javascript合并相同的数组元素

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

Answer 1

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

Answer 2

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;
}

Answer 3

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]] */