比较2个对象数组并删除重复项
比较2个对象数组并删除重复项
提问于 2019-01-11 15:31:08
我在JavaScript中有两个对象数组，想要比较和合并内容，并按id对结果进行排序。具体地说，得到的排序数组应该包含第一个数组中的所有对象，以及第二个数组中id不在第一个数组中的所有对象。
下面的代码似乎可以工作(没有排序)。但必须有一种更好、更简洁的方法来实现这一点，特别是使用ES6的功能。我假设使用Set是可行的方法，但不确定如何实现。
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;
}
回答 7
Stack Overflow用户
回答已采纳
发布于 2019-01-11 15:35:30
具有O(N)复杂性的一种选择是在Set中创建id的cars1，然后将cars1和过滤后的cars2散布到输出数组中，使用过滤器测试在cars2中迭代的汽车中的id是否包括在集合中：
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);
对sort也是如此：
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);
Stack Overflow用户
发布于 2019-01-11 15:35:44
您可以使用concat、filter和map。
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
let ids = cars1.map(c => c.id);
cars1 = cars1.concat(cars2.filter(({id}) => !ids.includes(id)))
console.log(cars1);
Stack Overflow用户
发布于 2019-01-11 15:38:16
合并两个数组，将每个数组元素与它们的ids放在一个映射中，然后从映射值创建数组。
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},
];
cars = cars1.concat(cars2);
let foo = new Map();
for(const c of cars){
foo.set(c.id, c);
}
let final = [...foo.values()]
console.log(final)
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原文链接：
https://stackoverflow.com/questions/54142112
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