Compare two array of objects and push or pop objects from first array based on second array

Question

consider below 2 array of objects, say arr1 and arr2. based on id, if any new item exist in arr2 copy item to arr1 and remove item from arr1 if that item not present in arr2. Tried something like this,

const arr1 = [
  {name: "name1", id: 1},
  {name: "name2", id: 2},
  {name: "name3", id: 3}
];
    
const arr2 = [
  {name: "name1", id: 1},
  {name: "name2", id: 4},
  {name: "name3", id: 3}
];

const array1 = arr2.filter((val) => {
  return arr1.find((val2) => {
    const arr = [];
    if (!arr1.includes(val.id)) {
      arr.push(val);
    }
  });
});

console.log(array1)

required output

output = `[ 
{name: "name1", id: 1},
{name: "name3", id: 3},
{name: "name2", id: 4} 
]`

array1 = arr2.filter((val) => { return arr1.find((val2) => { const arr = []; if (!arr1.includes(val.id)) { arr.push(val); } }); }); i was trying something like this, but it is not feasible. — Praveen Kumar
– Praveen Kumar, Commented Aug 6, 2024 at 10:28

Haluk · Accepted Answer · 2024-08-06 10:19:35Z

1

EDIT:

Reduce method, based on "id" may be used but note that ordering does not comply with your expected result as commented below by Hao Wu

const arr1 = [
      {name: "name1", id: 1},
      {name: "name2", id: 2},
      {name: "name3", id: 3}
    ];

    const arr2 = [
      {name: "name1", id: 1},
      {name: "name2", id: 4},
      {name: "name3", id: 3}
    ];

let output = arr2.reduce((a,b) => {
    let a1 = arr1.find(e => e.id === b.id) || {};
    return a.concat(Object.assign(a1, b));
},[]);

console.log(output)

edited Aug 6, 2024 at 10:19

answered Aug 6, 2024 at 9:52

Haluk

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4 Comments

Hao Wu Over a year ago

OP wanted it to be based on id but you're comparing name, also the order is off.

Praveen Kumar Over a year ago

Thank you for the solution , this solution works for the problem statement i posted, how can i preserve the values of arr1 properties ? @Haluk

Haluk Over a year ago

What do you mean by "preserving properties", can you give an example?

Praveen Kumar Over a year ago

Sorry for late reply, above method replacing values of name from arr1 with arr2 values, ex: arr1 = [{name:"jhon", id:1}]; and arr2 = [{name:'', id:1}]; above function gives out put as [{name:"", id:1}]; i want to preserve values of arr1. @Haluk

Haluk · Accepted Answer · 2024-08-07 07:34:42Z

You can use below code:

    const arr1 = [
      { name: "name1", id: 1 },
      { name: "name2", id: 2 },
      { name: "name3", id: 3 }
    ];
    
    const arr2 = [
      { name: "name1", id: 1 },
      { name: "name2", id: 4 },
      { name: "name3", id: 3 }
    ];
    
    var idsInArr2 = new Set(arr2.map(item => item.id));
    
    // Remove items from arr1 that are not in arr2
    var updatedArr1 = arr1.filter(item => idsInArr2.has(item.id));
    
    // Create a set of IDs from updatedArr1 for quick lookup
    var idsInUpdatedArr1 = new Set(updatedArr1.map(item => item.id));
    
    // Add items to arr1 that are in arr2 but not in updatedArr1
    arr2.forEach(item => {
      if (!idsInUpdatedArr1.has(item.id)) {
        updatedArr1.push(item);
      }
    });
    
    console.log(updatedArr1);

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