比较对象javascript的两个数组并删除不相等的值

Question

I have two array of objects.

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}]

I want to compare these objects and my final result will be like

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

I tried

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

But am getting

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

How to achieve the desired output?

Answer 1

It looks like you should be filtering f , not e , because the result shows values from f , not from e .

For the least complexity, turn the e array's qId s into a Set for quick lookup first. ( Set s have O(1) lookup time, compared to O(N) complexity for .some )

 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

i hope you need to compare on qId..

 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

You can use Array.filter() and Array.some() in combination to get that result:

 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

You could check if array e has the same qId value for filtering 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); 

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