How to sum elements at the same index in array of arrays into a single array?

Question

Let's say that I have an array of arrays, like so:

[
  [0, 1, 3],
  [2, 4, 6],
  [5, 5, 7],
  [10, 0, 3]
]

How do I generate a new array that sums all of the values at each position of the inner arrays in javascript? In this case, the result would be: [17, 10, 19]. I need to be able to have a solution that works regardless of the length of the inner arrays. I think that this is possible using some combination of map and for-of, or possibly reduce, but I can't quite wrap my head around it. I've searched but can't find any examples that quite match this one.

What do you expect to do if some of the sub-arrays are different lengths. Do you want that to be an error? Or just use 0 for array elements in the shorter arrays that don't match up with the longer ones? — jfriend00
– jfriend00, Commented Dec 24, 2015 at 21:43
@jfriend00 that's a good question. Given that choice, I'd say use 0 for shorter arrays. — hmlee
– hmlee, Commented Dec 24, 2015 at 21:45

Nina Scholz · Accepted Answer · 2018-10-21 16:32:45Z

54

You can use Array.prototype.reduce() in combination with Array.prototype.forEach().

var array = [
        [0, 1, 3],
        [2, 4, 6],
        [5, 5, 7],
        [10, 0, 3]
    ],
    result = array.reduce(function (r, a) {
        a.forEach(function (b, i) {
            r[i] = (r[i] || 0) + b;
        });
        return r;
    }, []);
document.write('<pre>' + JSON.stringify(result, 0, 4) + '</pre>');

Update, a shorter approach by taking a map for reducing the array.

var array = [[0, 1, 3], [2, 4, 6], [5, 5, 7], [10, 0, 3]],
    result = array.reduce((r, a) => a.map((b, i) => (r[i] || 0) + b), []);
    
console.log(result);

edited Oct 21, 2018 at 16:32

answered Dec 24, 2015 at 21:44

Nina Scholz

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Comments

Tamas Hegedus · Accepted Answer · 2018-11-14 11:54:12Z

12

Using Lodash 4:

function sum_columns(data) {
  return _.map(_.unzip(data), _.sum);
}

var result = sum_columns([
  [1, 2],
  [4, 8, 16],
  [32]
]);

console.log(JSON.stringify(result));

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.js"></script>

For older Lodash versions and some remarks

Lodash 4 has changed the way _.unzipWith works, now the iteratee gets all the values passed as spread arguments at once, so we cant use the reducer style _.add anymore. With Lodash 3 the following example works just fine:

function sum_columns(data) {
  return _.unzipWith(data, _.add);
}

var result = sum_columns([
  [1, 2],
  [4, 8, 16],
  [32],
]);

console.log(JSON.stringify(result));

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/3.10.1/lodash.js"></script>

_.unzipWith will insert undefineds where the row is shorter than the others, and _.sum treats undefined values as 0. (as of Lodash 3)

If your input data can contain undefined and null items, and you want to treat those as 0, you can use this:

function sum_columns_safe(data) {
  return _.map(_.unzip(data), _.sum);
}

function sum_columns(data) {
  return _.unzipWith(data, _.add);
}

console.log(sum_columns_safe([[undefined]])); // [0]
console.log(sum_columns([[undefined]]));      // [undefined]

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/3.10.1/lodash.js"></script>

This snipet works with Lodash 3, unfortunately I didn't find a nice way of treating undefined as 0 in Lodash 4, as now sum is changed so _.sum([undefined]) === undefined

edited Nov 14, 2018 at 11:54

answered Dec 24, 2015 at 21:53

Tamas Hegedus

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9 Comments

hmlee Over a year ago

Thanks for this. I'm not all that familiar with lodash, but it looks like I should check it out more. Looking at the docs, they have an example that uses _.unzipWith and _.add, like: return _.unzipWith(data, _.add). Would that work too?

binki Over a year ago

Doesn't lodash have flatten()?

