基于 javascript/react 中的属性过滤数组

Question

您好，我需要过滤具有相同地址的公司（数组）并创建一个数组，例如：

 [{address:uniqueaddress1,organization:[company1,company2]},
  {address:uniqueaddress2,organization:[company3,company4]
  .....]   

我正在使用以下代码：

 var organizations = [];
 var  dataPoints = [];
 for (var i = 0; i < companies.length; i++) {
  for (var j = 0; j < companies.length; j++) {
     if (i === j) continue;
     if (companies[j].address === companies[i].address) {
        organizations.push(companies[j]);            
        companies[j].added = true;  //To mark it is added
    }
    dataPoints.push({address:companies[j].address, organizations: companies[j]});
   }   
 }

原始数组：

  0:Object
  added:true
  address:"sample address 1"
  id:258
  latitude:90.90227
  longitude:12.538208
  name:"Company name 1"
  postalCode:"90450"

Answer 1

您可以使用Map 以更好的方式实现此目的（尽管您也可以将这种方法与数组一起使用，就像您已经完成的那样）。您还可以删除内部循环，以便在更大的结果集上稍微提高效率，方法是在进行时添加，而不是为每个地址再次循环。

let original = [
    { address: '123 Example Street', id: 1 }, 
    { address: '123 Example Street', id: 2 },
    { address: '456 Example Street', id: 3 }
];

let grouped = new Map();

original.forEach(function(company) {

    let companies = grouped.get(company.address);

    // If we already have the key, then just push into the array.
    if (companies !== undefined) {
        companies.push(company);
    }
    else {
        // ...if not then create a new array.
        companies = [company];
    }

    grouped.set(company.address, companies);

});

Answer 2

如果您习惯使用ES6 语法，那么您可以使用filter 和map 方法来做到这一点。下面的代码过滤了数组comapanies，map方法创建了一个临时数组，然后我们使用indexOf方法检查我们的地图中是否可以有相同的对象。

let companies = [{
  "added": true,
  "address": "sample address 1",
  "id": 258,
  "latitude": 90.90227,
  "longitude": 12.538208,
  "name": "Company name 1"
}, {
  "added": true,
  "address": "sample address 1",
  "id": 258,
  "latitude": 90.90227,
  "longitude": 12.538208,
  "name": "Company name 1"
}, {
  "added": true,
  "address": "sample address 2",
  "id": 258,
  "latitude": 90.90227,
  "longitude": 12.538208,
  "name": "Company name 1"
}, {
  "added": true,
  "address": "sample address 2",
  "id": 258,
  "latitude": 90.90227,
  "longitude": 12.538208,
  "name": "Company name 1"
}]

function uniqueArray(array, prop) {
  return array.filter((obj, pos, arr) => {
    return arr.map(mapObj => mapObj[prop]).indexOf(obj[prop]) === pos;
  });
}
console.log(uniqueArray(companies, "address"))

Credits