合并一个属性相同的对象数组，同时将另一个属性的唯一值转移到数组中

Question

我有一个对象数组，其中大多数属性可能彼此重复，但有一个值可能不同:

const originalArray = [
{ id: 1, name: 'x', 'description': 'x', url: 'y' },
{ id: 2, name: 'x', 'description': 'x', url: 'z' }
]

我想根据名称的差异对这个数组进行重复数据删除，但保留URL的差异作为被重复数据删除数组中的数组:

const dedupedArray = [
{ id: 1, name: 'x', 'description': 'x', url: ['y','z'] },
]

function removeDuplicatesByProperty(keyFn, array) {
const mySet = new Set();
return array.filter(function (x) {
const key = keyFn(x),
isNew = !mySet.has(key);
if (isNew) mySet.add(key);
return isNew;
});
}
const dedupedArray = removeDuplicatesByProperty((x) => x.name, originalArray);

Answer 1

您可以使用Array.reduce:

const originalArray = [
{ id: 1, name: 'x', 'description': 'x', url: 'y' },
{ id: 2, name: 'x', 'description': 'x', url: 'z' }
]
const res = originalArray.reduce((a,b) => {
const found = a.find(e => e.name == b.name);
return found ? found.url.push(b.url) : a.push({...b, url:[b.url]}), a;
}, [])
console.log(res)

Answer 2

将数组缩减为Map，并使用mergeFn合并重复项。使用Array.from()或数组扩展将Map.values()迭代器转换回数组:

const originalArray = [
{ id: 1, name: 'x', 'description': 'x', url: 'y' },
{ id: 2, name: 'x', 'description': 'x', url: 'z' }
]
const dedupeMerge = (keyFn, mergeFn, array) => Array.from(
array.reduce((acc, o) => {
const key = keyFn(o);

if(!acc.has(key)) return acc.set(key, o);

return acc.set(key, mergeFn(acc.get(key), o));
}, new Map()).values()
);
const dedupedArray = dedupeMerge(
x => x.name, 
(o1, o2) => ({ ...o1, url: [o1.url, o2.url].flat() }),
originalArray
);
console.log(dedupedArray);