This is part of a bigger problem I try to solve in an exercise. It looks like this:
x is 10 times more likely to appear than y.
z appears 2x less often than y.
I solved this by calculating a single unit like this:
const x = 100;
const y = 10;
const z = 5;
const unit = 100 / (x + y + z);
unit equals 0.87
So when I do (0.87) + (0.87 * 10) + (0.87 * 5) I get 100%(almost)
Then I generate a random number between 0 and 1.
const randomNumber = Math.random();
function getValue() {
if (randomNumber <= 0.87) {
console.log('x');
} else if (randomNumber > 0.87 && randomNumber < 95.7) {
console.log('y');
} else console.log('z');
}
getValue();
If value<0.87 then I log out x, if value < 0.87+(0.087*10) I log y etc
Can anyone recommend a more logical and elegant way than this?
Your way looks clean for me except the fact that randomNumber > 0.87 is redundant.
if you store the value x, y and z in an array, you can probably write some cleaner code for example:
let prob = [100, 10, 5];
let sum = prob.reduce((a, b) => a + b, 0);
let normalizedProb = prob.map(p => p / sum);
let cummulativeProb = normalizedProb.map((cummulative => p => cummulative += p)(0));
for (let i = 0; i <= 50; i++) {
let r = Math.random();
console.log(cummulativeProb.filter(p => r >= p).length);
}
Also, you may want to read this post for faster implementation (in python though). However, the code will be more complicated for sure.
Since the weights are small integers, you can duplicate the x, y and z in an array, and just pick one random cell of the array:
let choices = "zyyxxxxxxxxxxxxxxxxxxxx";
console.log(choices[Math.floor(Math.random() * 23)]);
Here the magic number 23 is the number of choices, 1+2+20; and Math.floor(Math.random() * 23) is a random integer uniformly at random in range [0, 22] (both bounds included). See also:
Generating random whole numbers in JavaScript in a specific range?
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This is a sort of discrete math integration question - I need to fit a fixed quantity of items to a binomial distribution or bell curve over a fixed period.
Say I a total of M boxes being shipped over T days, where n is the number boxes that arrive on day t. I need a way to calculate n(t) for each day t, so that
The Sum ( n(t) ) 0 -> t = M
t is an integer, and n(t) is an integer, and
the shape of n(t) matches as closely as possible to a bell curve.
Edit
In case anyone does think this an SO-worthy question, here's the Javascript I cobbled together from the pointers in Yves Daoust answer.
/*
See https://stackoverflow.com/questions/5259421/
*/
function normal(x, mean, stdDev) {
return stdNormal(( x - mean ) / stdDev);
}
function stdNormal(z) {
// Power series is not stable at these extreme tail scenarios
if (z < -6) { return 0; }
if (z > 6) { return 1; }
let j, k ;
let m = 1; // m(k) == (2**k)/factorial(k)
let b = z; // b(k) == z ** (2*k + 1)
let z2 = z * z; // cache of z squared
let z4 = z2 * z2; // cache of z to the 4th
let values = [];
// Compute the power series in groups of two terms.
// This reduces floating point errors because the series
// alternates between positive and negative.
for (k=0; k<100; k+=2) {
const a = 2*k + 1;
let item = b / (a*m);
item *= (1 - (a*z2)/((a+1)*(a+2)));
values.push(item);
m *= (4*(k+1)*(k+2));
b *= z4;
}
// Add the smallest terms to the total first that
// way we minimize the floating point errors.
let total = 0;
for (k=49; k>=0; k--) {
total += values[k];
}
// Multiply total by 1/sqrt(2*PI)
// Then add 0.5 so that stdNormal(0) === 0.5
return 0.5 + 0.3989422804014327 * total;
}
/*
Compute the cdf of the binomial distribution between 0 and T, times M.
Round all values to integers. Let m(t) the numbers so obtained.
Use n(t) = m(t+1) - m(t). In doing so, you ensure Σ n(t) = n(T) - n(0) = M.
