Remove same Values from array of Object

Question

I want to remove same object from array by comparing 2 arrays.

Sample Data:

arr1 = [
  {id: 1, name: "a"},
  {id: 2, name: "b"},
  {id: 3, name: "c"},
  {id: 4, name: "d"},
];

arr2 = [
  {id: 1, name: "a"},
  {id: 4, name: "d"},
];

let newArray = []; // new array with with no same values it should be unique.
arr1.map((val, i)=>{
   arr2.map((val2)=>{
    if(val.id == val2.id){
       console.log('Matched At: '+ i) // do nothing
    }else{
      newArray.push(val);
    }
   })
})
console.log(newArray); // e.g: [{id: 2, name: "b"}, {id: 3, name: "c"},];

Does this answer your question? Remove duplicate values from JS array — Christos Lytras
– Christos Lytras, Commented Mar 16, 2020 at 14:59
what if arr2 contains some unique elements to be merged? for example arr2 = [ {id: 1, name: "a"}, {id: 4, name: "d"}, {id: 5, name: "x"}, ] — qiAlex
– qiAlex, Commented Mar 16, 2020 at 15:52

Keith · Accepted Answer · 2020-03-16 17:16:44Z

7

Array.filter combined with not Array.some.

The trick here is also to not some,..

const arr1 = [
  {id: 1, name: "a"},
  {id: 2, name: "b"},
  {id: 3, name: "c"},
  {id: 4, name: "d"},
], arr2 = [
  {id: 1, name: "a"},
  {id: 4, name: "d"},
];

const newArray=arr1.filter(a=>!arr2.some(s=>s.id===a.id));

console.log(newArray);

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

As mentioned in comments the question could be interpreted slightly differently. If you also want the unqiue items from arr2, you basically just do it twice and join. IOW: check what not in arr2 is in arr1, and then check what not in arr1 that's in arr2.

eg..

const notIn=(a,b)=>a.filter(f=>!b.some(s=>f.id===s.id));
const newArray=[...notIn(arr1, arr2), ...notIn(arr2, arr1)];

Update 2: Time complexity, as mentioned by qiAlex there is loops within loops. Although some will short circuit on finding a match, if the dataset gets large things could slow down. This is were Set and Map comes in.

So to fix this using a Set.

const notIn=(a,b)=>a.filter(a=>!b.has(a.id));
const newArray=[
  ...notIn(arr1, new Set(arr2.map(m=>m.id))),
  ...notIn(arr2, new Set(arr1.map(m=>m.id)))
];

edited Mar 16, 2020 at 17:16

answered Mar 16, 2020 at 14:58

Keith

24.4k2 gold badges32 silver badges48 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

qiAlex Over a year ago

what if arr2 contains some unique elements to be merged?

Keith Over a year ago

@qiAlex You mean like a unique merge? would need to confirm with OP on that one, for me I interpreted the question as return one array that does not contain another.

qiAlex Over a year ago

At first I undesrtand a question as like as you did, it is a fair assumption, but wasn't clearly specified by OP. it is also possible to understand a question as 2 arrays that should be transformed into a one excluding duplicates, but theoretically both arr1 and arr2 can contain unique items.

Keith Over a year ago

@qiAlex You can basically do it the same way, just do it twice and join. I'll update answer with this info.

Najam Us Saqib Over a year ago

In my scenario I didn't have unique values in arr2 so the answer worked perfectly.

Rob Brander · Accepted Answer · 2020-03-16 14:55:04Z

2

const isInArray = (arr, id, name) => arr.reduce((result, curr) => ((curr.name === name && curr.id === id) || result), false)

const newArray = arr1.reduce((result, curr) => (isInArray(arr2, curr.id, curr.name) ? result : result.concat(curr)), [])

answered Mar 16, 2020 at 14:55

Rob Brander

3,7831 gold badge22 silver badges33 bronze badges

Comments

palaѕн · Accepted Answer · 2020-03-16 14:51:37Z

1

You can update you code using filter() method, instead of using .map() method like:

const arr1 = [
  {id: 1, name: "a"},
  {id: 2, name: "b"},
  {id: 3, name: "c"},
  {id: 4, name: "d"},
], arr2 = [
  {id: 1, name: "a"},
  {id: 4, name: "d"},
];

let newArray = []; // new array with with no same values it should be unique.
newArray = arr1.filter(function(a) {
    for(var i=0; i < arr2.length; i++){
      if(a.id == arr2[i].id) return false;
    }
    return true;
});
console.log(newArray);

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

answered Mar 16, 2020 at 14:51

palaѕн

74.2k17 gold badges123 silver badges140 bronze badges

Comments

Kunal Mukherjee · Accepted Answer · 2020-03-16 15:00:40Z

1

You check each element in first array whether its id lies in the second array by using Array.prototype.some. If the element is not present then only yield it.

