從數組中刪除重復項僅返回一個對象

Question

我正在嘗試從此json刪除重復的條目，但它僅返回一個我不明白我要去哪里的對象。

代碼如下。

// exemplary array of objects (id 'NewLive' occurs twice)
var arr = [
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
    obj = {}, new_arr = [];

// in the end the last unique object will be considered
arr.forEach(function(v){
    obj[v['id']] = v;
   console.log(JSON.stringify(new_arr));
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });

console.log(JSON.stringify(new_arr));

我與此同時附加codepen。

https://codepen.io/anon/pen/oQXJWK

Answer 1

您的代碼返回單個元素的原因是因為您使用的是v['id']但是對象上沒有id屬性，因此在整個循環中，您需要反復設置obj[undefined] 。

在您的jsfiddle代碼中，這看起來是正確的，並且代碼似乎按預期工作。

如果有人遇到這個問題，以了解如何從javascript中的數組中刪除重復項 ，則有以下幾種選擇：

經典方式：好循環

本質上，這是您使用的解決方案，遍歷數組，檢查鍵是否已添加到結果數組中，如果不存在，則將元素添加到結果中。

例：

const result = [];
const knownIDs = new Set();
for (const item of input) {
  if (!knownIDs.has(item.jobcodeid.S)) {
    result.push(item);
    knownIDs.add(item.jobcodeid.S);
  }
}

前往地圖並返回

要過濾重復項，可以將元素轉換為鍵->值的Map ，然后再轉換回數組。 之所以可行，是因為鍵在Map是唯一的，並且重復項將自動消除。 這種方法的主要優點是，由於代碼的簡單性，它將減少錯誤。

console.log(
  Array.from(
    new Map(
      input.map(i => [i.jobcodeid.S, i])
    ).values()
  )
)

過濾並設置

另一種選擇是使用Set記錄已知ID，並使用filter刪除具有已知ID的項目。 此方法的優點是，由於意圖很明確，因此可能更易於閱讀。 而且，與轉換為Map並返回相比，此方法的性能更高。

const knownKeys = new Set();
console.log(
  input.filter(i => {
    if (!knownKeys.has(i.jobcodeid.S)) {
      knownKeys.add(i.jobcodeid.S);
      return true;
   }
  })
);

看到他們的行動：

 const input = [{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}]; // Classic for loop const result = []; const knownIDs = new Set(); for (const item of input) { if (!knownIDs.has(item.jobcodeid.S)) { result.push(item); knownIDs.add(item.jobcodeid.S); } } console.log(result.map(r => r.jobcodeid.S)); // To Map and back console.log( Array.from( new Map( input.map(i => [i.jobcodeid.S, i]) ).values() ) ) // filter and set const knownKeys = new Set(); console.log( input.filter(i => { if (!knownKeys.has(i.jobcodeid.S)) { knownKeys.add(i.jobcodeid.S); return true; } }) ); 

作為記錄，我對公認的解決方案，我的產品以及雅克的答案所帶來的性能改進進行了基准測試

accepted solution x 1,892,585 ops/sec ±3.48% (89 runs sampled)
Map and back x 495,116 ops/sec ±2.27% (90 runs sampled)
Set and filter x 1,600,833 ops/sec ±1.98% (90 runs sampled)
Jacques x 2,110,510 ops/sec ±0.98% (92 runs sampled)
Fastest is Jacques

如您所見， Jacques的解決方案的確速度快一倍，因此，如果您打算過濾大型陣列，或者如果性能是關鍵，則絕對應該選擇它！

Answer 2

首先，您必須使用obj[v['jobcodeid']] = v; 而不是obj[v['id']] = v; 。

但是由於v[jobcodeid]是一個對象，因此js會將其轉換為字符串，即[object Object]並且最終數組中將只有一個元素。

 // exemplary array of objects (id 'NewLive' occurs twice) var arr=[{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}], obj = {}, new_arr = []; // in the end the last unique object will be considered arr.forEach(function(v){ obj[v['jobcodeid']] = v; }); new_arr = Object.keys(obj).map(function(id) { return obj[id]; }); console.log(JSON.stringify(new_arr)); 

