filter out duplicate objects by value name and place into new array

Question

I am trying to filter out duplicates objects in array by object value name and place into new array called finalResult.

Basically i want to only have objects with unique country names in my new array.

example arrays:

var result = [
   {country: 'united states', numofdistributors: 5},
   {country: 'united states', numofdistributors: 5},
   {country: 'brazil', numofdistributors: 2},
   {country: 'Germany', numofdistributors: 1},
   {country: 'india', numofdistributors: 6},
   {country: 'united states', numofdistributors: 5},
   {country: 'Egypt', numofdistributors: 1},
   {country: 'Germany', numofdistributors: 1},
];

This is what i want my array to turn out like:

var finalResult = [
   {country: 'united states', numofdistributors: 5},
   {country: 'brazil', numofdistributors: 2},
   {country: 'Germany', numofdistributors: 1},
   {country: 'india', numofdistributors: 6},
   {country: 'Egypt', numofdistributors: 1},
];

Here is my attempt to solve with no luck and only frustration:

var finalResults = [];
result.forEach(function(country){
    if(finalResults.indexOf(country.country) != -1){
        console.log('already added');
    }else {
       //console.log('Does not exist');
       finalResults.push(country);
});
console.log(finalResults);

I made a few changes on your code and updated my answer

Mojtaba
– Mojtaba

2016-05-02 20:56:04 +00:00
Commented May 2, 2016 at 20:56 — Mojtaba
– Mojtaba, Commented May 2, 2016 at 20:56

Nina Scholz · Accepted Answer · 2016-05-02 20:47:19Z

2

You could filter the data with Array#filter and use an object for already inserted items.

var result = [{ country: 'united states', numofdistributors: 5 }, { country: 'united states', numofdistributors: 5 }, { country: 'brazil', numofdistributors: 2 }, { country: 'Germany', numofdistributors: 1 }, { country: 'india', numofdistributors: 6 }, { country: 'united states', numofdistributors: 5 }, { country: 'Egypt', numofdistributors: 1 }, { country: 'Germany', numofdistributors: 1 }, ],
    finalResult = result.filter(function (a) {
        var key = a.country + '|' + a.numofdistributors;
        if (!this[key]) {
            this[key] = true;
            return true;
        }
    }, Object.create(null));

document.write('<pre>' + JSON.stringify(finalResult, 0, 4) + '</pre>');

answered May 2, 2016 at 20:47

Nina Scholz

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2 Comments

Роман Парадеев Over a year ago

Wow, nice idea to provide an empty object as a shared storage.

Travis Michael Heller Over a year ago

@Nina Scholz, thank you so much! saved me much headache:)

Mojtaba · Accepted Answer · 2016-05-02 20:55:36Z

1

You already done it. Just a simple mistake:

var result = [
   {country: 'united states', numofdistributors: 5},
   {country: 'united states', numofdistributors: 5},
   {country: 'brazil', numofdistributors: 2},
   {country: 'Germany', numofdistributors: 1},
   {country: 'india', numofdistributors: 6},
   {country: 'united states', numofdistributors: 5},
   {country: 'Egypt', numofdistributors: 1},
   {country: 'Germany', numofdistributors: 1},
];

var finalResult = [
   {country: 'united states', numofdistributors: 5},
   {country: 'brazil', numofdistributors: 2},
   {country: 'Germany', numofdistributors: 1},
   {country: 'india', numofdistributors: 6},
   {country: 'Egypt', numofdistributors: 1},
];

var finalResults = [];
var countries = [];
result.forEach(function(country){
    if(countries.indexOf(country.country) != -1){
        console.log('already added');
    }else {
       //console.log('Does not exist');
       finalResults.push(country);
       countries.push(country.country);
}
}
);
console.log(finalResults);

edited May 2, 2016 at 20:55

answered May 2, 2016 at 20:48

Mojtaba

5,0175 gold badges24 silver badges40 bronze badges

2 Comments

Travis Michael Heller Over a year ago

Wow, probably should of considered stepping away from my computer a bit. Thank you for pointing that out.

Mojtaba Over a year ago

@TravisMichaelHeller, Yea. Sometimes we need a break and fresh up the brain

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