还有另一个答案，但似乎没有人提到jsperf，他们比较了几种算法和技术支持:https://jsperf.com/array-difference-javascript似乎使用过滤器得到了最好的结果。谢谢

ES2015的函数方法

计算两个数组之间的差值是Set操作之一。这个术语已经表明应该使用本机Set类型，以便提高查找速度。不管怎样，当你计算两个集合之间的差值时，有三种排列:

[+left difference] [-intersection] [-right difference]
[-left difference] [-intersection] [+right difference]
[+left difference] [-intersection] [+right difference]

下面是反映这些排列的功能性解决方案。

离开的区别:

// small, reusable auxiliary functions const apply = f => x => f(x); const flip = f => y => x => f(x) (y); const createSet = xs => new Set(xs); const filter = f => xs => xs.filter(apply(f)); // left difference const differencel = xs => ys => { const zs = createSet(ys); return filter(x => zs.has(x) ? false : true ) (xs); }; // mock data const xs = [1,2,2,3,4,5]; const ys = [0,1,2,3,3,3,6,7,8,9]; // run the computation console.log( differencel(xs) (ys) );

正确的区别:

差异是微不足道的。这与翻转的参数不同。为了方便，你可以写一个函数:const differencer = flip(difference)。这是所有!

对称的区别:

现在我们有了左边和右边，实现对称的差异也变得微不足道:

// small, reusable auxiliary functions const apply = f => x => f(x); const flip = f => y => x => f(x) (y); const concat = y => xs => xs.concat(y); const createSet = xs => new Set(xs); const filter = f => xs => xs.filter(apply(f)); // left difference const differencel = xs => ys => { const zs = createSet(ys); return filter(x => zs.has(x) ? false : true ) (xs); }; // symmetric difference const difference = ys => xs => concat(differencel(xs) (ys)) (flip(differencel) (xs) (ys)); // mock data const xs = [1,2,2,3,4,5]; const ys = [0,1,2,3,3,3,6,7,8,9]; // run the computation console.log( difference(xs) (ys) );

我想这个例子是一个很好的起点，可以让你了解函数式编程的含义:

使用可以以许多不同方式组合在一起的构建块进行编程。

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;
    });
}

对我来说，把它作为部分函数处理比较容易。很惊讶没有看到函数式编程的解决方案，这是我在ES6中的:

const arrayDiff = (a, b) => {
  return diff(b)(a);
}

const contains = (needle) => (array) => {
  for (let i=0; i < array.length; i++) {
    if (array[i] == needle) return true;
  }

  return false;
}

const diff = (compare) => {
    return (array) => array.filter((elem) => !contains(elem)(compare))
}

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' ]

一个衬垫

const unique = (a) => [...new Set(a)]; const uniqueBy = (x,f)=>Object.values(x.reduce((a,b)=>((a[f(b)]=b),a),{})); const intersection = (a, b) => a.filter((v) => b.includes(v)); const diff = (a, b) => a.filter((v) => !b.includes(v)); const symDiff = (a, b) => diff(a, b).concat(diff(b, a)); const union = (a, b) => diff(a, b).concat(b); const a = unique([1, 2, 3, 4, 5, 5]); console.log(a); const b = [4, 5, 6, 7, 8]; console.log(intersection(a, b), diff(a, b), symDiff(a, b), union(a, b)); console.log(uniqueBy( [ { id: 1, name: "abc" }, { id: 2, name: "xyz" }, { id: 1, name: "abc" }, ], (v) => v.id )); const intersectionBy = (a, b, f) => a.filter((v) => b.some((u) => f(v, u))); console.log(intersectionBy( [ { id: 1, name: "abc" }, { id: 2, name: "xyz" }, ], [ { id: 1, name: "abc" }, { id: 3, name: "pqr" }, ], (v, u) => v.id === u.id )); const diffBy = (a, b, f) => a.filter((v) => !b.some((u) => f(v, u))); console.log(diffBy( [ { id: 1, name: "abc" }, { id: 2, name: "xyz" }, ], [ { id: 1, name: "abc" }, { id: 3, name: "pqr" }, ], (v, u) => v.id === u.id ));

打印稿

操场上的链接

const unique = <T>(array: T[]) => [...new Set(array)];


const intersection = <T>(array1: T[], array2: T[]) =>
  array1.filter((v) => array2.includes(v));


const diff = <T>(array1: T[], array2: T[]) =>
  array1.filter((v) => !array2.includes(v));


const symDiff = <T>(array1: T[], array2: T[]) =>
  diff(array1, array2).concat(diff(array2, array1));


const union = <T>(array1: T[], array2: T[]) =>
  diff(array1, array2).concat(array2);


const intersectionBy = <T>(
  array1: T[],
  array2: T[],
  predicate: (array1Value: T, array2Value: T) => boolean
) => array1.filter((v) => array2.some((u) => predicate(v, u)));


const diffBy = <T>(
  array1: T[],
  array2: T[],
  predicate: (array1Value: T, array2Value: T) => boolean
) => array1.filter((v) => !array2.some((u) => predicate(v, u)));


const uniqueBy = <T>(
  array: T[],
  predicate: (v: T, i: number, a: T[]) => string
) =>
  Object.values(
    array.reduce((acc, value, index) => {
      acc[predicate(value, index, array)] = value;
      return acc;
    }, {} as { [key: string]: T })
  );