Spread the object and omit keys

Lastly I needed to create a shallow copy of an object but omit some keys. To make the code more compact I didn't want to use the delete operator. ES6 spread operator allows to do it.

Below is an example original object. I want to copy it but omit the city and country keys.

const user = {
    age: 20,
    name: 'Garry',
    city: 'Copenhagen',
    country: 'Denmark'
};

const { city, country, ...userCopy } = user;

console.log(userCopy);

/**
{
    "age": 20,
    "name": "Garry"
}
**/

However, this solution creates new, unused variables (city and country) with their values:

console.log(city) // 'Copenhagen'
console.log(country) // 'Denmark'

Ideally we don't want to create any side effects. Here's the solution.

function omitKeys(keys, obj) {
    const excluded = new Set(keys);
    return Object.fromEntries(Object.entries(obj).filter(e => !excluded.has(e[0])))
}

console.log(omitKeys(['city, country'], user));
/**
{
    "age": 20,
    "name": "Garry"
}
**/

console.log(city) // city is not defined;
console.log(country) // country is not defined;