在两个字典数组中查找重复键以使用新数组字典更新旧数组字典

Posted 2023-03-12

【中文标题】在两个字典数组中查找重复键以使用新数组字典更新旧数组字典

【英文标题】：Find duplicate keys in Two Arrays of Dictionaries to update old Array dictionary with new Array dictionary

【发布时间】：2017-03-29 18:28:55

【问题描述】：

我想用新词典更新已保存的词典。如果不存在值，则需要将新值附加到字典中。还使用字典的新值更新旧的现有值。

我保存的旧字典数组：

 var savedDict = [
             ["id":"1","pic":"Alice.png","name":"Alice Smith","position":"Nurse"],
             ["id":"2","pic":"brad.png","name":"Brad Smith MD","position":"Primary Doctor"],
             ["id":"12","pic":"bob.png","name":"Bob Smith PhD","position":"Hospital Coordinator"],
            ]

新下载的字典数组：

let newDict =   [
             ["id":"1","pic":"alice.png","name":"Alice Smith","position":"Nurse"],
             ["id":"2","pic":"brad.png","name":"Brad Smith MD","position":"Primary Doctor"],
             ["id":"3","pic":"user.png","name":"Dr. Quam","position":"Immunologist"],
             ["id":"4","pic":"jennifer.jpg","name":"Jennifer Johnson","position":"Case Manager"],
             ["id":"5","pic":"user.png","name":"John Banks MD","position":"Cardiologist"],
             ["id":"6","pic":"tammie.png","name":"Tammie Summers","position":"Case Manager"]
            ]

我想： 1.将“Alice.png”更新为“alice.png”，“id”=1， 2.ignore "id" = 2 因为两个字典数组相同， 3. 将 newDict 中的所有新项目追加到 savedDict

最终的字典应该是：

 upToDateDict = [
             ["id":"1","pic":"alice.png","name":"Alice Smith","position":"Nurse"],
             ["id":"2","pic":"brad.png","name":"Brad Smith MD","position":"Primary Doctor"],
             ["id":"12","pic":"bob.png","name":"Bob Smith PhD","position":"Hospital Coordinator"],
             ["id":"3","pic":"user.png","name":"Dr. Quam","position":"Immunologist"],
             ["id":"4","pic":"jennifer.jpg","name":"Jennifer Johnson","position":"Case Manager"],
             ["id":"5","pic":"user.png","name":"John Banks MD","position":"Cardiologist"],
             ["id":"6","pic":"tammie.png","name":"Tammie Summers","position":"Case Manager"]
            ]

到目前为止我已经试过了：

 import UIKit
 import Foundation

 func arrayContains(array:[[String:String]], value:[String:String]) -> Bool 
     for item in array 
         if item == value 
             return true
         
     
     return false
 

 var upToDateDict = savedDict//Array<Dictionary<String,String>>()

 //Save all of the dictionaries from the 1st array (savedDict) that aren't in the 2nd array (newDict)
 for item in newDict 
     if !arrayContains(array: savedDict, value: item) 
         upToDateDict.append(item)
     
 

 print(upToDateDict)

 //find duplicate keys and check if need to update saved dict with new dict

 let key = "id"

 for dict1 in savedDict 
     if let value = dict1[key] 
         for dict2 in newDict 
             if dict2[key] == value  //if true duplicate "id" key's  found

                 if dict1 != dict2  //not duplicate for all keys, so update saved with new dictionary

                     print("found \(key):\(value) in both arrays")
                     print("dict1:\(dict1)")
                     print("dict2:\(dict2)")

                     savedDict[dict1] = dict2 //what I want to do but will not build 

                  else //all keys duplicate in dict1 and dict2

             
         
     
 

我只想用 new 覆盖这个字典，所以我将 ["id":"1","pic":"Alice.png"... 更新为 ["id":"1","pic": “alice.png”...但这不起作用：

 savedDict[dict1] = dict2 

【问题讨论】：

你应该创建自定义的Class 而不是这样的字典。之后一切都会容易得多。

自定义类会是什么样子？

检查我的答案

【参考方案1】：

你应该创造

class Employee 
    var id: String
    var pic: String
    var name: String
    var position: String

    init(id: String, pic: String, name: String, position: String) 
        self.id = id
        self.pic = pic
        self.name = name
        self.position = position
    

那么你就有了

var oldEmployes: [Employee] = [Employee(id: "1", pic: "pic.png", name: "Name", position: "Position"), 
                               Employee(id: "2", pic: "pic2.png", name: "Name2", position: "Position2")]
var newEmployes: [Employee] = [Employee(id: "2", pic: "pic21.png", name: "Name21", position: "Position2"), 
                               Employee(id: "3", pic: "pic3.png", name: "Name3", position: "Position3")]

更新数组做一些类似的事情（使用高级函数可以更美观，但我认为这样看起来很清楚）

for employee in newEmployes 
     var isNew = true
     for oldEmployee in oldEmployes 
           if oldEmployee.id == employee.id 
               oldEmployee.name = employee.name
               oldEmployee.pic = employee.pic
               oldEmployee.position = employee.position
               isNew = false
           
     
     if isNew 
         oldEmployes.append(employee)
     

【讨论】：

你应该使Employee符合compere协议，这样比较员工实例或将它们放入数组中会更容易

如果您决定将class 更改为struct，请小心，因为这样您就不能只更新oldEmployee 对象 - 因为struct 是按值处理的，而不是按引用处理的

谢谢，这看起来不错。你能告诉我一个如何用上面的数据初始化 oldEmployes 的例子吗？

该协议被命名为Equatable 我在第一条评论中提到（developer.apple.com/reference/swift/equatable）

@Brian: 看看init(id: String, pic: String, name: String, position: String) 哪里可以let emp1 = Employee(id: "1", pic: "abc.jpg", name: "def", position: "efg")