Javascript 2d array indexOf

phoeberesnik picture phoeberesnik · Jul 24, 2014 · Viewed 62.5k times · Source

I have a 2d array like this:

var arr = [[2,3],[5,8],[1,1],[0,9],[5,7]];

Each index stores an inner array containing the coordinates of some element.

How can I use Array.indexOf() to check if the newly generated set of coordinates is already contained in arr? I want to push into arr if only the coordinate is NOT a duplicate.

Here is my attempt that didn't work:

if (arr.indexOf([x, y]) == -1) {
    arr.push([x, y]);
}

It looks like indexOf() doesn't work for 2d arrays...

Answer

user654628 picture user654628 · Jul 24, 2014

You cannot use indexOf to do complicated arrays (unless you serialize it making everything each coordinate into strings), you will need to use a for loop (or while) to search for that coordinate in that array assuming you know the format of the array (in this case it is 2d).

var arr = [[2,3],[5,8],[1,1],[0,9],[5,7]];
var coor1 = [0, 9];
var coor2 = [1, 2];

function isItemInArray(array, item) {
    for (var i = 0; i < array.length; i++) {
        // This if statement depends on the format of your array
        if (array[i][0] == item[0] && array[i][1] == item[1]) {
            return true;   // Found it
        }
    }
    return false;   // Not found
}

// Test coor1
console.log("Is it in there? [0, 9]", isItemInArray(arr, coor1));   // True

// Test coor2
console.log("Is it in there? [1, 2]", isItemInArray(arr, coor2));   // False

// Then
if (!isItemInArray(arr, [x, y])) {
   arr.push([x, y]);
}

This implementation loops and grabs every value. If you care about performance you can do more complicated things like sorting the original array by the first index and then using binary search on the first index.

Another way is to bucket the first coordinate of each item in the array in an object (like a hashtable) and bucket the second value in each of those buckets to reduce search times; more info here http://en.wikipedia.org/wiki/Bucket_sort.

Otherwise this is probably good enough for what you need.