替换数组中的对象

问题描述 投票：0回答：18

我有这个javascript对象：

var arr1 = [{id:'124',name:'qqq'}, 
           {id:'589',name:'www'}, 
           {id:'45',name:'eee'},
           {id:'567',name:'rrr'}]

var arr2 = [{id:'124',name:'ttt'}, 
           {id:'45',name:'yyy'}]

我需要将 arr1 中的对象替换为 arr2 中具有相同 id 的项目。

所以这是我想要得到的结果：

var arr1 = [{id:'124',name:'ttt'}, 
           {id:'589',name:'www'}, 
           {id:'45',name:'yyy'},
           {id:'567',name:'rrr'}]

如何使用javascript实现它？

javascript lodash
18个回答
281
投票

您可以将 

Array#map
与 
Array#find
 一起使用。

arr1.map(obj => arr2.find(o => o.id === obj.id) || obj);


var arr1 = [{
    id: '124',
    name: 'qqq'
}, {
    id: '589',
    name: 'www'
}, {
    id: '45',
    name: 'eee'
}, {
    id: '567',
    name: 'rrr'
}];

var arr2 = [{
    id: '124',
    name: 'ttt'
}, {
    id: '45',
    name: 'yyy'
}];

var res = arr1.map(obj => arr2.find(o => o.id === obj.id) || obj);

console.log(res);

这里，如果在 

arr2.find(o => o.id === obj.id)
中找到 
arr2
，
id
将返回来自 
arr2
的元素，即对象。如果不是，则返回 
arr1
中的相同元素，即 
obj
。

10
投票

关于时间与空间总是会有一场很好的争论，但是这些天我发现从长远来看使用空间更好。抛开数学不谈，让我们看一下使用散列图、字典或关联数组是您想要标记的任何简单数据结构..

    var marr2 = new Map(arr2.map(e => [e.id, e]));
    arr1.map(obj => marr2.has(obj.id) ? marr2.get(obj.id) : obj);

我喜欢这种方法，因为尽管您可能会与数字较小的数组争论，但您却在浪费空间，因为像 @Tushar 方法这样的内联方法的性能与此方法几乎没有区别。不过，我运行了一些测试，图表显示了两种方法在 n 0 - 1000 范围内的性能（以毫秒为单位）。您可以根据您的情况决定哪种方法最适合您，但根据我的经验，用户不太关心小空间，但他们确实关心小速度。

这是我为数据源运行的性能测试

var n = 1000;
var graph = new Array();
for( var x = 0; x < n; x++){
  var arr1s = [...Array(x).keys()];
  var arr2s = arr1s.filter( e => Math.random() > .5);
  var arr1 = arr1s.map(e => {return {id: e, name: 'bill'}});
  var arr2 = arr2s.map(e => {return {id: e, name: 'larry'}});
  // Map 1
  performance.mark('p1s');
  var marr2 = new Map(arr2.map(e => [e.id, e]));
  arr1.map(obj => marr2.has(obj.id) ? marr2.get(obj.id) : obj);
  performance.mark('p1e');
  // Map 2
  performance.mark('p2s');
  arr1.map(obj => arr2.find(o => o.id === obj.id) || obj);
  performance.mark('p2e');
  graph.push({ x: x, r1: performance.measure('HashMap Method', 'p1s', 'p1e').duration, r2: performance.measure('Inner Find', 'p2s','p2e').duration});
}
9
投票

有什么问题

Object.assign(target, source)
？

数组在Javascript中仍然是类型对象，因此只要找到匹配的键，使用分配仍然应该重新分配由运算符解析的任何匹配的键，对吗？

8
投票

我想建议另一种解决方案：

const objectToReplace = this.array.find(arrayItem => arrayItem.id === requiredItem.id);
Object.assign(objectToReplace, newObject);
5
投票

