通過對象的id在javascript數組中查找並移動對象

Question

我有2個對象數組。每個對象都有一個Id屬性。現在，如果我有第三個只有Ids的數組，那麼根據這些Ids並將它們移動到數組2，從array1中查找對象的更好更快的方法是什麼。

非常感謝回答..

示例代碼：

Person = function(id, fn, ln) { 
    this.id = id, 
    this.firstName = fn, 
    this.lastName = ln 
} 

array1 = new Array(); 
// add 500 new Person objects to this array 

array2 = new Array(); 
// add some other new Person objects to this array 

function moveArrayItems(ids) { 
    // ids is an array of ids e.g. [1,2,3,4,5,6,...] 
    // Now I want to find all the person objects from array1 whose ids 
    // match with the ids array passed into this method. Then move them to array2. 
    // What is the best way to achive this? 
} 

Answer 1

如果你真的有每個陣列中500+的對象，你可能會更好過使用散列存儲的對象，通過ID鍵：

var people = { 
       1: {id:1, name:"George Washington"}, 
       2: {id:2, name:"John Adams"}, 
       3: {id:3, name:"Thomas Jefferson"}, // ... 
      } 

var people2 = {} 

現在是微不足道的（和很多

function moveArrayItems(ids) { 
    var i,id; 
    for (i=0; i<ids.length; i++){ 
     id = ids[i]; 
     if (people1[id]) { 
      people2[id] = people1[id]; 
      delete people1[id]; 
     } 
    } 
} 

Answer 2

只是一些未經檢驗的快速僞代碼。這給出了一個O（n^2）算法，因此它可能不是最好的。

function moveArrayItems(ids) { 
    // ids is an array of ids e.g. [1,2,3,4,5,6,...] 
    //Now I want to find all the person objects from array1 whose ids match with the ids array passed into this method. Then move them to array2. 
    //What is the best way to achive this? 
    for(i = 0;i < ids.length;i++){ 
     var found = false; 
     var j = 0; 
     while(!found && j < array1.length){ 
      if(array1[j].id = ids[i]){ 
      array2.push(array1[j]); 
      found = true; 
      }    
      j++; 
     } 
    } 
} 

Answer 3

首先一點功能通過John Resig

// Array Remove - By John Resig (MIT Licensed) 
Array.prototype.remove = function(from, to) { 
    var rest = this.slice((to || from) + 1 || this.length); 
    this.length = from < 0 ? this.length + from : from; 
    return this.push.apply(this, rest); 
}; 

：更快）圍繞由ID移動的東西

然後，合併文森特的解決方案

function moveArrayItems(ids) 
{ 
    // ids is an array of ids e.g. [1,2,3,4,5,6,...] 
    // Now I want to find all the person objects from array1 
    // whose ids match with the ids array passed into 
    // this method. Then move them to array2. 
    // What is the best way to achive this? 

    for(i = 0; i < ids.length; i++) 
    {  
    var found = false; 
    var j = 0; 
    while(!found && j < array1.length) 
    { 
     if(array1[j].id = ids[i]) 
     { 
     array2.push(array1[j]); 
     array1.remove(i); 
     found = true; 
     }     
     j++; 
    } 
    } 
} 

Answer 4

好問題。這實際上讓我回去參考基礎知識。關於JS陣列的關鍵是它的稀疏。您可以創建一個數組併爲任何索引分配值（例如：10和23）。基於這一事實

array1 = new Array(); 

array1[person1.id] = person1; 
array1[person2.id] = person2; 
.... goes till N 

function moveArrayItems(ids) { 
    for(index in ids) { 
     array2.push(array1[ids[index]]); 
     delete array1[ids[index]]; 
    } 
} 

注：我假設Person.id是整數並且小於2^32 - 1.參見JS文件，當ID是大於或浮點數。 JS Array實現不是連續的塊，所以不要認爲爲索引12345分配值需要12345個連續的內存塊。