javascript merge array elements that are the same

Question

I have this array: I was wondering how can I remove the duplicate arrays elements regardless of the position of the elements inside the array

example
array1 = [[0, 1], [1, 0], [2, 3], [3, 2], [4, 5], [5, 4]];
//the [0,1],[1,0] will be counted as the same element, so as the others..
//to
array1 = [[0, 1], [2, 3], [4, 5]]

thanks in advance

sorry i have this so far

newArrayVal = [];

arr1=  [[0, 1], [1, 0], [2, 3], [3, 2], [4, 5], [5, 4]];
for(i=0; i<arr1.length; i++){
  newArrayVal[i] = arr1[i].sort();
}
console.log(newArrayVal);
//thanks frenchie I think i am near to the solution, i just need to filter this now, hehe thanks

I would start with a loop. Edit: I misread, you can probably just use .filter. — Evan Davis
– Evan Davis, Commented Mar 7, 2015 at 4:53
I'd first change the inner arrays so they're all sorted internally. Then, loop through the array and at each iteration you reloop through the array and remove any element identical to the one you're on. — frenchie
– frenchie, Commented Mar 7, 2015 at 5:03
thanks everyone all the answers works, i'll just mark as answer the first one thanks again — user4006175
– user4006175, Commented Mar 7, 2015 at 5:39

tim-we · Accepted Answer · 2015-03-09 01:04:12Z

3

You could do that with a loop in a loop:

array1 = [[0, 1], [1, 0], [2, 3], [3, 2], [4, 5], [5, 4]];

function equal(a1, a2) {
   return (a1[0]==a2[0] && a1[1]==a2[1]) || (a1[0]==a2[1] && a1[1]==a2[0]);
}

for(var i=0; i<array1.length; i++) {
    for(var j=i+1; j<array1.length; j++) {
        if(equal(array1[i], array1[j])) {
            array1.splice(j,1);
        }
    }
}

// array1 = [[0,1],[2,3],[4,5]]

This should do the job, hope it helps :) I put it on jsfiddle: http://jsfiddle.net/eu1v8eso/

In this example equal() is the function checking if the conditions apply.

Note that this solution also works with an array containing strings instead of numbers as I'm not using a sort function here (which Sly_cardinal does in his solution)

I think kennebec's solution is the better answer to this question as it is much cleaner and uses the new Array.prototype.filter function

edited Mar 9, 2015 at 1:04

answered Mar 7, 2015 at 5:12

tim-we

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Sly_cardinal · Accepted Answer · 2015-03-07 05:14:29Z

Based on Mathletics and frenchie's comments, here is a solution:

/**
 * Returns the unique sub-lists of the given list:
 * 
 * e.g. 
 * given:
 *     getUnique([[0, 1], [1, 0], [2, 3], [3, 2], [4, 5], [5, 4]])
 * 
 * returns:
 * 
 *     [[0, 1], [2, 3], [4, 5]]
 * 
 * @param  {number[][]} list List of array of numbers
 * @return {number[][]} Returns a list of unique sub-arrays
 */
function getUnique(list){

    var numSort = function(a, b){
        return a - b;
    };

    var getItemId = function(list){
        return list.join(',');
    };

    // Map each sub-list to a unique ID.
    // Keep track of whether we've seen that
    // ID before and only keep the first list
    // with that ID.
    var uniqueMap = {};
    var resultList = list.filter(function(item, index){
        var keepItem = false;
        var itemId = getItemId(item.sort(numSort));
        if (!uniqueMap[itemId]){
            uniqueMap[itemId] = true;
            keepItem = true;
        }
        return keepItem;
    });

    return resultList;
}

kennebec · Accepted Answer · 2015-03-07 06:34:27Z

2

You can sort and join the interior arrays for an easier lookup.

function uniqueDeepArray(A){
    var next, b= A.map(function(itm){
        return itm.sort().join('');
    });
    return A.filter(function(itm, i){
        return b.indexOf(b[i])== i;
    });
}



var a1= [[0, 1], [1, 0], [2, 3], [3, 2], [4, 5], [5, 4]],
a2= uniqueDeepArray(a1);

/* a2= (Array) [[0,1],[2,3],[4,5]] */

edited Mar 7, 2015 at 6:34

answered Mar 7, 2015 at 6:16

kennebec

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