是否有一种方法可以在JavaScript中返回两个数组之间的差异?

例如:

var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];

// need ["c", "d"]

当前回答

我就遇到了这个问题,就是求两个简单数组的差值

var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];

// need ["c", "d"]

我不明白为什么不使用基本的for循环:

for(var i=0; i < a1.length; i++) {
  for(var j=0; j < a2.length; j++) {
    if(a1[i] == a2[j]) {
      a2.splice(j, 1);
    }
  }
}

这将返回所需的["c", "d"]

[编辑]提议正上方,看到晚了。

不管怎样,有什么好的理由来避免这个简单的解决方案吗?

其他回答

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)

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;
    });
}

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}

Array.prototype.diff =函数(a) { return this.filter(function(i) {return a.f index_of (i) < 0;}); }; ////////////// //示例// ////////////// Const dif1 =[1,2,3,4,5,6]。Diff ([3,4,5]); console.log (dif1);// => [1,2,6] const dif2 =[“test1”、“test2”,“test3”,“test4”,“test5”,“test6”].diff([“test1”、“test2”,“test3”、“test4”)); console.log (dif2);// => ["test5", "test6"]

注意:. indexof()和.filter()在IE9之前是不可用的。

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' ]