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How do you do random?

David J Eddy on October 05, 2018

I was watching a VueJS tutorial and was surprised at what it takes in ECMAscript to generate a random integer between 0 and 10. Math.floor(Math....

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Tetri Mesquita • Oct 5 '18

const random = 7;

Fran González • Oct 6 '18

xkcd.com/221/

Mihail Malo • Oct 8 '18

Pfff, everyone knows random is 4 not 7.

Tetri Mesquita • Oct 8 '18

Pfff, you need update your mindset 🤨

Mihail Malo • Oct 8 '18 • Edited

Updates just break everything.
How can you know that no consumer of your code assumes it will be 4?
Sure, you claimed it is random, in the "documentation", but it's not like that ever works.

There are probably children out there holding down spacebar to stay warm in the winter! YOUR UPDATE MURDERS CHILDREN.

Elliot Derhay • Oct 6 '18

🤣

Yaser Al-Najjar • Oct 6 '18

You made my day 🤣🤣

JavaScript Joel • Oct 5 '18 • Edited

Math.random() produces a random float (all JavaScript numbers are of type float) from 0 to 1.

Adding the requirement of 0-10 means you have to multiply by 10 or * 10.

The last requirement was the number must be an Integer, which is the math.floor.

Of course, if you only wanted a random float between 0 and 1, it would be just one call.

And in PHP if you wanted a random float between 0 and 1, you would also have to do more work: Random Float between 0 and 1 in PHP

So the complexity actually comes from the requirements you give it.

But asking for a random integer is a common task. So it's handy to create function and keep it in your library. Don't go sprinkling Math.floor(Math.random() * 10) randomly around your codebase :D

const pseudoRandomInteger = (start, end) =>
  start + Math.floor(Math.random() * (end - start))

P.S. I prefixed this with pseudo because there is no true Random in JavaScript. This is very important when it comes to cryptography. Google PRNG if you want to learn more.

For a cryptographically strong random value, you might want to look into Crypto.getRandomValues()

Cheers!

David J Eddy • Oct 5 '18

...the complexity actually comes from the requirement...

I wish PM's understood this more.

Klemen Slavič • Oct 6 '18

There's a bug in your example; if the output is supposed to be [start, end), then the correct implementation is:

const pseudoRandomInteger = (start, end) =>
  start + Math.floor(Math.random() * (end - start));

✌😉

JavaScript Joel • Oct 6 '18

I probably should have run it once. lol.

I gotta stop typing code directly into editors.

Good catch!

Jochem Stoel • Oct 5 '18

const between = (min, max) => Math.floor(Math.random() * max) + min

Klemen Slavič • Oct 6 '18 • Edited

There is a subtle bug to consider with your code, though. If you want to guarantee an even spread among the numbers in the set of [0, 10], you'll have problems with 10 never showing up in your histogram. The reason for this is that the random value will almost never be exactly 1, as most PRNGs will actually output something in the range of [0, 1) (note the non-inclusiveness of 1 in the return values).

To correct that problem, you have to be a little smarter about how you map the [0,1] interval to buckets of integers:

const rand = (max) => Math.floor(Math.random() * (max + 1 - Number.EPSILON));

In the case of 10, you'll be multiplying a number in the [0,1] range to something that's just under 11. Number.EPSILON is the smallest float fraction that JavaScript numbers can express, so it's as close as the JavaScript number precision gets. And because the number 11 is never a value that appears in the output, Math.floor will never return anything above 10, guaranteed.

Of course, if your PRNG never returns 1 in its output, it's safe to just do max + 1. :)

Here's an illustration:

repl.it/@KrofDrakula/safeRandom

Elliot Derhay • Oct 6 '18

What about using Math.round() in place of Math.floor() in the original?

Klemen Slavič • Oct 6 '18

In that case, 0 and 10 would appear with about 50% the frequency of every other element.

If you imagine the number line after multiplying by 10, you'll understand what I mean:

0                    1                    2
|--------------------|--------------------|---....

<---------0---------><---------1---------><---....   Math.floor()
0---------><---------1---------><---------2---....   Math.round()
0<--------1----------><--------2----------><--....   Math.ceil()

This diagram shows you which floats map to which integer with different functions. As you can see, in the case of Math.round(), 0 has only half the length of every other number, so it would appear half as often. Same goes for 10 at the end of the interval:

9                    10
|--------------------|

9---------><--------10    Math.round()
<----------9--------->    Math.floor()

While 10 would appear in the resulting histogram, its frequency would be about half the frequency of every other non-zero integer.

