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"}
使用ES6“reduce”和“find”数组助手方法的简单解决方案
工作效率高,非常好!
"use strict";
var things = new Object();
things.thing = new Array();
things.thing.push({
place: "here",
name: "stuff"
});
things.thing.push({
place: "there",
name: "morestuff"
});
things.thing.push({
place: "there",
name: "morestuff"
});
// the logic is here
function removeDup(something) {
return something.thing.reduce(function (prev, ele) {
var found = prev.find(function (fele) {
return ele.place === fele.place && ele.name === fele.name;
});
if (!found) {
prev.push(ele);
}
return prev;
}, []);
}
console.log(removeDup(things));
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);
}