抓取数组中没有任何重复项的对象的最有效方法

Question

我有2个脚本对象数组：arr1=[{'id':'A'},{'id':'B'}, {'id':'C'}]; arr2=[{'id':'A'},{'id':'C'}, {'id':'D'}, {'id': 'E'}];如何仅获取没有重复项的元素？谢谢。编辑我不想删除重复项。我只想获取唯一对象。output = [{'id':'B'}, {'id':'D'}, {'id':'E'}]不output = [{'id':'A'},{'id':'B'}, {'id':'C'}, {'id':'D'}, {'id':'E'}]抱歉，如果我不清楚

Answer 1

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

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

const out = Object.entries(arr1.concat(arr2).reduce((a, c) => {

  a[c.id] = a[c.id] || 0;

  a[c.id]++;

  return a;

}, {})).filter(([k, v]) => v === 1).map(([k]) => ({id: k}));

console.log(out);

Answer 2

有点乱，但它有效。首先查找仅出现一次的 id，然后筛选组合数组以仅包含唯一 ID。

arr1 = [{ id: "A" }, { id: "B" }, { id: "C" }];

arr2 = [{ id: "A" }, { id: "C" }, { id: "D" }, { id: "E" }];

function getCount(array, elem) {

    return array.filter((e) => e === elem).length;

}

uniqueIds = [...arr1, ...arr2]

    .map((e) => e.id)

    .filter((e, i, a) => getCount(a, e) === 1);

let uniques = [...arr1, ...arr2].filter((e) => uniqueIds.includes(e.id));

console.log(uniques);

Answer 3

好吧，所以也许不是。以下是一种像您要求的那样检索输出的方法：

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

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

function getUniques(arr) {

  let counter = arr.reduce((acc, val) =>

    (acc[val.id] = (acc[val.id] || 0) + 1, acc), {});

  return Object.keys(counter)

    .filter(k => counter[k] === 1)

    .map(k => ({id: k}));

}

console.log(getUniques([...arr1, ...arr2]));

Answer 4

只是另一种方法：

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

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

 var res = [];

 var newArr = arr1.concat(arr2);

 var reverseArr = newArr.slice().reverse();

 newArr.forEach((el,idx)=> {

 let iex = newArr.findIndex((e,i)=> e.id == el.id );

 if (iex === idx && iex === newArr.length - (reverseArr.findIndex((e,i)=> e.id == 

   el.id )+ 1) ){

     res.push(el)

   }

});

console.log(res)