Thanks to Yves Daoust
*/
function distributeItems(itemsToPlace = 100, steps = 7) {
const mean = Math.floor((steps - 1) / 2);
const stdDev = mean / 2.96; /// for 'standard' std dev ;-)
const m = [0]; // cdf
const n = [0]; // items
for (var step = 1; step <= steps; step++) {
m.push(Math.round(normal(step, mean, stdDev) * itemsToPlace,0));
n.push( m[step] - m[step - 1] );
}
const interimCount = n.reduce(function (sum, elt) { return sum+elt; }, 0);
const discrepancy = itemsToPlace - interimCount;
if (discrepancy !==0) {
n[n.length-1] += discrepancy;
}
return n;
}
const n = distributeItems(40,7);
console.log('Items: ',n, 'Total: ',n.reduce(function (sum, elt) { return sum+elt; }, 0))
// Result
// [ 0, 1, 5, 14, 14, 5, 1, 0 ] 40
Maybe not an optimal solution, but should not be far.
Compute the cdf of the binomial distribution between 0 and T, times M.
Round all values to integers. Let m(t) the numbers so obtained.
Use n(t) = m(t+1) - m(t). In doing so, you ensure Σ n(t) = n(T) - n(0) = M.
In the picture below, the blue curve is an exact binomial N=7, p=0.3, adjusted for M=20. The orange curve is obtained by the above procedure. As you can check, 2+5+6+4+2+1+0+0=20.
For example I have 3 points: (y:0, x:0), (y:100, x:10), (y:50, x:100). So I need to get 10 points among this polyline the distance between those is equal. I know how to get points between 2 ones, but I defenitely don't know how to receive among several points.
For receiving distance between 2 points I use this:
function getDistance(y1,x1,y2,x2){
return Math.sqrt(Math.pow(y2-y1, 2) + Math.pow(x2-x1, 2))
}
For computing single points I use it:
function interpolate(a, b, frac){
return {
y: a.y+(b.y-a.y)*frac,
x: a.x+(b.x-a.x)*frac
};
}
Can anyone help me?
This is working fine (for the example I'm using 3 points on the same line, but it should work for every combination)
function getDistance({y: y1, x:x1}, {y:y2, x:x2}){
return Math.sqrt(Math.pow(y2-y1, 2) + Math.pow(x2-x1, 2))
}
function interpolate(a, b, frac){
return {
x: a.x+(b.x-a.x)*frac,
y: a.y+(b.y-a.y)*frac,
};
}
//generate N point in a line
function generateOnLineEveryDistance(a, b, n, distance){
let res = [];
for(let i = 0; i < n ; i++){
// add a point interpolating on a line after (i + 1) * distance from the beginning point (i+1 to avoid the starting point 0,0)
res.push(interpolate(a, b, (i + 1) * distance))
}
return res;
}
function generate(points, n){
// calculate total distance to find out how distant we have to place the points
let totalDistance = 0;
for(let i = 1; i < points.length; i++){
totalDistance += getDistance(points[i - 1], points[i])
}
// distance to place the points
const pointDistance = totalDistance / (n - 1);
// the first point is always included
let res = [points[0]];
// now, we will consider a segment as point[i - 1] & point[i], and we will consider only the piece where we can fit point:
// eg. in a segment long 10 ([x:0, y:0], [x:0, y:10]), and a pointDistance of 4, we will consider only [x:0, y:0], [x:0, y:8]
// and the remainder is 2... this remainder will be used in the next segment, "appending it at the beginning"
// let's say the second segment is [x:0, y:10], [x:0, y:20], with the remainder it will be [x:0, y:8], [x:0, y:20]
let remainder = 0;
for(let i = 1; i < points.length ; i++){
// add the remainder if exists at the beginning of the current segment (point[i-1], point[i])
// source https://stackoverflow.com/questions/7740507/extend-a-line-segment-a-specific-distance
if(remainder > 0){
let a = points[i];
let b = points[i - 1];
let lengthAB = Math.sqrt(Math.pow(a.x - b.x, 2) + Math.pow(a.y - b.y, 2))
points[i - 1].x = b.x + (b.x - a.x) / lengthAB * remainder;
points[i - 1].y = b.y + (b.y - a.y) / lengthAB * remainder;
}
// points we need to generate
let nPoints = Math.floor(getDistance(points[i - 1], points[i]) / pointDistance)
// remainder to add to the next iteration
remainder = getDistance(points[i - 1], points[i]) - nPoints * pointDistance;
// add to the result the points
res = [
...res, // previous result
...generateOnLineEveryDistance( // new points
points[i - 1], // first point to consider
points[i], // second point to consider
