ios - 如何从快速字典数组中获取所选键数组的值及其复杂性

Question

假设我收到了以下回复。我想从下面的操作中了解两件事。

1.) 如何在 swift 中使用高阶函数改进/优化以下代码。

2.) 还想了解代码当前的复杂性以及您可能建议的任何优化代码的复杂性。

在下面的响应中，我只想检查 keyToBeChecked 中描述的某些键的值，并且对于该特定键，我需要执行操作。操作完成后，我想向响应添加一个新 key (key6)，如下所示。以下操作对我来说效果很好，这就是我打算做的。我正在寻找上面提到的两件事

var response = [["key1": 1, "key2": 0, "name": "John", "key3": 1, "key4": 1, "place": "Newyork", "key5": 0],
                ["key1": 0, "key2": 1, "name": "Mike", "key3": 1, "key4": 0, "place": "California", "key5": 1],
                ["key1": 1, "key2": 0, "name": "John", "key3": 0, "key4": 1, "place": "Boston", "key5": 1]]

let keysToBeChecked = ["key1", "key2", "key3", "key4", "key5"]

for var item in response{
var dict = [String: String]()
   for(key, value) in item{
       if keysToBeChecked.contains(key){
           dict[key] = "\(value)"
           if dict[key] == "1"{
               //perform required operations
               output
           }
       }
   }

   item["key6"] = output
   response.append(item)
}
print(response)//should print the below


My expected output is

response = [["key1": 1, "key2": 0, "name": "John", "key3": 1, "key4": 1, "place": "Newyork", "key5": 0, "key6": "output"],
   ["key1": 0, "key2": 1, "name": "Mike", "key3": 1, "key4": 0, "place": "California", "key5": 1, "key6": "output"],
   ["key1": 1, "key2": 0, "name": "John", "key3": 0, "key4": 1, "place": "Boston", "key5": 1, "key6": "output"]]

Answer 1

对于高阶函数，你可以尝试，

let result = response.map { (dict) -> [String:Any] in
    var filteredDict = [String:Any]()
    dict.forEach({ (key, value) in
        if keysToBeChecked.contains(key) {
            filteredDict[key] = value
            if (value as? Int) == 1 {
                filteredDict["key6"] = "output"
            }
        }
    })
    return filteredDict
}