您應該使用v.jobcodeid.S作為對象的鍵。

 // exemplary array of objects (id 'NewLive' occurs twice) var arr=[{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}], obj = {}, new_arr = []; // in the end the last unique object will be considered arr.forEach(function(v){ obj[v.jobcodeid.S] = v; }); new_arr = Object.keys(obj).map(function(id) { return obj[id]; }); console.log(JSON.stringify(new_arr)); 

Answer 3

發布答案以顯示另一種更高效率的方法。

 var arr = [ {"jobcodeid":{"S":"Etc_new"} }, {"jobcodeid":{"S":"NewLive"} }, {"jobcodeid":{"S":"NewLiveVid"}}, {"jobcodeid":{"S":"New_Live"}}, {"jobcodeid":{"S":"New_Live_Vid"}}, {"jobcodeid":{"S":"Newest"}}, {"jobcodeid":{"S":"NewestLive"}}, {"jobcodeid":{"S":"NewestLiveVid"}}, {"jobcodeid":{"S":"Very_New_Vid"}}, {"jobcodeid":{"S":"Etc_new"}}, {"jobcodeid":{"S":"NewLive"}}, {"jobcodeid":{"S":"NewLiveVid"}}, {"jobcodeid":{"S":"New_Live"}}, {"jobcodeid":{"S":"New_Live_Vid"}}, {"jobcodeid":{"S":"Newest"}}, {"jobcodeid":{"S":"NewestLive"}}, {"jobcodeid":{"S":"NewestLiveVid"}}, {"jobcodeid":{"S":"Very_New_Vid"}} ], obj = {}, new_arr = []; // in the end the last unique object will be considered for (const job of arr) { if (!obj[job.jobcodeid.S]) { obj[job.jobcodeid.S] = true; new_arr.push(job); } } console.log(JSON.stringify(new_arr)); 

這個答案總是運行N次迭代。 設置唯一值后在鍵之間循環時，它最多可以運行2N次迭代。 （從談論大O /復雜性改為更清楚）

Answer 4

您只需要使用Set ！

const arr = [
    { jobcodeid: { S: "Etc_new" } },
    { jobcodeid: { S: "NewLive" } },
    { jobcodeid: { S: "NewLiveVid" } },
    { jobcodeid: { S: "New_Live" } },
    { jobcodeid: { S: "New_Live_Vid" } },
    { jobcodeid: { S: "Newest" } },
    { jobcodeid: { S: "NewestLive" } },
    { jobcodeid: { S: "NewestLiveVid" } },
    { jobcodeid: { S: "Very_New_Vid" } },
    { jobcodeid: { S: "Etc_new" } },
    { jobcodeid: { S: "NewLive" } },
    { jobcodeid: { S: "NewLiveVid" } },
    { jobcodeid: { S: "New_Live" } },
    { jobcodeid: { S: "New_Live_Vid" } },
    { jobcodeid: { S: "Newest" } },
    { jobcodeid: { S: "NewestLive" } },
    { jobcodeid: { S: "NewestLiveVid" } },
    { jobcodeid: { S: "Very_New_Vid" } }
];

const uniqueItems = [...new Set(arr.map(i => i.jobcodeid.S))]

Answer 5

也可以嘗試此方法..解決此問題的另一種方法

var arr = [
  {"jobcodeid":{"S":"Etc_new"}
  },
  {"jobcodeid":{"S":"NewLive"}
  },
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
    obj = {}, new_arr = [];


arr.forEach(function(v){
    obj[v['id']] = v;

    for(var i=0;i< new_arr.length;i++){
      if(new_arr[i].jobcodeid.S == v.jobcodeid.S)       {
         return;
      }
    } 
      new_arr.push(v);     
});

console.log(new_arr);