比较 2 个对象数组并删除重复项

Question

我在 JavaScript 中有 2 个对象数组，我想比较和合并内容并按 id 对结果进行排序。具体来说，生成的排序数组应包含第一个数组中的所有对象，以及第二个数组中具有不在第一个数组中的 id 的所有对象。

以下代码似乎有效（减去排序）。但必须有更好、更简洁的方法来做到这一点，尤其是使用 ES6 的特性。我假设使用 Set 是可行的方法，但不确定具体如何实现。

     var cars1 = [
        {id: 2, make: "Honda", model: "Civic", year: 2001},
        {id: 1, make: "Ford",  model: "F150",  year: 2002},
        {id: 3, make: "Chevy", model: "Tahoe", year: 2003},
    ];

    var cars2 = [
        {id: 3, make: "Kia",    model: "Optima",  year: 2001},
        {id: 4, make: "Nissan", model: "Sentra",  year: 1982},
        {id: 2, make: "Toyota", model: "Corolla", year: 1980},
    ];

    // Resulting cars1 contains all cars from cars1 plus unique cars from cars2
    cars1 = removeDuplicates(cars2);
    console.log(cars1);

    function removeDuplicates(cars2){
        for (entry in cars2) {
            var keep = true;

            for (c in cars1) {
                if (cars1[c].id === cars2[entry].id) {
                    keep = false;
                }
            }

            if (keep) {
                cars1.push({
                    id:cars2[entry].id,
                    make:cars2[entry].make,
                    model:cars2[entry].model,
                    year:cars2[entry].year
                })
            }
        }
        return cars1;
    }

原文由 Woodchuck 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

One option with O(N) complexity would be to make a Set of the id s in cars1 , then spread cars1 和过滤后的 cars2 进入输出数组，过滤器测试 id 在汽车中被迭代 cars2 是否包含在 Set:– 中

 var cars1 = [
    {id: 2, make: "Honda", model: "Civic", year: 2001},
    {id: 1, make: "Ford",  model: "F150",  year: 2002},
    {id: 3, make: "Chevy", model: "Tahoe", year: 2003},
];

var cars2 = [
    {id: 3, make: "Kia",    model: "Optima",  year: 2001},
    {id: 4, make: "Nissan", model: "Sentra",  year: 1982},
    {id: 2, make: "Toyota", model: "Corolla", year: 1980},
];
const cars1IDs = new Set(cars1.map(({ id }) => id));
const combined = [
  ...cars1,
  ...cars2.filter(({ id }) => !cars1IDs.has(id))
];
console.log(combined);

至 sort 以及：

 combined.sort(({ id: aId }, {id: bId }) => aId - bId);

 var cars1 = [
    {id: 2, make: "Honda", model: "Civic", year: 2001},
    {id: 1, make: "Ford",  model: "F150",  year: 2002},
    {id: 3, make: "Chevy", model: "Tahoe", year: 2003},
];

var cars2 = [
    {id: 3, make: "Kia",    model: "Optima",  year: 2001},
    {id: 4, make: "Nissan", model: "Sentra",  year: 1982},
    {id: 2, make: "Toyota", model: "Corolla", year: 1980},
];
const cars1IDs = new Set(cars1.map(({ id }) => id));
const combined = [
  ...cars1,
  ...cars2.filter(({ id }) => !cars1IDs.has(id))
];
combined.sort(({ id: aId }, {id: bId }) => aId - bId);
console.log(combined);

原文由 CertainPerformance 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 2

您可以拿走 Map 并先拿走其中的地图或实际汽车。

 var cars1 = [{ id: 2, make: "Honda", model: "Civic", year: 2001 }, { id: 1, make: "Ford",  model: "F150",  year: 2002 }, { id: 3, make: "Chevy", model: "Tahoe", year: 2003 }],
    cars2 = [{ id: 3, make: "Kia",    model: "Optima",  year: 2001 }, { id: 4, make: "Nissan", model: "Sentra",  year: 1982 }, { id: 2, make: "Toyota", model: "Corolla", year: 1980 }],
    result = Array
        .from(
            [...cars1, ...cars2]
                .reduce((m, c) => m.set(c.id, m.get(c.id) || c), new Map)
                .values()
        )
        .sort((a, b) => a.id - b.id);

console.log(result);

原文由 Nina Scholz 发布，翻译遵循 CC BY-SA 4.0 许可协议