由于您使用的是 Lodash，您可以使用 

_.map
和 
_.find
来确保支持主要浏览器。

最后我会选择这样的东西： 

function mergeById(arr) {
  return {
    with: function(arr2) {
      return _.map(arr, item => {
        return _.find(arr2, obj => obj.id === item.id) || item
      })
    }
  }
}

var result = mergeById([{id:'124',name:'qqq'}, 
           {id:'589',name:'www'}, 
           {id:'45',name:'eee'},
           {id:'567',name:'rrr'}])
    .with([{id:'124',name:'ttt'}, {id:'45',name:'yyy'}])

console.log(result);
<script src="https://raw.githubusercontent.com/lodash/lodash/4.13.1/dist/lodash.js"></script>

5
投票

我喜欢用 

arr2
遍历 
foreach()
并使用 
findIndex()
检查 
arr1
中是否出现：

var arr1 = [{id:'124',name:'qqq'}, 
           {id:'589',name:'www'}, 
           {id:'45',name:'eee'},
           {id:'567',name:'rrr'}]

var arr2 = [{id:'124',name:'ttt'}, 
           {id:'45',name:'yyy'}]

arr2.forEach(element => {
            const itemIndex = arr1.findIndex(o => o.id === element.id);
            if(itemIndex > -1) {
                arr1[itemIndex] = element;
            } else {
                arr1 = arr1.push(element);
            }       
        });
    
console.log(arr1)

4
投票

感谢 ES6，我们可以用简单的方法实现它 -> 例如在 util.js 模块上 ;)))。

  1. 合并2个实体数组

    export const mergeArrays = (arr1, arr2) => 
   arr1 && arr1.map(obj => arr2 && arr2.find(p => p.id === obj.id) || obj);

获取 2 个数组并将其合并.. Arr1 是主数组，优先级是 高合并过程

  1. 将数组与相同类型的实体合并

    export const mergeArrayWithObject = (arr, obj) => arr && arr.map(t => t.id === obj.id ? obj : t);

它将相同类型的数组与某种类型合并为

示例：人员数组 ->

[{id:1, name:"Bir"},{id:2, name: "Iki"},{id:3, name:"Uc"}]   
second param Person {id:3, name: "Name changed"}

结果是

[{id:1, name:"Bir"},{id:2, name: "Iki"},{id:3, name:"Name changed"}]
3
投票

考虑到接受的答案对于大型数组 O(nm) 可能效率较低，我通常更喜欢这种方法，O(2n + 2m)：

function mergeArrays(arr1 = [], arr2 = []){
    //Creates an object map of id to object in arr1
    const arr1Map = arr1.reduce((acc, o) => {
        acc[o.id] = o;
        return acc;
    }, {});
    //Updates the object with corresponding id in arr1Map from arr2, 
    //creates a new object if none exists (upsert)
    arr2.forEach(o => {
        arr1Map[o.id] = o;
    });

    //Return the merged values in arr1Map as an array
    return Object.values(arr1Map);
}

单元测试：

it('Merges two arrays using id as the key', () => {
   var arr1 = [{id:'124',name:'qqq'}, {id:'589',name:'www'}, {id:'45',name:'eee'}, {id:'567',name:'rrr'}];
   var arr2 = [{id:'124',name:'ttt'}, {id:'45',name:'yyy'}];
   const actual = mergeArrays(arr1, arr2);
   const expected = [{id:'124',name:'ttt'}, {id:'589',name:'www'}, {id:'45',name:'yyy'}, {id:'567',name:'rrr'}];
   expect(actual.sort((a, b) => (a.id < b.id)? -1: 1)).toEqual(expected.sort((a, b) => (a.id < b.id)? -1: 1));
})
2
投票
// here find all the items that are not it the arr1
const temp = arr1.filter(obj1 => !arr2.some(obj2 => obj1.id === obj2.id))
// then just concat it
arr1 = [...temp, ...arr2]
2
投票

这是一种更透明的方法。我发现单行代码更难阅读，也更难调试。

export class List {
    static replace = (object, list) => {
        let newList = [];
        list.forEach(function (item) {
            if (item.id === object.id) {
                newList.push(object);
            } else {
                newList.push(item);
            }
        });
        return newList;
    }
}
0
投票

