2 then there should be no bias...Garmaise
We just discovered a genius bug with this. Because checksum returns an int, and the range of an int is -2^31 (-2,147,483,648) to 2^31-1 (2,147,483,647), the abs() function can return an overflow error if the result happens to be exactly -2,147,483,648! The chances are obviously very low, about 1 in 4 billion, however we were running it over a ~1.8b row table every day, so it was happening about once a week! Fix is to cast the checksum to bigint before the abs.Feebleminded
I was hitting @Feebleminded 's bug with a much smaller dataset - only 350k rows. The trigger was wrapping an outer select around it and asking for AVG() of the random number column. Also instead of a static Mod (e.g. 14) I was using COUNT ( * ) OVER ( PARTITION BY 1 ). Anyway @Feebleminded your fix rocks! Thanks and stay Evil. My code is now: ABS ( CAST ( CHECKSUM ( NewId() ) AS bigint ) ) % COUNT ( * ) OVER ( PARTITION BY 1 ) + 1 AS [Random Row Number]Boys
@EvilPuppetMaster, @MikeHoney Instead of casting to bigint, you could just perform the % operation first (it returns the same magnitude remainder whether positive or negative), then perform the abs operation: abs(checksum(newid()) % 14). This should give the same result without the cast.Marmawke
I think that this should say "a uniform distribution" not "normalized distribution" -- each number is equally likely, it's not a bell curve."Normalized" has a specific mathematical meaning.Deuno
Note that if used in a WHERE clause, the expression is still only evaluated once for the entire row set, as is the original issue that OP faced. For example, instead of omitting 90% of records in the table, 90% of the time it omits 100% of records (Anchorman movie reference) with this query: select * from @Test where ABS(CHECKSUM(NEWID())) % 100 >= 90Jacklin
In that case you might want to try the SELECT TOP(@n) PERCENT * syntax where @n is a value from 0 to 100 that you want as a percent inclusion. But you then need to provide an ORDER BY clause (hint: use ORDER BY NEWID()). Also note that, unless @n is exactly 0, at least one row will always be returned.Jacklin
It's actually not a uniform distribution, because the number of 32-bit numbers is not evenly divisible by 14. Some numbers are more probable than others.Izaguirre
K
116

When called multiple times in a single batch, rand() returns the same number.

I'd suggest using convert(varbinary,newid()) as the seed argument:

SELECT table_name, 1.0 + floor(14 * RAND(convert(varbinary, newid()))) magic_number 
FROM information_schema.tables

newid() is guaranteed to return a different value each time it's called, even within the same batch, so using it as a seed will prompt rand() to give a different value each time.

Edited to get a random whole number from 1 to 14.

Kobold answered 25/6, 2009 at 17:20 Comment(6)
How do you get a number out of a guid or varbinary? I'll update the question to indicate I'm hoping for an integer.Condolence
You multiply it by a number and floor it :) so if you want five digits, multiply by 100000, and convert to an int. Ugly, but simple enough to do.Kobold
As a further addendum - that will give you up to five digits - if you want to zero-pad it, you'll have to use a char datatype, and use replicate to zero-pad up to 5 digits.Kobold
If you use the ceiling function instead of floor, you don't have to add 1.Everest
Even when I use this, there are times that RAND() always gives me the same result. Even stranger, there are times that it jumps from a correct to an incorrect behaviour depending on the number of times I am using it. I am trying to implement a RANDOM INNER JOIN and if I ask for more than 19 (!!!) rows, it starts giving me always the same result...Affirmative
@JohannesWentu do you think in possibility of Sql Server is maybe reusing the previous plans, so you get the same results? Maybe you can use a Sql Hint to force avoid this.Dordrecht
B
99
RAND(CHECKSUM(NEWID()))

The above will generate a (pseudo-) random number between 0 and 1, exclusive. If used in a select, because the seed value changes for each row, it will generate a new random number for each row (it is not guaranteed to generate a unique number per row however).

