Sum over variable in array using reduce in javascript

Question

I want to sum over the number of calls in my array 'data'. I've found the 'reduce' function, but don't know how to select the calls part of the array. Here's my attempt at doing it:

data = {
        links: [
                  {source: 0,target: 1, calls: 20, texts:0},
                  {source: 0,target: 2, calls: 5, texts:0},
                  {source: 0,target: 3, calls: 8, texts:0},
                  {source: 0,target: 4, calls: 3, texts:0},
                  {source: 0,target: 5, calls: 2, texts:0},
                  {source: 0,target: 6, calls: 3, texts:0},
                  {source: 0,target: 7, calls: 5, texts:0},
                  {source: 0,target: 8, calls: 2, texts:0}
                ]
        }

var total_calls = data.links.calls.reduce(function(a, b) {
  return a + b;
});

thefourtheye · Accepted Answer · 2015-03-03 11:08:39Z

3

You need to iterate over the data.links array, like this

var total_calls = data.links.reduce(function(result, currentObject) {
  return result + currentObject.calls;
}, 0);
console.log(total_calls);
// 48

answered Mar 3, 2015 at 11:08

thefourtheye

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georg · Accepted Answer · 2015-03-03 11:18:04Z

1

How about making this a little bit more reusable?

data = {
        links: [
                  {source: 0,target: 1, calls: 20, texts:0},
                  {source: 0,target: 2, calls: 5, texts:0},
                  {source: 0,target: 3, calls: 8, texts:0},
                  {source: 0,target: 4, calls: 3, texts:0},
                  {source: 0,target: 5, calls: 2, texts:0},
                  {source: 0,target: 6, calls: 3, texts:0},
                  {source: 0,target: 7, calls: 5, texts:0},
                  {source: 0,target: 8, calls: 2, texts:0}
                ]
        }

pluck = function(ary, prop) {
  return ary.map(function(x) { return x[prop] });
}

sum = function(ary) {
  return ary.reduce(function(a, b) { return a + b }, 0);
}

result = sum(pluck(data.links, 'calls'))
document.write(result)

answered Mar 3, 2015 at 11:18

georg

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1 Comment

pir Over a year ago

Both solutions are quite nice, but for my purposes it's actually perfect that's reusable.

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