nPoints, // number of points to generate
pointDistance / getDistance(points[i - 1], points[i]) // the ratio we want to use to generate them
)
]
}
// small fix in case of .333333 decimals that will "miss" the last point because it's .9999 and not 1.00000
if(res.length < n){
res = [...res, points[points.length - 1]]
}
return res;
}
const points = [
{
x: 0,
y: 0
} , {
x: 0,
y: 10
} , {
x: 0,
y: 20
}
]
console.log(generate(points, 4))
however, I can see this library already doing it, I've not checked it out, but maybe is worth checking it out because the code I'm providing is pretty untested and unreadable
UPDATE:
I've tested it against the examples they are providing and it's giving back the same result, so GG
I have a array of n elements. I am at a given position say x and I need to move to position y, I need to find out what is the difference or number of steps if I traverse array by going forward and by going backward.
const foo = Array.from(Array(15).keys());
const length = foo.length;
const currentItem = 2;
const next = 12;
let diff = currentItem - next;
if (diff < 0) {
if (next > currentItem) {
diff = next - currentItem;
console.log((diff));
} else {
diff = length - Math.abs(diff);
console.log(-diff);
}
} else {
if (next < currentItem) {
console.log(-(diff));
} else {
console.log(diff);
}
}
I am trying to fin in above code if I need to move forward or backward. In above example I expect answer as -6 but I get answer 10. I am getting bit confused in the loops. Any better and smarter way to do this?
Here is how I'd do it:
// The % operator in JavaScript is the remainder operator.
// The following function defines the modulo operator.
const mod = (x, y) => (x % y + y) % y;
// Law: mod(start + shortestPath(start, end, length), length) === end
const shortestPath = (start, end, length) => {
const forward = mod(end - start, length);
const backward = mod(start - end, length);
return forward > backward ? -backward : forward;
};
const shortestPathLaw = (start, end, length) =>
mod(start + shortestPath(start, end, length), length) === end;
console.assert(shortestPathLaw(4, 6, 10));
console.assert(shortestPathLaw(8, 6, 10));
console.log(shortestPath(4, 6, 10)); // 2
console.log(shortestPath(8, 6, 10)); // -2
We use the modulo operator to calculate the forward distance from point A to point B (i.e. B - A). The backward distance from point A to point B is the same as the forward distance from point B to point A (i.e. A - B). Then we pick the shorter of the two distances, and if the distance we picked is the backward distance, then we negate it to show that we're traveling backward from A to reach B.
Is it possible to get a random number between 1-100 and keep the results mainly within the 40-60 range? I mean, it will go out of that range rarely, but I want it to be mainly within that range... Is it possible with JavaScript/jQuery?
Right now I'm just using the basic Math.random() * 100 + 1.
The simplest way would be to generate two random numbers from 0-50 and add them together.
This gives a distribution biased towards 50, in the same way rolling two dice biases towards 7.
In fact, by using a larger number of "dice" (as #Falco suggests), you can make a closer approximation to a bell-curve:
function weightedRandom(max, numDice) {
let num = 0;
for (let i = 0; i < numDice; i++) {
num += Math.random() * (max/numDice);
}
return num;
}
JSFiddle: http://jsfiddle.net/797qhcza/1/
You have some good answers here that give specific solutions; let me describe for you the general solution. The problem is:
I have a source of more-or-less uniformly distributed random numbers between 0 and 1.
I wish to produce a sequence of random numbers that follow a different distribution.
The general solution to this problem is to work out the quantile function of your desired distribution, and then apply the quantile function to the output of your uniform source.
The quantile function is the inverse of the integral of your desired distribution function. The distribution function is the function where the area under a portion of the curve is equal to the probability that the randomly-chosen item will be in that portion.
