获取条件输出的最有效方法

Question

我有2个数组json对象

arr1=[{'id':'A'},{'id':'B'}, {'id':'C'}];
arr2=[{'id':'A'},{'id':'C'}, {'id':'D'}, {'id': 'E'}];

我如何获取所有唯一的json对象？我希望我的输出成为

output = [{'id':'B'}, {'id':'D'}, {'id':'E'}]

谢谢

编辑

我不想删除重复项。我只想得到独一无二的对象。

output = [{'id':'B'}, {'id':'D'}, {'id':'E'}]

NOT

output = [{'id':'A'},{'id':'B'}, {'id':'C'}, {'id':'D'}, {'id':'E'}]

对不起，如果我不清楚的话

Answer 1

这个问题被问到一个[[lot，但是在这种特殊情况下，可以这样解决：

arr1=[{'id':'A'},{'id':'B'}, {'id':'C'}];
arr2=[{'id':'A'},{'id':'C'}, {'id':'D'}, {'id': 'E'}];

var uniques = [...arr1, ...arr2]
    .filter((v, i, a) => a.find(el => el.id === v.id) === v );

console.log(uniques);

Answer 2

arr1=[{'id':'A'},{'id':'B'}, {'id':'C'}];
arr2=[{'id':'A'},{'id':'C'}, {'id':'D'}, {'id': 'E'}];

function uniqById(...args) {
  const attr = 'id';
  const seen = {};
  return arr1.concat(arr2).filter(o => seen[o[attr]] ? false : seen[o[attr]] = true, true);
}

console.log(uniqById(arr1, arr2));