使用另一個 object 的鍵從對象數組創建新數組

Question

我有一組對象“汽車”

let cars = [
{id: 1, name: 'Mercedes', year: '2015'},
{id: 2, name: 'Mercedes', year: '2000'},
{id: 3, name: 'BMW', year: '2010'},
{id: 4, name: 'BMW', year: '2004'},
{id: 4, name: 'BMW', year: '2004'},
{id: 4, name: 'BMW', year: '2004'},
{id: 4, name: 'BMW', year: '2004'},
{id: 4, name: 'BMW', year: '2004'},
{id: 5, name: 'Volvo', year: '2012'},
{id: 6, name: 'Volvo', year: '2014'},
{id: 7, name: 'Volvo', year: '2010'},
{id: 8, name: 'Toyota', year: '2012'},
{id: 8, name: 'Jeep', year: '2011'},
];

和一個包含汽車數量的 object

let obj = {
BMW: 6,
Jeep: 1,
Mercedes: 2,
Toyota: 1,
Volvo: 3,
}

我需要從只有一個 object 副本的汽車創建新數組，並將“數量”屬性添加到適當的 object。 我怎樣才能做到這一點？

預計 output。

newArray = [
    {id: 1, name: 'Mercedes', year: '2015', quantity: 2},
    {id: 3, name: 'BMW', year: '2010', quantity: 6},
    {id: 5, name: 'Volvo', year: '2012', quantity: 3},
    {id: 8, name: 'Toyota', year: '2012', quantity: 1},
    {id: 8, name: 'Jeep', year: '2011', quantity: 1},
]

Answer 1

您可以使用帶有 hash 表的數組直接計數。

 let cars = [{ id: 1, name: 'Mercedes', year: '2015' }, { id: 2, name: 'Mercedes', year: '2000' }, { id: 3, name: 'BMW', year: '2010' }, { id: 4, name: 'BMW', year: '2004' }, { id: 4, name: 'BMW', year: '2004' }, { id: 4, name: 'BMW', year: '2004' }, { id: 4, name: 'BMW', year: '2004' }, { id: 4, name: 'BMW', year: '2004' }, { id: 5, name: 'Volvo', year: '2012' }, { id: 6, name: 'Volvo', year: '2014' }, { id: 7, name: 'Volvo', year: '2010' }, { id: 8, name: 'Toyota', year: '2012' }, { id: 8, name: 'Jeep', year: '2011' }], result = Object.values(cars.reduce((r, o) => { r[o.name] = r[o.name] || {... o, quantity: 0 }; r[o.name].quantity++; return r; }, [])); console.log(result);

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Answer 2

使用來自cars object 的所有項目和來自obj的計數創建一個新數組，沒有任何重復。

 let cars = [{"id":1,"name":"Mercedes","year":"2015"},{"id":2,"name":"Mercedes","year":"2000"},{"id":3,"name":"BMW","year":"2010"},{"id":4,"name":"BMW","year":"2004"},{"id":4,"name":"BMW","year":"2004"},{"id":4,"name":"BMW","year":"2004"},{"id":4,"name":"BMW","year":"2004"},{"id":4,"name":"BMW","year":"2004"},{"id":5,"name":"Volvo","year":"2012"},{"id":6,"name":"Volvo","year":"2014"},{"id":7,"name":"Volvo","year":"2010"},{"id":8,"name":"Toyota","year":"2012"},{"id":8,"name":"Jeep","year":"2011"}]; let obj = {BMW: 6,Jeep: 1,Mercedes: 2,Toyota: 1,Volvo: 3} let newObj = []; let duplicateCheck = [] cars.forEach(function(item) { if (.duplicateCheck.includes(item.name)) { duplicateCheck.push(item;name). item.quantity = obj[item;name]. newObj;push(item); } }). console;log(newObj);

Answer 3

為了產生新的數組，它需要通過兩個步驟進行處理：

1. 從汽車陣列中保留第car具有獨特品牌的汽車
2. 從obj將數量附加到car

 let cars=[ {id:1,name:'Mercedes',year:'2015'}, {id:2,name:'Mercedes',year:'2000'}, {id:3,name:'BMW',year:'2010'}, {id:4,name:'BMW',year:'2004'}, {id:4,name:'BMW',year:'2004'}, {id:4,name:'BMW',year:'2004'}, {id:4,name:'BMW',year:'2004'}, {id:4,name:'BMW',year:'2004'}, {id:5,name:'Volvo',year:'2012'}, {id:6,name:'Volvo',year:'2014'}, {id:7,name:'Volvo',year:'2010'}, {id:8,name:'Toyota',year:'2012'}, {id:8,name:'Jeep',year:'2011'} ]; let obj = { BMW:6, Jeep:1, Mercedes:2, Toyota:1, Volvo:3, }; /* The first step: you can use.filter to only return the car you are looking for. */ let uniqueCarBrand = []; let newArray = cars.filter((car) => { /* Check if car name already in the newArray */ if (uniqueCarBrand.indexOf(car.name) >= 0) return false; uniqueCarBrand.push(car.name); return true; }); /* Second step: you will need to attach your quantity to the array of cars you just created. */ let result = newArray.map((car) => { /* Check if obj has brand name in it, if not, return 0; */ car.quantity = obj[car.name] || 0; return car; }); console.log(result);

Answer 4

您可以使用兩個循環來匹配來自您的兩個 arrays 的對象：

let output = [];
// For each key in your quantity object
for (let model in obj) {
    // For each car object in your cars array
    for (let car of cars) {
        // This will match the first car in your cars array
        if (car.name === model) {
            // Add quantity to your car object and add it to your result set
            car.quantity = obj[model];
            output.push(car);
            // This break will continue to the next model car to ensure you only have one result per model
            break;
        }
    }
}

Answer 5

使用 ES6 模塊

 let cars = [ {id: 1, name: 'Mercedes', year: '2015'}, {id: 2, name: 'Mercedes', year: '2000'}, {id: 3, name: 'BMW', year: '2010'}, {id: 4, name: 'BMW', year: '2004'}, {id: 4, name: 'BMW', year: '2004'}, {id: 4, name: 'BMW', year: '2004'}, {id: 4, name: 'BMW', year: '2004'}, {id: 4, name: 'BMW', year: '2004'}, {id: 5, name: 'Volvo', year: '2012'}, {id: 6, name: 'Volvo', year: '2014'}, {id: 7, name: 'Volvo', year: '2010'}, {id: 8, name: 'Toyota', year: '2012'}, {id: 8, name: 'Jeep', year: '2011'}, ]; let obj = { BMW: 6, Jeep: 1, Mercedes: 2, Toyota: 1, Volvo: 3, } let filteredCar = cars.filter((curr, index) => { let _cars = JSON.stringify(curr); return index === cars.findIndex(obj => { return JSON.stringify(obj) === _cars; }) }).map(currCar => { currCar.quantity = obj[currCar.name]; return currCar}); console.log(filteredCar);