Sum arrays in javascript

Question

I have these arrays:

// I want to calculate these arrays each other.
var a = [1,2,3]
var b = [1,2,3]
var c = [3,4,5]
// ->
var result = [5, 8, 11]

I can calculate two array with below method but how can I sum n arrays?

Array.prototype.Sum = function (arr) {
    var s = [];
    if (arr != null && this.length == arr.length) {
        for (var i = 0; i < arr.length; i++) {
            var c = this[i] + arr[i]
            s.push(c);
        }
    }
    return s;
}

a.sum(b).sum(c)? Note lower case s, Sum is not a constructor. ;-) — RobG
– RobG, Commented Dec 18, 2014 at 0:49
@Lazy—if you have that many, you should find the looping answers much faster than those using built–in methods, see http://jsperf.com/summing-arrays-again where loops are 10 times faster than built–ins. Don't judge a solution by the lines of code. ;-) — RobG
– RobG, Commented Dec 18, 2014 at 3:38

Oriol · Accepted Answer · 2014-12-18 01:04:17Z

5

Assuming

var arrays = [
    [1,2,3],
    [1,2,3],
    [3,4,5]
];

Using your Sum with EcmaScript 5 array methods,

var result = arrays.reduce(function(arr1, arr2){
    return arr1.Sum(arr2);
});

Alternatively, consider not polluting Array.prototype and just using something like (ES5)

var result = arrays.reduce(function(arr1, arr2) {
    return arr1.map(function(num, index){
        return num+arr2[index];
    });
});

Or, simplifying using EcmaScript 6 arrow functions,

var result = arrays.reduce(
    (arr1, arr2) => arr1.map(
        (num, index) => num+arr2[index]
    )
);

edited Dec 18, 2014 at 1:04

answered Dec 18, 2014 at 0:50

Oriol

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5 Comments

elclanrs Over a year ago

This is not it, result gives [6,6,12] when it should be [5,8,11]

Oriol Over a year ago

@elclanrs True, fixed.

Witiko Over a year ago

Be aware that the reduce and map functions are parts of the ECMAScript 5 specification, which is only supported by modern browsers. If that's an issue for you, you will have to use a polyfill, which adds the functionality to older browsers.

RobG Over a year ago

Note that for loops are nearly always faster than methods like map and reduce. Slightly more code, but faster.

RobG Over a year ago

Oh, and if the built–ins have been pollyfilled, they will be very, very much slower than plain loops.

Witiko · Accepted Answer · 2014-12-18 01:12:22Z

3

A plain old ECMAScript 3 solution:

var arrays = [
    [1, 2, 3],
    [1, 2, 3],
    [3, 4, 5]
];

function sum() {
  var sum = [];
  for(var i = 0; i < arguments.length; ++i) {
    for(var j = 0; j < arguments[i].length; ++j)
      sum[j] = (sum[j] || 0) + arguments[i][j];
  } return sum;
}

sum.apply(this, arrays);              // [ 5, 8, 11 ]
sum(arrays[0], arrays[1], arrays[2]); // [ 5, 8, 11 ]

answered Dec 18, 2014 at 1:12

Witiko

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2 Comments

RobG Over a year ago

Loops are fast, you might find sum.push(...) faster than sum[j] = ....

Witiko Over a year ago

That wouldn't preserve the semantics, though. I'm accessing each sum[j] arguments.length times (given the input arrays have the same length).

Hilder Vitor Lima Pereira · Accepted Answer · 2014-12-18 01:13:56Z

3

With this function, you can sum several arrays and you don't need to "encapsulate" your arrays into another array...

Also, it works on old browsers.

function Sum(){
  if (arguments.length == 0){
    return null;
  }
  var s = arguments[0];
  for(var j = 1; j < arguments.length; j++) {
     var arr = arguments[j];
     for (var i = 0; i < arr.length; i++) {
         s[i] = s[i] + arr[i];
     }
  }
  return s;
}

Just call Sum(a, b) or Sum(a, b, c) ...

You can put it in your prototype too if you want (and make little changes to use this if you want to sum in place).

edited Dec 18, 2014 at 1:13

answered Dec 18, 2014 at 1:06

Hilder Vitor Lima Pereira

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1 Comment

RobG Over a year ago

The first if block is redundant. If no arguments are provided, s will be undefined, the outer for loop will not run so undefined will be returned. Or you could do return s || null;, but undefined makes more sense to me. ;-)

Chris Franklin · Accepted Answer · 2014-12-18 01:01:35Z

-1

a.sum(b).sum(c);

That should do it.

edited Dec 18, 2014 at 1:01

answered Dec 18, 2014 at 0:50

Chris Franklin

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