How to merge two arrays of objects, and sum the values of duplicate object keys in JavaScript?

Question

So, I have an indeterminate amount of arrays in an object, and I need to merge the objects in them. The arrays are guaranteed to be the same length, and the objects are guaranteed to have the same keys.

I've tried iterating through it, though for some reason, it sums the objects in every key in the array.

Example:

var demo = {
  "key1": [{"a": 4, "b": 3, "c": 2, "d": 1}, {"a": 2, "b": 3, "c": 4, "d": 5}, {"a": 1, "b": 4, "c": 2, "d": 9}]
  "key2": [{"a": 3, "b": 5, "c": 3, "d": 4}, {"a": 2, "b": 9, "c": 1, "d": 3}, {"a": 2, "b": 2, "c": 2, "d": 3}]
};

mergeArrays(demo);

/* Returns:
  {
    "arbitraryname": [{"a": 7, "b": 8, "c": 5, "d": 5}, {"a": 4, "b": 12, "c": 5, "d": 8}, {"a": 3, "b": 6, "c": 4, "d": 12}]
  }
*/

My current implementation attempts to do something like this:

var skeleton = {
  "a": 0,
  "b": 0,
  "c": 0,
  "d": 0
};

function mergeArrays = function (obj) {
  var flat = {};
  var keys = _.keys(prod);
  var len = obj[keys[0]].length;
  flat["arbitraryname"] = [];
  for (var i = 0; i < len; i++) {
    flat["arbitraryname"].push(skeleton);
  }
  flat["arbitraryname"].forEach(function (v, k) {
    _.forIn(v, function (key, value) {
      flat["arbitraryname"][k][key] += value;
    });
  });
  return flat;
}

Is there any easier way to do that? It also appears to return an array right now that's entirely identical. E.g.:

[{"a": 14, "b": 26, "c": 14, "d": 25}, {"a": 14, "b": 26, "c": 14, "d": 25}, {"a": 14, "b": 26, "c": 14, "d": 25}]

Where each object is a sum of ALL elements in the array. What am I doing wrong? Is there an easier way?

what are expected results? That identical set of objects doesn't make sense — charlietfl
– charlietfl, Commented Jan 16, 2016 at 5:25
I didn't clarify this very well. I had an issue where it was pushing the same value for each key in the array. The reason this happened was because assigning the skeleton to the object didn't break the reference to the original object, causing it to update, and cascade down to each key. I fixed this by running JSON.parse(JSON.stringify(skeleton)). I also found a solution that works that's similar, but not exact. I'll try and post it tonight. — Brandon Anzaldi
– Brandon Anzaldi, Commented Jan 18, 2016 at 21:06

charlietfl · Accepted Answer · 2016-01-16 05:31:58Z

3

The following matches the commented array at top of question. Expected results are a little unclear

Using Object.keys() makes it fairly easy.

var demo = {
    "key1": [{"a": 4, "b": 3, "c": 2, "d": 1}, {"a": 2, "b": 3, "c": 4, "d": 5}, {"a": 1, "b": 4, "c": 2, "d": 9}],
    "key2": [{"a": 3, "b": 5, "c": 3, "d": 4}, {"a": 2, "b": 9, "c": 1, "d": 3}, {"a": 2, "b": 2, "c": 2, "d": 3}]
};

// iterate and map using first array
var res = demo.key1.map(function(item, idx) { 
    //  loop over keys of each object in array and return 
    //  new objects with sums of matching index and keys
    return Object.keys(item).reduce(function(obj,key) {
        obj[key] = item[key] + demo.key2[idx][key];
        return obj;
    },{});  
});

// print "res" for demo
document.getElementById('pre').innerHTML =JSON.stringify(res, null, 4)

<pre id="pre"></pre>

edited Jan 16, 2016 at 5:31

answered Jan 16, 2016 at 5:18

charlietfl

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Comments

Matthew Herbst · Accepted Answer · 2016-01-16 05:24:57Z

1

I mean, you could just write it yourself - it's not overly complicated since your rules make it well-defined (plus vanilla is cross-browser compatible):

var resultArray = [];

for(var key in demo) {
  if(demo.hasOwnProperty(key)) {
    demo[key].forEach(function(obj, i) {
      if(!resultArray[i]) {
        resultArray[i] = {};
      }

      for(var letterKey in obj) {
        if(obj.hasOwnProperty(letterKey)) {
          if(resultArray[i][letterKey]) {
            resultArray[i][letterKey] += obj[letterKey];
          } else {
            resultArray[i][letterKey] = obj[letterKey];
          } 
        }
      }
    });
  }
}

That should work.

answered Jan 16, 2016 at 5:24

Matthew Herbst

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