How to print extra object in an array of objects after comparison?

Question

const array1 = [
  {id: 1, Q_type: "AL"},
  {id: 2, Q_type: "BL"},
  {id: 3, Q_type: "CL"},
  {id: 4, Q_type: "DL"}
]
const array2 = [
  {id: 2, Q_type: "BL"},
  {id: 3, Q_type: "CL"},
  {id: 4, Q_type: "DL"}
]

const arrAfterComparison = array1.filter(val => !array2.includes(val))

I am trying to compare array1 and array2 and getting the object which is not present in both of these arrays

Expected output

arrAfterComparison = [{id:1,Q_type:"AL"}]

@Andreas added my unsuccefful attempt.. i have tried many solutions but i am stuck here :( — Ali
– Ali, Commented Feb 5, 2021 at 14:29
@Phil expected o/p would be this [{id:4,Q_type:"AL"}] or [{id:1,Q_type:"AL"}] ? — Rajneesh
– Rajneesh, Commented Feb 5, 2021 at 14:31

topdev87 · Accepted Answer · 2021-02-05 14:29:51Z

1

Use Array.some() inside Array.filter() method callback.

const array1 = [
  {id: 1, Q_type: "AL"},
  {id: 2, Q_type: "BL"},
  {id: 3, Q_type: "CL"},
  {id: 4, Q_type: "DL"}
]
const array2 = [
  {id: 2, Q_type: "BL"},
  {id: 3, Q_type: "CL"},
  {id: 1, Q_type: "DL"}
]

const output = array1.filter(item => !array2.some(a => a.Q_type === item.Q_type))
console.log(output);

answered Feb 5, 2021 at 14:29

topdev87

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2 Comments

Ali Over a year ago

This solves it flawlessly.. been stuck here for a while .. never had to use the some() method before i will read the docx thanks

Andreas Over a year ago

Don't you think there will be another question anywhere on SO that already solves this problem?

The Bomb Squad · Accepted Answer · 2021-02-05 14:42:12Z

When you compare, actually 2 objects are not similar.. look closely(it isn't an error)

Both the ones with id:1 have different values for Q_type
The one with the id of 4(we all see this one)

const array1 = [
  {id: 1, Q_type: "AL"},
  {id: 2, Q_type: "BL"},
  {id: 3, Q_type: "CL"},
  {id: 4, Q_type: "DL"}
]
const array2 = [
  {id: 2, Q_type: "BL"},
  {id: 3, Q_type: "CL"},
  {id: 1, Q_type: "DL"}
]

function oddIndexesOut(_arr1,_arr2){
  //objects may "look" the same but if they don't point to the same thing, they're not equal.. however if i turn it into a string, things that "look" equal are equal
  _arr1=_arr1.map(a=>JSON.stringify(a))
  _arr2=_arr2.map(a=>JSON.stringify(a))
  
  //comparison function(if things "look" similar)
  function compare(arr1,arr2){
    var x=arr1.filter(a=>!arr2.includes(a))
    return x.map(a=>JSON.parse(a))
  }
  
  //the longest array is used(so that checking can be full)
  if(_arr1.length>_arr2.length){
    return compare(_arr1,_arr2)
  }
  return compare(_arr2,_arr1)
}

console.log(oddIndexesOut(array1,array2))

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