javascript - generate a new random number - Stack Overflow

admin管理员组

文章数量:1305167

I have a variable that has a number between 1-3.

I need to randomly generate a new number between 1-3 but it must not be the same as the last one.

It happens in a loop hundreds of times.

What is the most efficient way of doing this?

I have a variable that has a number between 1-3.

I need to randomly generate a new number between 1-3 but it must not be the same as the last one.

It happens in a loop hundreds of times.

What is the most efficient way of doing this?

• javascript

Share Improve this question Follow asked Apr 17, 2012 at 10:24 Moshe ShahamMoshe Shaham 16k2424 gold badges8181 silver badges122122 bronze badges 7

• 1 is it between 1-3, are we talking whole numbers? so options are 1, 2 or 3? – filype Commented Apr 17, 2012 at 10:26
• 2 You realise that generating a number from 1-3 without repeating is a) not random and b) functionally equivalent to selecting one of two numbers, i.e. probability 0.5? – Widor Commented Apr 17, 2012 at 10:26
• 1 ok i understand b) but why a)? – Moshe Shaham Commented Apr 17, 2012 at 10:28
• What have you tried so far? The efficiency of the loop depends on many things: Are you using this random in the loop only, should it be passed forward as an argument etc. – Teemu Commented Apr 17, 2012 at 10:39
• @Moshe because for it to be truly random, each number would have to be independent of the last. As you have the condition that there must be no repeats, the future numbers are affected by the past. – Widor Commented Apr 17, 2012 at 10:41

 |  Show 2 more ments

6 Answers 6

Sorted by: Reset to default 5

May the powers of modular arithmetic help you!!

This function does what you want using the modulo operator:

/**
 * generate(1) will produce 2 or 3 with probablity .5
 * generate(2) will produce 1 or 3 with probablity .5
 * ... you get the idea.
 */
function generate(nb) {
    rnd = Math.round(Math.random())
    return 1 + (nb + rnd) % 3
}

if you want to avoid a function call, you can inline the code.

Here is a jsFiddle that solves your problem : http://jsfiddle/AsMWG/

I've created an array containing 1,2,3 and first I select any number and swap it with the last element. Then I only pick elements from position 0 and 1, and swap them with last element.

var x = 1; // or 2 or 3
// this generates a new x out of [1,2,3] which is != x
x = (Math.floor(2*Math.random())+x) % 3 + 1;

You can randomly generate numbers with the random number generator built in to javascript. You need to use Math.random().

If you're push()-ing into an array, you can always check if the previously inserted one is the same number, thus you regenerate the number. Here is an example:

var randomArr = [];
var count = 100;
var max = 3;
var min = 1;

while (randomArr.length < count) {
    var r = Math.floor(Math.random() * (max - min) + min);

    if (randomArr.length == 0) {
        // start condition
        randomArr.push(r); 
    } else if (randomArr[randomArr.length-1] !== r) { 
        // if the previous value is not the same 
        // then push that value into the array
        randomArr.push(r);
    }
}

As Widor mented generating such a number is equivalent to generating a number with probability 0.5. So you can try something like this (not tested):

var x; /* your starting number: 1,2 or 3 */
var y = Math.round(Math.random()); /* generates 0 or 1 */

var i = 0;
var res = i+1;
while (i < y) {
   res = i+1;
   i++;
   if (i+1 == x) i++;
}

The code is tested and it does for what you are after.

var RandomNumber = {

    lastSelected: 0,

    generate: function() {

        var random = Math.floor(Math.random()*3)+1;

        if(random == this.lastSelected) {
            generateNumber();
        }
        else {
            this.lastSelected = random;
            return random;
        }
    }
}


RandomNumber.generate();

本文标签： javascriptgenerate a new random numberStack Overflow