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how to add random items to an array using a function that generates random numbers

I have this problem I'm not really sure how to do it. I'm having trouble figuring out how to add items to an array using a function that generates random numbers.
First I have to build a function that takes three parameters. one is the minimum random number, the other one is the maximum, the third parameter is the length of the array.
I've tried some stuff like removing the variable x setting it to an empty array.
var newArray = [generateRandomArray(10, 1, 10)]
function getRandomInt(min, max) {
return Math.floor(Math.random() * (
max + 1 - min)) + min;
}
var generateRandomArray = function(max, min, length) {
var x;
for (y = 1; y >= length; y++) {
x = [x.push(getRandomInt(min, max))]
console.log(x)
}
return x;
}
var newArray = [generateRandomArray(10, 1, 10)]
returns [5, 7, 9, 6, 6, 10, 8, 10, 9, 7]
I just get []

You were attempting to call generateRandomArray in the snippet before the function existed, and your for loop in generateRandomArray would never run as y started as 1 and you were comparing it as >= length:
function getRandomInt(min, max) {
return Math.floor(Math.random() * (
max + 1 - min)) + min;
}
var generateRandomArray = function(max, min, length) {
var x = [];
for (var y = 0; y < length; y++) {
x.push(getRandomInt(min,max));
}
return x;
}
var newArray = generateRandomArray(10, 1, 10);
console.log(newArray);

Comments in code to show fixes.
Also just as a side note, I would recommend switching the min and max around on one of the two functions that use them for consistency. Just a Good Practice.
//moved generateRandomArray function to beginning so it is defined before it is called
var generateRandomArray = function(max, min, length) {
var x = []; //initialise array
for (y = 1; y <= length; y++) { //fix the incorrect sign so the loop works
x.push(getRandomInt(min, max)); //fix the fact you kept overriding x and instead just push to the array
console.log(x)
}
return x;
}
var newArray = [generateRandomArray(10, 1, 10)]
console.log("newArray", newArray);
function getRandomInt(min, max) {
return Math.floor(Math.random() * (
max + 1 - min)) + min;
}

Understanding formula for generating random number in interval [duplicate]

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?

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... ... 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.

var randomnumber = Math.floor(Math.random() * (maximum - minimum + 1)) + minimum;

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.

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.

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).

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)

function randomRange(min, max) {
return ~~(Math.random() * (max - min + 1)) + min
}
Alternative if you are using Underscore.js you can use
_.random(min, max)

If you need a variable between 0 and max, you can use:
Math.floor(Math.random() * max);

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

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))

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). 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.

Use this function to get random numbers in a given range:
function rnd(min, max) {
return Math.floor(Math.random()*(max - min + 1) + min);
}

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

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);
}

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).

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"> [' + (i + min - 1) + ']: '+array[i]+'</span> ');
}
document.write('<hr/>');
</script>
</head>
<body>
</body>
</html>

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;
}

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;

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.

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;
}

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.

I found this simple method on W3Schools:
Math.floor((Math.random() * max) + min);

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]

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;
}

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.

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);

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.

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.

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;
};

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;
}

How to get percent from an binary int, with every bit representing x percent?

I have an binary bitmap with 10bits. Every bit represents 10%. Is there any simple math function to get the sum of percent from the bitmap?
Sample
0000000000 = 0%
0000000001 = 10%
1000000000 = 10%
0000100000 = 10%
1000000001 = 20%
0000000011 = 20%
0000110000 = 20%
0010000010 = 20%
1010000010 = 30%
Be aware this is just an example of how the bits are activated. The number i have actually is an int such as 0,1 to 1023.

You don't have to use a loop. You don't have to do math. Just do it like this:
var number = 1000100010;
alert(number.toString().split("1").length - 1);
//A little more deep:
var number2 = 1100100000;
alert((number2.toString().split("1").length - 1) * 10 + "%");

Use a loop:
function count_1bits(n, bitsize) {
var count = 0;
for (var i = 0; i < bitsize; i++) {
count += (n & 1); // test low-order bit
n >>= 1; // shift number down 1 bit
}
return count;
}
var pct = count_1bits(bitmap, 10)*100;

You can count the bits with a function like this, then just multiply by 10 to get it as percent:
function countBits(n, cnt) {
if (cnt == 1) return n & 1;
var half = Math.floor(cnt / 2);
return countBits(n >> half, cnt - half) + countBits(n, half);
}
n = 1023;
document.write(countBits(n, 10) * 10 + '%');

Try doing something like
x = '0000100011';
function binTopre(x) {
c = (x.match(/1/g)||[]).length;
return (c*10) + '%';
}
console.log(binTopre(x));//30%

random percentage that fluctuates javascript/jquery

So I'm trying to pick a random number based on a percentage. 0-5
0 - 25% (25/100)
1 - 25% (25/100)
2 - 20% (20/100)
3 - 15% (15/100)
4 - 10% (10/100)
5 - 5% (5/100)
But, sometimes one or more of those values will need to be omitted from the picking. So you could encounter something like this
0 - 25% (25/65)
2 - 20% (20/65)
3 - 15% (15/65)
5 - 5% (5/65)
Which means that the percentage has to scale with the other numbers there.
I thought I could use a random function to do something like this
var1 = Math.floor((Math.random() * 100) + 1);
//0 - 25% random(1-25)
//1 - 25% random(26-50)
//2 - 20% random(51-70)
//3 - 15% random(71-85)
//4 - 10% random(86-95)
//5 - 5% random(96-100)
if(var1 <= 25){
var2 = 0;
}else if(var1 <= 50){
var2 = 1;
}else if(var1 <= 70){
var2 = 2;
}else if(var1 <= 85){
var2 = 3;
}else if(var1 <= 95){
var2 = 4;
}else if(var1 <= 100){
var2 = 5;
}else{
// error
}
But I ran into a problem when one or more of the variables (0-5) is omitted. It is possible to setup a bunch of if/else statements but it's not a very practical solution.
Does anybody know a better way I could possibly do this?
Also, if you're unsure of what my question is, please say so. I will try to be more specific and clear on my intentions.

