有几种方法可以做到。你可以使用地图。合并功能:

let mergedMap = map1.merge(map2);

注意:如果任何Map的键是相同的，将使用最后一个要合并的Map中的重复键的值。

更多信息请点击这里:https://untangled.io/immutable-js-6-ways-to-merge-maps-with-full-live-examples/#:~:text=merge()，merged%20in%20will%20be%20used。

集:

var merged = new Set([...set1, ...set2, ...set3])

地图:

var merged = new Map([...map1, ...map2, ...map3])

注意，如果多个映射具有相同的键，则合并映射的值将是具有该键的最后一个合并映射的值。

编辑:

I benchmarked my original solution against other solutions suggests here and found that it is very inefficient. The benchmark itself is very interesting (link) It compares 3 solutions (higher is better): @fregante (formerly called @bfred.it) solution, which adds values one by one (14,955 op/sec) @jameslk's solution, which uses a self invoking generator (5,089 op/sec) my own, which uses reduce & spread (3,434 op/sec) As you can see, @fregante's solution is definitely the winner. Performance + Immutability With that in mind, here's a slightly modified version which doesn't mutates the original set and excepts a variable number of iterables to combine as arguments: function union(...iterables) { const set = new Set(); for (const iterable of iterables) { for (const item of iterable) { set.add(item); } } return set; } Usage: const a = new Set([1, 2, 3]); const b = new Set([1, 3, 5]); const c = new Set([4, 5, 6]); union(a,b,c) // {1, 2, 3, 4, 5, 6}

原来的答案

我想建议另一种方法，使用reduce和spread运算符:

实现

function union (sets) {
  return sets.reduce((combined, list) => {
    return new Set([...combined, ...list]);
  }, new Set());
}

用法:

const a = new Set([1, 2, 3]);
const b = new Set([1, 3, 5]);
const c = new Set([4, 5, 6]);

union([a, b, c]) // {1, 2, 3, 4, 5, 6}

Tip:

我们还可以使用rest操作符来使界面更好:

function union (...sets) {
  return sets.reduce((combined, list) => {
    return new Set([...combined, ...list]);
  }, new Set());
}

现在，我们不再传递一个集合数组，而是可以传递任意数量的集合参数:

union(a, b, c) // {1, 2, 3, 4, 5, 6}

我已经创建了一个小片段，使用ES6中的一个函数合并任意数量的set。你可以改变“设置”为“地图”，让它与地图一起工作。

const mergeSets = (...args) => {
    return new Set(args.reduce((acc, current) => {
        return [...acc, ...current];
    }, []));
};

const foo = new Set([1, 2, 3]);
const bar = new Set([1, 3, 4, 5]);

mergeSets(foo, bar); // Set(5) {1, 2, 3, 4, 5}
mergeSets(foo, bar, new Set([6])); // Set(6) {1, 2, 3, 4, 5, 6}

以下是我使用生成器的解决方案:

地图:

let map1 = new Map(), map2 = new Map();

map1.set('a', 'foo');
map1.set('b', 'bar');
map2.set('b', 'baz');
map2.set('c', 'bazz');

let map3 = new Map(function*() { yield* map1; yield* map2; }());

console.log(Array.from(map3)); // Result: [ [ 'a', 'foo' ], [ 'b', 'baz' ], [ 'c', 'bazz' ] ]

集:

let set1 = new Set(['foo', 'bar']), set2 = new Set(['bar', 'baz']);

let set3 = new Set(function*() { yield* set1; yield* set2; }());

console.log(Array.from(set3)); // Result: [ 'foo', 'bar', 'baz' ]

被认可的答案很好，但每次都会创建一个新的集合。

如果要更改现有对象，请使用helper函数

在不久的将来，你可以只使用setA.union(setB)

Set

function concatSets(set, ...iterables) {
    for (const iterable of iterables) {
        for (const item of iterable) {
            set.add(item);
        }
    }
}

用法:

const setA = new Set([1, 2, 3]);
const setB = new Set([4, 5, 6]);
const setC = new Set([7, 8, 9]);
concatSets(setA, setB, setC);
// setA will have items 1, 2, 3, 4, 5, 6, 7, 8, 9

Map

function concatMaps(map, ...iterables) {
    for (const iterable of iterables) {
        for (const item of iterable) {
            map.set(...item);
        }
    }
}

用法:

const mapA = new Map().set('S', 1).set('P', 2);
const mapB = new Map().set('Q', 3).set('R', 4);
concatMaps(mapA, mapB);
// mapA will have items ['S', 1], ['P', 2], ['Q', 3], ['R', 4]