Example when combined with an upper limit of 10 (produces numbers 1 - 10):

CAST(RAND(CHECKSUM(NEWID())) * 10 as INT) + 1

Transact-SQL Documentation:

  1. CAST(): https://learn.microsoft.com/en-us/sql/t-sql/functions/cast-and-convert-transact-sql
  2. RAND(): http://msdn.microsoft.com/en-us/library/ms177610.aspx
  3. CHECKSUM(): http://msdn.microsoft.com/en-us/library/ms189788.aspx
  4. NEWID(): https://learn.microsoft.com/en-us/sql/t-sql/functions/newid-transact-sql
Boating answered 27/1, 2012 at 20:30 Comment(0)
F
49

Random number generation between 1000 and 9999 inclusive:

FLOOR(RAND(CHECKSUM(NEWID()))*(9999-1000+1)+1000)

"+1" - to include upper bound values(9999 for previous example)

Ferocity answered 5/8, 2015 at 11:27 Comment(1)
The upper bound is exclusive with this method, so if you want to include the top number you would need to do FLOOR(RAND(CHECKSUM(NEWID()))*(10000-1000)+1000)Grantham
E
30

Answering the old question, but this answer has not been provided previously, and hopefully this will be useful for someone finding this results through a search engine.

With SQL Server 2008, a new function has been introduced, CRYPT_GEN_RANDOM(8), which uses CryptoAPI to produce a cryptographically strong random number, returned as VARBINARY(8000). Here's the documentation page: https://learn.microsoft.com/en-us/sql/t-sql/functions/crypt-gen-random-transact-sql

So to get a random number, you can simply call the function and cast it to the necessary type:

select CAST(CRYPT_GEN_RANDOM(8) AS bigint)

or to get a float between -1 and +1, you could do something like this:

select CAST(CRYPT_GEN_RANDOM(8) AS bigint) % 1000000000 / 1000000000.0
Empson answered 14/6, 2017 at 20:13 Comment(3)
Nice actually, when you’re worried about NEWID() running low or being semantically unfitting, or just need a positive integer from ABS(CAST(CRYPT_GEN_RANDOM(4)AS INT)).Kuo
NOTE: CRYPT_GEN_RANDOM cannot be called from within a db function. "Invalid use of a side-effecting operator 'Crypt_Gen_Random' within a function." Perhaps it writes to some system table.Melba
That is also true for NEWID() and RAND().Melba
R
15

The Rand() function will generate the same random number, if used in a table SELECT query. Same applies if you use a seed to the Rand function. An alternative way to do it, is using this:

SELECT ABS(CAST(CAST(NEWID() AS VARBINARY) AS INT)) AS [RandomNumber]

Got the information from here, which explains the problem very well.

Restive answered 25/6, 2009 at 17:23 Comment(0)
O
7

Do you have an integer value in each row that you could pass as a seed to the RAND function?

To get an integer between 1 and 14 I believe this would work:

FLOOR( RAND(<yourseed>) * 14) + 1
Overshoe answered 25/6, 2009 at 17:23 Comment(2)
This works in theory, but in practice I've found the RAND(<seed>) doesn't appear to be very random for minor changes in <seed>. For example a quick test I did: I let <seed> be 184380, 184383, 184386, and the corresponding RAND(<seed>) values were: 0.14912, 0.14917, 0.14923.Chariot
Maybe to get some more "seemingly" random results, try something like: RAND(<seed>)*100000) - FLOOR(RAND(<seed>)*100000)Chariot
A
7
select round(rand(checksum(newid()))*(10)+20,2)

Here the random number will come in between 20 and 30. round will give two decimal place maximum.

If you want negative numbers you can do it with

select round(rand(checksum(newid()))*(10)-60,2)

Then the min value will be -60 and max will be -50.