I think, it is exactly what you want. Example
function RangeArray(diffs) {
this.diffs = diffs;
}
// Omit range by index
RangeArray.prototype.omitAt = function(index) {
var arr = this.diffs;
// move value to the next item in array
arr[index + 1] += arr[index];
arr[index] = 0;
}
// Find index of range for value
RangeArray.prototype.indexOf = function(value) {
var arr = this.diffs;
var sum = 0;
for (var index = 0; index < arr.length; ++index) {
sum += arr[index];
if (value < sum) {
return index;
}
}
return -1;
}
// ------- tests ----------
// array of ranges created using your percentage steps
var ranges = new RangeArray([25,25,20,15,10,5]);
// your random values
var values = [1, 26, 51, 70, 86, 99];
// test resutls: indexes of ranges for values
var indexes;
console.log('ranges: 1-25, 26-50, 51-70, 71-85, 86-95, 96-100');
console.log('random values: ' + values.join(', '));
// for your random values indexOf should return 0, 1, 2, 3, 4, 5 accordingly
indexes = values.map(function(x) { return ranges.indexOf(x); });
console.log('test 1 results: ' + indexes.join(', '));
// omit range at index 1 and 4
ranges.omitAt(1);
ranges.omitAt(4);
// for your random values indexOf should return 0, 2, 2, 3, 5, 5 accordingly
indexes = values.map(function(x) { return ranges.indexOf(x); });
console.log('test 2 results: ' + indexes.join(', '));

You store your ranges in an array with a the percentage of each block then you can check them. Be sure that your array sums up to 100.
You can use a simple loop.
var1 = Math.floor((Math.random() * 100) + 1);
var2 =0;
results = [25,25,20,15,10,5];
total = 0;
for (i = 0;i <results.length; i++ ) {
total += results[i];
if (var1 < total) {
var2 = i;
break;
}
}
console.log(var1)
console.log(var2)
example fiddle: http://jsfiddle.net/auo6nc7p/
So in the array above the first item has 25% probablility, the second 25% and so on. You can then remove the element from the array if you do not want them to be picked up.
Is that answering your question ?

Generate incremented random numbers between 0 to 100 every second where new number should be greater than the previous nymber

I need a javascript where with every second call i get a random number between 0-100 where current number should be greater than previous number. I wrote a code to get random numbers every second but stuck at generating it with increment.
<script>
var span = document.getElementsByTagName("span")[3];
var i = 100;
(function randNumber() {
var a = Math.floor((Math.random() * i) + 1);
span.innerHTML = "RandNumber: " + a;
setTimeout( randNumber, 1000);
})();
Note : The numbers should generate randomly.
example result may be : 2,5,7,8,22,23,34,56,78,88.....

You should not create a new random number between zero and your maximum (i), but just between the last created number (lastNumber) and max (i).
Also you might want to stop, when the random numbers reached the maximum.
var span = document.getElementsByTagName("span")[3],
i = 100,
lastNumber = 0;
function randNumber() {
lastNumber = lastNumber + Math.floor( Math.random() * (i - lastNumber) + 1 );
span.innerHTML = lastNumber;
if( lastNumber < i ) {
setTimeout( randNumber, 1000 );
}
}
randNumber();
AS for the comments and the requirement of a minimum amount of steps until reaching the maximum:
In each iteration, just increase you number by a random value between 1 and ((max - min) / steps. It is pretty much the same code as above.
var span = document.getElementsByTagName("span")[3],
max = 100,
min = 0,
lastNumber = 0,
minSteps = 30;
// how wide can the average step be at most?
var stepWidth = (max - min) / minSteps;
function randNumber() {
lastNumber = lastNumber + Math.floor( Math.random() * stepWidth + 1 );
span.innerHTML = lastNumber;
if( lastNumber < max ) {
setTimeout( randNumber, 1000 );
}
}
randNumber();

If you extract the "between two numbers" logic into its own function like this, this becomes a lot easier. This way, you just generate a number between your last generated number and your maximum.
var max = 100
var last_number = 1
var interval_id = setInterval(function(){
last_number = get_random_number_greater_between(last_number, max)
span.innerHTML = "RandNumber: " + last_number;
if(last_number >= max){
window.clearInterval(interval_id)
}
}, 1000)
function get_random_number_greater_between(low, high){
return Math.floor(Math.random() * (high - low + 1) + low);
}

Try this:
// generate a random number from 0 to max - 1.
function rand(max) {
return Math.floor(max * Math.random());
}
// generate a random number from min to max.
function range(min, max) {
return min + rand(1 + max - min);
}
// the loop function
function next(n, callback) {
var m = range(n, 100);
if (m < 100) setTimeout(next, 1000, m + 1, callback);
callback(m);
}
var span = document.getElementById("rand");
next(0, function (m) {
span.innerHTML = "Random number: " + m;
});
<span id="rand"></span>