typescript 获取对象数组中重复对象的列表

Get list of duplicate objects in an array of objects

提问by ua_boaz

I am trying to get duplicate objects within an array of objects. Let's say the object is like below.

我试图在对象数组中获取重复的对象。假设对象如下所示。

values = [
  { id: 10, name: 'someName1' },
  { id: 10, name: 'someName2' },
  { id: 11, name: 'someName3' },
  { id: 12, name: 'someName4' }
];

Duplicate objects should return like below:

重复的对象应该返回如下：

duplicate = [
  { id: 10, name: 'someName1' },
  { id: 10, name: 'someName2' }
];

回答by ggorlen

You can use Array#reduceto make a counter lookup table based on the idkey, then use Array#filterto remove any items that appeared only once in the lookup table. Time complexity is O(n).

您可以使用Array#reduce基于id键的计数器查找表，然后用于Array#filter删除查找表中仅出现一次的任何项目。时间复杂度为 O(n)。

const values = [{id: 10, name: 'someName1'}, {id: 10, name: 'someName2'}, {id: 11, name:'someName3'}, {id: 12, name: 'someName4'}];

const lookup = values.reduce((a, e) => {
  a[e.id] = ++a[e.id] || 0;
  return a;
}, {});

console.log(values.filter(e => lookup[e.id]));

回答by Aljosha Novakovic

You haven't clarified whether two objects with different ids, but the same "name" count as a duplicate. I will assume those do not count as a duplicate; in other words, only objects with the same id will count as duplicate.

您还没有澄清两个具有不同 id 的对象，但相同的“名称”是否算作重复。我会假设那些不算作重复；换句话说，只有具有相同 id 的对象才算作重复。

let ids = {};
let dups = [];

values.forEach((val)=> {
  if (ids[val.id]) {
    // we have already found this same id
    dups.push(val)
  } else {
    ids[val.id] = true;
  }
})
return dups;

回答by HMagdy

Let's say you have:

假设你有：

arr = [
    { id:10, name: 'someName1' },
    { id:10, name: 'someName2' },
    { id:11, name: 'someName3' },
    { id:12, name: 'someName4' }
]

So, to get unique items:

因此，要获得独特的物品：

unique = arr
     .map(e => e['id'])
     .map((e, i, final) => final.indexOf(e) === i && i)
     .filter(obj=> arr[obj])
     .map(e => arr[e]);

Then, result will be

那么，结果将是

unique = [
     { id:10, name: 'someName1' },
     { id:11, name: 'someName3' },
     { id:12, name: 'someName4' }
]

And, to get duplicate ids:

并且，要获得重复的 ID：

duplicateIds = arr
     .map(e => e['id'])
     .map((e, i, final) => final.indexOf(e) !== i && i)
     .filter(obj=> arr[obj])
     .map(e => arr[e]["id"])

List of IDs will be

ID 列表将是

duplicateIds = [10]

Thus, to get duplicates objects:

因此，要获取重复的对象：

duplicate = arr.filter(obj=> dublicateIds.includes(obj.id));

Now you have it:

现在你拥有了：

duplicate = [
    { id:10, name: 'someName1' },
    { id:10, name: 'someName2' }
]

Thanks https://reactgo.com/removeduplicateobjects/

谢谢https://reactgo.com/removeduplicateobjects/

回答by Ori Drori

With lodash you can use _.groupBy()to group elements by their id. Than _.filter()out groups that have less than two members, and _.flatten()the results:

随着lodash可以使用_.groupBy()由他们的组元素id。比_.filter()少于两个成员的组，_.flatten()结果：

const values = [{id: 10, name: 'someName1'}, {id: 10, name: 'someName2'}, {id: 11, name:'someName3'}, {id: 12, name: 'someName4'}];

const result = _.flow([
  arr => _.groupBy(arr, 'id'), // group elements by id
  g => _.filter(g, o => o.length > 1), // remove groups that have less than two members
  _.flatten // flatten the results to a single array
])(values);

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>

回答by Akrion

With lodash you can solve this with filterand countByfor complexity of O(n):

随着lodash你可以解决这个问题filter，并countBy为复杂性O(n)：

const data = [{ id: 10,name: 'someName1' }, { id: 10,name: 'someName2' }, { id: 11,name: 'someName3' }, { id: 12,name: 'someName4' } ]

const counts = _.countBy(data, 'id')
console.log(_.filter(data, x => counts[x.id] > 1))

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>

You could do the same with ES6 like so:

你可以像这样对 ES6 做同样的事情：

const data = [{ id: 10,name: 'someName1' }, { id: 10,name: 'someName2' }, { id: 11,name: 'someName3' }, { id: 12,name: 'someName4' } ]

const countBy = (d, id) => d.reduce((r,{id},i,a) => (r[id] = a.filter(x => x.id == id).length, r),{})
const counts = countBy(data, 'id')
console.log(data.filter(x => [x.id] > 1))

回答by Saurabh

Try this

试试这个

function checkDuplicateInObject(propertyName, inputArray) {
  var seenDuplicate = false,
  testObject = {};

  inputArray.map(function(item) {
  var itemPropertyName = item[propertyName];    
  if (itemPropertyName in testObject) {
  testObject[itemPropertyName].duplicate = true;
  item.duplicate = true;
  seenDuplicate = true;
 }
 else {
   testObject[itemPropertyName] = item;
   delete item.duplicate;
 }
});

 return seenDuplicate;
}

referred from : http://www.competa.com/blog/lets-find-duplicate-property-values-in-an-array-of-objects-in-javascript/

引用自：http: //www.competa.com/blog/lets-find-duplicate-property-values-in-an-array-of-objects-in-javascript/

回答by ControlAltDel

create a function, objectsEqual, and then use a filter

创建一个函数，objectsEqual然后使用过滤器

values.filter(function(num, elem) {
  for (var i = 0; i < values.length; i++) {
    if (i == num) continue;
    if (objectsEqual(values[num], values[i])) return true;
  }
  return false;
}

回答by Niall

You can use an array to store unique elements and use filter on values to only return duplicates.

您可以使用数组来存储唯一元素，并对值使用过滤器以仅返回重复项。

const unique = []

const duplicates = values.filter(o => {

   if(unique.find(i => i.id === o.id && i.name === o.name)) {
     return true
   }

   unique.push(o)
   return false;
})