How to get sum of an array object keys in underscore.js?

Question

I have the following array:

var items = [
   {price1: 100, price2: 200, price3: 150},
   {price1: 10, price2: 50},
   {price1: 20, price2: 20, price3: 13},
]

I need to get object with sum of all keys like the following:

var result = {price1: 130, price2: 270, price3: 163};

I know I may to use just loop but I'm looking for a approach in underscore style :)

Have you tried their documentation? If so, have you tried to accomplish this for yourself? — user2437417
– user2437417, Commented Jun 27, 2013 at 18:28
@ArturNowak: I'll go ahead and clean up most of my comments since they no longer make sense now that the rest are gone. It was a fruitless conversation anyway. — user2437417
– user2437417, Commented Jun 27, 2013 at 20:22

Artur Nowak · Accepted Answer · 2013-06-27 18:48:48Z

5

Not very pretty, but I think the fastest method is to do it like this

_(items).reduce(function(acc, obj) {
  _(obj).each(function(value, key) { acc[key] = (acc[key] ? acc[key] : 0) + value });
  return acc;
}, {});

Or, to go really over the top (I think it will can be faster than above one, if you use lazy.js instead of underscore):

_(items).chain()
  .map(function(it) { return _(it).pairs() })
  .flatten(true)
  .groupBy("0") // groups by the first index of the nested arrays
  .map(function(v, k) { 
    return [k, _(v).reduce(function(acc, v) { return acc + v[1] }, 0)]     
  })
  .object()
  .value()

edited Jun 27, 2013 at 18:48

answered Jun 27, 2013 at 18:34

Artur Nowak

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6 Comments

user2437417 Over a year ago

Should use .each() or .forEach() (whatever underscore names it) instead of .map() since you're not building an Array.

user2437417 Over a year ago

@Erik: I thought you wanted "better than loop"? This uses two loops.

user2437417 Over a year ago

@Erik: I already read it. Read my first comment under your question again.

Erik Over a year ago

I'm looking for underscore approach. Where here is the loops like for, for/in in Artur answer?

user2437417 Over a year ago

@Erik: They're called .reduce() and .map()

|

Bergi · Accepted Answer · 2013-06-27 19:44:53Z

1

For aggregating I'd recommend reduce:

_.reduce(items, function(acc, o) {
    for (var p in acc) acc[p] += o[p] || 0;
    return acc;
}, {price1:0, price2:0, price3:0});

Or better

_.reduce(items, function(acc, o) {
    for (var p in o)
        acc[p] = (p in acc ? acc[p] : 0) + o[p];
    return acc;
}, {});

edited Jun 27, 2013 at 19:44

answered Jun 27, 2013 at 18:37

Bergi

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4 Comments

Artur Nowak Over a year ago

What if items have different keys than price1, price2 and price3?

Bergi Over a year ago

Well, this solution lets you choose which properties you want to have summed up. Of course you might also compute that list dynamically, or just use for (var p in o) acc[p] = (p in acc ? acc[p] : 0) + o[p]; instead (especially if each item has different properties)

YD1m Over a year ago

Returns {price1: 130, price2: 270, price3: NaN}

Bergi Over a year ago

@YD1m: Thanks for the hint, I first believed every item had the same set of properties. Solution adapted (and integrated second one)

YD1m · Accepted Answer · 2013-06-27 19:35:33Z

0

Js from hell:

var names = _.chain(items).map(function(n) {  return _.keys(n);  }).flatten().unique().value();    
console.log(_.map(names, function(n) {              
      return  n + ': ' + eval((_.chain(items).pluck(n).compact().value()).join('+')); 
}).join(', '));

jsfiddle

answered Jun 27, 2013 at 19:35

YD1m

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