Apodaca answered 7/11, 2017 at 10:37 Comment(0)
U
6

If you need to preserve your seed so that it generates the "same" random data every time, you can do the following:

1. Create a view that returns select rand()

if object_id('cr_sample_randView') is not null
begin
    drop view cr_sample_randView
end
go

create view cr_sample_randView
as
select rand() as random_number
go

2. Create a UDF that selects the value from the view.

if object_id('cr_sample_fnPerRowRand') is not null
begin
    drop function cr_sample_fnPerRowRand
end
go

create function cr_sample_fnPerRowRand()
returns float
as
begin
    declare @returnValue float
    select @returnValue = random_number from cr_sample_randView
    return @returnValue
end
go

3. Before selecting your data, seed the rand() function, and then use the UDF in your select statement.

select rand(200);   -- see the rand() function
with cte(id) as
(select row_number() over(order by object_id) from sys.all_objects)
select 
    id,
    dbo.cr_sample_fnPerRowRand()
from cte
where id <= 1000    -- limit the results to 1000 random numbers
Usually answered 11/5, 2015 at 17:15 Comment(0)
R
5

If you don't need it to be an integer, but any random unique identifier, you can use newid()

SELECT table_name, newid() magic_number 
FROM information_schema.tables
Residence answered 25/6, 2009 at 17:22 Comment(0)
S
5

try using a seed value in the RAND(seedInt). RAND() will only execute once per statement that is why you see the same number each time.

Singultus answered 25/6, 2009 at 17:22 Comment(1)
Simplest! Though the values seem a lot more scattered, using digits from the middle of that, like RIGHT(CONVERT(BIGINT, RAND(RecNo) * 1000000000000), 2) (note: I'm seeing RIGHT implicitly convert the BIGINT to CHAR, but to be rigorous, you'd have another CONVERT in there).Corona
V
5

You would need to call RAND() for each row. Here is a good example

https://web.archive.org/web/20090216200320/http://dotnet.org.za/calmyourself/archive/2007/04/13/sql-rand-trap-same-value-per-row.aspx

Vannoy answered 25/6, 2009 at 17:22 Comment(3)
Dead link :( Any copies that could be included into the answer?Thermostatics
He puts RAND() into a view, puts a SELECT of that view into a function, and then calls the function from anywhere. Clever.Corona
I posted a solution that solves the problem in exactly the same way as in the linked article, but here in this blog directly as an answer five posts ago! No one called me clever envy face heheUsually
O
4

The problem I sometimes have with the selected "Answer" is that the distribution isn't always even. If you need a very even distribution of random 1 - 14 among lots of rows, you can do something like this (my database has 511 tables, so this works. If you have less rows than you do random number span, this does not work well):

SELECT table_name, ntile(14) over(order by newId()) randomNumber 
FROM information_schema.tables

This kind of does the opposite of normal random solutions in the sense that it keeps the numbers sequenced and randomizes the other column.

Remember, I have 511 tables in my database (which is pertinent only b/c we're selecting from the information_schema). If I take the previous query and put it into a temp table #X, and then run this query on the resulting data:

select randomNumber, count(*) ct from #X
group by randomNumber

I get this result, showing me that my random number is VERY evenly distributed among the many rows:

enter image description here

Ozone answered 4/4, 2015 at 2:17 Comment(0)
G
4

It's as easy as:

DECLARE @rv FLOAT;
SELECT @rv = rand();

And this will put a random number between 0-99 into a table:

CREATE TABLE R
(
    Number int
)

DECLARE @rv FLOAT;
SELECT @rv = rand();

INSERT INTO dbo.R
(Number)
    values((@rv * 100));

SELECT * FROM R
Guyot answered 30/10, 2018 at 1:35 Comment(0)
P
3

Use newid()

select newid()

or possibly this

select binary_checksum(newid())
Peroration answered 25/6, 2009 at 17:22 Comment(0)
D
3
select ABS(CAST(CAST(NEWID() AS VARBINARY) AS INT)) as [Randomizer]

has always worked for me

Discoloration answered 24/10, 2017 at 20:31 Comment(0)
S
3

If you want to generate a random number between 1 and 14 inclusive.