I give an example of how to do so here:
http://ericlippert.com/2012/02/21/generating-random-non-uniform-data/
The code in there is in C#, but the principles apply to any language; it should be straightforward to adapt the solution to JavaScript.
Taking arrays of numbers, etc. isn't efficient. You should take a mapping which takes a random number between 0 to 100 and maps to the distribution you need. So in your case, you could take f(x)=-(1/25)x2+4x to get a distribution with the most values in the middle of your range.
I might do something like setup a "chance" for the number to be allowed to go "out of bounds". In this example, a 20% chance the number will be 1-100, otherwise, 40-60:
$(function () {
$('button').click(function () {
var outOfBoundsChance = .2;
var num = 0;
if (Math.random() <= outOfBoundsChance) {
num = getRandomInt(1, 100);
} else {
num = getRandomInt(40, 60);
}
$('#out').text(num);
});
function getRandomInt(min, max) {
return Math.floor(Math.random() * (max - min + 1)) + min;
}
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<button>Generate</button>
<div id="out"></div>
fiddle: http://jsfiddle.net/kbv39s9w/
I needed to solve this problem a few years ago and my solution was easier than any of the other answers.
I generated 3 randoms between the bounds and averaged them. This pulls the result towards the centre but leaves it completely possible to reach the extremities.
It looks stupid but you can use rand twice:
var choice = Math.random() * 3;
var result;
if (choice < 2){
result = Math.random() * 20 + 40; //you have 2/3 chance to go there
}
else {
result = Math.random() * 100 + 1;
}
Sure it is possible. Make a random 1-100. If the number is <30 then generate number in range 1-100 if not generate in range 40-60.
There is a lot of different ways to generate such random numbers. One way to do it is to compute the sum of multiple uniformly random numbers. How many random numbers you sum and what their range is will determine how the final distribution will look.
The more numbers you sum up, the more it will be biased towards the center. Using the sum of 1 random number was already proposed in your question, but as you notice is not biased towards the center of the range. Other answers have propose using the sum of 2 random numbers or the sum of 3 random numbers.
You can get even more bias towards the center of the range by taking the sum of more random numbers. At the extreme you could take the sum of 99 random numbers which each were either 0 or 1. That would be a binomial distribution. (Binomial distributions can in some sense be seen as the discrete version of normal distributions). This can still in theory cover the full range, but it has so much bias towards the center that you should never expect to see it reach the endpoints.
This approach means you can tweak just how much bias you want.
What about using something like this:
var loops = 10;
var tries = 10;
var div = $("#results").html(random());
function random() {
var values = "";
for(var i=0; i < loops; i++) {
var numTries = tries;
do {
var num = Math.floor((Math.random() * 100) + 1);
numTries--;
}
while((num < 40 || num >60) && numTries > 1)
values += num + "<br/>";
}
return values;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<div id="results"></div>
The way I've coded it allows you to set a couple of variables:
loops = number of results
tries = number of times the function will try to get a number between 40-60 before it stops running through the while loop
Added bonus: It uses do while!!! Awesomeness at its best
You can write a function that maps random values between [0, 1) to [1, 100] according to weight. Consider this example:
Here, the value 0.95 maps to value between [61, 100].
In fact we have .05 / .1 = 0.5, which, when mapped to [61, 100], yields 81.
