当向现有集合中添加多个元素(来自数组或另一个集合)时，调用new Set(…anArrayOrSet)没有任何意义。

我在reduce函数中使用了这个，它只是简单的愚蠢。即使你有…数组展开运算符可用，在这种情况下不应该使用它，因为它浪费处理器、内存和时间资源。

// Add any Map or Set to another
function addAll(target, source) {
  if (target instanceof Map) {
    Array.from(source.entries()).forEach(it => target.set(it[0], it[1]))
  } else if (target instanceof Set) {
    source.forEach(it => target.add(it))
  }
}

演示片段

// Add any Map or Set to another function addAll(target, source) { if (target instanceof Map) { Array.from(source.entries()).forEach(it => target.set(it[0], it[1])) } else if (target instanceof Set) { source.forEach(it => target.add(it)) } } const items1 = ['a', 'b', 'c'] const items2 = ['a', 'b', 'c', 'd'] const items3 = ['d', 'e'] let set set = new Set(items1) addAll(set, items2) addAll(set, items3) console.log('adding array to set', Array.from(set)) set = new Set(items1) addAll(set, new Set(items2)) addAll(set, new Set(items3)) console.log('adding set to set', Array.from(set)) const map1 = [ ['a', 1], ['b', 2], ['c', 3] ] const map2 = [ ['a', 1], ['b', 2], ['c', 3], ['d', 4] ] const map3 = [ ['d', 4], ['e', 5] ] const map = new Map(map1) addAll(map, new Map(map2)) addAll(map, new Map(map3)) console.log('adding map to map', 'keys', Array.from(map.keys()), 'values', Array.from(map.values()))

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}

一个很好的解决方案，无论你是否有两个或更多的映射合并是将它们作为一个数组，并使用以下:

Array.prototype.merge = function () {
  return this.reduce((p, c) => Object.assign(c, p), {});
};

当向现有集合中添加多个元素(来自数组或另一个集合)时，调用new Set(…anArrayOrSet)没有任何意义。

我在reduce函数中使用了这个，它只是简单的愚蠢。即使你有…数组展开运算符可用，在这种情况下不应该使用它，因为它浪费处理器、内存和时间资源。

// Add any Map or Set to another
function addAll(target, source) {
  if (target instanceof Map) {
    Array.from(source.entries()).forEach(it => target.set(it[0], it[1]))
  } else if (target instanceof Set) {
    source.forEach(it => target.add(it))
  }
}

演示片段

// Add any Map or Set to another function addAll(target, source) { if (target instanceof Map) { Array.from(source.entries()).forEach(it => target.set(it[0], it[1])) } else if (target instanceof Set) { source.forEach(it => target.add(it)) } } const items1 = ['a', 'b', 'c'] const items2 = ['a', 'b', 'c', 'd'] const items3 = ['d', 'e'] let set set = new Set(items1) addAll(set, items2) addAll(set, items3) console.log('adding array to set', Array.from(set)) set = new Set(items1) addAll(set, new Set(items2)) addAll(set, new Set(items3)) console.log('adding set to set', Array.from(set)) const map1 = [ ['a', 1], ['b', 2], ['c', 3] ] const map2 = [ ['a', 1], ['b', 2], ['c', 3], ['d', 4] ] const map3 = [ ['d', 4], ['e', 5] ] const map = new Map(map1) addAll(map, new Map(map2)) addAll(map, new Map(map3)) console.log('adding map to map', 'keys', Array.from(map.keys()), 'values', Array.from(map.values()))

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}

将集合转换为数组，将它们平直，最后构造函数将惟一化。

const union = (...sets) => new Set(sets.map(s => [...s]).flat());