Generating random number in Array without repeatition - JavaScript

Question

so I know this questing has been asked, but all the answers that were given are already known to me. I don't want to make a variable of all the posible numbers (that was always the answer). So to go to the question, I want to make a random number generator, that will generate me 7 numbers, that must not be the same. For example, I get random numbers: "5,16,12,5,21,37,2" ... But I don't want the number 5 to be used again, so I want different numbers. I made a code for the generation, but I can not think of any good method/way to do this. I was thinking that maybe check if the number is already in array and if it is, then generate another number, but as I'm amateur in JavaScript, I don't know how to do this. So here is my JavaScript code:

// JavaScript Document

function TableOn()
{
    document.write("<table border='1'>");
}
function TableOff()
{
    document.write("</table>");
}

function RandNum()
{
    var n = new Array();
    for(var i=0;i<7;i++)
    {
        n[i] = Math.round((1+(Math.random()*40)));
    }
    TableOn();
    for(var c=0;c<7;c=c+1)
    {
        document.write("<tr><td>"+n[c]+"</td></tr>");
    }
    TableOff();
}

In HTML I just have a button, that is onclick="RandNum()" ... Pardon for my English.

Answer 1

I would do it like this:

var nums = [], numsLen = 5, maxNum = 100, num;
while (nums.length < numsLen) {
    num = Math.round(Math.random() * maxNum);
    if (nums.indexOf(num) === -1) {
        nums.push(num);
    }
}

This generates an array with 5 random numbers in the range 0..100. (numsLen cannot be greater than maxNum.)

Answer 2

These commands can be used to check if a value is/is not in your array:

if ( !!~n.indexOf(someVal) ) {
    // someVal is in array "n"
}
if ( !~n.indexOf(someVal) ) {
    // someVal is not in array "n"
}

Answer 3

    var n = new Array(),num;
function TableOn()
{
    document.write("<table border='1'>");
}
function TableOff()
{
    document.write("</table>");
}
function check_repition()
{
num=Math.round((1+(Math.random()*40)))
if(n.indexOf(num)==-1)
return true;
else 
return false;
}

 function RandNum()
    {

        for(var i=0;i<7;i++)
        {
if(check_repition())
            n[i] =num;
else // keep checking
{
    check_repition()
     n[i] =num;
}
        }
        TableOn();
        for(var c=0;c<7;c=c+1)
        {
            document.write("<tr><td>"+n[c]+"</td></tr>");
        }
        TableOff();
    }
    RandNum()

Answer 4

I'd use a string, storing the generated random numbers with a divider. Then check if the newly generated number is in that string.

Something like this

generated = "";
for(var i=0;i<7;i++)
{
    generate = Math.round((1+(Math.random()*40))); //generate = 5
    while (generated.indexOf("[" + generate + "]") != -1) { //checking if the [5] is already in the generated string, and loop until it's a different number
        generate = Math.round((1+(Math.random()*40))); //get a new random number
    }
    generated += "[" + generate + "]";
    n[i] = generate;
}

or you can take another longer approach

for(var i=0;i<7;i++)
{
    repeated = true;
    while (repeated) { 
        repeated = false;
        generate = Math.round((1+(Math.random()*40)));
        for (var a=0; a < i, a++) {
            if (generate == n[a]) { repeated = true; }
        }
    }
    n[i] = generate;
}

Answer 5

Not sure if OP's requirement of non-repeating is really needed, but here's a fiddle of something that could work if your number range isn't too big:

http://jsfiddle.net/6rEDV/1/

function range(start, end, step) {
    if (typeof step === 'undefined') {
        step = 1;
    }

    if (typeof start === 'undefined' || typeof end === 'undefined') {
        throw TypeError('Must have start and end');
    }

    var ret = [];
    for (var i = start; i <= end; i += step) {
        ret.push(i);
    }

    return ret;
};

// source: http://stackoverflow.com/a/6274381/520857
function shuffle(o) { //v1.0
    for(var j, x, i = o.length; i; j = Math.floor(Math.random() * i), x = o[--i], o[i] = o[j], o[j] = x);
    return o;
};

function getNonRepeatingRand(min, max, num) {
    var arr = shuffle(range(min, max));
    return arr.slice(0, num-1);
}

// Get 7 random numbers between and including 1 and 1000 that will *not* repeat
console.log(getNonRepeatingRand(1,1000,7));

A possibly slower, but less memory intensive method:

http://jsfiddle.net/Qnd8Q/

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

function getNonRepeatingRand(min, max, num) {
    var ret = [];

    for (var i = 0; i < num; i++) {
        var n = rand(min, max);
        if (ret.indexOf(n) == -1) {
            ret.push(n);
        } else {
            i--;
        }
    }
    return ret;
}

console.log(getNonRepeatingRand(1,5,5));

Answer 6

Here's a function to generate an array of n unrepeated random numbers in [min, max):

function rands(n, min, max) {
  var range = max - min;
  if (range < n)
    throw new RangeError("Specified number range smaller than count requested");

  function shuffle() {
    var deck = [], p, t;
    for (var i = 0; i < range; ++i)
      deck[i] = i + min;

    for (i = range - 1; i > 0; --i) {
      p = Math.floor(Math.random() * i);
      t = deck[i];
      deck[i] = deck[p];
      deck[p] = t;
    }
    return deck.slice(0, n);
  }

  function find() {
    var used = {}, rv = [], r;
    while (rv.length < n) {
      r = Math.floor(Math.random() * range + min);
      if (!used[r]) {
        used[r] = true;
        rv.push(r);
      }
    }
    return rv;
  }

  return range < 3 * n ? shuffle() : find();
}

This code checks the range of possible values and compares it to the number of random values requested. If the range is less than three times the number of values requested, the code uses the shuffle to avoid the terrible performance of the lookup approach. If the range is large, however, the lookup approach is used instead.