獲取字典數組中的重復值。 迅速

Question

我有這樣的數組

[
    ["itm_id": 4, "itm_name": Chicken],
    ["itm_id": 4, "itm_name": Chicken],
    ["itm_id": 4, "itm_name": Chicken],
    ["itm_id": 7, "itm_name": Cat]
]

我有此字典數組，我正在嘗試按字典中的值對它們進行分組。 因此，在上面的示例中，我想知道要創建一個字典來知道我有多少個重復鍵的字典：

[["item_id" : 4, count: 3], ["item_id" : 7, count: 1]]

itm_id: 4重復3次，因此計數為3， itm_id: 7僅重復一次。

我該如何實現

Answer 1

我建議您創建項目數組的struct ，而不是像這樣的字典

struct Item{
    var itemID : Int
    var name : String

    init(dictionary:[String:Any]) {
        itemID = dictionary["itm_id"] as? Int ?? 0
        name = dictionary["itm_name"] as? String ?? ""
    }
}

一旦有了Items數組，就可以將特定項ID的元素映射為數組，以獲取計數並將其從數組中刪除。 看看下面的代碼。 這不是最干凈的實現，但是它將幫助您解決問題。

func countDuplicates(){
    let dictionary = [["itm_id": 4, "itm_name": "Chicken"],["itm_id": 4, "itm_name": "Chicken"],["itm_id": 4, "itm_name": "Chicken"],["itm_id": 7, "itm_name": "Cat"]]
    var items = [Item]()
    var countArray = [[String:Any]]()
    dictionary.forEach{
        items.append(Item(dictionary: $0))
    }
    while items.count > 0 {
        if let firstItem = items.first{
            let duplicateItems = items.filter{$0.itemID == firstItem.itemID}
            var countDictionary = [String:Any]()
            countDictionary["itm_id"] = firstItem.itemID
            countDictionary["count"] = duplicateItems.count
            countArray.append(countDictionary)
            items = items.filter{$0.itemID != firstItem.itemID}
        }
    }
    print(countArray)
}

這將打印[["itm_id": 4, "count": 3], ["itm_id": 7, "count": 1]]

我以為“ Chicken ，“ Cat是弦樂器。 如果它們不是字符串而是類類型，則可以將Item結構重寫為這樣的形式

class Animal{}

class Chicken:Animal{}

class Cat:Animal{}

struct Item<T:Animal>{
    var itemID : Int
    var name : String
    var animal : Animal

    init(dictionary:[String:Any],animal:T) {
        itemID = dictionary["itm_id"] as? Int ?? 0
        name = dictionary["itm_name"] as? String ?? ""
        self.animal = animal
    }
}

然后您可以像這樣初始化Item

yourItem = Item(dictionary:yourDictionary,animal:Cat())

Answer 2

// Swift 3.1

選項1：使用字典

func countRepeats() -> [[String: Int]] {
    var repeats = [Int: Int]()
    arr.forEach { item in
        if let id = item["itm_id"] as? Int {
            repeats[id] = repeats[id] == nil ? 1 : repeats[id]! + 1
        }
    }
    return repeats.map {["item_id" : $0.key, "count": $0.value]}
}

print(countRepeats()) // [["item_id": 4, "count": 3], ["item_id": 7, "count": 1]]

選項2：建議使用Struct代替Dictionary

public struct ItemCount {
    let id: Int
    let count: Int
}

func countRepeats() -> [ItemCount] {
    var repeats = [Int: Int]()
    arr.forEach { item in
        if let id = item["itm_id"] as? Int {
            repeats[id] = repeats[id] == nil ? 1 : repeats[id]! + 1
        }
    }
    return repeats.map {ItemCount(id:$0.key, count: $0.value)}
}

print(countRepeats())

Answer 3

我建議在類（或結構）中處理數據，但是如果必須使用字典數組，則這是獲取計數的一種方法：

// extract all the "itm_id" values from all dictionaries in the array
// use a set to identify unique values
// map each unique values to its count in the array

let ids = dictionaries.flatMap{$0["itm_id"] as? Int}
let idCounts = Set(ids).map{ id in (id, ids.filter{$0==id}.count) }

// You can use map() to turn this into yet another array of dictionaries

let idCountDict = idCounts.map{["itm_id":$0 ,"count":$1]}

print(idCounts)     // [(4, 3), (7, 1)]
print(idCountDict)  // [["itm_id": 4, "count": 3], ["itm_id": 7, "count": 1]]