Find n highest values in an object JavaScript

Let's say, we have an object that describes various qualities of a football player like this ?

const qualities = {
    defence: 82,
    attack: 92,
    heading: 91,
    pace: 96,
    dribbling: 88,
    tenacity: 97,
    vision: 91,
    passing: 95,
    shooting: 90
};

We wish to write a function that takes in such object and a number n (n ? no. of keys in object) and returns an object with n highest key value pairs.

Like for n = 2

Output should be ?

{
    tenacity: 97,
    pace: 96
}

Implementation

The complete code for this function will be ?

const qualities = {
    defence: 82,
    attack: 92,
    heading: 91,
    pace: 96,
    dribbling: 88,
    tenacity: 97,
    vision: 91,
    passing: 95,
    shooting: 90
};

const pickHighest = (obj, num = 1) => {
    const requiredObj = {};
    if(num > Object.keys(obj).length){
        return false;
    };
    Object.keys(obj).sort((a, b) => obj[b] - obj[a]).forEach((key, ind) =>
    {
        if(ind < num){
            requiredObj[key] = obj[key];
        }
    });
    return requiredObj;
};

console.log("Top 3 qualities:", pickHighest(qualities, 3));
console.log("Top 1 quality:", pickHighest(qualities, 1));
console.log("Top 5 qualities:", pickHighest(qualities, 5));

Output

The output in the console will be ?

Top 3 qualities: { tenacity: 97, pace: 96, passing: 95 }
Top 1 quality: { tenacity: 97 }
Top 5 qualities: { tenacity: 97, pace: 96, passing: 95, attack: 92, heading: 91 }

How It Works

The function uses Object.keys() to get all property names, then sorts them by their values in descending order using obj[b] - obj[a]. The forEach loop adds only the first num properties to the result object.

Alternative Approach Using Object.entries()

const pickHighestAlternative = (obj, num = 1) => {
    if(num > Object.keys(obj).length) return false;
    
    return Object.fromEntries(
        Object.entries(obj)
            .sort(([,a], [,b]) => b - a)
            .slice(0, num)
    );
};

console.log("Alternative approach:", pickHighestAlternative(qualities, 3));

Alternative approach: { tenacity: 97, pace: 96, passing: 95 }

Conclusion

Both approaches sort object entries by values and return the top n key-value pairs. The second method is more concise using Object.entries() and Object.fromEntries().