有没有一个简单的方法来合并ES6映射在一起(像Object.assign)?说到这里,ES6集合(比如Array.concat)呢?
当前回答
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()))
其他回答
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' ]
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}
你可以使用spread语法将它们合并在一起:
const map1 = {a: 1, b: 2}
const map2 = {b: 1, c: 2, a: 5}
const mergedMap = {...a, ...b}
=> {a: 5, b: 1, c: 2}
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。
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]