从对象数组中删除相同的值

Question

我想通过比较 2 个数组从数组中删除相同的对象。

样本数据：

arr1 = [
  {id: 1, name: "a"},
  {id: 2, name: "b"},
  {id: 3, name: "c"},
  {id: 4, name: "d"},
];

arr2 = [
  {id: 1, name: "a"},
  {id: 4, name: "d"},
];

let newArray = []; // new array with with no same values it should be unique.
arr1.map((val, i)=>{
   arr2.map((val2)=>{
    if(val.id == val2.id){
       console.log('Matched At: '+ i) // do nothing
    }else{
      newArray.push(val);
    }
   })
})
console.log(newArray); // e.g: [{id: 2, name: "b"}, {id: 3, name: "c"},];

Answer 1

您可以使用filter() 方法更新您的代码，而不是使用.map() 方法，例如：

const arr1 = [
  {id: 1, name: "a"},
  {id: 2, name: "b"},
  {id: 3, name: "c"},
  {id: 4, name: "d"},
], arr2 = [
  {id: 1, name: "a"},
  {id: 4, name: "d"},
];

let newArray = []; // new array with with no same values it should be unique.
newArray = arr1.filter(function(a) {
    for(var i=0; i < arr2.length; i++){
      if(a.id == arr2[i].id) return false;
    }
    return true;
});
console.log(newArray);

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Answer 2

const isInArray = (arr, id, name) => arr.reduce((result, curr) => ((curr.name === name && curr.id === id) || result), false)

const newArray = arr1.reduce((result, curr) => (isInArray(arr2, curr.id, curr.name) ? result : result.concat(curr)), [])

Answer 3

Array.filter 与非Array.some 结合。

这里的诀窍也是不要some,..

const arr1 = [
  {id: 1, name: "a"},
  {id: 2, name: "b"},
  {id: 3, name: "c"},
  {id: 4, name: "d"},
], arr2 = [
  {id: 1, name: "a"},
  {id: 4, name: "d"},
];

const newArray=arr1.filter(a=>!arr2.some(s=>s.id===a.id));

console.log(newArray);

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如 cmets 中所述，该问题的解释可能略有不同。如果您还想要 arr2 中的 unqiue 项目，您基本上只需执行两次并加入。 IOW：检查 arr1 中没有 arr2 的内容，然后检查 arr2 中没有 arr1 的内容。

例如..

const notIn=(a,b)=>a.filter(f=>!b.some(s=>f.id===s.id));
const newArray=[...notIn(arr1, arr2), ...notIn(arr2, arr1)];

更新 2： 时间复杂度，正如 qiAlex 所提到的，循环中存在循环。尽管some 会在找到匹配项时短路，但如果数据集变大，事情可能会变慢。这是Set 和Map 进来的。

所以要使用Set 来解决这个问题。

const notIn=(a,b)=>a.filter(a=>!b.has(a.id));
const newArray=[
  ...notIn(arr1, new Set(arr2.map(m=>m.id))),
  ...notIn(arr2, new Set(arr1.map(m=>m.id)))
];

Answer 4

您使用Array.prototype.some 检查第一个数组中的每个元素是否其id 位于第二个数组中。如果元素不存在，则只生成它。

const arr1 = [
  {id: 1, name: "a"},
  {id: 2, name: "b"},
  {id: 3, name: "c"},
  {id: 4, name: "d"},
];

const arr2 = [
  {id: 1, name: "a"},
  {id: 4, name: "d"},
];

const result = arr1.filter(x => !arr2.some(y => y.id === x.id));

console.log(result);

Answer 5

我认为一个简单的比较器可以用于获取差异然后将它们连接起来。 使用这种方法，您不需要检查哪个数组更大。

arr1 = [  {id: 1, name: "a"},  {id: 2, name: "b"},  {id: 3, name: "c"},  {id: 4, name: "d"}];

arr2 = [  {id: 1, name: "a"},  {id: 4, name: "d"},];

function localComparer(b){
  return function(a){
    return b.filter(
    function(item){
      return item.id == a.id && item.name == a.name
    }).length == 0;
  }
}

var onlyInArr1 = arr1.filter(localComparer(arr2));
var onlyInArr2 = arr2.filter(localComparer(arr1));

console.log(onlyInArr1.concat(onlyInArr2));

Answer 6

我们可以通过检查some元素是否不包含在当前数组中来过滤值：

const result = arr1.reduce((a, c) => {
  if (!arr2.some(a2 => a2.id === c.id))
      a.push(c);
  return a;
}, [])

一个例子：

let arr1 = [
  {id: 1, name: "a"},
  {id: 2, name: "b"},
  {id: 3, name: "c"},
  {id: 4, name: "d"},
];

let arr2 = [
  {id: 1, name: "a"},
  {id: 4, name: "d"},
];

const result = arr1.reduce((a, c) => {
  if (!arr2.some(a2 => a2.id === c.id))
      a.push(c);
  return a;
}, [])

console.log(result);

Answer 7

试试这个 -

const arr1 = [
  {id: 1, name: "a"},
  {id: 2, name: "b"},
  {id: 3, name: "c"},
  {id: 4, name: "d"},
];

const arr2 = [
  {id: 1, name: "a"},
  {id: 4, name: "d"},
];

const arr3 = [...arr1, ...arr2];
const mySubArray = _.uniq(arr3, 'id');
console.log(mySubArray);

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Answer 8

每个答案都有很多循环。

代码的复杂度我的答案是2N，

想法是：

1. 合并数组。

2. 第一个循环 - 以某种方式标记重复项

3. 第二个循环 - 过滤掉重复项

arr1 = [
  {id: 1, name: "a"},
  {id: 2, name: "b"},
  {id: 3, name: "c"},
  {id: 4, name: "d"},
];

arr2 = [
  {id: 1, name: "a"},
  {id: 4, name: "d"},
];

let newArray = [...arr1, ...arr2].reduce((acc, item, index) => {
  acc.items.push(item);

  if (typeof acc.map[item.id] !== 'undefined') {
    acc.items[acc.map[item.id]] = null;
    acc.items[index] = null;
  }
  acc.map[item.id] = index;
  
  return acc
},  {map: {}, items: []}).items.filter(item => !!item)


console.log(newArray);