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
对于一个可读且简单的解决方案搜索者，她是我的版本：
function removeDupplicationsFromArrayByProp(originalArray, prop) {
let results = {};
for(let i=0; i<originalArray.length;i++){
results[originalArray[i][prop]] = originalArray[i];
}
return Object.values(results);
}
其他回答
TypeScript解决方案
这将删除重复的对象，并保留对象的类型。
function removeDuplicateObjects(array: any[]) {
return [...new Set(array.map(s => JSON.stringify(s)))]
.map(s => JSON.parse(s));
}
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 }]
这里是ES6的解决方案，您只想保留最后一项。该解决方案功能强大，符合Airbnb风格。
const things = {
thing: [
{ place: 'here', name: 'stuff' },
{ place: 'there', name: 'morestuff1' },
{ place: 'there', name: 'morestuff2' },
],
};
const removeDuplicates = (array, key) => {
return array.reduce((arr, item) => {
const removed = arr.filter(i => i[key] !== item[key]);
return [...removed, item];
}, []);
};
console.log(removeDuplicates(things.thing, 'place'));
// > [{ place: 'here', name: 'stuff' }, { place: 'there', name: 'morestuff2' }]
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]));
};
这是我的解决方案，它基于object.prop搜索重复的对象，当找到重复的对象时，它会将array1中的值替换为array2值
function mergeSecondArrayIntoFirstArrayByProperty(array1, array2) {
for (var i = 0; i < array2.length; i++) {
var found = false;
for (var j = 0; j < array1.length; j++) {
if (array2[i].prop === array1[j].prop) { // if item exist in array1
array1[j] = array2[i]; // replace it in array1 with array2 value
found = true;
}
}
if (!found) // if item in array2 not found in array1, add it to array1
array1.push(array2[i]);
}
return array1;
}
