Array(96) [ {…}, {…}, {…}, {…}, {…}, {…}, {…}, {…}, {…}, {…}, … ]
?
0: Object { id: 1, name: "PrimShal01", period: 3, … }
1: Object { id: 61, name: "TertDeep01", period: 1, … }
2: Object { id: 37, name: "SecoDeep01", period: 2, … }
3: Object { id: 49, name: "TertShal01", period: 1, … } ?
4: Object { id: 13, name: "PrimDeep01", period: 3, … }
5: Object { id: 61, name: "TertDeep01", period: 1, … }
當我嘗試下面的代碼時，我只得到唯一的 id，但我想要物件：
const uniques = [new Set(all_filter_ids.map(pos => pos.id))]
當我嘗試以下代碼時，我得到與以前相同的結果：
const uniques = [new Set(all_filter_ids)]
uj5u.com熱心網友回復：
將它們轉換為由 ID 索引的 Map（一個鍵只能存在一個物件），然后將 Map 的值轉換回陣列。
const map = new Map(all_filter_ids.map(pos => [pos.id, pos]));
const uniques = [...map.values()];
uj5u.com熱心網友回復：
另一種解決方案：
const arr = [{ id: 1, name: "PrimShal01", period: 3},{ id: 61, name: "TertDeep01", period: 1},{ id: 37, name: "SecoDeep01", period: 2},{ id: 49, name: "TertShal01", period: 1},{ id: 13, name: "PrimDeep01", period: 3},{ id: 61, name: "TertDeep01", period: 1}]
const result = Object.values(
arr.reduce((acc, obj) => ({ ...acc, [obj.id]: obj }), {})
);
console.log(result);.as-console-wrapper{min-height: 100%!important; top: 0}uj5u.com熱心網友回復：
維護一個set用于跟蹤 ID 和使用的filter
const uniq = (arr, track = new Set()) =>
arr.filter(({ id }) => (track.has(id) ? false : track.add(id)));
const arr = [
{ id: 1, name: "PrimShal01", period: 3 },
{ id: 61, name: "TertDeep01", period: 1 },
{ id: 37, name: "SecoDeep01", period: 2 },
{ id: 49, name: "TertShal01", period: 1 },
{ id: 13, name: "PrimDeep01", period: 3 },
{ id: 61, name: "TertDeep01", period: 1 },
];
console.log(uniq(arr))轉載請註明出處，本文鏈接：https://www.uj5u.com/shujuku/395628.html
標籤：javascript 数组 目的 ecmascript-6 独特的
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