I have an app that allows you to rate media 0-5 stars as integers using Core Data. I have the aggregate fetch request using propertiesToGroupBy to return a count of each media type and rating. The print() result of the request is

[
    ["rating": 0, "count": 17, "type": movie], 
    ["count": 10, "rating": 0, "type": show], 
    ["count": 8, "type": movie, "rating": 1], 
    ["rating": 2, "count": 36, "type": movie], 
    ["rating": 2, "count": 15, "type": show], 
    ["type": movie, "count": 161, "rating": 3], 
    ["rating": 3, "type": show, "count": 26], 
    ["rating": 4, "count": 217, "type": movie], 
    ["count": 48, "rating": 4, "type": show], 
    ["type": movie, "rating": 5, "count": 114], 
    ["count": 63, "rating": 5, "type": show],
    ["count": 8, "rating": 4, "type": web]
]

returned as an NSDictionary/dictionaryResultType. I would like to transform this result into separate integer arrays for each type, with the count for each rating 5-0, even if there is no result for the combination of type and rating. Something like:

ratings = [5, 4, 3, 2, 1, 0]

and then

movie = [114, 217, 161, 36, 8, 17]
show = [63, 48, 26, 15, 0, 10]
web = [0, 8, 0, 0, 0, 0]

Is there an efficient way to do this, like with mapping or grouped dictionaries?

hi Brian,

what you show is an array of dictionaries; and if you have defined the values movie, show, and web to be, say integers, you have something of type [[String : Int]]. so this code works:

let movie = 1 // my assumptions
let show = 2
let web = 3

let myArray = [
    ["rating": 0, "count": 17, "type": movie],
  //
  // and other data as shown above ...
  //
    ["count": 8, "rating": 4, "type": web]
]

// finds the one entry, if any, for the given type and rating and returns 
// the count, else returns 0
func count(type: Int, rating: Int) -> Int {
    let dictionaryEntry = myArray.first(where: { $0["type"] ==  type && $0["rating"] == rating })
    return dictionaryEntry?["count"] ?? 0
}

let ratings = [5,4,3,2,1,0]
print(ratings.map({ count(type: movie, rating: $0) }))
print(ratings.map({ count(type: show, rating: $0) }))
print(ratings.map({ count(type: web, rating: $0) }))

hope that helps,

DMG