Here is the function:
/*
* Function that returns a function that maps random number to value according to map of probability
*/
function createDistributionFunction(data) {
// cache data + some pre-calculations
var cache = [];
var i;
for (i = 0; i < data.length; i++) {
cache[i] = {};
cache[i].valueMin = data[i].values[0];
cache[i].valueMax = data[i].values[1];
cache[i].rangeMin = i === 0 ? 0 : cache[i - 1].rangeMax;
cache[i].rangeMax = cache[i].rangeMin + data[i].weight;
}
return function(random) {
var value;
for (i = 0; i < cache.length; i++) {
// this maps random number to the bracket and the value inside that bracket
if (cache[i].rangeMin <= random && random < cache[i].rangeMax) {
value = (random - cache[i].rangeMin) / (cache[i].rangeMax - cache[i].rangeMin);
value *= cache[i].valueMax - cache[i].valueMin + 1;
value += cache[i].valueMin;
return Math.floor(value);
}
}
};
}
/*
* Example usage
*/
var distributionFunction = createDistributionFunction([
{ weight: 0.1, values: [1, 40] },
{ weight: 0.8, values: [41, 60] },
{ weight: 0.1, values: [61, 100] }
]);
/*
* Test the example and draw results using Google charts API
*/
function testAndDrawResult() {
var counts = [];
var i;
var value;
// run the function in a loop and count the number of occurrences of each value
for (i = 0; i < 10000; i++) {
value = distributionFunction(Math.random());
counts[value] = (counts[value] || 0) + 1;
}
// convert results to datatable and display
var data = new google.visualization.DataTable();
data.addColumn("number", "Value");
data.addColumn("number", "Count");
for (value = 0; value < counts.length; value++) {
if (counts[value] !== undefined) {
data.addRow([value, counts[value]]);
}
}
var chart = new google.visualization.ColumnChart(document.getElementById("chart"));
chart.draw(data);
}
google.load("visualization", "1", { packages: ["corechart"] });
google.setOnLoadCallback(testAndDrawResult);
<script src="https://www.google.com/jsapi"></script>
<div id="chart"></div>
Here's a weighted solution at 3/4 40-60 and 1/4 outside that range.
function weighted() {
var w = 4;
// number 1 to w
var r = Math.floor(Math.random() * w) + 1;
if (r === 1) { // 1/w goes to outside 40-60
var n = Math.floor(Math.random() * 80) + 1;
if (n >= 40 && n <= 60) n += 40;
return n
}
// w-1/w goes to 40-60 range.
return Math.floor(Math.random() * 21) + 40;
}
function test() {
var counts = [];
for (var i = 0; i < 2000; i++) {
var n = weighted();
if (!counts[n]) counts[n] = 0;
counts[n] ++;
}
var output = document.getElementById('output');
var o = "";
for (var i = 1; i <= 100; i++) {
o += i + " - " + (counts[i] | 0) + "\n";
}
output.innerHTML = o;
}
test();
<pre id="output"></pre>
Ok, so I decided to add another answer because I felt like my last answer, as well as most answers here, use some sort of half-statistical way of obtaining a bell-curve type result return. The code I provide below works the same way as when you roll a dice. Therefore, it is hardest to get 1 or 99, but easiest to get 50.
var loops = 10; //Number of numbers generated
var min = 1,
max = 50;
var div = $("#results").html(random());
function random() {
var values = "";
for (var i = 0; i < loops; i++) {
var one = generate();
var two = generate();
var ans = one + two - 1;
var num = values += ans + "<br/>";
}
return values;
}
function generate() {
return Math.floor((Math.random() * (max - min + 1)) + min);
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<div id="results"></div>
I'd recommend using the beta distribution to generate a number between 0-1, then scale it up. It's quite flexible and can create many different shapes of distributions.
Here's a quick and dirty sampler:
rbeta = function(alpha, beta) {
var a = 0
for(var i = 0; i < alpha; i++)
a -= Math.log(Math.random())
var b = 0
for(var i = 0; i < beta; i++)
b -= Math.log(Math.random())
return Math.ceil(100 * a / (a+b))
}
var randNum;
// generate random number from 1-5
var freq = Math.floor(Math.random() * (6 - 1) + 1);
// focus on 40-60 if the number is odd (1,3, or 5)
// this should happen %60 of the time
if (freq % 2){
randNum = Math.floor(Math.random() * (60 - 40) + 40);
}
else {
randNum = Math.floor(Math.random() * (100 - 1) + 1);
}
The best solution targeting this very problem is the one proposed by BlueRaja - Danny Pflughoeft but I think a somewhat faster and more general solution is also worth mentioning.
When I have to generate random numbers (strings, coordinate pairs, etc.) satisfying the two requirements of
The result set is quite small. (not larger than 16K numbers)
The result set is discreet. (like integer numbers only)
I usually start by creating an array of numbers (strings, coordinate pairs, etc.) fulfilling the requirement (In your case: an array of numbers containing the more probable ones multiple times.), then choose a random item of that array. This way, you only have to call the expensive random function once per item.
