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"}
当前回答
如果您严格希望基于一个属性删除重复项，则可以基于place属性将数组缩减为和对象，因为对象只能具有唯一的键，因此只需获取值即可返回数组：
const unique = Object.values(things.thing.reduce((o, t) => ({ ...o, [t.place]: t }), {}))
其他回答
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);
}
let data = [
{
'name': 'Amir',
'surname': 'Rahnama'
},
{
'name': 'Amir',
'surname': 'Stevens'
}
];
let non_duplicated_data = _.uniqBy(data, 'name');
如果不想指定财产列表：
function removeDuplicates(myArr) {
var props = Object.keys(myArr[0])
return myArr.filter((item, index, self) =>
index === self.findIndex((t) => (
props.every(prop => {
return t[prop] === item[prop]
})
))
)
}
再见！与IE11不兼容。
var testArray= ['a','b','c','d','e','b','c','d'];
function removeDuplicatesFromArray(arr){
var obj={};
var uniqueArr=[];
for(var i=0;i<arr.length;i++){
if(!obj.hasOwnProperty(arr[i])){
obj[arr[i]] = arr[i];
uniqueArr.push(arr[i]);
}
}
return uniqueArr;
}
var newArr = removeDuplicatesFromArray(testArray);
console.log(newArr);
Output:- [ 'a', 'b', 'c', 'd', 'e' ]
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"
}]