如果您不关心数组的顺序，那么您可能想使用 

differenceBy()
获得 
arr1
和 
arr2
之间的差异，然后只需使用 concat() 即可追加所有更新的对象。 id





var result = _(arr1).differenceBy(arr2, 'id').concat(arr2).value();



var arr1 = [{
  id: '124',
  name: 'qqq'
}, {
  id: '589',
  name: 'www'
}, {
  id: '45',
  name: 'eee'
}, {
  id: '567',
  name: 'rrr'
}]

var arr2 = [{
  id: '124',
  name: 'ttt'
}, {
  id: '45',
  name: 'yyy'
}];

var result = _(arr1).differenceBy(arr2, 'id').concat(arr2).value();

console.log(result);





    





      

      
      
      

      

        

          

            

              
0
投票

              
            

          

        

        



话虽如此，为了便于阅读，我将问题分解为两个函数。



<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.13.1/lodash.js"></script>



jsfiddle

    




      

      
      
      

      

        

          

            

              
0
投票

              
            

          

        

        

    



      

      
      
      

      

        

          

            

              
0
投票

              
            

          

        

        



function getMatch(elem) {
    function action(ele, val) {
        if(ele === val){ 
            elem = arr2[i]; 
        }
    }

    for (var i = 0; i < arr2.length; i++) {
        action(elem.id, Object.values(arr2[i])[0]);
    }
    return elem;
}

var modified = arr1.map(getMatch);


    



      

      
      
      

      

        

          

            

              
0
投票

              
            

          

        

        


masterData = [{id: 1, name: "aaaaaaaaaaa"}, 
        {id: 2, name: "Bill"},
        {id: 3, name: "ccccccccc"}];

updatedData = [{id: 3, name: "Cat"},
               {id: 1, name: "Apple"}];

updatedData.forEach(updatedObj=> {
       // For every updatedData object (dataObj), find the array index in masterData where the IDs match.
       let indexInMasterData = masterData.map(masterDataObj => masterDataObj.id).indexOf(updatedObj.id); // First make an array of IDs, to use indexOf().
       // If there is a matching ID (and thus an index), replace the existing object in masterData with the updatedData's object.
       if (indexInMasterData !== undefined) masterData.splice(indexInMasterData, 1, updatedObj);
});

/* masterData becomes [{id: 1, name: "Apple"}, 
                       {id: 2, name: "Bill"},
                       {id: 3, name: "Cat"}];  as you want.`*/


    



      

      
      
      

      

        

          

            

              
0
投票

              
            

          

        

        




//newArr contains the mapped array from arr2 to arr1. 
//arr1 still contains original value

var newArr = arr1.map(obj => arr2.find(o => o.id === obj.id) || obj);






    





      

      
      
      

      

        

          

            

              
0
投票

              
            

          

        

        
使用 key 合并和更新记录的高效 JavaScript 代码

论坛上已经有一些很好的答案，但我仍在发布我在 JavaScript 中合并和更新记录的改进解决方案。






Array.prototype.update = function(...args) {
  return this.map(x=>args.find((c)=>{return c.id===x.id})  || x)    
}

const result = 
        [
            {id:'1',name:'test1'}, 
            {id:'2',name:'test2'}, 
            {id:'3',name:'test3'},
            {id:'4',name:'test4'}
        ]
        .update({id:'1',name:'test1.1'}, {id:'3',name:'test3.3'})

console.log(result)





    





      

      
      
      

      

        

          

            

              
-3
投票

              
            

          

        

        



const existingItems = [
  { id: 1, name: 'Item A', category: 'Category 1' },
  { id: 2, name: 'Item B', category: 'Category 2' },
  { id: 3, name: 'Item C', category: 'Category 1' },
  // Add more source items as needed
];

// Usually new data from the server
const newItems = [
  { id: 3, name: 'Item C', category: 'Category 3' },
  { id: 4, name: 'Item D', category: 'Category 4' },
  // Add more new items as needed
];

let key = "id";

let newList = [...new Map([...existingItems, ...newItems].map((item) => [item[key], item])).values()];

console.log(newList);

    

      

      
    

  

  

    
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