js completely scrambles array - Stack Overflow

Question

var arr = [0,1,2,3,4,5,6,7]

Completely disrupted, it is required that the position of each element must be changed

Answer 1

First of all, the question is not about traditional array reordering. If it is traditional array reordering, of course Fisher-Yates is the first choice. The black technology of Math.random() cannot guarantee consistent probability. For details, you can see my previous post Analysis https://github.com/hanzichi/u...

If every position needs to be changed, just write a piece of code and it will barely work:

function shuffle(a) {
  let len = a.length;
  let shuffled = Array(len);

  for (let i = len; i--; ) {
    let rand = ~~(Math.random() * i);
    shuffled[i] = a[rand];
    a[rand] = a[i];
  }

  return shuffled;
}

Created a new array and polluted the original array, which is very inelegant. The subject can improve it by himself

Answer 2

Putting the last one first will realize that every element is moved, but what does it mean to completely disrupt it?

var arr = [0,1,2,3,4,5,6,7], last = arr[arr.length - 1];
arr.splice(arr.length - 1, 1);
arr.unshift(last) 

Answer 3

It is impossible to guarantee that every position will change randomly.

If you want to completely disrupt it, you can’t guarantee that the positions will all change.

The simpler method is to scramble it once and then compare it with the array until the conditions are met.

Answer 4

刚没看清 , 这个可以吗?
function upset(arr){
     let newArr = arr.slice().sort(() => Math.random()>0.5 ? 1 : -1)
     let result = arr.some((item,index)=>{
         return item === newArr[index]
     })
     if(!result){
         return newArr
     }else{
         return upset(newArr)
     }
}

Answer 5

    function compare(a , b){
        Math.random > 0.5 ? return -1 ; return 1
    } 
    arr.sort(compare);
    

Answer 6

function shuffle (arr) {

    var _floor = Math.floor,
        _random = Math.random,
        len = arr.length;
    for(var i = 0; i < len - 1; ++i){
        var rand_pos = _floor(_random() * (len - i)) + i;
        if(rand_pos != i){
            var tmp = arr[i];
            arr[i] = arr[rand_pos];
            arr[rand_pos] = tmp;
        }
    }
    return arr;
};

Answer 7

Random panning, Caesar encryption is booming...

function shuffle (arr) {
  if (!Array.isArray(arr)) { return [] }

  var len = Math.floor(Math.random() * (arr.length - 1)) + 1
  return arr.slice(len).concat(arr.slice(0, len))
}

Answer 8

All the simplest positions can be moved

> arr = [0,1,2,3,4,5,6,7]
> arr.unshift(arr.pop())
> arr
[ 7, 0, 1, 2, 3, 4, 5, 6 ]
> arr.sort((a , b)=>Math.random() > 0.5 ? -1:1)
[ 6, 5, 4, 0, 1, 3, 7, 2 ]
> arr.sort((a , b)=>Math.random() > 0.5 ? -1:1)
[ 6, 0, 5, 3, 4, 7, 1, 2 ]

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All the positions are movedIn fact, it is not the most chaotic;

The most chaotic thing is probably shuffling the cards randomly, with a certain probability of keeping a certain number unchanged, so there is no pattern to follow.