Javascript Array Sorting Error

Question

I have array of cars and an array of price corresponding to each car.

I want to implement the function highest3 which will return an array of 3 cars in order by their price (highest to lowest).

However, if the price is equal, then it returns the cars in alphabetical order.

In this example below, "Hummer" would be the first entry.

• Question: What am I doing wrong?

Original Code

var cars = ["Ferrari", "Lamborghini", "Jaguar", "Hummer", "Toyota"];
var price = [12, 34.5, 3.54, 45.9, 3.44];

result == ["Hummer", "Lamborghini", "Ferrari"];

function highest3 (cars, price) {

//Enter Code Here

}

What I had Tried So Far

var cars = ["Ferrari", "Lamborghini", "Jaguar", "Hummer", "Toyota"];
var price = [12, 34.5, 3.54, 45.9, 3.44];

result == ["Hummer", "Lamborghini", "Ferrari"];

function highest3 (cars, price) {
  for(var i=0; i<cars.length; i++) {
    console.log(price[i], cars[i]);
    result[i] = cars[price[i]];
  }
  return result;

alert(highest3(cars, price));
}

JSFiddle at http://jsfiddle.net/d5vTc/4/ UPDATED

Answer 1

You are probably better off reformatted the data to something easier, then using a custom sort comparer. The below sorts by price (high to low), as well as alphabetical for the identical price.

http://jsfiddle.net/T8qb4/

var cars = [{name:"Ferrari",price:12}, 
            {name:"Lamborghini", price:34.5},
            {name:"Jaguar", price:3.54},
            {name:"Honda", price:3.54},
            {name:"Hummer", price:45.9}, 
            {name:"Toyota", price:3.44}];

function compare(a,b) {
  if (a.price > b.price)
     return -1;
  if (a.price < b.price)
    return 1;
  if(a.price == b.price){
      if(a.name < b.name){
          return -1;
      } else {
          return 1;
      }
  }
  return 0;
}

cars.sort(compare);

Answer 2

You will need to do something like this:

1. combine the car with the price in an array of objects.
2. sort the combined array first for price but if equal then alphabetically.
3. convert combined array back to an array of cars as per sort function (2).
4. take the first three cars from the new array created in (3).

Javascript

var cars = ["Ferrari", "Lamborghini", "Jaguar", "Hummer", "Toyota", "Renault"],
    price = [12, 34.5, 3.54, 45.9, 3.44, 3.44],
    combined = cars.map(function (car, index) {
        return {
            car: car,
            price: price[index]
        };
    }),
    result;

combined.sort(function (a, b) {
    if (a.price === b.price) {
        if (a.car === b.car) {
            return 0;
        }

        if (a.car < b.car) {
            return -1;
        }

        return 1;
    }

    if (a.price > b.price) {
        return -1
    }

    return 1;
});

result = combined.map(function (element) {
    return element.car;
}).slice(0, 3);

console.log(result);

Output

["Hummer", "Lamborghini", "Ferrari"] 

On jsFiddle

Or if you prefer you can chain all of those methods

var cars = ["Ferrari", "Lamborghini", "Jaguar", "Hummer", "Toyota", "Renault"],
    price = [12, 34.5, 3.54, 45.9, 3.44, 3.44],
    result = cars.map(function (car, index) {
        return {
            car: car,
            price: price[index]
        };
    }).sort(function (a, b) {
        if (a.price === b.price) {
            if (a.car === b.car) {
                return 0;
            }

            if (a.car < b.car) {
                return -1;
            }

            return 1;
        }

        if (a.price > b.price) {
            return -1
        }

        return 1;
    }).map(function (element) {
        return element.car;
    }).slice(0, 3);

console.log(result);

Answer 3

You can use sort with a suitable function:

var cars = ["Ferrari", "Lamborghini", "Jaguar", "Hummer", "Toyota"];
var price = [12, 34.5, 3.54, 45.9, 3.44];

function topN(data, values, n) {
    var d = data.slice(0);
    d.sort(function (a, b) {
             return values[data.indexOf(b)] - values[data.indexOf(a)];
           });
    return d.slice(0,n);
}

console.log(topN(cars, price, 3)); // Hummer,Lamborghini,Ferrari

You will need a shim for Array.prototype.indexOf for hosts that don't have it.

Edit

To also sort alphabetically where the values are the same:

function topN(data, values, n) {
    var d = data.slice(0);
    function qSort(a, b){return a<b? -1 : a>b? 1 : 0;}
    d.sort(function (a, b) {
             var aval = values[data.indexOf(a)];
             var bval = values[data.indexOf(b)];
             return aval == bval? qSort(a, b) : bval - aval;
           });
    return d.slice(0,n);
}

Answer 4

I would recommend making an array that contains objects which contain the car make and price as properties. Then use the sort function to return the top 3 cars in the array by price:

var cars = [
    {make:"Ferrari",price:12}, 
    {make:"Lamborghini", price: 34.5},
    {make:"Jaguar", price:3.54},
    {make:"Hummer", price:45.9}, 
    {make:"Toyota", price:3.44}
];


function top(cars, num) {
    return cars.slice(0).sort(function(a,b){
        return b.price - a.price;
    }).slice(0,num);
}

//usage:
top(cars, 3);

JS Fiddle: http://jsfiddle.net/d5vTc/2/