symmetric difference for arrays with objects JavaScript

Question

let arr1 = [{ countryCode: "ITA", index: 2, name: "Italy"}, { countryCode: "NLD", index: 1, name: "Netherlands"}];

let arr2 = [{ countryCode: "NLD", index: 1, name: "Netherlands"}, { countryCode: "BEL", index: 3, name: "Belgium"}];

I want it to return the symmetric difference, so it should return:

[{ countryCode: "ITA", index: 2, name: "Italy"},  {countryCode: "BEL", index: 3, name: "Belgium"}]

How can I accomplish this in Javascript? I tried to do the following:

let difference = arr1
                 .filter(x => !arr2.includes(x))
                 .concat(arr2.filter(x => !arr1.includes(x)));

But this doesn't seem to work for arrays with objects in them.

The reason is because equality checks on objects checks only the reference to the object, not each and every one of its key:value pairs. — Randy Casburn
– Randy Casburn, Commented Jan 21, 2021 at 21:19

Nina Scholz · Accepted Answer · 2021-01-21 21:23:15Z

2

You could take a Map and add or delete the item, if it is in the map.

const
    array1 = [{ countryCode: "ITA", index: 2, name: "Italy"}, { countryCode: "NLD", index: 1, name: "Netherlands" }],
    array2 = [{ countryCode: "NLD", index: 1, name: "Netherlands" }, { countryCode: "BEL", index: 3, name: "Belgium" }],
    map = new Map,
    cb = o => map.delete(o.countryCode) || map.set(o.countryCode, o);

array1.forEach(cb);
array2.forEach(cb);

console.log(Array.from(map.values()));

answered Jan 21, 2021 at 21:23

Nina Scholz

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CertainPerformance · Accepted Answer · 2021-01-21 21:16:33Z

1

I'd collect one of the unique values in the objects (like countryCode) into an array of just those values, then you can use your original approach:

const arr1 = [{ countryCode: "ITA", index: 2, name: "Italy"}, { countryCode: "NLD", index: 1, name: "Netherlands"}];
const arr2 = [{ countryCode: "NLD", index: 1, name: "Netherlands"}, { countryCode: "BEL", index: 3, name: "Belgium"}];

const arr1Codes = arr1.map(a => a.countryCode);
const arr2Codes = arr2.map(a => a.countryCode);

const result = arr1
  .filter(a => !arr2Codes.includes(a.countryCode))
  .concat(arr2.filter(a => !arr1Codes.includes(a.countryCode)));
console.log(result);

answered Jan 21, 2021 at 21:16

CertainPerformance

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Majed Badawi · Accepted Answer · 2021-01-21 21:35:10Z

1

You can first get the set of the unique country codes and then get the unique elements in each array using .filter:

const arr1 = [
  { countryCode: "ITA", index: 2, name: "Italy"}, 
  { countryCode: "NLD", index: 1, name: "Netherlands"}
];
const arr2 = [
  { countryCode: "NLD", index: 1, name: "Netherlands"}, 
  { countryCode: "BEL", index: 3, name: "Belgium"}
];

const arr = [...arr1, ...arr2];
const countryCodes = arr.map(e => e.countryCode);

const uniqueCountryCodes = new Set(
  countryCodes.filter(e => countryCodes.indexOf(e) === countryCodes.lastIndexOf(e))
);

const difference = arr.filter(x => uniqueCountryCodes.has(x.countryCode));

console.log(difference);

edited Jan 21, 2021 at 21:35

answered Jan 21, 2021 at 21:23

Majed Badawi

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