使用findIndex查找和更新對象中深層嵌套數組的性能

Question

我有以下數據結構：

const myData = [
        {
            "trips": [
                {
                    "destination": "Hungary",
                    "id": "34547",
                    "stars": 0
                },
                {
                    "destination": "Hungary",
                    "id": "14542",
                    "stars": 0
                },
                {
                    "destination": "Hungary",
                    "id": "88247",
                    "stars": 0
                },
                {
                    "destination": "Hungary",
                    "id": "11447",
                    "stars": 0
                },
            ],
            "descr": "Holidays",
            "id": "243567"
        },
    ]

假設我們有N個具有唯一ID的對象：

給定項id ，行程id和替換行程對象，找到並替換行程對象。

例：

const itemId = 243567;
const tripId = 14542;

const replacement = { 
  destination: Beijing
  id: 14542
  stars: 4
};;

我的解決方案如下：

 const myData = [{ "trips": [{ "destination": "Hungary", "id": "34547", "stars": 0 }, { "destination": "Hungary", "id": "14542", "stars": 0 }, { "destination": "Hungary", "id": "88247", "stars": 0 }, { "destination": "Hungary", "id": "11447", "stars": 0 }, ], "descr": "Holidays", "id": "243567" }]; const itemId = 243567; const tripId = 14542; const replacement = { destination: "Beijing" id: 14542 stars: 4 }; const itemIndex = myData .findIndex(element => element.id === itemId); const tripIndex = myData[itemIndex].trips .findIndex(element => element.id === tripId); Object.assign(myData[itemIndex].trips[tripIndex], replacement); 

該解決方案將如何執行？是否有更快的實施方法？

Answer 1

如果您只需要對給定的數據集執行一次這樣的查找和變異，那么您當前的工作就可以了。

但是，如果您將在同一數據集中執行多次查找和更改（因此在重新加載之前），則應按ID和Trip ID鍵入數據。 為此，您可以使用此函數，在加載數據集后應調用一次：

 function hashData(myData) { const result = {}; for (const row of myData) { const obj = result[row.id] = {}; for (const trip of row.trips) { obj[trip.id] = trip; } } return result; } // Sample data const myData = [{ "trips": [{"destination": "Hungary", "id": "34547", "stars": 0 },{"destination": "Hungary", "id": "14542", "stars": 0 },{"destination": "Hungary", "id": "88247", "stars": 0 },{"destination": "Hungary", "id": "11447", "stars": 0}], "descr": "Holidays", "id": "243567"}]; // Key it by id and trip id: const hash = hashData(myData); // Mutate one particular entry: Object.assign(hash[243567][88247], { destination: 'PARADISE', id: "9999", stars: 5 }); // Display result console.log(myData); 

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如果您不介意更多冗長的代碼，那么使用單獨的分配替換Object.assign將在當前瀏覽器中提供更好的性能：

const obj = hash[243567][88247];
obj.destination = 'PARADISE';
obj.id = "9999";
obj.stars = 5;