Tamas Hegedus Over a year ago

@hmlee yes, that's even better. The only difference is _.map(_.unzip([[undefined]]), _.sum) -> [0], but _.unzipWith([[undefined]], _.add) === [undefined]

Tamas Hegedus Over a year ago

@binki it does, but it doesn't help in this case

Penny Liu Over a year ago

@TamasHegedus How can I use lodash in v4, which executes with (...group).

|

isvforall · Accepted Answer · 2016-04-29 13:26:08Z

5

One-liner in ES6, with map and reduce

var a = [ [0, 1, 3], [2, 4, 6], [5, 5, 7], [10, 0, 3] ];

var sum = a[0].map((_, i) => a.reduce((p, _, j) => p + a[j][i], 0));

document.write(sum);

answered Apr 29, 2016 at 13:26

isvforall

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Comments

epascarello · Accepted Answer · 2015-12-24 22:10:49Z

0

Assuming that the nested arrays will always have the same lengths, concat and reduce can be used.

    function totalIt (arr) {
        var lng = arr[0].length;
       return [].concat.apply([],arr)  //flatten the array
                .reduce( function(arr, val, ind){ //loop over and create a new array
                    var i = ind%lng;  //get the column
                    arr[i] = (arr[i] || 0) + val; //update total for column
                    return arr;  //return the updated array
                }, []);  //the new array used by reduce
    }
    
    var arr = [
      [0, 1, 3],
      [2, 4, 6],
      [5, 5, 7],
      [10, 0, 3]
    ];
    console.log(totalIt(arr));  //[17, 10, 19]

answered Dec 24, 2015 at 22:10

epascarello

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Comments

Abhishek Gurjar · Accepted Answer · 2017-04-09 13:21:44Z

0

Assuming array is static as op showned.

a = [
      [0, 1, 3],
      [2, 4, 6],
      [5, 5, 7],
      [10, 0, 3]
    ]

b = []
     	
for(i = 0; i < a[0].length; i++){						
  count = 0
  for(j = 0; j < a.length; j++){				
     count += a[j][i]			
   }
   b.push(count)
}		
console.log(b)

answered Apr 9, 2017 at 13:21

Abhishek Gurjar

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Comments

David Lukas · Accepted Answer · 2022-01-04 18:36:14Z

So far, no answer using the for ... of mentioned in the question.
I've used a conditional statement for different lengths of inner arrays.

var a = [
  [0, 1, 3],
  [2, 4, 6],
  [5, 5, 7],
  [10, 0, 3]
];
i = 0;
r = []
for (const inner of a) {
  j = 0;
  for (const num of inner) {
    if (j == r.length) r.push(num)
    else r[j] += num
    j++;
  }
  i++;
}
console.log(r);

True, in this case, the classic for cycle fits better than for ... of.
The following snippet uses a conditional (ternary) operator.

var a = [
  [0, 1, 3],
  [2, 4, 6],
  [5, 5, 7],
  [10, 0, 3]
];

r = [];
for (i = 0; i < a.length; i++) {
  for (j = 0; j < a[i].length; j++) {
    j==r.length ? r.push(a[i][j]) : r[j]+=a[i][j]
  }
}

console.log(r);

A solution using maps and reductions, adding elements from different lengths of arrays.

var array = [
  [0],
  [2, 4],
  [5, 5, 7, 10, 20, 30],
  [10, 0]
];
b = Array(array.reduce((a, b) => Math.max(a, b.length), 0)).fill(0);
result = array.reduce((r, a) => b.map((_, i) => (a[i] || 0) + (r[i] || 0)), []);

console.log(result);

pelak · Accepted Answer · 2017-05-09 09:54:22Z

-2

const ar = [
  [0, 1, 3],
  [2, 4, 6],
  [5, 5, 7],
  [10, 0, 3]
]   

ar.map( item => item.reduce( (memo, value)=> memo+= value, 0 ) )
//result-> [4, 12, 17, 13]

answered May 9, 2017 at 9:54

pelak

685 bronze badges

2 Comments

Donald Duck is with Ukraine Over a year ago

While this code may answer the question, providing additional context regarding why and/or how this code answers the question improves its long-term value.

JJP Over a year ago

This does not answer the question, it sums every array.

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