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
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}
其他回答
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>
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']
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()))
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' ]
要合并数组集合中的集合,您可以执行
var Sets = [set1, set2, set3];
var merged = new Set([].concat(...Sets.map(set => Array.from(set))));
对我来说有点神秘的是,为什么下面这些应该是等价的,但至少在巴别塔失败了:
var merged = new Set([].concat(...Sets.map(Array.from)));