Remove duplicates from a multidimensional array

Question

I have found questions that kind of touch on the issue I'm having, but I haven't found a solution that works for me yet. I have this array: [[1, red], [2, green], [3, red], [3, blue], [5, green]] and I need it to return [[1, red], [2, green], [3, blue]. What I need the code to do is go through the array and find ONLY colors that match, not numbers, and get rid of that entire index.

I have tried something like this

var uniqueArray = colors.filter(function(item, pos) {
return colors.indexOf(item) == pos;
});

I'm thinking that this code is searching for a complete match, and I only require a partial match. So basically, how would I modify .filter() to get rid of partial duplicates (only matching the colors)?

Please let me know if I need to provide any more information.

your question is a bit confusing. Why do you need the 1, 2 and 3 in the result you're searching for? Could these not be determined as answer.indexOf('blue') + 1 or something of the like? — Julie
– Julie, Commented Sep 23, 2016 at 21:49

user5522551 · Accepted Answer · 2016-09-23 21:59:25Z

1

// Parameter marr: multidimensional array
function removeSameColors(marr){
    var carr = [];
    var rarr = [];
    var j = -1;

    for(var i = 0, l = marr.length; i < l; i++){
        if(carr[marr[i][1]] !== true){
            carr[marr[i][1]] = true;
            rarr[++j] = marr[i];
        }
    }

    return rarr;
}

That should solve your problem with very low execution time.

answered Sep 23, 2016 at 21:59

user5522551

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Comments

Nina Scholz · Accepted Answer · 2016-09-23 21:54:33Z

1

You could use a hash table with the color and use Array#filter for the wanted items.

var data = [[1, 'red'], [2, 'green'], [3, 'red'], [3, 'blue'], [5, 'green']],
    result = data.filter(function (a) {
        if (!this[a[1]]) {
            return this[a[1]] = true;                    
        }
    }, Object.create(null));

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

answered Sep 23, 2016 at 21:54

Nina Scholz

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The One and Only ChemistryBlob · Accepted Answer · 2016-09-26 13:20:06Z

1

I would use a for loop to populate a new, unique array:

var old_array = [[1, red], [2, green], [3, red], [3, blue], [5, green]],
    old_count = old_array.length,
    unique_array = [], check_array = [], i = 0;

for(; i < old_count; i++) {
    if (check_array.indexOf(old_array[i][1]) === -1) {
        check_array.push(old_array[i][1]);// this array is filled with new colors, and used for checking
        unique_array.push(old_array[i]);
    }
}

edited Sep 26, 2016 at 13:20

answered Sep 23, 2016 at 21:53

The One and Only ChemistryBlob

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2 Comments

Andrea Over a year ago

I ran this on fiddle, and the output was an empty array jsfiddle.net/sxrapxrv/1

The One and Only ChemistryBlob Over a year ago

My apologies, I had 1 mistake I didn't catch. I updated my code and now it works fine...see fiddle here ---> jsfiddle.net/sxrapxrv/2

Sherif · Accepted Answer · 2016-09-23 22:02:14Z

0

I would just keep track of the unique colors in an object passed into the filter, because as a hash it's guaranteed to be unique. If the object doesn't have a property by that color name, it returns it from the filter. Otherwise if it does it ignores the item.

var colors = [[1, "red"], [2, "green"], [3, "red"], [3, "blue"], [5, "green"]];

var uniqueArray = colors.filter(function(item, pos) {
    if (!this.hasOwnProperty(item[1])) {
        return this[item[1]] = true;
    }
    return false;
}, {});

This gives you uniqueArray = [[1,"red"],[2,"green"],[3,"blue"]], as expected.

answered Sep 23, 2016 at 22:02

Sherif

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