Aggregating object values of JavaScript arrays?

Question

In JavaScript, given n number of arrays as input in this format: (n=2)

array1:
[{x: 1, y: 5},{x: 2, y: 3},{x: 3, y: 6}]

array2:
[{x: 1, y: 2},{x: 2, y: 6},{x: 3, y: 2}]

How do I aggregate the Y-values easily and get this resulting array:

arrayOutput:
[{x: 1, y: 7},{x: 2, y: 9},{x: 3, y: 8}]

Thank you.

Answer 1

Benchmark Example

Note that the mixed code is faster followed by loops followed by native/underscore.

function createOutput(arr1, arr2, arr3 /*, ... */) {
    return Array.prototype.reduce.call(arguments, function (prev, curr) {
        return prev.map(function(val, index) {
            return {
                x: val.x,
                y: val.y + curr[index].y
            }
        });
    });
}

Assumes arrays are sorted and contain all values of x in order 1..n with no gaps.

Kind of requires ES5. This can be swapped out for _ which gives this kind of functionality in a cross browser manner.

With underscore it is

function createOutput() {
    return _.reduce(arguments, function (memo, arr) {
        return _.map(memo, function(val, index) {
            return { x: val.x, y: val.y + arr[index].y };
        });
    });
}

Answer 2

Update: The additional comment about the x values and their positions in the arrays makes the below irrelevant.

There's no particular trick, you just loop through the arrays and build up your result. It's nothing more than a nested loop. If you're trying to be maximally efficient across a broad range of JavaScript engines, avoid unnecessary function calls.

Something along the lines of:

function sumYValues(arrays) {
    var outer, inner, array, entry, sum, result, x;

    // Create our result array with a copy of the first array
    result = [];
    if (arrays.length > 0) {
        array = arrays[0];
        for (inner = 0; inner < array.length; ++inner) {
            entry = array[inner];
            result[inner] = {x: entry.x, y: entry.y};
        }

        // Add in the remaining values
        for (outer = 1; outer < arrays.length; ++outer) {
            array = arrays[outer];
            // You might want an assert here verifying that result.length == array.length
            for (inner = 0; inner < array.length; ++inner) {
                entry = array[inner];
                // You might want an assert here verifying that result[inner].x == entry.x
                result[inner].y += entry.y;
            }
        }
    }

    return result;
}

Those loops count from 0 (or 1) to array.length - 1. You might profile whether going backwards (array.length - 1 to 0 (or 1)) is any faster, mostly the "down to 0" one. I used to assume it was because it was in C when I was a fresh-faced youth (comparisons to 0 are faster than comparisons to another variable), but that assumption may or may not be valid in JavaScript.

---

There's no particular shortcut, you just loop through the arrays, do your comparisons, and build up your result.

If the x values will be unique in each array, it may be easier to keep track of your ongoing sum by using an object rather than an array and using x values as keys, and then converting it into an array when you're done. E.g.:

function sumYValues(arrays) {
    var outer, inner, ar, entry, sum, result, x;

    sum = {};
    for (outer = 0; outer < arrays.length; ++outer) {
        ar = arrays[outer];
        for (inner = 0; inner < arrays.length; ++inner) {
            entry = ar[inner];
            sum[entry.x] = (sum[entry.x] || 0) + entry.y;
        }
    }

    result = [];
    for (x in sum) {
        result.push({x: x, y: sum[x]});
    }

    return result;
}

The above is mostly just to demonstrate using sum, an object, as a map of x => y values, although it does implement at least some of the summing logic as well.

This line may need some explanation:

            sum[entry.x] = (sum[entry.x] || 0) + entry.y;

If sum doesn't have an entry for that x value, sum[entry.x] will be undefined, which is a "falsey" value. So we use the curiously-powerful || operator to either get the value for that x from sum or 0, and then add the current entry's y to it and store the result.