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"}
您也可以使用地图:
const dedupThings = Array.from(things.thing.reduce((m, t) => m.set(t.place, t), new Map()).values());
完整样本:
const 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"});
const dedupThings = Array.from(things.thing.reduce((m, t) => m.set(t.place, t), new Map()).values());
console.log(JSON.stringify(dedupThings, null, 4));
结果:
[
{
"place": "here",
"name": "stuff"
},
{
"place": "there",
"name": "morestuff"
}
]
可以使用for循环和条件使其唯一
const data = [
{ id: 1 },
{ id: 2 },
{ id: 3 },
{ id: 4 },
{ id: 5 },
{ id: 6 },
{ id: 6 },
{ id: 6 },
{ id: 7 },
{ id: 8 },
{ id: 8 },
{ id: 8 },
{ id: 8 }
];
const filtered= []
for(let i=0; i<data.length; i++ ){
let isHasNotEqual = true
for(let j=0; j<filtered.length; j++ ){
if (filtered[j].id===data[i].id){
isHasNotEqual=false
}
}
if (isHasNotEqual){
filtered.push(data[i])
}
}
console.log(filtered);
/*
output
[ { id: 1 },
{ id: 2 },
{ id: 3 },
{ id: 4 },
{ id: 5 },
{ id: 6 },
{ id: 7 },
{ id: 8 } ]
*/
向列表中再添加一个。将ES6和Array.reduce与Array.find一起使用。在此示例中,根据guid属性筛选对象。
let filtered = array.reduce((accumulator, current) => {
if (! accumulator.find(({guid}) => guid === current.guid)) {
accumulator.push(current);
}
return accumulator;
}, []);
扩展此选项以允许选择属性并将其压缩为一行:
const uniqify = (array, key) => array.reduce((prev, curr) => prev.find(a => a[key] === curr[key]) ? prev : prev.push(curr) && prev, []);
要使用它,请将对象数组和要进行重复数据消除的键的名称作为字符串值传递:
const result = uniqify(myArrayOfObjects, 'guid')