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スプレッド演算子を使用してオブジェクト配列から重複を削除する方法

IDが一意のキーである以下のオブジェクト配列があります。 "

var test = [
  {id: 1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: ""},
  {id: 2, PlaceRef: "*00022", Component: "BAXI10R", SubLocCode: "KIT", BarCode:""},
  {id: 1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: ""},
  {id: 3, PlaceRef: "*00011", Component: "ANR190", SubLocCode: "B1", BarCode: ""}
]

これから、spreadオペレーターを使用して一意のオブジェクトを取得したいので、以下のコードを使用してみました：

const uniKeys = [...(new Set(test.map(({ id }) => id)))];

Idのみを取得できますが、spread演算子を使用して一意のオブジェクトを取得するにはどうすればよいですか。また、新しいES6機能の実装も役立ちます。

javascriptangularecmascript-6typescript2.0
10
tracer

mapメソッドを使用してオブジェクトの配列にfind戻ることができます。これにより、そのIDを持つ最初のオブジェクトが返されます。

var test = [{id:1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: ""},{id:2, PlaceRef: "*00022", Component: "BAXI10R", SubLocCode: "KIT", BarCode:""},{id:1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: ""},{id:3, PlaceRef: "*00011", Component: "ANR190", SubLocCode: "B1", BarCode: ""}]
var uniq = [...new Set(test.map(({id}) => id))].map(e => test.find(({id}) => id == e));        
console.log(uniq)

代わりにfilterメソッドを使用することもできます。

var test = [{id:1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: ""},{id:2, PlaceRef: "*00022", Component: "BAXI10R", SubLocCode: "KIT", BarCode:""},{id:1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: ""},{id:3, PlaceRef: "*00011", Component: "ANR190", SubLocCode: "B1", BarCode: ""}]

var uniq = test.filter(function({id}) {
  return !this[id] && (this[id] = id)
}, {})
        
console.log(uniq)
7
Nenad Vracar

Set を使用して、不明なidでフィルタリングできます。

var test = [{ id: 1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: "" }, { id: 2, PlaceRef: "*00022", Component: "BAXI10R", SubLocCode: "KIT", BarCode: "" }, { id: 1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: "" }, { id: 3, PlaceRef: "*00011", Component: "ANR190", SubLocCode: "B1", BarCode: "" }],
    unique = test.filter((s => ({ id }) => !s.has(id) && s.add(id))(new Set));
    
console.log(unique);
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4
Nina Scholz

IDでマップを作成し、値を抽出できます。 [...new Map(test.map(item => [item.id, item])).values()]

var test = [{id:1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: ""},
{id:2, PlaceRef: "*00022", Component: "BAXI10R", SubLocCode: "KIT", BarCode:""},
{id:1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: ""},
{id:3, PlaceRef: "*00011", Component: "ANR190", SubLocCode: "B1", BarCode: ""}]

console.log([
  ...new Map(test.map(item => [item.id, item])).values()
])
2
Yury Tarabanko
var test = [{id:1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: ""},
{id:2, PlaceRef: "*00022", Component: "BAXI10R", SubLocCode: "KIT", BarCode:""},
{id:1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: ""},
{id:3, PlaceRef: "*00011", Component: "ANR190", SubLocCode: "B1", BarCode: ""}];


var uniqArray = Array.from(new Map(test.map(e=>[e.id, e])).values());
console.log(uniqArray)
1
Shishir Arora