Dictionary(grouping: people, by: with 2 items in swift 4

Question

I have a data like this

 "geoData": [
  {

    "lat": 50.26877,
    "long": 19.034,
    "msg": "TEST 3",
    "geoRadius": 30.48,
    "pri" : 1
  },
  {
    "lat": 50.26877,
    "long": 19.034,
    "msg": "TEST 2",
    "geoRadius": 30.48,
    "pri" : 2
  },
  {
    "lat": 50.26877,
    "long": 10.034,
    "msg": "TEST 1",
    "geoRadius": 30.48,
    "pri" : 1
  }
  {
    "lat": 50.26877,
    "long": 10.034,
    "msg": "TEST 4",
    "geoRadius": 30.48,
    "pri" : 2
  }
]

Now i want to have a result dictionary as group by latitude and longitude and sort by pri

I tried something like this

        let result = Dictionary(grouping: geoNotifications,
                            by: { (geoNotification)->Bool in
                                return geoNotification.latitude && geoNotification.longitude })

But it doesn't work Can someone share the logic for the same?

Answer 1

In the code you have given, I assume geoNotifications is an array of struct/objects similar to the below code:

struct GeoNotification
{
    var latitude: Double
    var longitude: Double
    var msg: String
    var geoRadius: String
    var pri: Int

    init(dict: [String:Any])
    {
        self.latitude = dict["lat"] as! Double
        self.longitude = dict["long"] as! Double
        self.geoRadius = dict["geoRadius"] as! String
        self.msg = dict["msg"] as! String
        self.pri = dict["pri"] as! Int
    }
}

Having an array of GeoNotification objects, we can achieve both of your requirements as:

1. Grouping using longitude

var dict = Dictionary(grouping: geoNotifications) { $0.longitude }

Result:

["10.034000000000001": [geoNotification1, geoNotification2] , "19.034": [geoNotification3, geoNotification4]]

2. Sorting with pri

for (key, value) in dict
{
    dict[key] = value.sorted(by: { $0.pri < $1.pri })
}

In the above code, I've picked every key-value pair from the dict we got earlier and sorted the value(it's an array of GeoNotification objects) according to pri.