CODEWITHSUNDEEP
javascriptbackbone.jsunderscore.js
Erik
Erik

Reputation: 14770

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

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 :)

Upvotes: 6

Views: 11523

Answers (3)

Bergi
Bergi

Reputation: 664936

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;
}, {});

Upvotes: 1

YD1m
YD1m

Reputation: 5895

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

Upvotes: 0

Artur Nowak
Artur Nowak

Reputation: 5354

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()

Upvotes: 5

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