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
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' ]
其他回答
不,它们没有内置操作,但你可以很容易地创建自己的操作:
Map.prototype.assign = function(...maps) {
for (const m of maps)
for (const kv of m)
this.add(...kv);
return this;
};
Set.prototype.concat = function(...sets) {
const c = this.constructor;
let res = new (c[Symbol.species] || c)();
for (const set of [this, ...sets])
for (const v of set)
res.add(v);
return res;
};
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}
例子
const mergedMaps = (...maps) => {
const dataMap = new Map([])
for (const map of maps) {
for (const [key, value] of map) {
dataMap.set(key, value)
}
}
return dataMap
}
使用
const map = mergedMaps(new Map([[1, false]]), new Map([['foo', 'bar']]), new Map([['lat', 1241.173512]]))
Array.from(map.keys()) // [1, 'foo', 'lat']
根据Asaf Katz的回答,以下是一个打字版本:
export function union<T> (...iterables: Array<Set<T>>): Set<T> {
const set = new Set<T>()
iterables.forEach(iterable => {
iterable.forEach(item => set.add(item))
})
return set
}
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' ]