查找重複的對象數組，並選擇其中一個比較它們的值

-3

var array1 = [ 
    { 
    "_id" : ObjectId("5972dbe185c5a00906fbefa7"), 
    "name" : "user1", 
    "score" : 2.0 
    }, 
    { 
    "_id" : ObjectId("59994cf90b79bc4e8233ed1d"), 
    "name" : "user2", 
    "score" : 1.5 
    }, 
    { 
    "_id" : ObjectId("59880bbaa329573db1180acd"), 
    "name" : "user3", 
    "score" : 0.5 
    }, 
    { 
    "_id" : ObjectId("5972fa945ad42b149da788b2"), 
    "name" : "user4", 
    "score" : 3.0 
    } 
] 

var array2 = [ 
    { 
    "_id" : ObjectId("5972dbe185c5a00906fbefa7"), 
    "name" : "user1", 
    "score" : 1.5 
    }, 
    { 
    "_id" : ObjectId("599d2d8fe9feeb290c699535"), 
    "name" : "user5", 
    "score" : 1.5 
    }, 
    { 
    "_id" : ObjectId("59880bbaa329573db1180acd"), 
    "name" : "user3", 
    "score" : 3.0 
    }, 
    { 
    "_id" : ObjectId("8072fa945ad42b149da799cc"), 
    "name" : "user8", 
    "score" : 2.5 
    } 
] 

array3 = _.unionWith(array1, array2, _.isEqual);

我有用戶對象的兩個數組從兩個不同的搜索查詢返回，現在有可能重複的用戶對象，我沒有刪除使用lodash的重複項._union具有函數，但是我未能通過向要返回的對象的score屬性添加另一個重複對象分數來選擇並返回其中一個重複對象。我如何以更有效的方式來做到這一點。感謝查找重複的對象數組，並選擇其中一個比較它們的值

來源

2017-09-15 Javid Sapand

發佈您的代碼將是一個良好的開端。 – TheCog

歡迎來到SO。請訪問[幫助]，看看有什麼和如何問。建議：投入努力和代碼 – mplungjan

哦對不起！因爲這是我的第一個問題，我認爲我的問題描述足以被理解。 –

您可以將哈希表作爲對象的引用並減少數組。

var array1 = [{ _id: "5972dbe185c5a00906fbefa7", name: "user1", score: 2.0 }, { _id: "59994cf90b79bc4e8233ed1d", name: "user2", score: 1.5 }, { _id: "59880bbaa329573db1180acd", name: "user3", score: 0.5 }, { _id: "5972fa945ad42b149da788b2", name: "user4", score: 3.0 }], 
 
    array2 = [{ _id: "5972dbe185c5a00906fbefa7", name: "user1", score: 1.5 }, { _id: "599d2d8fe9feeb290c699535", name: "user5", score: 1.5 }, { _id: "59880bbaa329573db1180acd", name: "user3", score: 3.0 }, { _id: "8072fa945ad42b149da799cc", name: "user8", score: 2.5 }], 
 
    hash = Object.create(null), 
 
    result = [array1, array2].reduce(function (r, a) { 
 
     a.forEach(function (o) { 
 
      if (!hash[o._id]) { 
 
       hash[o._id] = o; 
 
       r.push(hash[o._id]); 
 
       return; 
 
      } 
 
      hash[o._id].score += o.score; 
 
     }); 
 
     return r; 
 
    }, []); 
 

 
console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

來源

2017-09-15 19:21:39

使用循環，你可以做這樣的事情：

var array1 = [ 
 
    { 
 
    "_id" : "5972dbe185c5a00906fbefa7", 
 
    "name" : "user1", 
 
    "score" : 2.0 
 
    }, 
 
    { 
 
    "_id" : "59994cf90b79bc4e8233ed1d", 
 
    "name" : "user2", 
 
    "score" : 1.5 
 
    }, 
 
    { 
 
    "_id" : "59880bbaa329573db1180acd", 
 
    "name" : "user3", 
 
    "score" : 0.5 
 
    }, 
 
    { 
 
    "_id" : "5972fa945ad42b149da788b2", 
 
    "name" : "user4", 
 
    "score" : 3.0 
 
    } 
 
] 
 

 
var array2 = [ 
 
    { 
 
    "_id" : "5972dbe185c5a00906fbefa7", 
 
    "name" : "user1", 
 
    "score" : 1.5 
 
    }, 
 
    { 
 
    "_id" : "599d2d8fe9feeb290c699535", 
 
    "name" : "user5", 
 
    "score" : 1.5 
 
    }, 
 
    { 
 
    "_id" : "59880bbaa329573db1180acd", 
 
    "name" : "user3", 
 
    "score" : 3.0 
 
    }, 
 
    { 
 
    "_id" : "8072fa945ad42b149da799cc", 
 
    "name" : "user8", 
 
    "score" : 2.5 
 
    } 
 
] 
 

 

 

 
array1.forEach((item)=>{ 
 
    array2.forEach((item2) => { 
 
    if(item._id === item2._id){ 
 
     item2.score = item2.score +item.score;//add score from first array dup to second 
 
    } 
 
    
 
    }); 
 
    }); 
 
console.log(array2)

現在數組2具有用於陣列1所有重複

注比分：我刪除的ObjectId得到它在代碼段中工作

來源

2017-09-15 18:36:14 rjustin

查找重複的對象數組，並選擇其中一個比較它們的值

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