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"}
如果您发现需要经常基于特定字段从数组中删除重复的对象,那么创建一个可以从项目中任何位置导入的独特(数组、谓词)函数可能是值得的。这看起来像
const things = [{place:"here",name:"stuff"}, ...];
const distinctThings = distinct(things, thing => thing.place);
不同的函数可以使用上面许多好答案中给出的任何实现。最简单的方法是使用findIndex:
const distinct = (items, predicate) => items.filter((uniqueItem, index) =>
items.findIndex(item =>
predicate(item) === predicate(uniqueItem)) === index);
const uniqueElements = (arr, fn) => arr.reduce((acc, v) => {
if (!acc.some(x => fn(v, x))) { acc.push(v); }
return acc;
}, []);
const stuff = [
{place:"here",name:"stuff"},
{place:"there",name:"morestuff"},
{place:"there",name:"morestuff"},
];
const unique = uniqueElements(stuff, (a,b) => a.place === b.place && a.name === b.name );
//console.log( unique );
[{
"place": "here",
"name": "stuff"
},
{
"place": "there",
"name": "morestuff"
}]
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