const arr1 = [
  {id: 1, name: "a"},
  {id: 2, name: "b"},
  {id: 3, name: "c"},
  {id: 4, name: "d"},
];

const arr2 = [
  {id: 1, name: "a"},
  {id: 4, name: "d"},
];

const result = arr1.filter(x => !arr2.some(y => y.id === x.id));

console.log(result);

answered Mar 16, 2020 at 15:00

Kunal Mukherjee

5,8733 gold badges29 silver badges55 bronze badges

Comments

AbdurrahmanY · Accepted Answer · 2020-03-16 15:02:29Z

1

I think a simple comparer can works for getting differences and then concat them. with this method you dont need to check which array is bigger.

arr1 = [  {id: 1, name: "a"},  {id: 2, name: "b"},  {id: 3, name: "c"},  {id: 4, name: "d"}];

arr2 = [  {id: 1, name: "a"},  {id: 4, name: "d"},];

function localComparer(b){
  return function(a){
    return b.filter(
    function(item){
      return item.id == a.id && item.name == a.name
    }).length == 0;
  }
}

var onlyInArr1 = arr1.filter(localComparer(arr2));
var onlyInArr2 = arr2.filter(localComparer(arr1));

console.log(onlyInArr1.concat(onlyInArr2));

answered Mar 16, 2020 at 15:02

AbdurrahmanY

90513 silver badges23 bronze badges

Comments

StepUp · Accepted Answer · 2020-03-16 15:06:33Z

1

We can filter values by checking whether some element is not contained in current array:

const result = arr1.reduce((a, c) => {
  if (!arr2.some(a2 => a2.id === c.id))
      a.push(c);
  return a;
}, [])

An example:

let arr1 = [
  {id: 1, name: "a"},
  {id: 2, name: "b"},
  {id: 3, name: "c"},
  {id: 4, name: "d"},
];

let arr2 = [
  {id: 1, name: "a"},
  {id: 4, name: "d"},
];

const result = arr1.reduce((a, c) => {
  if (!arr2.some(a2 => a2.id === c.id))
      a.push(c);
  return a;
}, [])

console.log(result);

answered Mar 16, 2020 at 15:06

StepUp

38.5k16 gold badges95 silver badges159 bronze badges

Comments

Kunal Mukherjee · Accepted Answer · 2020-03-16 15:18:58Z

1

Try this one -

const arr1 = [
  {id: 1, name: "a"},
  {id: 2, name: "b"},
  {id: 3, name: "c"},
  {id: 4, name: "d"},
];

const arr2 = [
  {id: 1, name: "a"},
  {id: 4, name: "d"},
];

const arr3 = [...arr1, ...arr2];
const mySubArray = _.uniq(arr3, 'id');
console.log(mySubArray);

<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.9.1/underscore-min.js"></script>

edited Mar 16, 2020 at 15:18

Kunal Mukherjee

5,8733 gold badges29 silver badges55 bronze badges

answered Mar 16, 2020 at 15:14

Ankit Kumar Rajpoot

5,6542 gold badges41 silver badges33 bronze badges

Comments

qiAlex · Accepted Answer · 2020-03-16 16:09:59Z

1

So many loops in every answer.

Complexity of the code my answer is 2N,

Idea is:

to merge arrays.
first loop - mark duplicates somehow
second loop - filter duplicates out

arr1 = [
  {id: 1, name: "a"},
  {id: 2, name: "b"},
  {id: 3, name: "c"},
  {id: 4, name: "d"},
];

arr2 = [
  {id: 1, name: "a"},
  {id: 4, name: "d"},
];

let newArray = [...arr1, ...arr2].reduce((acc, item, index) => {
  acc.items.push(item);

  if (typeof acc.map[item.id] !== 'undefined') {
    acc.items[acc.map[item.id]] = null;
    acc.items[index] = null;
  }
  acc.map[item.id] = index;
  
  return acc
},  {map: {}, items: []}).items.filter(item => !!item)


console.log(newArray);

edited Mar 16, 2020 at 16:09

answered Mar 16, 2020 at 15:24

qiAlex

4,3562 gold badges22 silver badges35 bronze badges

1 Comment

Keith Over a year ago

Complexity of the code my answer is 2N, Good point, if the dataset gets large, and performance is an issue doing this would be a good idea. Currently on this small dataset it's about 8X slower than a simple filter & some, but I don't think it would take much bigger dataset's before this gets faster. I'm up-voting this one for that reason. I suppose if performance is really paramount you could even do both by checking the lengths.. :)

Collectives™ on Stack Overflow

Remove same Values from array of Object

8 Answers 8

5 Comments

Comments

Comments

Comments

Comments

Comments

Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

8 Answers 8

5 Comments

Comments

Comments

Comments

Comments

Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related