Comparing 2 arrays and pushing missing values to an Array - Javascript

Question

Its quite a simple problem but I am unable to figure out the logic.

I have Arr1 = [1,2,3,4] and Arr2 = [3,4,5,6]

I would like compare Arr2 with Arr1 and add missing values to MissingArr = [];

my code

let missing = [];  
 arrOfNum.forEach( (number) => {

  // Arr2 = 3,4,5,6
  console.log("number: "+number);
  for(let i = 0; i < this.numbers.length; i++){

      // Arr1 = 1,2,3,4
      console.log("this.numbers: "+this.numbers[i]['number']); 

      if(number == this.numbers[i]['number'])
      {
        //Do Nothing
      }else{
        missing.push(this.numbers[i]['number']);
      }
  }
});

I am not sure how to do a check of Arr2 to Arr1 (complete entire forloop) and only if Arr2 value does not exist in Arr1 then add it to missing array.

Update

Arr1 = [{id:1, number:1}, {id:2, number:2},{id:3, number:3}, {id:4, number:4}]
Arr2 = [3,4,5,6]

so missing array should be: Missing= [5,6]

@NenadVracar I forgot to mention my Arr2 is just arr of numbers whereas Arr1 is Array of objects. will this still work? !Arr1['numbers'].includes(e) — Murlidhar Fichadia
– Murlidhar Fichadia, Commented Jan 27, 2018 at 15:10
@NenadVracar I have posted actual data to my updated section of the question — Murlidhar Fichadia
– Murlidhar Fichadia, Commented Jan 27, 2018 at 15:18

Nenad Vracar · Accepted Answer · 2018-01-27 15:20:52Z

3

You can do this with filter and some methods.

const Arr1 = [{id:1, number:1}, {id:2, number:2},{id:3, number:3}, {id:4, number:4}],
Arr2 = [3,4,5,6]

const result = Arr2.filter(e => !Arr1.some(o => o.id == e));
console.log(result)

answered Jan 27, 2018 at 15:20

Nenad Vracar

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the_makeshift_dev · Accepted Answer · 2018-01-27 15:10:56Z

0

You can use Array.filter to achieve this. Array.filter returns an array containing items matching the predicate.

var arr1 = [{key: 1},{key: 2},{key: 3},{key: 4}];
var arr2 = [3,4,5,6];

//This will return all values in array2 missing in array1
var missing = arr1.filter(function(arr1_item){
    return arr2.indexOf(arr1_item.key) === -1;
});

answered Jan 27, 2018 at 15:10

the_makeshift_dev

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