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
考虑lodash.uniqWith
const objects = [{ 'x': 1, 'y': 2 }, { 'x': 2, 'y': 1 }, { 'x': 1, 'y': 2 }];
_.uniqWith(objects, _.isEqual);
// => [{ 'x': 1, 'y': 2 }, { 'x': 2, 'y': 1 }]
其他回答
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')
let data = [
{
'name': 'Amir',
'surname': 'Rahnama'
},
{
'name': 'Amir',
'surname': 'Stevens'
}
];
let non_duplicated_data = _.uniqBy(data, 'name');
基本方法是：
const obj = {};
for (let i = 0, len = things.thing.length; i < len; i++) {
obj[things.thing[i]['place']] = things.thing[i];
}
things.thing = new Array();
for (const key in obj) {
things.thing.push(obj[key]);
}
这个呢
function dedupe(arr, compFn){
let res = [];
if (!compFn) compFn = (a, b) => { return a === b };
arr.map(a => {if(!res.find(b => compFn(a, b))) res.push(a)});
return res;
}
es6魔术在一条线上。。。在那时候可读！
// returns the union of two arrays where duplicate objects with the same 'prop' are removed
const removeDuplicatesWith = (a, b, prop) => {
a.filter(x => !b.find(y => x[prop] === y[prop]));
};
