Javascript ES6 calculate the sum of an array of arrays of objects

Question

I have multiple arrays of objects with a string property "CountType" and a number property "ItemCount".

The objects must be added up to calculate the total of all items by "CountType" property.

For example, I need to know the total number of Volumes, Sheets, Photos, etc.

Attempt #1: This code I attempted returns 0. Possibly that is because I need to loop through the outer array to do the comparison across the nested arrays.

function sumByProperty(items, prop) {
  if (items == null) {
      return 0;
  }
  return items.reduce(function (a, b) {
      return b[prop] == null ? a : a + b[prop];
  }, 0);
}

INPUT

Item count

[
    [
      {"CountType":"Volumes","ItemCount":3},
      {"CountType":"Sheets","ItemCount":6},
      {"CountType":"Photos","ItemCount":3},
      {"CountType":"Boxes","ItemCount":1},
      {"CountType":"Other","ItemCount":2}
    ],
    [
      {"CountType":"Volumes","ItemCount":1},
      {"CountType":"Sheets","ItemCount":1},
      {"CountType":"Photos","ItemCount":3},
      {"CountType":"Boxes","ItemCount":0},
      {"CountType":"Other","ItemCount":1}
    ],
    [
      {"CountType":"Volumes","ItemCount":1},
      {"CountType":"Sheets","ItemCount":0},
      {"CountType":"Photos","ItemCount":3},
      {"CountType":"Boxes","ItemCount":4},
      {"CountType":"Other","ItemCount":5}
    ]
]

DEISRED OUTPUT

Total count

[
  {"CountType":"Volumes","ItemCount":5},
  {"CountType":"Sheets","ItemCount":7},
  {"CountType":"Photos","ItemCount":9},
  {"CountType":"Boxes","ItemCount":5},
  {"CountType":"Other","ItemCount":8}
]

UPDATE: Here is how I am trying to run the function:

https://jsfiddle.net/yd051o76/2/

Your sumByProperty function seems fine. Please show us how you are calling it on the input — Bergi
– Bergi, Commented Jul 22, 2019 at 19:14

Jonas Wilms · Accepted Answer · 2019-07-22 19:13:33Z

4

Possibly that is because I need to loop through the outer array to do the comparison across the nested arrays.

Indeed, you could flatten the array:

 function sumByProperty(items, prop) {
   return items.flat().reduce(function (a, b) {
     return b[prop] == null ? a : a + b[prop];
   }, 0);
 }

 const result = [
   { CountType: "Volumes", ItemCount: sumByProperty(input, "Volumes"), },
   //...
 ];

But I'd rather group dynamically using a hashtable, that way you only have to iterate once and you don't need to name all properties:

  const hash = new Map();

   for(const { CountType, ItemCount } of input.flat())
      hash.set(CountType, (hash.get(CountType) || 0) + ItemCount);

  const result = [...hash.entries()].map(([CountType, ItemCount]) => ({ CountType, ItemCount }));

answered Jul 22, 2019 at 19:13

Jonas Wilms

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5 Comments

Kamil Kiełczewski Over a year ago

Heh - why Map ? :)

Jonas Wilms Over a year ago

Because it guarantees O(1) lookuptime and is thus the appropriate datastructure.

Jonas Wilms Over a year ago

That doesnt guarantee O(1) lookup time (most engines will choose a hashtable though), but I prefer the explicitness of Map.

Kamil Kiełczewski Over a year ago

interesting information (+1) - any proof (or link to proof) ?

Tyler Roper Over a year ago

For what it's worth, MDN 's Map documentation has a section Objects and Maps compared.

mwilson · Accepted Answer · 2019-07-22 21:48:58Z

Here's an example using 2 reduce functions with a for ... in ... to merge the results back into the parent reducer.

The trick is to make sure you initialize your accumulator as null so you can define your own. Then, it's just a matter of reducing the array of arrays and having another reducer to run the array of objects. Once done, you just need to merge the result of the inner array back into the parent. (Might be a better way to do this part)

Fiddle: https://jsfiddle.net/mswilson4040/jcp9x6ba/26/

const json = [
  [
    {"CountType":"Volumes","ItemCount":3},
    {"CountType":"Sheets","ItemCount":6},
    {"CountType":"Photos","ItemCount":3},
    {"CountType":"Boxes","ItemCount":1},
    {"CountType":"Other","ItemCount":2}
  ],
  [
    {"CountType":"Volumes","ItemCount":1},
    {"CountType":"Sheets","ItemCount":1},
    {"CountType":"Photos","ItemCount":3},
    {"CountType":"Boxes","ItemCount":0},
    {"CountType":"Other","ItemCount":1}
  ],
  [
    {"CountType":"Volumes","ItemCount":1},
    {"CountType":"Sheets","ItemCount":0},
    {"CountType":"Photos","ItemCount":3},
    {"CountType":"Boxes","ItemCount":4},
    {"CountType":"Other","ItemCount":5}
  ]
];

const counts = json.reduce( (acc, currentArray) => {
	acc = acc ? acc : {};
  const arrCounts = currentArray.reduce( (_acc, _item) => {
  	_acc = _acc ? _acc : {};
    _acc[_item.CountType] = _acc[_item.CountType] ? _acc[_item.CountType] + _item.ItemCount : _item.ItemCount;
    return _acc;
  }, null);
  for (const item in arrCounts) {
  	acc[item] = acc[item] ? acc[item] + arrCounts[item] : arrCounts[item];
  }
  return acc;
}, null);


console.log(counts);

Kamil Kiełczewski · Accepted Answer · 2019-07-22 19:17:46Z

0

Try (data is input, h = {}, the r is result);

data.map(a=> a.map(x=> h[x.CountType]=x.ItemCount+(h[x.CountType]||0) ))

r= Object.keys(h).map(k=> ({CountType:k, ItemCount:h[k] }) );

let data = [[
  {"CountType":"Volumes","ItemCount":3},
  {"CountType":"Sheets","ItemCount":6},
  {"CountType":"Photos","ItemCount":3},
  {"CountType":"Boxes","ItemCount":1},
  {"CountType":"Other","ItemCount":2}
],
[
  {"CountType":"Volumes","ItemCount":1},
  {"CountType":"Sheets","ItemCount":1},
  {"CountType":"Photos","ItemCount":3},
  {"CountType":"Boxes","ItemCount":0},
  {"CountType":"Other","ItemCount":1}
],
[
  {"CountType":"Volumes","ItemCount":1},
  {"CountType":"Sheets","ItemCount":0},
  {"CountType":"Photos","ItemCount":3},
  {"CountType":"Boxes","ItemCount":4},
  {"CountType":"Other","ItemCount":5}
]];

let r,h = {};
data.map(a=> a.map(x=> h[x.CountType]=x.ItemCount+(h[x.CountType]||0) ))
r= Object.keys(h).map(k=> ({CountType:k, ItemCount:h[k] }) );

console.log(r);

answered Jul 22, 2019 at 19:17

Kamil Kiełczewski

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2 Comments

Jonas Wilms Over a year ago

This isn't code golf. Why h, r ... ? That makes things unreadable

Jonas Wilms Over a year ago

Also why .map ?

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