Non volatile random numbers between

[Dicken]

Hi, 
is there a way to create a stablel random list bwtween, two numbers;     so a staable list  1 to 10  could be 

= List.Transform( 
 List.Random( 20,1), (x)=> 
   Number.IntegerDivide( ( x * 10  ) ,1 ) 
)

 the problem is this method always starts at  1   and i would like to cntrol the min and max vaulues,
what i woud like is stable version of   

 List.Transform( {1..100},(x)=>
let n = 
 Number.IntegerDivide( Number.RandomBetween( 97, 122),1)
 in 
 Character.FromNumber(n) )

 any suggestions, 

Richard.

[anilelmastasi]

Hello @Dicken ,

Can you try List.Random function? 

https://community.fabric.microsoft.com/t5/Power-BI-Community-Blog/List-RandomBetween-function-in-Pow...

You can check this.

If this solved your issue, please mark it as the accepted solution. ✅

[ibarrau]

Hi. Do you mean a list of random that even after a refresh will keep same values?

You could build it like the following code, specifying a seed to make it deterministic. Add a New blank query and create the function, you can call it getRandom:

(min as number, max as number, count as number, seed as number) =>
let
    range = max - min + 1,

    rnd =
        List.Transform(
            List.Random(count, seed),
            each Number.IntegerDivide(_ * range, 1) + min
        )
in
    rnd

Then you can call it any where, for example using your nros as example:

getRandom(97,122,100,1)

That would generate 100 estable numbers between 97 and 122.

I hope that helps,

[Dicken]

Not sure i get this ;

let
    range = max - min + 1,

    rnd =
        List.Transform(
            List.Random(count, seed),
            each Number.IntegerDivide(_ * range, 1) + min
        )
in
    rnd

it does not seem very different to what I posted, and haveing tried, i still have a range starting at 0 ,
can you show an actual example,  where a  random list stable List is produced between a min andn max valess  say   10 an 20 ?   I have tried the abvoe 

let 
range = (122 - 97)+1
in
    List.Transform( 
         List.Random( 20, 1) , (x)=> Number.IntegerDivide( x * range , 1)
    )

and it just takes gvies range  of 1  - 26  ? 

Richard.