比較對象javascript的兩個數組並刪除不相等的值

Question

我有兩個對象數組。

e = [{uniqueId:'',active:'a',qId:10},{uniqueId:'',active:'a',qId:11}]
f = [{uniqueId:50,active:'a',qId:10},{uniqueId:51,active:'a',qId:11},{uniqueId:52,active:'a',qId:13}]

我想比較這些對象，我的最終結果將是

result = [{uniqueId:50,active:'a',qId:10},{uniqueId:51,active:'a',qId:11}]

我試過了

let result = e.filter(o1 => f.some(o2 => o1.qId != o2.qId));

但是越來越

[{uniqueId:50,active:'a',qId:10},{uniqueId:51,active:'a',qId:11},{uniqueId:52,active:'a',qId:13}]

如何達到期望的輸出？

Answer 1

看起來您應該過濾f而不是e ，因為result顯示的是f而不是e 。

為了使復雜度最小，請首先將e數組的qId轉換為Set以便快速查找。 （與.some O(N)復雜度相比， Set具有O(1)查找時間）

 const e = [{uniqueId:'',active:'a',qId:10},{uniqueId:'',active:'a',qId:11}] const f = [{uniqueId:50,active:'a',qId:10},{uniqueId:51,active:'a',qId:11},{uniqueId:52,active:'a',qId:13}] const qIds = new Set(e.map(({ qId }) => qId)); console.log(f.filter(({ qId }) => qIds.has(qId))); 

Answer 2

我希望您需要在qId上進行比較。

 let e = [{uniqueId:'',active:'a',qId:10},{uniqueId:'',active:'a',qId:11}] let f = [{uniqueId:50,active:'a',qId:10},{uniqueId:51,active:'a',qId:11},{uniqueId:52,active:'a',qId:13}] let res = []; f.forEach(fo => { e.forEach(eo => { if(fo.qId === eo.qId){ res.push(fo) } }) }) console.log(res) 

Answer 3

您可以結合使用Array.filter()和Array.some()獲得該結果：

 e = [{uniqueId:'',active:'a',qId:10},{uniqueId:'',active:'a',qId:11}] f = [{uniqueId:50,active:'a',qId:10},{uniqueId:51,active:'a',qId:11},{uniqueId:52,active:'a',qId:13}]; var res = f.filter(fItem => e.some(({qId}) => fItem.qId === qId)); console.log(res); 

Answer 4

您可以檢查數組e是否具有相同的qId值來過濾f 。

 var e = [{ uniqueId: '', active: 'a', qId: 10 }, { uniqueId: '', active: 'a', qId: 11 }], f = [{ uniqueId: 50, active: 'a', qId: 10 }, { uniqueId: 51, active: 'a', qId: 11 }, { uniqueId: 52, active: 'a', qId: 13 }], result = f.filter(({ qId }) => e.some(o => o.qId === qId)); console.log(result); 

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