这个怎么样:

Array.prototype.contains = function(needle){
  for (var i=0; i<this.length; i++)
    if (this[i] == needle) return true;

  return false;
} 

Array.prototype.diff = function(compare) {
    return this.filter(function(elem) {return !compare.contains(elem);})
}

var a = new Array(1,4,7, 9);
var b = new Array(4, 8, 7);
alert(a.diff(b));

这样你就可以用array1。diff(array2)来得到它们的区别算法的时间复杂度很可怕- O(array1。长度(我相信)

类似于Ian Grainger的解决方案(但在打字稿中):

function findDiffs(arrayOne: string[], arrayTwo: string[]) {

    let onlyInArrayOne = []
    let onlyInArrayTwo = []
    let share = []
    let [arrayOneCopy, arrayTwoCopy] = [[...arrayOne], [...arrayTwo]]

    arrayOneCopy.sort(); arrayTwoCopy.sort()

    while (arrayOneCopy.length !== 0 && arrayTwoCopy.length !== 0) {
        if (arrayOneCopy[0] == arrayTwoCopy[0]) {
            share.push(arrayOneCopy[0])
            arrayOneCopy.splice(0, 1)
            arrayTwoCopy.splice(0, 1)
        }
        if (arrayOneCopy[0] < arrayTwoCopy[0]) {
            onlyInArrayOne.push(arrayOneCopy[0])
            arrayOneCopy.splice(0, 1)
        }
        if (arrayOneCopy[0] > arrayTwoCopy[0]) {
            onlyInArrayTwo.push(arrayTwoCopy[0])
            arrayTwoCopy.splice(0, 1)
        }
    }
    onlyInArrayTwo = onlyInArrayTwo.concat(arrayTwoCopy)
    onlyInArrayOne = onlyInArrayOne.concat(arrayOneCopy)

    return {
        onlyInArrayOne,
        onlyInArrayTwo,
        share,
        diff: onlyInArrayOne.concat(onlyInArrayTwo)
    }
}

// arrayOne: [ 'a', 'b', 'c', 'm', 'y' ] 
// arrayTwo: [ 'c', 'b', 'f', 'h' ]
//
// Results: 
// { 
//    onlyInArrayOne: [ 'a', 'm', 'y' ],
//    onlyInArrayTwo: [ 'f', 'h' ],
//    share: [ 'b', 'c' ],
//    diff: [ 'a', 'm', 'y', 'f', 'h' ] 
// }

所选的答案只对了一半。您必须比较数组的两种方式才能得到完整的答案。

const ids_exist = [
   '1234',
   '5678',
   'abcd',
]

const ids_new = [
  '1234',
  '5678',
  'efjk',
  '9999',
]

function __uniq_Filter (__array_1, __array_2) {
  const one_not_in_two = __array_1.filter(function (obj) {
    return __array_2.indexOf(obj) == -1
  })
  const two_not_in_one = __array_2.filter(function (obj) {
    return __array_1.indexOf(obj) == -1
  })
  return one_not_in_two.concat(two_not_in_one)
}

let uniq_filter = __uniq_Filter(ids_exist, ids_new)

console.log('uniq_filter', uniq_filter) // => [ 'abcd', 'efjk', '9999' ]

困难的方法(如果你想做一些比.indexOf更奇特的东西)

var difference = function (source, target) {
    return source.reduce(function (diff, current) { 
        if (target.indexOf(current) === -1) { 
            diff.push(current); 
        }

        return diff; 
    }, []);
}

简单的方法

var difference = function (source, target) {
    return source.filter(function (current) {
        return target.indexOf(current) === -1;
    });
}

求两个没有重复项的数组的差值:

function difference(arr1, arr2){

  let setA = new Set(arr1);
  let differenceSet = new Set(arr2.filter(ele => !setA.has(ele)));
  return [...differenceSet ];

}

1.difference([2,2,3,4]，[2,3,3,4])将返回[]

2.difference([1,2,3]，[4,5,6])将返回[4,5,6]

3.difference([1,2,3,4]，[1,2])返回[]

4.difference([1,2]，[1,2,3,4])将返回[3,4]

注意:上述解决方案要求始终将较大的数组作为第二个参数发送。要找到绝对差值，首先需要找到两者的较大数组，然后对它们进行处理。

求两个不存在重复项的数组的绝对差值:

function absDifference(arr1, arr2){

  const {larger, smaller} = arr1.length > arr2.length ? 
  {larger: arr1, smaller: arr2} : {larger: arr2, smaller: arr1}
  
  let setA = new Set(smaller);
  let absDifferenceSet = new Set(larger.filter(ele => !setA.has(ele)));
  return [...absDifferenceSet ];

}

1. absdifference((2, 2, 3, 4),[2、3、3、4])将返回[]

2. absdifference([1, 2, 3],[4、5、6])将返回(4、5、6)

3. absdifference([1、2、3、4],[1,2])将返回(3、4)

4. absdifference([1, 2],[1、2、3、4])将返回(3、4)

请注意这两个解决方案中的示例3

如果你想找出两个object数组之间的差异，你可以这样做:

let arrObj = [{id: 1}，{id: 2}，{id: 3}] let arrObj2 = [{id: 1}，{id: 3}] let result = arrObj。filter(x => arrObj2.)每个(x2 => x2。Id == x.id) console.log(结果)