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
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']
其他回答
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());
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' ]
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}
我创建了一个helper方法来合并映射,并以所需的任何成对方式处理重复键的值:
const mergeMaps = (map1, map2, combineValuesOfDuplicateKeys) => {
const mapCopy1 = new Map(map1);
const mapCopy2 = new Map(map2);
mapCopy1.forEach((value, key) => {
if (!mapCopy2.has(key)) {
mapCopy2.set(key, value);
} else {
const newValue = combineValuesOfDuplicateKeys
? combineValuesOfDuplicateKeys(value, mapCopy2.get(key))
: mapCopy2.get(key);
mapCopy2.set(key, newValue);
mapCopy1.delete(key);
}
});
return new Map([...mapCopy1, ...mapCopy2]);
};
const mergeMaps = (map1, map2, combineValuesOfDuplicateKeys) => { const mapCopy1 = new Map(map1); const mapCopy2 = new Map(map2); mapCopy1.forEach((value, key) => { if (!mapCopy2.has(key)) { mapCopy2.set(key, value); } else { const newValue = combineValuesOfDuplicateKeys ? combineValuesOfDuplicateKeys(value, mapCopy2.get(key)) : mapCopy2.get(key); mapCopy2.set(key, newValue); mapCopy1.delete(key); } }); return new Map([...mapCopy1, ...mapCopy2]); }; const map1 = new Map([ ["key1", 1], ["key2", 2] ]); const map2 = new Map([ ["key2", 3], ["key4", 4] ]); const show = (object) => { return JSON.stringify(Array.from(object), null, 2) } document.getElementById("app").innerHTML = ` <h1>Maps are awesome!</h1> <div>map1 = ${show(map1)}</div> <div>map2 = ${show(map2)}</div><br> <div>Set value of last duplicate key:<br>merged map = ${show(mergeMaps(map1, map2))}</div><br> <div>Set value of pair-wise summated duplicate keys:<br>merged map = ${show(mergeMaps(map1, map2, (value1, value2) => value1 + value2))}</div><br> <div>Set value of pair-wise difference of duplicate keys:<br>merged map = ${show(mergeMaps(map1, map2, (value1, value2) => value1 - value2))}</div><br> <div>Set value of pair-wise multiplication of duplicate keys:<br>merged map = ${show(mergeMaps(map1, map2, (value1, value2) => value1 * value2))}</div><br> <div>Set value of pair-wise quotient of duplicate keys:<br>merged map = ${show(mergeMaps(map1, map2, (value1, value2) => value1 / value2))}</div><br> <div>Set value of pair-wise power of duplicate keys:<br>merged map = ${show(mergeMaps(map1, map2, (value1, value2) => Math.pow(value1, value2)))}</div><br> `; <!DOCTYPE html> <html> <head> <title>Parcel Sandbox</title> <meta charset="UTF-8" /> </head> <body> <div id="app"></div> <script src="src/index.js"> </script> </body> </html>
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}