How can I generate a random number within a range

2020-02-05 04:00发布

Basically I pick a random number between 0-24:

Math.floor(Math.random() * myArray.length); // myArray contains 25 items

Lets say it comes out to be 8. Now I want to get another number in the same range 0-24 but this time, I do not want an 8. The next time, I might roll a 15. Now I want to roll again but I don't want an 8 or 15. The way I am handling this now is by using do while loops and if the number comes out the same, I just reroll.

This is a small portion of my homework and I, in fact, have it working to meet all the requirements so I guess you could say this is for my own personal benefit so I can write this properly and not end up on "the daily wtf".

8条回答
孤傲高冷的网名
2楼-- · 2020-02-05 04:12

<div id="number" style="color: red; margin-left: 200px;">array</div>
<div id="arr" style="color: red; margin-left: 200px;">length</div>
<script>
  var arrayOfIndexesToExclude = new Array();
  function getRandomWithManyExclusions(){
    var rand = null;
      
		  	do{
		    	rand = Math.round(Math.random() * ( 9));
		    	if(arrayOfIndexesToExclude.length >= 10){
		 			  arrayOfIndexesToExclude.length = 0;	   			
		    	}
				}while(arrayOfIndexesToExclude.includes(rand));
			
			arrayOfIndexesToExclude.push(rand);  
    
    document.getElementById("number").innerHTML = arrayOfIndexesToExclude;
    document.getElementById("arr").innerHTML = arrayOfIndexesToExclude.length;
  }
</script>

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▲ chillily
3楼-- · 2020-02-05 04:13

This is easy guys. You do not want recursion for this one. These answers are really bad. Ideally you do not want to hardcode the array, either.

function getRandomWithOneExclusion(lengthOfArray,indexToExclude){

  var rand = null;  //an integer

    while(rand === null || rand === indexToExclude){
       rand = Math.round(Math.random() * (lengthOfArray - 1));
    }

  return rand;
}

now use the value returned from the above function to choose an element from whatever array you want, just like so:

var arr = [];
var random = getRandomWithOneExclusion(arr.length,5);  //array has length x, we want to exclude the 5th element
var elem = arr[random];

that's it. if you wanted to exclude more than value, then you would have to make this more sophisticated, but for excluding one value, this works well. A recursive solution for this is overkill and a bad idea.

I haven't tested this, but to exclude more than one element, try this:

function getRandomWithManyExclusions(originalArray,arrayOfIndexesToExclude){

   var rand = null;

   while(rand === null || arrayOfIndexesToExclude.includes(rand)){
         rand = Math.round(Math.random() * (originalArray.length - 1));
    }
     return rand;
  }

The above method does not sound too different from the OP's original method. This method works properly because it does not sample in a biased way from the array.

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smile是对你的礼貌
4楼-- · 2020-02-05 04:22

step 1> create an array CHECK_ARRAY fill the array with value which is out of the range of your random number [fill it with 26 if you want to generate number within 0-25]

step2-> generate a random number and add it to RANDOM_ARRAY and also add it to the CHECK_ARRAY that is

i=0;
CHECK_ARRAY[i]=random;
i++;

step3-> generate a new random number and go though the CHECK_ARRAY, if you found 26 then ignore, else if you found duplicate then re-generate a random number and continue step 3 again until you found an unique random number !

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男人必须洒脱
5楼-- · 2020-02-05 04:25

Hmz :-? Fastest way to randomly get items from an array and ensure they're all unique would be:

var array = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24];

Array.prototype.shuffle = function shuffle(){
    var tempSlot;
    var randomNumber;
    for(var i =0; i != this.length; i++){
        randomNumber = Math.floor(Math.random() * this.length);
        tempSlot = this[i]; 
        this[i] = this[randomNumber]; 
        this[randomNumber] = tempSlot;
    }
}

while(array.length!=0){
    array.shuffle();
    alert(array.pop());    
}
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该账号已被封号
6楼-- · 2020-02-05 04:25

Suppose you need to choose a random number from the range 1...5 and exclude the values 2, 4 then:

  • Pick a random number from the range 1...3
  • Sort excluded number list
  • For each excluded number less than the random number: add one to the random number

function getRandomExcept(min, max, except) {
  except.sort(function(a, b) {
    return a - b;
  });
  var random = Math.floor(Math.random() * (max - min + 1 - except.length)) + min;
  var i;
  for (i = 0; i < except.length; i++) {
    if (except[i] > random) {
      break;
    }
    random++;
  }
  return random;
}

/*
 * Test iterations. Make sure that:
 * excluded numbers are skipped 
 * numbers are equally distributed
 */
(function(min, max, except) {
  var iterations = 1000000;
  var i;
  var random;
  var results = {};
  for (i = 0; i < iterations; i++) {
    random = getRandomExcept(min, max, except);
    results[random] = (results[random] || 0) + 1;
  }
  for (random in results) {
    console.log("value: " + random + ", count: " + results[random] + ", percent: " + results[random] * 100 / iterations + "%");
  }
})(1, 5, [2, 4]);

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乱世女痞
7楼-- · 2020-02-05 04:33

Set an array with all the values (this is only a valid option if you're only doing small numbers, like the 25 in your example), like this:

var array = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24];

then, pick a random number between 0 and the array length:

var num = Math.floor(Math.random() * array.length);

remove that index number from the array:

var roll = array.splice(num, 1);

Javascript splice() removes indexed items from an array and returns the item(s) as an array. Perfect for your use.

Grab the first index from the roll, since we only cut 1 out anyway:

var yourNumber = roll[ 0 ];

Keep doing for as many rolls as you want. Also, you might want to store the original array as a copy so that you can "reset" the numbers easily.

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