只是想…为了一个挑战;-)这个工作…(对于字符串，数字等的基本数组)没有嵌套数组

function diffArrays(arr1, arr2, returnUnion){
  var ret = [];
  var test = {};
  var bigArray, smallArray, key;
  if(arr1.length >= arr2.length){
    bigArray = arr1;
    smallArray = arr2;
  } else {
    bigArray = arr2;
    smallArray = arr1;
  }
  for(var i=0;i<bigArray.length;i++){
    key = bigArray[i];
    test[key] = true;
  }
  if(!returnUnion){
    //diffing
    for(var i=0;i<smallArray.length;i++){
      key = smallArray[i];
      if(!test[key]){
        test[key] = null;
      }
    }
  } else {
    //union
    for(var i=0;i<smallArray.length;i++){
      key = smallArray[i];
      if(!test[key]){
        test[key] = true;
      }
    }
  }
  for(var i in test){
    ret.push(i);
  }
  return ret;
}

array1 = "test1", "test2","test3", "test4", "test7"
array2 = "test1", "test2","test3","test4", "test5", "test6"
diffArray = diffArrays(array1, array2);
//returns ["test5","test6","test7"]

diffArray = diffArrays(array1, array2, true);
//returns ["test1", "test2","test3","test4", "test5", "test6","test7"]

注意排序可能不会像上面提到的那样…但是如果需要的话，可以在数组上调用.sort()对其进行排序。

我在这里读到的答案有很多问题，使得它们在实际编程应用中价值有限。

First and foremost, you're going to want to have a way to control what it means for two items in the array to be "equal". The === comparison is not going to cut it if you're trying to figure out whether to update an array of objects based on an ID or something like that, which frankly is probably one of the most likely scenarios in which you will want a diff function. It also limits you to arrays of things that can be compared with the === operator, i.e. strings, ints, etc, and that's pretty much unacceptable for grown-ups.

其次，diff操作有三种状态结果:

在第一个数组中但不在第二个数组中的元素 两个数组共用的元素 在第二个数组中但不在第一个数组中的元素

我认为这意味着你需要不少于2个循环，但我愿意接受肮脏的技巧，如果有人知道如何将其减少到一个。

这里是我拼凑的一些东西，我想强调的是，我绝对不在乎它在旧版本的Microshaft浏览器中不起作用。如果您在IE这样的较差的编码环境中工作，那么您就可以自行修改它，使其在您无法满意的限制范围内工作。

Array.defaultValueComparison = function(a, b) {
    return (a === b);
};

Array.prototype.diff = function(arr, fnCompare) {

    // validate params

    if (!(arr instanceof Array))
        arr = [arr];

    fnCompare = fnCompare || Array.defaultValueComparison;

    var original = this, exists, storage, 
        result = { common: [], removed: [], inserted: [] };

    original.forEach(function(existingItem) {

        // Finds common elements and elements that 
        // do not exist in the original array

        exists = arr.some(function(newItem) {
            return fnCompare(existingItem, newItem);
        });

        storage = (exists) ? result.common : result.removed;
        storage.push(existingItem);

    });

    arr.forEach(function(newItem) {

        exists = original.some(function(existingItem) {
            return fnCompare(existingItem, newItem);
        });

        if (!exists)
            result.inserted.push(newItem);

    });

    return result;

};

function diff(arr1, arr2) {
  var filteredArr1 = arr1.filter(function(ele) {
    return arr2.indexOf(ele) == -1;
  });

  var filteredArr2 = arr2.filter(function(ele) {
    return arr1.indexOf(ele) == -1;
  });
  return filteredArr1.concat(filteredArr2);
}

diff([1, "calf", 3, "piglet"], [1, "calf", 3, 4]); // Log ["piglet",4]

在这种情况下，您可以使用Set。它针对这种操作(并、交、差)进行了优化。

确保它适用于你的案例，一旦它不允许重复。

var a = new JS.Set([1,2,3,4,5,6,7,8,9]);
var b = new JS.Set([2,4,6,8]);

a.difference(b)
// -> Set{1,3,5,7,9}

var result = [];
var arr1 = [1,2,3,4];
var arr2 = [2,3];
arr1.forEach(function(el, idx) {
    function unEqual(element, index, array) {
        var a = el;
        return (element!=a);
    }
    if (arr2.every(unEqual)) {
        result.push(el);
    };
});
alert(result);

使用ES7有一个更好的方法:

十字路口

 let intersection = arr1.filter(x => arr2.includes(x));

对于[1,2,3][2,3]，它将生成[2,3]。另一方面，for[1,2,3][2,3,5]将返回相同的结果。

区别

let difference = arr1.filter(x => !arr2.includes(x));

对于[1,2,3][2,3]，它将生成[1]。另一方面，for[1,2,3][2,3,5]将返回相同的结果。

对于对称差分，你可以这样做:

let difference = arr1
                 .filter(x => !arr2.includes(x))
                 .concat(arr2.filter(x => !arr1.includes(x)));

通过这种方式，您将得到一个包含arr1中所有不在arr2中的元素的数组，反之亦然

正如@Joshaven Potter在他的回答中指出的，你可以把这个添加到Array中。原型，所以它可以这样使用:

Array.prototype.diff = function(arr2) { return this.filter(x => !arr2.includes(x)); }
[1, 2, 3].diff([2, 3])