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

另一个选项是创建一个自定义indexOf函数,该函数比较每个对象所选属性的值,并将其包装在reduce函数中。

var uniq = redundant_array.reduce(function(a,b){
      function indexOfProperty (a, b){
          for (var i=0;i<a.length;i++){
              if(a[i].property == b.property){
                   return i;
               }
          }
         return -1;
      }

      if (indexOfProperty(a,b) < 0 ) a.push(b);
        return a;
    },[]);
let data = [
  {
    'name': 'Amir',
    'surname': 'Rahnama'
  }, 
  {
    'name': 'Amir',
    'surname': 'Stevens'
  }
];
let non_duplicated_data = _.uniqBy(data, 'name');
function filterDuplicateQueries(queries){
    let uniqueQueries = [];
     queries.forEach((l, i)=>{
        let alreadyExist = false;
        if(uniqueQueries.length>0){
            uniqueQueries.forEach((k, j)=>{
                if(k.query == l.query){
                    alreadyExist = true;
                }
            });
        }
        if(!alreadyExist){
           uniqueQueries.push(l)
        }
    });

这是我的两分钱。如果您知道财产的顺序相同,则可以将元素串接起来,并从数组中删除重复项,然后再次解析数组。类似于:

var things=新对象();things.thing=新数组();thing.thing.push({place:“here”,name:“stuff”});things.thing.push({place:“there”,name:“morestuff”});things.thing.push({place:“there”,name:“morestuff”});let-stringified=things.thing.map(i=>JSON.sringify(i));let unique=stringified.filter((k,idx)=>stringified.indexOf(k)==idx).map(j=>JSON.parse(j))console.log(唯一);

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')