Distribution
5% for [ 0,39]
90% for [40,59]
5% for [60,99]
Solution
var f = Math.random();
if (f < 0.05) return random(0,39);
else if (f < 0.95) return random(40,59);
else return random(60,99);
Generic Solution
random_choose([series(0,39),series(40,59),series(60,99)],[0.05,0.90,0.05]);
function random_choose (collections,probabilities)
{
var acc = 0.00;
var r1 = Math.random();
var r2 = Math.random();
for (var i = 0; i < probabilities.length; i++)
{
acc += probabilities[i];
if (r1 < acc)
return collections[i][Math.floor(r2*collections[i].length)];
}
return (-1);
}
function series(min,max)
{
var i = min; var s = [];
while (s[s.length-1] < max) s[s.length]=i++;
return s;
}
You can use a helper random number to whether generate random numbers in 40-60 or 1-100:
// 90% of random numbers should be between 40 to 60.
var weight_percentage = 90;
var focuse_on_center = ( (Math.random() * 100) < weight_percentage );
if(focuse_on_center)
{
// generate a random number within the 40-60 range.
alert (40 + Math.random() * 20 + 1);
}
else
{
// generate a random number within the 1-100 range.
alert (Math.random() * 100 + 1);
}
If you can use the gaussian function, use it. This function returns normal number with average 0 and sigma 1.
95% of this number are within average +/- 2*sigma. Your average = 50, and sigma = 5 so
randomNumber = 50 + 5*gaussian()
The best way to do that is generating a random number that is distributed equally in a certain set of numbers, and then apply a projection function to the set between 0 and a 100 where the projection is more likely to hit the numbers you want.
Typically the mathematical way of achieving this is plotting a probability function of the numbers you want. We could use the bell curve, but let's for the sake of easier calculation just work with a flipped parabola.
Let's make a parabola such that its roots are at 0 and 100 without skewing it. We get the following equation:
f(x) = -(x-0)(x-100) = -x * (x-100) = -x^2 + 100x
Now, all the area under the curve between 0 and 100 is representative of our first set where we want the numbers generated. There, the generation is completely random. So, all we need to do is find the bounds of our first set.
The lower bound is, of course, 0. The upper bound is the integral of our function at 100, which is
F(x) = -x^3/3 + 50x^2
F(100) = 500,000/3 = 166,666.66666 (let's just use 166,666, because rounding up would make the target out of bounds)
So we know that we need to generate a number somewhere between 0 and 166,666. Then, we simply need to take that number and project it to our second set, which is between 0 and 100.
We know that the random number we generated is some integral of our parabola with an input x between 0 and 100. That means that we simply have to assume that the random number is the result of F(x), and solve for x.
In this case, F(x) is a cubic equation, and in the form F(x) = ax^3 + bx^2 + cx + d = 0, the following statements are true:
a = -1/3
b = 50
c = 0
d = -1 * (your random number)
Solving this for x yields you the actual random number your are looking for, which is guaranteed to be in the [0, 100] range and a much higher likelihood to be close to the center than the edges.
This answer is really good. But I would like to post implementation instructions (I'm not into JavaScript, so I hope you will understand) for different situation.
Assume you have ranges and weights for every range:
ranges - [1, 20], [21, 40], [41, 60], [61, 100]
weights - {1, 2, 100, 5}
Initial Static Information, could be cached:
Sum of all weights (108 in sample)
Range selection boundaries. It basically this formula: Boundary[n] = Boundary[n - 1] + weigh[n - 1] and Boundary[0] = 0. Sample has Boundary = {0, 1, 3, 103, 108}
Number generation:
Generate random number N from range [0, Sum of all weights).
for (i = 0; i < size(Boundary) && N > Boundary[i + 1]; ++i)
Take ith range and generate random number in that range.
Additional note for performance optimizations. Ranges don't have to be ordered neither ascending nor descending order, so for faster range look-up range that has highest weight should go first and one with lowest weight should go last.