CODEWITHSUNDEEP
javascriptarraystreeflatten
Nitin
Nitin

Reputation: 165

Better Solution for Unflattening Array of Objects

For the following array of Objects

[
 {"Corp": "ABC", "T1": "HW A/V", "T2": "A/V System","T3": "Audio","Prod": "Audio System","Mfr": "One"},
 {"Corp": "ABC", "T1": "HW A/V", "T2": "A/V System","T3": "Audio","Prod": "Audio System","Mfr": "Two"},
 {"Corp": "ABC", "T1": "HW A/V", "T2": "A/V System","T3": "Video","Prod": "Video System","Mfr": "Other"}
]

I intend to get a unflatten object like the one below :

[{
"Corp": "ABC",
"List": [{
    "T1": "HW A/V",
    "List": [{
        "T2": "A/V System",
        "List": [{
            "T3": "Audio",
            "List": [{
                "Prod": "Audio System",
                "List": [
                    {"Mfr": "One"},
                    {"Mfr": "Two"}
                ]
            }]
        },
        {
            "T3": "Video",
            "List": [{
                "Prod": "Video System",
                "List": [
                    {"Mfr": "Other"}
                ]
            }]
        }]
    }]
}]

}] I did get the result I intend to get as described above. I used underscore to get the result. Following code snippet did the job for me :

var items = _.map(_.groupBy(itemList, 'Corp'), function (a) {
        return _.extend(_.pick(a[0], 'Corp'), {
            List: _.map(_.groupBy(a, 'T1'), function (b) {
                return _.extend(_.pick(b[0], 'T1'), {
                    List: _.map(_.groupBy(b, 'T2'), function (c) {
                        return _.extend(_.pick(c[0], 'T2'), {
                            List: _.map(_.groupBy(c, 'T3'), function (d) {
                                return _.extend(_.pick(d[0], 'T3'), {
                                    List: _.map(_.groupBy(d, 'Prod'), function (e) {
                                        return _.extend(_.pick(e[0], 'Prod'), {
                                            List: _.map(e, function (elem) {
                                                return _.pick(elem, 'Mfr')
                                            })
                                        });
                                    })
                                });
                            })
                        });
                    })
                });
            })
        });
    });

Now, So All I am looking for, is if someone can enhance my solution. I want to optimize both space and time for this process.

PS: In morning, I had asked a similar question requesting for the solution, and that question was marked as TOO BROAD and was put on HOLD, so I have added my solution with this question, now all I am looking for is a better solution.

Thanks

Upvotes: 2

Views: 494

Answers (1)

Arnauld
Arnauld

Reputation: 6130

In order to avoid imbricated statements, you may want to define a list of 'expand keys' and iterate on them.

It's half-tempting to automatically extract the keys with something like:

expandKeys = _.keys(itemList[0]);

But since Javascript doesn't guarantee the order of the keys in an object, you really should define this list explicitly instead.

Below is some example code.

var itemList = [
  {"Corp": "ABC", "T1": "HW A/V", "T2": "A/V System", "T3": "Audio", "Prod": "Audio System", "Mfr": "One"},
  {"Corp": "ABC", "T1": "HW A/V", "T2": "A/V System", "T3": "Audio", "Prod": "Audio System", "Mfr": "Two"},
  {"Corp": "ABC", "T1": "HW A/V", "T2": "A/V System", "T3": "Video", "Prod": "Video System", "Mfr": "Other"}
];

var expandKeys = [ 'Corp', 'T1', 'T2', 'T3', 'Prod', 'Mfr' ];

function expandList(list, keys) {
  var node, obj, root = {};

  _.each(list, function(item) {
    obj = root;
    _.each(keys, function(key) {
      obj = (obj.List = obj.List || []);
      node = _.find(obj, function(i) { return i[key] == item[key]; });

      if(node === undefined) {
        obj.push(node = {});
        node[key] = item[key];
      }
      obj = node;
    });
  });
  return root.List;
}

var res = expandList(itemList, expandKeys);
console.log(res);
<script src="http://underscorejs.org/underscore-min.js"></script>

Upvotes: 2

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