Question

不确定是否需要减少，排序，过滤或使用较新的独特方法。我什至尝试了Equatable解决方案。

我需要自动识别具有匹配值的键，并仅采用那些键并将其放入新的字典或数组中。

var userDB: [String:Any] = ["userID1":"gameIDa", "userID2":"gameIDa", "userID3":"gameIDc", "userID4":"gameIDd", "userID5":"gameIDe"]

如您所见，只有这两个ID具有相同的gameIDa值。我需要此结果的输出。

// [userID1, userID2]

谢谢

Answer 1

您可以使用Dictionary(grouping:by:)轻松实现此目的，然后将其简化为仅包含具有多个值的条目的字典。

var userDB: [String: String] = ["userID1":"gameIDa", "userID2":"gameIDb", "userID3":"gameIDc", "userID4":"gameIDa", "userID5":"gameIDe"]

let dict = Dictionary(grouping: userDB.keys) { userDB[$0] ?? "" }.reduce(into: [String:[String]](), { (result, element) in
    if element.value.count > 1 {
        result[element.key] = element.value
    }
})

dict

[“ gameIDa”：[“ userID1”，“ userID4”]]

Answer 2

首先，为了能够比较值，Dictionary Value类型必须是Equatable类型。完成此操作后，您可以轻松过滤保存查询值的键：

extension Dictionary where Value: Equatable {
    func keysWithCommonValues() -> [Key] {
        // go over all keys and keep ones for which the dictionary has one another key with the same value
        return keys.filter { key in contains { $0.key != key && $0.value == self[key] } }
    }
}

// userDB is not inferred as [String:String]
var userDB = ["userID1":"gameIDa", "userID2":"gameIDa", "userID3":"gameIDc", "userID4":"gameIDd", "userID5":"gameIDe"]
print(userDB.keysWithCommonValues()) // ["userID1", "userID2"]

筛选/识别具有匹配值的字典键

2 个答案: