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Generating random whole numbers in JavaScript in a specific range

How can I generate random whole numbers between two specified variables in JavaScript, e.g. x = 4 and y = 8 would output any of 4, 5, 6, 7, 8?

here is a useful gist: gist.github.com/kerimdzhanov/7529623
As a side note: for those using npm and looking for a quick, reliable and ready-made solution there's lodash.random that can be easily required with a super small footprint (it will import just the method itself and not the whole lodash).
Can you be explicit in the question about the number range? In particular, zero. What about negative numbers? ("Texts that exclude zero from the natural numbers sometimes refer to the natural numbers together with zero as the whole numbers"). (But without "Edit:", "Update:", or similar - the question should appear as if it was written today.)
Many answers here answer some different question (they are not real answers). It is like some users only read "Generating random whole numbers" and never get to the "in a specific range" part (or even the body with the [4; 8] example).

u
user229044

There are some examples on the Mozilla Developer Network page:

/**
 * Returns a random number between min (inclusive) and max (exclusive)
 */
function getRandomArbitrary(min, max) {
    return Math.random() * (max - min) + min;
}

/**
 * Returns a random integer between min (inclusive) and max (inclusive).
 * The value is no lower than min (or the next integer greater than min
 * if min isn't an integer) and no greater than max (or the next integer
 * lower than max if max isn't an integer).
 * Using Math.round() will give you a non-uniform distribution!
 */
function getRandomInt(min, max) {
    min = Math.ceil(min);
    max = Math.floor(max);
    return Math.floor(Math.random() * (max - min + 1)) + min;
}

Here's the logic behind it. It's a simple rule of three:

Math.random() returns a Number between 0 (inclusive) and 1 (exclusive). So we have an interval like this:

[0 .................................... 1)

Now, we'd like a number between min (inclusive) and max (exclusive):

[0 .................................... 1)
[min .................................. max)

We can use the Math.random to get the correspondent in the [min, max) interval. But, first we should factor a little bit the problem by subtracting min from the second interval:

[0 .................................... 1)
[min - min ............................ max - min)

This gives:

[0 .................................... 1)
[0 .................................... max - min)

We may now apply Math.random and then calculate the correspondent. Let's choose a random number:

                Math.random()
                    |
[0 .................................... 1)
[0 .................................... max - min)
                    |
                    x (what we need)

So, in order to find x, we would do:

x = Math.random() * (max - min);

Don't forget to add min back, so that we get a number in the [min, max) interval:

x = Math.random() * (max - min) + min;

That was the first function from MDN. The second one, returns an integer between min and max, both inclusive.

Now for getting integers, you could use round, ceil or floor.

You could use Math.round(Math.random() * (max - min)) + min, this however gives a non-even distribution. Both, min and max only have approximately half the chance to roll:

min...min+0.5...min+1...min+1.5   ...    max-0.5....max
└───┬───┘└────────┬───────┘└───── ... ─────┘└───┬──┘   ← Math.round()
   min          min+1                          max

With max excluded from the interval, it has an even less chance to roll than min.

With Math.floor(Math.random() * (max - min +1)) + min you have a perfectly even distribution.

min.... min+1... min+2 ... max-1... max.... max+1 (is excluded from interval)
|        |        |         |        |        |
└───┬───┘└───┬───┘└─── ... ┘└───┬───┘└───┬───┘   ← Math.floor()
   min     min+1               max-1    max

You can't use ceil() and -1 in that equation because max now had a slightly less chance to roll, but you can roll the (unwanted) min-1 result too.


It's only doing that because it's calling floor, which rounds down.
@thezachperson31 You could use round, but then both, min and max only had half the chance to roll like the other numbers do. You could also substract one and take ceil. This however leaves the max number with a minimal less chance to roll due to the [0,1) Interval.
I've created a JSFiddle if anyone wants to test the distribution of this method: jsfiddle.net/F9UTG/1
@JackFrost yeah, that's right. You're not dumb, you're just learning :)
This question is old, but understanding this answer took me way too much time O.o, I think expanding math.random on next JavaScript version would be kind of useful
D
Darin Dimitrov
var randomnumber = Math.floor(Math.random() * (maximum - minimum + 1)) + minimum;

I know this is a VERY old answer, but using (Math.random() * (maximum - minimum + 1) ) << 0 is faster.
@IsmaelMiguel Using binary operators (x << 0, x | 0, ~~x) instead of Math.floor() converts x into a two-complement with much smaller range than Number.MAX_SAFE_INTEGER (2³²⁻¹ vs. 2⁵³), thus you have to use it with caution!
@IsmaelMiguel Yo I just tried your method in the console and randomly got a negative value! Math.randRange = (minimum, maximum) => (Math.random() * (maximum - minimum + 1) ) << 0 Math.randRange(2,657348096152) -1407373159
@bluejayke Because 657348096152 (1001100100001100111111111111000010011000 in binary) has 40 bits, while bitwise arithmetics use 32 bits. If you do 657348096152|0 you get 218099864 (1100111111111111000010011000 in binary).
This is a smart answer. Making the range internally [min, max+1) actually achieves the desired result of [min, max] being both inclusive. Thank you! :)
C
Community

Math.random()

Returns an integer random number between min (included) and max (included):

function randomInteger(min, max) {
  return Math.floor(Math.random() * (max - min + 1)) + min;
}

Or any random number between min (included) and max (not included):

function randomNumber(min, max) {
  return Math.random() * (max - min) + min;
}

Useful examples (integers):

// 0 -> 10
Math.floor(Math.random() * 11);

// 1 -> 10
Math.floor(Math.random() * 10) + 1;

// 5 -> 20
Math.floor(Math.random() * 16) + 5;

// -10 -> (-2)
Math.floor(Math.random() * 9) - 10;

** And always nice to be reminded (Mozilla):

Math.random() does not provide cryptographically secure random numbers. Do not use them for anything related to security. Use the Web Crypto API instead, and more precisely the window.crypto.getRandomValues() method.


If you take Math.ceil, then +1 can be spared
P
Peter Mortensen

Use:

function getRandomizer(bottom, top) {
    return function() {
        return Math.floor( Math.random() * ( 1 + top - bottom ) ) + bottom;
    }
}

Usage:

var rollDie = getRandomizer( 1, 6 );

var results = ""
for ( var i = 0; i<1000; i++ ) {
    results += rollDie() + " ";    // Make a string filled with 1000 random numbers in the range 1-6.
}

Breakdown:

We are returning a function (borrowing from functional programming) that when called, will return a random integer between the the values bottom and top, inclusive. We say 'inclusive' because we want to include both bottom and top in the range of numbers that can be returned. This way, getRandomizer( 1, 6 ) will return either 1, 2, 3, 4, 5, or 6.

('bottom' is the lower number, and 'top' is the greater number)

Math.random() * ( 1 + top - bottom )

Math.random() returns a random double between 0 and 1, and if we multiply it by one plus the difference between top and bottom, we'll get a double somewhere between 0 and 1+b-a.

Math.floor( Math.random() * ( 1 + top - bottom ) )

Math.floor rounds the number down to the nearest integer. So we now have all the integers between 0 and top-bottom. The 1 looks confusing, but it needs to be there because we are always rounding down, so the top number will never actually be reached without it. The random decimal we generate needs to be in the range 0 to (1+top-bottom) so we can round down and get an integer in the range 0 to top-bottom:

Math.floor( Math.random() * ( 1 + top - bottom ) ) + bottom

The code in the previous example gave us an integer in the range 0 and top-bottom, so all we need to do now is add bottom to that result to get an integer in the range bottom and top inclusive. :D

NOTE: If you pass in a non-integer value or the greater number first you'll get undesirable behavior, but unless anyone requests it I am not going to delve into the argument checking code as it’s rather far from the intent of the original question.


I realize this is about 2½ years later, but with the input 1 and 6 your function returns values 1,2,3,4 and 5, but never a 6, as it would if it was "inclusive".
@some, It could be worse, I am 2½ years + 1 day later ^^
+1, I tested your code, it appears to create a correct value. Creative structure to handle fixed scenarios that might be repeated a lot in the code.
Why do you have a function within a function for this?
@Alph.Dev To decouple the logic that uses the random number generator from the logic that decides exactly what random number distribution to use. When the code that uses the random number generator accepts it as a parameter (a 0-argument function that always returns a new random number) it can work with any sort of random number generator.
P
Peter Mortensen

All these solutions are using way too much firepower. You only need to call one function: Math.random();

Math.random() * max | 0;

This returns a random integer between 0 (inclusive) and max (non-inclusive).


This is so clean. Thanks!
The OP was asking about a RANGE between 4 & 8, not 8 and 0
Then it doesn't work. Math.random() * 10 | 5 outputs only 5 | 7 | 13
Beware: This answer is not reliable. Max: 5 & Min: 1 returns: 1, 3, 5.
There's nothing wrong with the answer. People clearly have no clue what the | bitwise-OR operator does. As it is stated, this solution holds true for numbers between lower bound 0 and non-inclusive upper bound max.
n
nbro

Return a random number between 1 and 10:

Math.floor((Math.random()*10) + 1); 

Return a random number between 1 and 100:

Math.floor((Math.random()*100) + 1)

is your "between" inclusive or exclusive? i.e. is it [1,10], [1,10), (1,10], or (1,10)?
It is partialy inclusive: [1, *)
what is the need of + 1 at the end of the function? It works perfectly I guess.
@Shachi: It is the lower bound (a badly chosen example). 4, as in the question, would be better.
1 is too special. This will break down for other numbers, e.g., 4 and 8 as in the question (the range will (approximately) be [4;12], not [4;8]).
C
Codler
function randomRange(min, max) {
  return ~~(Math.random() * (max - min + 1)) + min
}

Alternative if you are using Underscore.js you can use

_.random(min, max)

Underscore actually provides a _.uniqueId() function you can call for client side models.
Using binary operators (x << 0, x | 0, ~~x) instead of Math.floor() converts x into a two-complement with much smaller range than Number.MAX_SAFE_INTEGER (2³²⁻¹ vs. 2⁵³), thus you have to use it with caution!
P
Peter Mortensen

If you need a variable between 0 and max, you can use:

Math.floor(Math.random() *  max);

Is max inclusive or exclusive>?
@Tree using Math.floor max is exclusive. If you want max to be inclusive you could use Math.round.
Like many other answers here, this doesn't answer the question (my emphasis): "How can I generate random whole numbers between two specified variables in JavaScript, e.g. x = 4 and y = 8 would output any of 4, 5, 6, 7, 8?". In other words, a random number in a specified range/closed interval ([4; 8] in the example). Even the title says "in a specific range". This belongs in comments.
S
Starkers

The other answers don't account for the perfectly reasonable parameters of 0 and 1. Instead you should use the round instead of ceil or floor:

function randomNumber(minimum, maximum){
    return Math.round( Math.random() * (maximum - minimum) + minimum);
}

console.log(randomNumber(0,1));  # 0 1 1 0 1 0
console.log(randomNumber(5,6));  # 5 6 6 5 5 6
console.log(randomNumber(3,-1)); # 1 3 1 -1 -1 -1

Your answer is true but I think your example is wrong.. console.log(randomNumber(5,6)); # 9 6 6 5 7 7 Do 9 & 7 come between 5 & 6? ...... you should correct it or explain..
The last example could be considered an empty range. For example, invalid input parameters. With an empty result, an error thrown, or similar.
P
Peter Mortensen

Cryptographically strong

To get a cryptographically strong random integer number in the range [x,y], try:

let cs = (x,y) => x + (y - x + 1)*crypto.getRandomValues(new Uint32Array(1))[0]/2**32 | 0 console.log(cs(4, 8))


this one is fascinating
I'd recommend this
I went down a rabbit hole on this one trying to learn what cryptographically secure even meant. Ended up here: crypto.stackexchange.com/questions/39186/…
+1, this is the best one! However, I used (x, y) => x + crypto.getRandomValues(new Uint32Array(1))[0] % (y - x + 1) (integer modulo rather than floating point division)
Would it be possible to get an explanation of e.g. when cryptographic strength is important and why this answer is different from the Math.random() ones?
3
3 revs, 2 users 54%

Use this function to get random numbers in a given range:

function rnd(min, max) {
    return Math.floor(Math.random()*(max - min + 1) + min);
}

How is this different from previous answers?
P
Peter Mortensen

Here is the Microsoft .NET Implementation of the Random class in JavaScript—

var Random = (function () {
    function Random(Seed) {
        if (!Seed) {
            Seed = this.milliseconds();
        }
        this.SeedArray = [];
        for (var i = 0; i < 56; i++)
            this.SeedArray.push(0);
        var num = (Seed == -2147483648) ? 2147483647 : Math.abs(Seed);
        var num2 = 161803398 - num;
        this.SeedArray[55] = num2;
        var num3 = 1;
        for (var i_1 = 1; i_1 < 55; i_1++) {
            var num4 = 21 * i_1 % 55;
            this.SeedArray[num4] = num3;
            num3 = num2 - num3;
            if (num3 < 0) {
                num3 += 2147483647;
            }
            num2 = this.SeedArray[num4];
        }
        for (var j = 1; j < 5; j++) {
            for (var k = 1; k < 56; k++) {
                this.SeedArray[k] -= this.SeedArray[1 + (k + 30) % 55];
                if (this.SeedArray[k] < 0) {
                    this.SeedArray[k] += 2147483647;
                }
            }
        }
        this.inext = 0;
        this.inextp = 21;
        Seed = 1;
    }

    Random.prototype.milliseconds = function () {
        var str = new Date().valueOf().toString();
        return parseInt(str.substr(str.length - 6));
    };

    Random.prototype.InternalSample = function () {
        var num = this.inext;
        var num2 = this.inextp;
        if (++num >= 56) {
            num = 1;
        }
        if (++num2 >= 56) {
            num2 = 1;
        }
        var num3 = this.SeedArray[num] - this.SeedArray[num2];
        if (num3 == 2147483647) {
            num3--;
        }
        if (num3 < 0) {
            num3 += 2147483647;
        }
        this.SeedArray[num] = num3;
        this.inext = num;
        this.inextp = num2;
        return num3;
    };

    Random.prototype.Sample = function () {
        return this.InternalSample() * 4.6566128752457969E-10;
    };

    Random.prototype.GetSampleForLargeRange = function () {
        var num = this.InternalSample();
        var flag = this.InternalSample() % 2 == 0;
        if (flag) {
            num = -num;
        }
        var num2 = num;
        num2 += 2147483646.0;
        return num2 / 4294967293.0;
    };

    Random.prototype.Next = function (minValue, maxValue) {
        if (!minValue && !maxValue)
            return this.InternalSample();
        var num = maxValue - minValue;
        if (num <= 2147483647) {
            return parseInt((this.Sample() * num + minValue).toFixed(0));
        }
        return this.GetSampleForLargeRange() * num + minValue;
    };

    Random.prototype.NextDouble = function () {
        return this.Sample();
    };

    Random.prototype.NextBytes = function (buffer) {
        for (var i = 0; i < buffer.length; i++) {
            buffer[i] = this.InternalSample() % 256;
        }
    };
    return Random;
}());

Use:

var r = new Random();
var nextInt = r.Next(1, 100); // Returns an integer between range
var nextDbl = r.NextDouble(); // Returns a random decimal

The MS DotNet Random class is under the Ms-RSL license, which means it is copyrighted. So be careful when using this derivated code, since it may form ground for copyright-infringement cases.
T
Travis

Here's what I use to generate random numbers.

function random(min,max) {
    return Math.floor((Math.random())*(max-min+1))+min;
}

Math.random() returns a number between 0 (inclusive) and 1 (exclusive). We multiply this number by the range (max-min). This results in a number between 0 (inclusive), and the range.

For example, take random(2,5)0. We multiply the random number 0≤x<1 by the range (5-2=3), so we now have a number, x where 0≤x<3.

In order to force the function to treat both the max and min as inclusive, we add 1 to our range calculation: Math.random()*(max-min+1). Now, we multiply the random number by the (5-2+1=4), resulting in an number, x, such that 0≤x<4. If we floor this calculation, we get an integer: 0≤x≤3, with an equal likelihood of each result (1/4).

Finally, we need to convert this into an integer between the requested values. Since we already have an integer between 0 and the (max-min), we can simply map the value into the correct range by adding the minimum value. In our example, we add 2 our integer between 0 and 3, resulting in an integer between 2 and 5.


As high is only used once, you may as well use high-low+1 rather than have the separate increment statement. Also, most users would expect the low parameter to come first.
random(2,5)0 doesn't appear to be valid.
@PeterMortensen Syntax Error. random(2,5) works fine. What is the 0 for?
P
Peter Mortensen

After generating a random number using a computer program, it is still considered as a random number if the picked number is a part or the full one of the initial one. But if it was changed, then mathematicians do not accept it as a random number and they can call it a biased number.

But if you are developing a program for a simple task, this will not be a case to consider. But if you are developing a program to generate a random number for a valuable stuff such as lottery program, or gambling game, then your program will be rejected by the management if you are not consider about the above case.

So for those kind of people, here is my suggestion:

Generate a random number using Math.random() (say this n):

Now for [0,10) ==>  n*10 (i.e. one digit) and for[10,100) ==> n*100 (i.e., two digits) and so on. Here square bracket indicates that the boundary is inclusive and a round bracket indicates the boundary is exclusive.

Then remove the rest after the decimal point. (i.e., get the floor) - using Math.floor(). This can be done.

If you know how to read the random number table to pick a random number, you know the above process (multiplying by 1, 10, 100 and so on) does not violate the one that I was mentioned at the beginning (because it changes only the place of the decimal point).

Study the following example and develop it to your needs.

If you need a sample [0,9] then the floor of n10 is your answer and if you need [0,99] then the floor of n100 is your answer and so on.

Now let’s enter into your role:

You've asked for numbers in a specific range. (In this case you are biased among that range. By taking a number from [1,6] by roll a die, then you are biased into [1,6], but still it is a random number if and only if the die is unbiased.)

So consider your range ==> [78, 247] number of elements of the range = 247 - 78 + 1 = 170; (since both the boundaries are inclusive).

/* Method 1: */
    var i = 78, j = 247, k = 170, a = [], b = [], c, d, e, f, l = 0;
    for(; i <= j; i++){ a.push(i); }
    while(l < 170){
        c = Math.random()*100; c = Math.floor(c);
        d = Math.random()*100; d = Math.floor(d);
        b.push(a[c]); e = c + d;
        if((b.length != k) && (e < k)){  b.push(a[e]); }
        l = b.length;
    }
    console.log('Method 1:');
    console.log(b);

/* Method 2: */

    var a, b, c, d = [], l = 0;
    while(l < 170){
        a = Math.random()*100; a = Math.floor(a);
        b = Math.random()*100; b = Math.floor(b);
        c = a + b;
        if(c <= 247 || c >= 78){ d.push(c); }else{ d.push(a); }
        l = d.length;
    }
    console.log('Method 2:');
    console.log(d);

Note: In method one, first I created an array which contains numbers that you need and then randomly put them into another array.

In method two, generate numbers randomly and check those are in the range that you need. Then put it into an array. Here I generated two random numbers and used the total of them to maximize the speed of the program by minimizing the failure rate that obtaining a useful number. However, adding generated numbers will also give some biasedness. So I would recommend my first method to generate random numbers within a specific range.

In both methods, your console will show the result (press F12 in Chrome to open the console).


"random" doesn't necessarily mean "uniformly distributed". "biased" doesn't imply "non-random". random means drawn from a probability distribution.
Can barely tell what this answer is trying to say. However, if you need random numbers for uses like lottery numbers and gambling. First you probably shouldn't be generating them on the client. Second, you need a cryptographically secure random number generator and the supplied algo is not sufficient. Calling random repeatedly does not make the result "more random". Author seems to be concerned about bias, but isn't providing a good algo for preventing it. In fact, the other short answers provided produce unbiased random numbers (assuming the underlying random generator is unbiased).
@JeffWalkerCodeRanger I think what he meant is that with the "normal" algorithm [i.e. Math.floor(Math.random() * (6 - 1 + 1) + 1)] the numbers 1 and 6 will necessarily be rolled fewer times than 2, 3, 4, and 5. However, the difference is basically insignificant.
P
Peter Mortensen

I wanted to explain using an example:

Function to generate random whole numbers in JavaScript within a range of 5 to 25

General Overview: (i) First convert it to the range - starting from 0. (ii) Then convert it to your desired range ( which then will be very easy to complete).

So basically, if you want to generate random whole numbers from 5 to 25 then:

First step: Converting it to range - starting from 0

Subtract "lower/minimum number" from both "max" and "min". i.e

(5-5) - (25-5)

So the range will be:

0-20 ...right?

Step two

Now if you want both numbers inclusive in range - i.e "both 0 and 20", the equation will be:

Mathematical equation: Math.floor((Math.random() * 21))

General equation: Math.floor((Math.random() * (max-min +1)))

Now if we add subtracted/minimum number (i.e., 5) to the range - then automatically we can get range from 0 to 20 => 5 to 25

Step three

Now add the difference you subtracted in equation (i.e., 5) and add "Math.floor" to the whole equation:

Mathematical equation: Math.floor((Math.random() * 21) + 5)

General equation: Math.floor((Math.random() * (max-min +1)) + min)

So finally the function will be:

function randomRange(min, max) {
   return Math.floor((Math.random() * (max - min + 1)) + min);
}

C
Chris
function getRandomInt(lower, upper)
{
    //to create an even sample distribution
    return Math.floor(lower + (Math.random() * (upper - lower + 1)));

    //to produce an uneven sample distribution
    //return Math.round(lower + (Math.random() * (upper - lower)));

    //to exclude the max value from the possible values
    //return Math.floor(lower + (Math.random() * (upper - lower)));
}

To test this function, and variations of this function, save the below HTML/JavaScript to a file and open with a browser. The code will produce a graph showing the distribution of one million function calls. The code will also record the edge cases, so if the the function produces a value greater than the max, or less than the min, you.will.know.about.it.

<html>
    <head>
        <script type="text/javascript">
        function getRandomInt(lower, upper)
        {
            //to create an even sample distribution
            return Math.floor(lower + (Math.random() * (upper - lower + 1)));

            //to produce an uneven sample distribution
            //return Math.round(lower + (Math.random() * (upper - lower)));

            //to exclude the max value from the possible values
            //return Math.floor(lower + (Math.random() * (upper - lower)));
        }

        var min = -5;
        var max = 5;

        var array = new Array();

        for(var i = 0; i <= (max - min) + 2; i++) {
          array.push(0);
        }

        for(var i = 0; i < 1000000; i++) {
            var random = getRandomInt(min, max);
            array[random - min + 1]++;
        }

        var maxSample = 0;
        for(var i = 0; i < max - min; i++) {
            maxSample = Math.max(maxSample, array[i]);
        }

        //create a bar graph to show the sample distribution
        var maxHeight = 500;
        for(var i = 0; i <= (max - min) + 2; i++) {
            var sampleHeight = (array[i]/maxSample) * maxHeight;

            document.write('<span style="display:inline-block;color:'+(sampleHeight == 0 ? 'black' : 'white')+';background-color:black;height:'+sampleHeight+'px">&nbsp;[' + (i + min - 1) + ']:&nbsp;'+array[i]+'</span>&nbsp;&nbsp;');
        }
        document.write('<hr/>');
        </script>
    </head>
    <body>

    </body>
</html>

n
nbro

For a random integer with a range, try:

function random(minimum, maximum) {
  var bool = true;

  while (bool) {
    var number = (Math.floor(Math.random() * maximum + 1) + minimum);
    if (number > 20) {
      bool = true;
    } else {
      bool = false;
    }
  }

  return number;
}

P
Peter Mortensen

To get a random number say between 1 and 6, first do:

0.5 + (Math.random() * ((6 - 1) + 1))

This multiplies a random number by 6 and then adds 0.5 to it. Next round the number to a positive integer by doing:

Math.round(0.5 + (Math.random() * ((6 - 1) + 1))

This round the number to the nearest whole number.

Or to make it more understandable do this:

var value = 0.5 + (Math.random() * ((6 - 1) + 1))
var roll = Math.round(value);
return roll;

In general, the code for doing this using variables is:

var value = (Min - 0.5) + (Math.random() * ((Max - Min) + 1))
var roll = Math.round(value);
return roll;

The reason for taking away 0.5 from the minimum value is because using the minimum value alone would allow you to get an integer that was one more than your maximum value. By taking away 0.5 from the minimum value you are essentially preventing the maximum value from being rounded up.


Make sense if you're excluding 0, then no need to "round down" range from 0 to 0.5.
P
Peter Mortensen

Using the following code, you can generate an array of random numbers, without repeating, in a given range.

function genRandomNumber(how_many_numbers, min, max) {

    // Parameters
    //
    //   how_many_numbers: How many numbers you want to
    //                     generate. For example, it is 5.
    //
    //   min (inclusive):  Minimum/low value of a range. It
    //                     must be any positive integer, but
    //                     less than max. I.e., 4.
    //
    //   max (inclusive):  Maximum value of a range. it must
    //                     be any positive integer. I.e., 50
    //
    //   Return type: array

    var random_number = [];
    for (var i = 0; i < how_many_numbers; i++) {
        var gen_num = parseInt((Math.random() * (max-min+1)) + min);
        do {
            var is_exist = random_number.indexOf(gen_num);
            if (is_exist >= 0) {
                gen_num = parseInt((Math.random() * (max-min+1)) + min);
            }
            else {
                random_number.push(gen_num);
                is_exist = -2;
            }
        }
        while (is_exist > -1);
    }
    document.getElementById('box').innerHTML = random_number;
}

P
Peter Mortensen

Random whole number between lowest and highest:

function randomRange(low, high) {
  var range = (high-low);
  var random = Math.floor(Math.random()*range);
  if (random === 0) {
    random += 1;
  }
  return low + random;
}

It is not the most elegant solution, but something quick.


The "+" in "+=" seems to be superfluous. The parentheses in (high-low) seems to be superfluous.
Considering the special case (if), what can be said about the distribution of the random numbers?
P
Peter Mortensen

I found this simple method on W3Schools:

Math.floor((Math.random() * max) + min);

Math.floor((Math.random() * 1) + 0); always gives 0
@madprops Because max number is exclusive. To get 0 or 1, you should set 2 as max number.
or you just add + 1 in the function that calls this method
Where did you find that? This will not output in the range [min; max]. For instance, if min = 3000 and max = 7000, it will (approximately) output in the range [3000; 10000] (not [3000; 7000]).
J
Jeremy

Math.random() is fast and suitable for many purposes, but it's not appropriate if you need cryptographically-secure values (it's not secure), or if you need integers from a completely uniform unbiased distribution (the multiplication approach used in others answers produces certain values slightly more often than others).

In such cases, we can use crypto.getRandomValues() to generate secure integers, and reject any generated values that we can't map uniformly into the target range. This will be slower, but it shouldn't be significant unless you're generating extremely large numbers of values.

To clarify the biased distribution concern, consider the case where we want to generate a value between 1 and 5, but we have a random number generator that produces values between 1 and 16 (a 4-bit value). We want to have the same number of generated values mapping to each output value, but 16 does not evenly divide by 5: it leaves a remainder of 1. So we need to reject 1 of the possible generated values, and only continue when we get one of the 15 lesser values that can be uniformly mapped into our target range. Our behaviour could look like this pseudocode:

Generate a 4-bit integer in the range 1-16.
If we generated  1,  6, or 11 then output 1.
If we generated  2,  7, or 12 then output 2.
If we generated  3,  8, or 13 then output 3.
If we generated  4,  9, or 14 then output 4.
If we generated  5, 10, or 15 then output 5.
If we generated 16 then reject it and try again.

The following code uses similar logic, but generates a 32-bit integer instead, because that's the largest common integer size that can be represented by JavaScript's standard number type. (This could be modified to use BigInts if you need a larger range.) Regardless of the chosen range, the fraction of generated values that are rejected will always be less than 0.5, so the expected number of rejections will always be less than 1.0 and usually close to 0.0; you don't need to worry about it looping forever.

const randomInteger = (min, max) => { const range = max - min; const maxGeneratedValue = 0xFFFFFFFF; const possibleResultValues = range + 1; const possibleGeneratedValues = maxGeneratedValue + 1; const remainder = possibleGeneratedValues % possibleResultValues; const maxUnbiased = maxGeneratedValue - remainder; if (!Number.isInteger(min) || !Number.isInteger(max) || max > Number.MAX_SAFE_INTEGER || min < Number.MIN_SAFE_INTEGER) { throw new Error('Arguments must be safe integers.'); } else if (range > maxGeneratedValue) { throw new Error(`Range of ${range} (from ${min} to ${max}) > ${maxGeneratedValue}.`); } else if (max < min) { throw new Error(`max (${max}) must be >= min (${min}).`); } else if (min === max) { return min; } let generated; do { generated = crypto.getRandomValues(new Uint32Array(1))[0]; } while (generated > maxUnbiased); return min + (generated % possibleResultValues); }; console.log(randomInteger(-8, 8)); // -2 console.log(randomInteger(0, 0)); // 0 console.log(randomInteger(0, 0xFFFFFFFF)); // 944450079 console.log(randomInteger(-1, 0xFFFFFFFF)); // Error: Range of 4294967296 covering -1 to 4294967295 is > 4294967295. console.log(new Array(12).fill().map(n => randomInteger(8, 12))); // [11, 8, 8, 11, 10, 8, 8, 12, 12, 12, 9, 9]


Is there a difference between crypto.getRandomValues (used here) and Crypto.getRandomValues?
P
Peter Mortensen

Here is an example of a JavaScript function that can generate a random number of any specified length without using Math.random():

function genRandom(length)
{
  const t1 = new Date().getMilliseconds();
  var min = "1", max = "9";
  var result;
  var numLength = length;
  if (numLength != 0)
  {
     for (var i = 1; i < numLength; i++)
     {
        min = min.toString() + "0";
        max = max.toString() + "9";
     }
  }
  else
  {
     min = 0;
     max = 0;
     return;
  }

  for (var i = min; i <= max; i++)
  {
       // Empty Loop
  }

  const t2 = new Date().getMilliseconds();
  console.log(t2);
  result = ((max - min)*t1)/t2;
  console.log(result);
  return result;
}

t1/t2 always very closer to 1. and hence your function returns same number when function is called repetitively .. jsbin.com/xogufacera/edit?js,console
check the jsbin url.. you will see the output yourself
It works great when length is between 4-10(As I have tested on my machine), because the values of constants T1 & T2 should have appropriate distance.
An explanation would be in order. E.g., what is the role of the time (timer/current date-time?)? What is the idea/gist? What is the purpose of the empty loop? To get some time to pass? What is some example output? From the Help Center: "...always explain why the solution you're presenting is appropriate and how it works". Please respond by editing (changing) your answer, not here in comments (without "Edit:", "Update:", or similar - the answer should appear as if it was written today).
P
Peter Mortensen

Use:

<!DOCTYPE html>
<html>
    <head>
        <meta charset="utf-8" />
    </head>

    <body>
        <script>
            /*
                Assuming that window.crypto.getRandomValues
                is available, the real range would be from
                0 to 1,998 instead of 0 to 2,000.

                See the JavaScript documentation
                for an explanation:

                  https://developer.mozilla.org/en-US/docs/Web/API/RandomSource/getRandomValues
            */
            var array = new Uint8Array(2);
            window.crypto.getRandomValues(array);
            console.log(array[0] + array[1]);
        </script>
    </body>
</html>

Uint8Array creates an array filled with a number up to three digits which would be a maximum of 999. This code is very short.


(The syntax highlighting of "Uint8Array" is truly weird.)
An explanation would be in order. E.g., what is the idea/gist? What is this window.crypto thingy? Why is it embedded in a web page? Would it work under Node.js? From the Help Center: "...always explain why the solution you're presenting is appropriate and how it works". Please respond by editing (changing) your answer, not here in comments (without "Edit:", "Update:", or similar - the answer should appear as if it was written today).
Like many other answers here, this doesn't answer the question. It answers some other question—(my emphasis) "How can I generate random whole numbers between two specified variables in JavaScript, e.g. x = 4 and y = 8 would output any of 4, 5, 6, 7, 8?". In other words, a random number in a specified range/closed interval ([4; 8] in the example). Even the title says "in a specific range".
P
Peter Mortensen

This is my take on a random number in a range, as in I wanted to get a random number within a range of base to exponent. E.g., base = 10, exponent = 2, gives a random number from 0 to 100, ideally, and so on.

If it helps using it, here it is:

// Get random number within provided base + exponent
// By Goran Biljetina --> 2012

function isEmpty(value) {
    return (typeof value === "undefined" || value === null);
}

var numSeq = new Array();

function add(num, seq) {
    var toAdd = new Object();
    toAdd.num = num;
    toAdd.seq = seq;
    numSeq[numSeq.length] = toAdd;
}

function fillNumSeq (num, seq) {
    var n;
    for(i=0; i<=seq; i++) {
        n = Math.pow(num, i);
        add(n, i);
    }
}

function getRandNum(base, exp) {
    if (isEmpty(base)) {
        console.log("Specify value for base parameter");
    }
    if (isEmpty(exp)) {
        console.log("Specify value for exponent parameter");
    }

    fillNumSeq(base, exp);

    var emax;
    var eseq;
    var nseed;
    var nspan;
    emax = (numSeq.length);
    eseq = Math.floor(Math.random()*emax) + 1;
    nseed = numSeq[eseq].num;
    nspan = Math.floor((Math.random())*(Math.random()*nseed)) + 1;
    return Math.floor(Math.random()*nspan) + 1;
}

console.log(getRandNum(10, 20), numSeq);

//Testing:
//getRandNum(-10, 20);
//console.log(getRandNum(-10, 20), numSeq);
//console.log(numSeq);

O
Oleg Valter is with Ukraine

Here is a function that generates a random number between min and max, both inclusive.

const randomInt = (max, min) => Math.round(Math.random() * (max - min)) + min;

How is that different from the previous answers? Does it work?
Yes, it works with both min and max being inclusive in the range using Math.round function.
I made the beginner mistake of posing more than one question. The capacity for more than one question is simply not there. Never ever do that. Always, always, always one question at a time.
S
Seb33300

This I guess, is the most simplified of all the contributions.

maxNum = 8,
minNum = 4

console.log(Math.floor(Math.random() * (maxNum - minNum) + minNum))

console.log(Math.floor(Math.random() * (8 - 4) + 4))

This will log random numbers between 4 and 8 into the console, 4 and 8 inclusive.


This doesnt work for maxNum = 2, minNum = 1, the outcome is always 1. In fact, I think it's the same with any min & max that are just 1 number appart; the lower boundary is always the result.
P
Peter Mortensen

Ionuț G. Stan wrote a great answer, but it was a bit too complex for me to grasp. So, I found an even simpler explanation of the same concepts at Math.floor( Math.random () * (max - min + 1)) + min) Explanation by Jason Anello.

Note: The only important thing you should know before reading Jason's explanation is a definition of "truncate". He uses that term when describing Math.floor(). Oxford dictionary defines "truncate" as:

Shorten (something) by cutting off the top or end.


P
Peter Mortensen

A function called randUpTo that accepts a number and returns a random whole number between 0 and that number:

var randUpTo = function(num) {
    return Math.floor(Math.random() * (num - 1) + 0);
};

A function called randBetween that accepts two numbers representing a range and returns a random whole number between those two numbers:

var randBetween = function (min, max) {
    return Math.floor(Math.random() * (max - min - 1)) + min;
};

A function called randFromTill that accepts two numbers representing a range and returns a random number between min (inclusive) and max (exclusive)

var randFromTill = function (min, max) {
    return Math.random() * (max - min) + min;
};

A function called randFromTo that accepts two numbers representing a range and returns a random integer between min (inclusive) and max (inclusive):

var randFromTo = function (min, max) {
    return Math.floor(Math.random() * (max - min + 1)) + min;
};

Beautiful, could you also note that randBetween is (exclusive) (exclusive)?
P
Peter Mortensen

You can you this code snippet,

let randomNumber = function(first, second) {
    let number = Math.floor(Math.random()*Math.floor(second));
    while(number < first) {

        number = Math.floor(Math.random()*Math.floor(second));
    }
    return number;
}

There is an unnecessary duplicated line here. Just use do - while instead of while
This will be very slow if first is 1,000,000,000 and second is 1,000,000,042. Perhaps add some caveats to your answer? Perhaps even some benchmarks to demonstrate when the performance implications become significant (to quantify it)? (But without "Edit:", "Update:", or similar - the answer should appear as if it was written today.)