CODEWITHSUNDEEP
javascriptunderscore.js
Vimalraj Selvam
Vimalraj Selvam

Reputation: 2245

Aggregate data using Javascript or Underscore

I have a raw data like this:

{
    "Team1": [
        {
            "priority": "P0",
            "status": "Open",
            "teamName": "Team1"
        },
        {
            "priority": "P1",
            "status": "Closed",
            "teamName": "Team1"
        },
        {
            "priority": "P0",
            "status": "Closed",
            "teamName": "Team1"
        }
    ],
    "Team2": [
        {
            "priority": "P1",
            "status": "Open",
            "teamName": "Team2"
        }
    ]
}

I would like to transform the data as follows:

{
    "Team1": {
        "teamName": "Team1",
        "Open": {
            "P0": 1,
            "P1": 0
        },
        "Closed": {
            "P0": 1,
            "P1": 1
        }
    },
    "Team2": {
        "teamName": "Team2",
        "Open": {
            "P0": 0,
            "P1": 1
        },
        "Closed": {
            "P0": 0,
            "P1": 1
        }
    }
}

I'm trying with Underscore.js, but not very successful since I'm novice in this.

I am till here:

_.groupBy(data, function(element) {
    element.teamName + '#' + element.status
});

But I'm clueless on how to proceed.

Upvotes: 2

Views: 2042

Answers (1)

Nina Scholz
Nina Scholz

Reputation: 386680

You could first iterate the object's keys and the arrays and get the included properties and then the same for the count.

var data = { "Team1": [{ "priority": "P0", "status": "Open" }, { "priority": "P1", "status": "Closed" }, { "priority": "P0", "status": "Closed" }], "Team2": [{ "priority": "P1", "status": "Open" }] },
    keys = Object.keys(data),
    priorities = new Set(),
    grouped = {};

keys.map(k => data[k].forEach(a => priorities.add(a.priority)));

keys.map(k => {
    grouped[k] = grouped[k] || {};
    data[k].forEach(a => {
        grouped[k][a.status] = grouped[k][a.status] || [...priorities].reduce((r, b) => (r[b] = 0, r), {});
        grouped[k][a.status][a.priority]++;
    });
});

console.log(grouped);
.as-console-wrapper { max-height: 100% !important; top: 0; }

Upvotes: 1

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