SELECT CONVERT(int, RAND() * (14 - 1) + 1)

OR

SELECT ABS(CHECKSUM(NewId())) % (14 -1) + 1
Sanyu answered 27/1, 2021 at 20:44 Comment(0)
G
2
    DROP VIEW IF EXISTS vwGetNewNumber;
    GO
    Create View vwGetNewNumber
    as
    Select CAST(RAND(CHECKSUM(NEWID())) * 62 as INT) + 1 as NextID,
    'abcdefghijklmnopqrstuvwxyz0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ'as alpha_num;

    ---------------CTDE_GENERATE_PUBLIC_KEY -----------------
    DROP FUNCTION IF EXISTS CTDE_GENERATE_PUBLIC_KEY;  
    GO
    create function CTDE_GENERATE_PUBLIC_KEY()
    RETURNS NVARCHAR(32)
    AS 
    BEGIN
        DECLARE @private_key NVARCHAR(32);
        set @private_key = dbo.CTDE_GENERATE_32_BIT_KEY();
        return @private_key;
    END;
    go

---------------CTDE_GENERATE_32_BIT_KEY -----------------
DROP FUNCTION IF EXISTS CTDE_GENERATE_32_BIT_KEY;  
GO
CREATE function CTDE_GENERATE_32_BIT_KEY()
RETURNS NVARCHAR(32)
AS 
BEGIN
    DECLARE @public_key NVARCHAR(32);
    DECLARE @alpha_num NVARCHAR(62);
    DECLARE @start_index INT = 0;
    DECLARE @i INT = 0;
    select top 1 @alpha_num = alpha_num from vwGetNewNumber;
        WHILE @i < 32
        BEGIN
          select top 1 @start_index = NextID from vwGetNewNumber;
          set @public_key = concat (substring(@alpha_num,@start_index,1),@public_key);
          set @i = @i + 1;
        END;
    return @public_key;
END;
    select dbo.CTDE_GENERATE_PUBLIC_KEY() public_key;
Gordy answered 27/6, 2018 at 12:33 Comment(3)
sorry @arnt if i did not explain well,Gordy
sorry @arnt , we have here two functions CTDE_GENERATE_32_BIT_KEY that generates a 32 bit alphanumeric key (can be extended to be more or less) and the other one called CTDE_GENERATE_PUBLIC_KEY that calls the first function and return back public key of 32 bit or you can return a private key of 16 bit ... you just need to call select dbo.CTDE_GENERATE_PUBLIC_KEY() as a public key; the logic behind is that we select one character from the alphanumeric character list 32 times and concatenate them together in order to get the random alphanumeric key. after research.Gordy
Nice. That explanation makes it a much better answer. (Someone flagged it for deletion; I voted to leave it open and left that comment for you.)Firestone
R
2
Update my_table set my_field = CEILING((RAND(CAST(NEWID() AS varbinary)) * 10))

Number between 1 and 10.

Rosabelle answered 18/4, 2019 at 18:22 Comment(0)
T
1

Try this:

SELECT RAND(convert(varbinary, newid()))*(b-a)+a magic_number 

Where a is the lower number and b is the upper number

Trowel answered 12/2, 2020 at 12:57 Comment(1)
Can you try to be more clear while answering a question?Pekin
T
1

If you need a specific number of random number you can use recursive CTE:

;WITH A AS (
        SELECT 1 X, RAND() R
    UNION ALL
        SELECT X + 1, RAND(R*100000) --Change the seed
        FROM A
        WHERE X < 1000 --How many random numbers you need
    )
SELECT
    X
    , RAND_BETWEEN_1_AND_14 = FLOOR(R * 14 + 1)
FROM A
OPTION (MAXRECURSION 0) --If you need more than 100 numbers
Thermopylae answered 14/4, 2021 at 19:15 Comment(0)

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