FYI, Math.round() is just:

const round = n => Math.floor(n + 0.5);

You're effectively just shifting the mapping from floats to integers, but you're not making room for the max number. To have even distribution, you must have all intervals of equal length, and all the intervals covered by the scaled random number interval.

Klemen Slavič • Oct 6 '18

Also, try out the above repl.it link, you can make your own random functions to see if your hunches are correct. :)

Elliot Derhay • Oct 6 '18

I'm on my smartphone at the moment, but thanks. Your other response makes a lot of sense. That's something I hadn't thought about before.

Dylan Miracle • Aug 1 '19

Typically don't people mean the set of numbers [0, 9] when they say they want a random int between 0 and 10? I would assume they want something out of 10 vs something out of 11

Klemen Slavič • Aug 1 '19

Well that is where your specification definition alarm bells should start ringing like crazy. We don't have a precise mathematical definition for ranges and sets like this in everyday language among lay folk, therefore you have to clarify what you mean. The example I gave above explicitly lays out the expectations, which the original text hadn't.

Technically speaking saying "between 0 and 10" could mean (0, 10), [0, 10], (0, 10] or [0, 10). In most conversations I've had, people tend to treat ranges as inclusive, so [0, 10] would probably be my best guess, but that's just anecdotal evidence on my part, and should always be checked.

Boris Jamot ✊ / • Oct 5 '18

cat /dev/random

Rishabh Gupta • Oct 6 '18

Going the OG route

rhymes • Oct 5 '18 • Edited

Python:

import random
random.randint(0, 10)

Ruby:

rand(11)

Go:

import "math/rand"

func main() {
    rand.Intn(11)
}

None of these are cryptographically secure though

Rishabh Gupta • Oct 6 '18

Python has a cryptographically secure library called "secrets", which I use a lot :-P

rhymes • Oct 6 '18

Yeah secrets is very useful and has a clear API

Will F • Oct 5 '18

Python has a few options, including the numpy module. Regardless of language, however, it often only takes a ~20 lines to define a pseudo-random (in numerical math we avoid saying "random") number generator method. Here is an implementation of the so-called Mersenne twister:

def rando(n, seed, S = [0, 10, 'integer']):
    i = int(16807)
    j = int(2147483647)
    u_list = []
    if S[2] == 'real':
        for k in range(n):
            seed = (i * seed) % j
            u = (S[1] - S[0]) * (float(seed) / float(j)) + S[0]
            u_list.append(u)
        return u_list
    elif S[2] == 'integer':
        for k in range(n):
            seed = (i * seed) % j
            u = int((S[1] - S[0] + 1) * (float(seed) / float(j)) + S[0])
            u_list.append(u)
        return u_list
    else:
        print('Set of numbers to draw from is undefined')

In this method, we declare two integers, i (16807 = 7^5), and j (the Mersenne prime). In specifying the arguments n (number of psuedo-random numbers desired), seed (for repeatability of the pseudo-random numbers), and S (min, max, types to be generated), what is repeatedly happening (depending on n) is that we are taking the remainder of (seed * i) and j, and dividing by j; over and over. This implementation of the Mersenne Twister is good for about 2³⁰ random numbers before it repeats itself. :)

Raunak Ramakrishnan • Oct 6 '18 • Edited

APL:

⍝ `? n` generates random number between 1 and n, both inclusive
(? 10) -1
⍝ If we need more than 1 number, use `⍴` function like:
(? 4⍴ 10) -1
⍝ In above code, x⍴y gives you an array of y's of length x

I am a beginner in this language but it is fascinating at how succinct and elegant some constructs are.

Adrian B.G. • Oct 5 '18

Others responded on-topic, so I would add an off-topic :)

I do not build a JS project without Lodash (or something similar), it covers the basic needs and lacks of JS, acting like a Standard library in other languages. A few of them are related to random

_.random([lower=0], [upper=1], [floating])

_.sample(collection) //get a random element
_.sampleSize(collection, [n=1])

But, statistically speaking, taking a float random and truncating it to just a small integer range it's like buying a Ferrari for commuting, just saying :D Most of the times is a waste of resources.

Devs should be more careful at the distributions, most of the cases the code will be used for a few values and a small range of values, which will lead to very not-so-random results.

If the random results will affect the User Experience you should think twice of using it, the screwed distribution will affect your sales and KPIs in unknown ways, most likely negative, and there are always better alternatives, that require more resources to implement of course.

