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javascript - One array taking random elements from another array

PHP中文網(wǎng) 2017-05-19 10:31:39

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There is an array a=[4,19,23,44,56,1], create a new array b, and b randomly selects one element at a time from a until it is exhausted. Implemented in JavaScript

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小葫蘆2017-05-19 10:33:39 7 floor

You can shuffle a with pseudo-randomness, exchange the numbers in any two positions, do this n times to achieve the shuffling effect, and then assign it to b.

Or just follow the steps. If code efficiency is not considered, the array operations provided by the lodash library can make the code more elegant:

var src = [4,19,23,44,56,1];

var shuffle = [];
while(src.length > 0){
    var random_index = Math.floor(Math.random() * src.length);
    shuffle.push(src[random_index]);
    
    src = src.filter(function(el, i){
        return i != random_index;
    });
}

I thought of a better ~~pseudo~~random method, which is directly sorted randomly, the code is simpler, and the operation efficiency is high:

var src = [4,19,23,44,56,1];

var shuffle = src;
shuffle.sort(function(){
    return Math.floor(Math.random() * 3) - 1;
});

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僅有的幸福2017-05-19 10:33:39 6 floor

var a=[4,19,23,44,56,1];
var b=[];
var c=[0,0,0,0,0,0];
var aLen = a.length;
while (c.indexOf(0)!=-1){
    var randomNum = Math.floor(Math.random()*aLen);
    if (c[randomNum] == 0){
        b.push(a[randomNum]);
        c[randomNum]=1
    }
}
console.log(b);

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世界只因有你2017-05-19 10:33:39 5 floor

The implementation is as follows:

var a = [4,19,23,44,56,1];
function randomSort(array){
    var newArray = [];
    for(let i = 0; i < array.length;){
        let j = Math.random() * array.length; // 獲取隨機(jī)的下標(biāo)值
        j = ~~j; // 取整
        let item = array.splice(j, 1)[0]; // 從原數(shù)組中剔除選中的元素
        newArray.push(item);
  }
  return newArray;
}
randomSort(a);

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漂亮男人2017-05-19 10:33:39 4 floor

My idea is to traverse this array, get random elements in this array, delete them from array a and push them into array b, and so on

var a = [4,19,23,44,56,1],
    aLen = a.length,
    b = [];
    
// 定義刪除數(shù)組中指定元素的方法
Array.prototype.removeByValue = function(val) {
  for(var i=0; i<this.length; i++) {
    if(this[i] == val) {
      this.splice(i, 1);
      break;
    }
  }
}

// 遍歷a數(shù)組
for(var i=0; i<aLen; i++){
    var cur = a[Math.floor(Math.random()*a.length)]; //獲取隨機(jī)的元素
    a.removeByValue(cur);    // 從a中刪除
    b.push(cur);            // 添加進(jìn)b
}

console.log(b)

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滿天的星座2017-05-19 10:33:39 3 floor

   var a=[4,19,23,44,56,1];
    function randomSelect(str){
        var leng= a.length;
        var b=[];
      while(leng!=0){
          b.push(a.splice(Math.random()*leng,1).join())
          leng--;
      }
       return b;
    }
    console.log(randomSelect(a));

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淡淡煙草味2017-05-19 10:33:39 2 floor

<!DOCTYPE html>
<html>
  <head>
    <meta charset="utf-8">
    <title></title>
  </head>
  <body>
    <script>
      //有一個(gè)數(shù)組a=[4,19,23,44,56,1]，新建一個(gè)數(shù)組b，b從a中一次隨機(jī)選取一個(gè)元素，取完為止。用JavaScript實(shí)現(xiàn)
      var a = [4,19,23,44,56,1];
      var b = [];

      function findInArr(num, arr){
        for(var i=0; i<arr.length; i++){
          if(num == arr[i]){
            return true;
          }
        }
        return false;
      }

      function rnd(m,n){
        return m + Math.floor(Math.random()*(n-m));
      }

      function randomPass(arr){
        var len = arr.length;
        var rndIndex = rnd(0,len-1);
        console.log(rndIndex);
        var temp = arr[rndIndex];
        if(!findInArr(temp,b)){
          b.push(temp);
          arr.splice(rndIndex,1);
          if(arr.length > 0){
            randomPass(arr);
          }
        }
      }

      randomPass(a);
      console.log(b); // [44, 56, 19, 23, 4, 1] (每次隨機(jī))
      console.log(a); // []

    </script>
  </body>
</html>

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曾經(jīng)蠟筆沒有小新2017-05-19 10:33:39 1 floor

I won’t write it down to the fourth floor. I just randomly select one from the array one at a time. After randomization, remove it from the array. Continue randomizing. Just push the removed ones into the new array and it will be ok

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