我有一个包含对象数组的对象。
obj = {};
obj.arr = new Array();
obj.arr.push({place:"here",name:"stuff"});
obj.arr.push({place:"there",name:"morestuff"});
obj.arr.push({place:"there",name:"morestuff"});
我想知道从数组中删除重复对象的最佳方法是什么。例如,obj.arr将变成。。。
{place:"here",name:"stuff"},
{place:"there",name:"morestuff"}
const objectsMap = new Map();
const placesName = [
{ place: "here", name: "stuff" },
{ place: "there", name: "morestuff" },
{ place: "there", name: "morestuff" },
];
placesName.forEach((object) => {
objectsMap.set(object.place, object);
});
console.log(objectsMap);
您还可以创建一个通用函数,该函数将根据传递给该函数的对象键过滤数组
function getUnique(arr, comp) {
return arr
.map(e => e[comp])
.map((e, i, final) => final.indexOf(e) === i && i) // store the keys of the unique objects
.filter(e => arr[e]).map(e => arr[e]); // eliminate the dead keys & store unique objects
}
你可以这样调用函数,
getUnique(things.thing,'name') // to filter on basis of name
getUnique(things.thing,'place') // to filter on basis of place
TypeScript函数将数组过滤到其唯一元素,其中唯一性由给定的谓词函数决定:
function uniqueByPredicate<T>(arr: T[], predicate: (a: T, b: T) => boolean): T[] {
return arr.filter((v1, i, a) => a.findIndex(v2 => predicate(v1, v2)) === i);
}
不打字员:
function uniqueByPredicate(arr, predicate) {
return l.filter((v1, i, a) => a.findIndex(v2 => predicate(v1, v2)) === i);
}