您可以使用一个公共对象并计算第一个数组中每个值的频率。对于第二个数组，减少公共对象中的值。然后遍历所有键并添加所有值大于1的键。

常量差值= (a1, a2) => { Var obj = {}; a1。forEach(obj[v] = (obj[v] || 0) + 1); a2。forEach(v => obj[v] = (obj[v] || 0) - 1); 返回对象 . keys (obj) .reduce((r,k) => { If (obj[k] > 0) r = r.concat (Array.from({长度:obj [k]}) .fill (k)); 返回r; }, []); }; const =结果不同([' a ', ' ', ' b ', ' c ', ' d '], [a, b]); console.log(结果);

根据之前的答案…取决于你是想要一个高效的还是“漂亮的联机”解决方案。

一般有三种方法……

"manual iterative" (using indexOf) - naive with O(n2) complexity (slow) var array_diff_naive = function(a,b){ var i, la = a.length, lb = b.length, res = []; if (!la) return b; else if (!lb) return a; for (i = 0; i < la; i++) { if (b.indexOf(a[i]) === -1) res.push(a[i]); } for (i = 0; i < lb; i++) { if (a.indexOf(b[i]) === -1) res.push(b[i]); } return res; } "abstract iterative" (using filter and concat library methods) - syntactic sugar for manual iterative (looks nicer, still sucks) var array_diff_modern = function(a1,a2){ return a1.filter(function(v) { return !a2.includes(v); } ) .concat(a2.filter(function(v) { return !a1.includes(v);})); } "using hashtable" (using object keys) - much more efficient - only O(n), but has slightly limited range of input array values var array_diff_hash = function(a1,a2){ var a = [], diff = []; for (var i = 0; i < a1.length; i++) { a[a1[i]] = true; } for (var i = 0; i < a2.length; i++) { if (a[a2[i]]) { delete a[a2[i]]; } else { a[a2[i]] = true; } } for (var k in a) { diff.push(k); } return diff; }

在jsperf上可以看到 https://jsperf.com/array-diff-algo

Samuel:“对于我的代码，我也需要删除副本，但我想这并不总是可取的。 我想主要的缺点是它可能会比较许多已经被拒绝的选择。”

当比较TWO列表、数组等，且元素小于1000时，行业标准中的 3GL世界是使用冒泡排序，以避免欺骗。

代码看起来像这样……(未经测试，但应该有效)

var Array01=new Array('A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P');
var Array02=new Array('X','B','F','W','Z','X','J','P','P','O','E','N','Q');
var Array03=Array01;

for(x=1; x<Array02.length; x++) {
 for(y=0; y<Array01.length-1; y++) {
  if (Array01[y]==Array02[x]) {Array03.splice(y,1);}}}

Array01=Array03;

要测试输出…

for(y=0; y<Array01.length; y++) {document.write(Array01[y])}

试一试。

var first = [ 1, 2, 3, 4, 5 ];
    var second = [ 4, 5, 6 ];
     
    var difference = first.filter(x => second.indexOf(x) === -1);
    console.log(difference);


Output: [ 1, 2, 3]

Var first = [1,2,3,4,5]; Var second = [4,5,6]; Var差=第一。filter(x => second.indexOf(x) === -1); console.log(差异);

下面是另一个可以返回差异的解决方案，就像git diff一样:(它已经用typescript编写，如果你不使用typescript版本，只需删除类型)

/**
 * util function to calculate the difference between two arrays (pay attention to 'from' and 'to'),
 * it would return the mutations from 'from' to 'to' 
 * @param { T[] } from
 * @param { T[] } to
 * @returns { { [x in string]: boolean } } it would return the stringified version of array element, true means added,
 * false means removed
 */
export function arrDiff<T>(from: T[], to: T[]): { [x in string]: boolean } {

  var diff: { [x in string]: boolean } = {};
  var newItems: T[] = []
  diff = from.reduce((a, e) => ({ ...a, [JSON.stringify(e)]: true }), {})

  for (var i = 0; i < to.length; i++) {
    if (diff[JSON.stringify(to[i])]) {
      delete diff[JSON.stringify(to[i])]
    } else {
      newItems.push(to[i])
    }
  }

  return {
    ...Object.keys(diff).reduce((a, e) => ({ ...a, [e]: false }), {}),
    ...newItems.reduce((a, e) => ({ ...a, [JSON.stringify(e)]: true }), {})
  }
}

下面是一个用法示例:

arrDiff(['a', 'b', 'c'], ['a', 'd', 'c', 'f']) //{"b": false, "d": true, "f": true}

我一直在寻找一个不涉及使用不同库的简单答案，我想出了我自己的答案，我想这里没有提到过。 我不知道它的效率如何，但它确实有效;

    function find_diff(arr1, arr2) {
      diff = [];
      joined = arr1.concat(arr2);
      for( i = 0; i <= joined.length; i++ ) {
        current = joined[i];
        if( joined.indexOf(current) == joined.lastIndexOf(current) ) {
          diff.push(current);
        }
      }
      return diff;
    }

对于我的代码，我也需要删除副本，但我想这并不总是可取的。

我想主要的缺点是它可能会比较许多已经被拒绝的选择。