根据 2 键从数组中删除重复对象答案

【问题标题】：remove duplicate object from array according 2 key根据 2 键从数组中删除重复对象
【发布时间】：2020-04-03 07:40:54
【问题描述】：

我有这样的对象数组，我想根据 2 个键删除重复的项目，在我的数据中，我想删除具有相同 room_id 和 mdate 的项目，在下面的代码对象 4 和 5 具有相同的 room_id 和 mdate 所以我应该删除其中一个，有什么解决方案吗？

我有删除重复对象的功能，但它可以检查一个键

function getUnique(arr, comp) {

    arr.map()
    const unique = arr
        .map(e => e[comp])

        // store the keys of the unique objects
        .map((e, i, final) => final.indexOf(e) === i && i)

        // eliminate the dead keys & store unique objects
        .filter(e => arr[e])
        .map(e => arr[e]);

    return unique;
}



arr = [{
cond: 2
extra_fee: 20000
kind: ""
mdate: "2019-12-15"
price: "100000"
room_id: 1
},
{
cond: 2
extra_fee: 20000
kind: ""
mdate: "2019-12-10"
price: "100000"
room_id: 1
},
{
cond: 2
extra_fee: 20000
kind: ""                       object 3 <--------------   is same - have same mdate and room_id
mdate: "2019-12-10"
price: "100000"
room_id: 2
},
{
cond: 2
extra_fee: 20000
kind: ""
mdate: "2019-12-10"
price: "100000"
room_id: 2                       object 4 <--------------   are same - have same mdate and room_id
},
{
cond: 2
extra_fee: 20000
kind: ""
mdate: "2019-12-10"
price: "100000"
room_id: 7
}
]

【问题讨论】：

标签： javascript ecmascript-6 ecmascript-5 ecmascript-2016

【解决方案1】：

这将删除重复项，

  let arr = [{
    cond: 2,
    extra_fee: 20000,
    kind: "",
    mdate: "2019-12-15",
    price: "100000",
    room_id: 1
  },
  {
    cond: 2,
    extra_fee: 20000,
    kind: "",
    mdate: "2019-12-10",
    price: "100000",
    room_id: 1
  },
  {
    cond: 2,
    extra_fee: 20000,
    kind: "",
    mdate: "2019-12-10",
    price: "100000",
    room_id: 2
  },
  {
    cond: 2,
    extra_fee: 20000,
    kind: "",
    mdate: "2019-12-10",
    price: "100000",
    room_id: 2
  },
  {
    cond: 2,
    extra_fee: 20000,
    kind: "",
    mdate: "2019-12-10",
    price: "100000",
    room_id: 7
  }
  ]
  
const arrayWithoutDuplicates = arr.filter((v,i,a)=>a.findIndex(t=>(t.room_id=== v.room_id && t.mdate===v.mdate))===i)
console.log(arrayWithoutDuplicates)

【讨论】：

【解决方案2】：

试试这个：

function getUnique(arr, comp) {
    let tmpObj = {};
    let unique = [];
    for (let i = 0; i < arr.length; i++) {
      if (tmpObj[arr[i][comp]] === undefined) {
        tmpObj[arr[i][comp]] = arr[i][comp];
        unique.push(arr[i]);
      }
    }
    return unique;
  }

  let arr = [{
    cond: 2,
    extra_fee: 20000,
    kind: "",
    mdate: "2019-12-15",
    price: "100000",
    room_id: 1
  },
  {
    cond: 2,
    extra_fee: 20000,
    kind: "",
    mdate: "2019-12-10",
    price: "100000",
    room_id: 1
  },
  {
    cond: 2,
    extra_fee: 20000,
    kind: "",
    mdate: "2019-12-10",
    price: "100000",
    room_id: 2
  },
  {
    cond: 2,
    extra_fee: 20000,
    kind: "",
    mdate: "2019-12-10",
    price: "100000",
    room_id: 2
  },
  {
    cond: 2,
    extra_fee: 20000,
    kind: "",
    mdate: "2019-12-10",
    price: "100000",
    room_id: 7
  }
  ]
  console.log(getUnique(arr, "room_id"));

【讨论】：

它会删除所有具有相同 room_id 的元素，如果 mdate 和 room_id 中的元素相同，我只想删除一个元素