Ross Henderson • Oct 8 '18

In PL/SQL:

A random number.

select DBMS_RANDOM.random from dual;
--Result: -860942438

A random round number.

select round(DBMS_RANDOM.random) from dual;
--Result: 1096364454

A random number between two values.

select round(DBMS_RANDOM.value(0, 100)) from dual;
--Result: 54

kip • Oct 5 '18

From a terminal can be(no for encrypt things):

> shuf -i 0-10 -n 1
4
>

David J Eddy • Oct 5 '18

They see me shuffling. LMFAO

Devin W. Leaman • Oct 6 '18

I originally found it on the Hey, Scripting Guy! blog, but it's pretty simple to use:

Get-Random -Count 2 -InputObject (1..10)

Where Count is how many random numbers you wish to generate, and InputObject is an array of the range of numbers you want to choose from.

Thorsten Hirsch • Oct 7 '18

Don't trust your own computer for randomness. Better trust a webservice. ;-)

wget -q -O - "https://www.random.org/integers/?num=1&min=1&max=9&col=1&base=10&format=plain&rnd=new"

Theqadiva • Oct 19 '18

I'm a tester and I use random in my automation. In groovy, I use:

import org.apache.commons.lang.RandomStringUtils as RandomStringUtils WebUI.setText(findTestObject('someWhereToPutNumbers'),RandomStringUtils.randomNumeric(4))

Christopher C. Johnson • May 27 '19 • Edited

The quick and dirty way in C# to get one random number is:

var randomNumber = new Random().Next(1, 10)

Keep in mind that Random() initializes with the current timestamp so you can't use it in rapid succession and expect a random result.

static void Main(string[] args)
{
     int randomSingle = GetSingleRandomNumber(min: 1, max: 10);

     Console.WriteLine(randomSingle);

     Console.WriteLine(Environment.NewLine);

     List<int> randomList = ListOfRandomNumbers
(min: 1, max: 10, numberOfItems: 5);

     foreach (var number in randomList)
     {
         Console.WriteLine(number);
     }

     Console.ReadKey();
 }

 static int GetSingleRandomNumber(int min, int max)
 {
     return new Random().Next(min, max);
 }

 static List<int> ListOfRandomNumbers
(int min, int max, int numberOfItems)
 {
      var output = new List<int>();

      for (int i = 0; i < numberOfItems; i++)
      {
         output.Add(new Random().Next(min, max));
      }

      return output;
 }

It's important to note what we're doing in the two functions. We're always adding a new Random() i.e

output.Add(new Random().Next(min, max));

Why? Because we get a new timestamp when we do so.

If instead, you did something like this:

var random = new Random();
for (int i = 0; i < numberOfItems; i++)
{
     output.Add(random.Next(min, max));
}

You'd end up with several repeating numbers as it's calculating off the same timestamp. This is fine for small sets of random numbers just don't go newing up Random() for a large set.

Jonathan Kuhl • May 11 '19

const random = (max, min=0) => Math.floor(Math.random() * (max - min)) + min

I wish JS had a more inuitive means of doing random, a built in RandInt function like Python has. Ah well.

Still beats having to make an entire object just for a random number Java.

Jonathan Kuhl • Jun 22 '19

It bothers me that JS has no decent random functions and I have to reach for this bizarre mathy statement to get a random integer from a range.

Really wish we had Python's random module.

I mean, it's not hard to implement, but it's inconvenient.

cout970 • Oct 6 '18

In kotlin

Random.nextInt(0..10)

Random.nextInt(from = 0, until = 11)

Thanks to the Random api in the stdlib this works in the JVM, Kotlin.js and Kotlin native

TMcSquared • Oct 6 '18 • Edited

RPL for HP48/50 series

<< RAND 10 * FLOOR >>

Lars Klopstra ⚡ • Dec 25 '18

Math.random() * 10 | 0;

Arit Developer • Oct 12 '18

rand 0..10

Ruby

Ash Grennan • Jun 12 '20

var random = new Random(); var i = random.Next(0, 11);

Less elegant than some examples I'll admit

Wojtek Bal • Oct 6 '18

rand(11)

CyberianNeptune • Mar 8 '19

import random
a = random.randint(0,10)
return a

Arie Beresteanu • Oct 6 '18

real(8) :: r
integer :: i

CALL RANDOM_NUMBER(r)
i = floor(r *10.0)

Gerard Klijs • Oct 6 '18

In clojure it's (rand-int 11) which you can also use in clojurescript.

APixel Visuals • Oct 5 '18

JS still works better than C to randomly generate and have the numbers actually be random 👀

Lea Rosema (she/her) • Oct 7 '18

I had a workshop about random numbers at the CodePenDay 2017 in Hamburg by Bullgit: github.com/bullgit/fair-random