如何只允许将数组推入数组一次？

Question

I have an array that has other arrays in it which have been pushed in. For an example:

const Arrays = [ [1,2,3], [4,5,6], [7,8,9], [2,1,3] ];
let myArr = [];

Arrays.map(arr => {
  if(myArr.indexOf(arr)){
   return
  }
  myArr.push(arr)
})

const myArr = [ [1,2,3], [4,5,6], [7,8,9], [2,1,3] ];

In this array you can see that there are two arrays with the same set of numbers 1 , 2 and 3 . I want to somehow set a condition saying:

If this array already exist then do not add this array in any order again to prevent this from happening. So that when it comes in the loop that this set of numbers comes up again it will just skip over it.

Answer 1

You can use some() and every() methods to check if same array already exists before push() .

 const myArr = [ [1,2,3], [4,5,6], [7,8,9] ]; let input = [2,1,3] function check(oldArr, newArr) { return oldArr.some(a => { return a.length == newArr.length && a.every(e => newArr.includes(e)) }) } if(!check(myArr, input)) myArr.push(input) console.log(myArr) 

Answer 2

You can make temp array with sorted element with joined and check by indexOf

 const myArr = [ [1,2,3], [4,5,6], [7,8,9], [2,1,3],[6,5,4] ]; var newArr = []; var temp = []; for(let i in myArr){ let t = myArr[i].sort().join(","); if(temp.indexOf(t) == -1){ temp.push(t); newArr.push(myArr[i]); } } console.log(newArr); 

Answer 3

The accepted answer does not respect the special case where only two values are in the array and the array has to check against two values in a different count like

[1, 1, 2]

and

[1, 2, 2]

which are different arrays.

For a working solution, I suggest to use a Map and count the occurences of same values of the first array and subtract the count of the values for the other arrray.

As result return the check if all elements of the Map are zero.

 function compare(a, b) { var map = new Map; a.forEach(v => map.set(v, (map.get(v) || 0) + 1)); b.forEach(v => map.set(v, (map.get(v) || 0) - 1)); return [...map.values()].every(v => !v); } var array = [[1, 2, 3], [4, 5, 6], [7, 8, 9], [2, 1, 3], [1, 1, 2], [1, 2, 2], [2, 1, 1], [2, 1, 2]], unique = array.reduce((r, a) => (r.some(b => compare(a, b)) || r.push(a), r), []); console.log(unique); 

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Answer 4

一种方法是使用.sort（）以数字方式对它们进行排序，然后比较它们。