如何在UITableView中按字母顺序对字典数组进行排序和分类为Swift 3.0中的部分

Question

我正在快速学习，并且有很多类似的字典，

[{“添加”：2017年12月24日，“ first_name”：阿卜杜拉，“电子邮件”：spaeker1@example.com，“ Last_name”：Jaleel，“ place”：印度，“称呼”：Mr}，{“添加“：2017-12-24，” first_name“：Catherine，”电子邮件“：spaeker1@example.com，”姓氏“：Rose，”地点“：印度，”敬礼“：Mrs}，{”添加“：2017- 12-24，“ first_name”：Alok，“ email”：spaeker1@example.com，“ Last_name”：Raj，“ place”：印度，“ salutation”：Mr}，{“添加”：2017-12-24， “ first_name”：达尔文，“电子邮件”：spaeker1@example.com，“ Last_name”：Jose，“ place”：印度，“ salutation”：Mr}]。

在这里，我想根据first_name按字母顺序列出UITableView中的所有发言人。 我可以按字母顺序对字典数组进行排序，但是问题是我想在UITableViewSection中显示字母，并且该部分仅包含仅以该字母开头的名称（例如iPhone的联系人屏幕）。

所以我可以按以下方式转换字典数组，

{a：[{“添加”：2017年12月24日，“ first_name”：阿卜杜拉，“电子邮件”：spaeker1@example.com，“ Last_name”：Jaleel，“ place”：印度，“ salutation”：Mr}， {“添加”：2017-12-24，“名字”：Alok，“电子邮件”：spaeker1@example.com，“姓氏”：Raj，“地点”：印度，“称呼”：先生}]，c：[ {“添加”：2017年12月24日，“ first_name”：凯瑟琳，“电子邮件”：spaeker1@example.com，“ Last_name”：Rose，“ place”：印度，“ salutation”：Mrs}]，d：[ {“添加”：2017-12-24，“名字”：达尔文，“电子邮件”：spaeker1@example.com，“姓氏”：Jose，“地点”：印度，“称呼”：先生}]}。

是否可以如上所述对字典数组进行分组？ 否则，如何在UITableViewSection中按字母顺序列出所有发言人。 请帮我。

Answer 1

您可以遍历字典数组，然后准备结果。

码

let dict = [ ["added": "2017-12-24", "first_name": "Abdullah", "email": "spaeker1@example.com", "Last_name": "Jaleel", "place": "India", "salutation":"Mr"], ["added": "2017-12-24", "first_name": "Catherine", "email": "spaeker1@example.com", "Last_name": "Rose", "place": "India", "salutation":"Mrs"], ["added": "2017-12-24", "first_name": "Alok", "email": "spaeker1@example.com", "Last_name": "Raj", "place": "India", "salutation":"Mr"], ["added": "2017-12-24", "first_name": "Darwin", "email": "spaeker1@example.com", "Last_name": "Jose", "place": "India", "salutation":"Mr"]]

var result: [String:[[String:String]]] = [:]
for d in dict {
   var tempContactArray: [[String:String]] = []
   let firstName = d["first_name"]?.lowercased()
   let firstChar = firstName![0]
   if let data = result[firstChar] {
       tempContactArray = data
   }
   tempContactArray.append(d)
   result[firstChar] = tempContactArray
}
let sortedKeysAndValues = result.sorted(by: { $0.0 < $1.0 })

print(sortedKeysAndValues)

将字符串的扩展写为

extension String {
subscript (i: Int) -> String {
    return String(self[index(startIndex, offsetBy: i)] as Character)
}

}

输出sortedKeysAndValues

[(key: "a", value: [["added": "2017-12-24", "Last_name": "Jaleel", "place": "India", "email": "spaeker1@example.com", "first_name": "Abdullah", "salutation": "Mr"], ["added": "2017-12-24", "Last_name": "Raj", "place": "India", "email": "spaeker1@example.com", "first_name": "Alok", "salutation": "Mr"]]), (key: "c", value: [["added": "2017-12-24", "Last_name": "Rose", "place": "India", "email": "spaeker1@example.com", "first_name": "Catherine", "salutation": "Mrs"]]), (key: "d", value: [["added": "2017-12-24", "Last_name": "Jose", "place": "India", "email": "spaeker1@example.com", "first_name": "Darwin", "salutation": "Mr"]])]

您无法对字典进行排序，因为字典是元素的无序集合。 因此，如果要排序，则必须为数组。

Answer 2

我想您可能会发现通过使用reduce + sorted组合可以发现有用，这样您可以按字母顺序对字典进行分组，并将first_name第一个字符添加为键：

斯威夫特3.0

let arrayOfDict = [ ["added": "2017-12-24", "first_name": "Abdullah", "email": "spaeker1@example.com", "Last_name": "Jaleel", "place": "India", "salutation":"Mr"],["added": "2017-12-24", "first_name": "Catherine", "email": "spaeker1@example.com", "Last_name": "Rose", "place": "India", "salutation":"Mrs"],["added": "2017-12-24", "first_name": "Alok", "email": "spaeker1@example.com", "Last_name": "Raj", "place": "India", "salutation":"Mr"],["added": "2017-12-24", "first_name": "Darwin", "email": "spaeker1@example.com", "Last_name": "Jose", "place": "India", "salutation":"Mr"]]

var result = arrayOfDict.reduce([String: [[String:Any]]]()) { res, element in
  var mutableRes = res
  if let firstLetter = element["first_name"]?.characters.prefix(0) {
    let initial = String(describing: firstLetter).lowercased()
    if mutableRes[initial] == nil {
      mutableRes[initial] = [[String:Any]]()
    }
    mutableRes[initial]?.append(element)
  }
  return mutableRes
}.sorted { return $0.key < $1.key }

迅捷4.0

let arrayOfDict = [ ["added": "2017-12-24", "first_name": "Abdullah", "email": "spaeker1@example.com", "Last_name": "Jaleel", "place": "India", "salutation":"Mr"],["added": "2017-12-24", "first_name": "Catherine", "email": "spaeker1@example.com", "Last_name": "Rose", "place": "India", "salutation":"Mrs"],["added": "2017-12-24", "first_name": "Alok", "email": "spaeker1@example.com", "Last_name": "Raj", "place": "India", "salutation":"Mr"],["added": "2017-12-24", "first_name": "Darwin", "email": "spaeker1@example.com", "Last_name": "Jose", "place": "India", "salutation":"Mr"]]

let result = arrayOfDict.reduce(into: [String: [[String:Any]]]()) { result, element in
  if let firstLetter = element["first_name"]?.first {
    let initial = String(describing: firstLetter).lowercased()
      result[initial, default: [[String:Any]]() ].append(element)
    }}.sorted { return $0.key < $1.key }
print(result)

最终结果将是：

[(key: "a", value: [["added": "2017-12-24", "Last_name": "Jaleel", "place": "India", "email": "spaeker1@example.com", "first_name": "Abdullah", "salutation": "Mr"], ["added": "2017-12-24", "Last_name": "Raj", "place": "India", "email": "spaeker1@example.com", "first_name": "Alok", "salutation": "Mr"]]), (key: "c", value: [["added": "2017-12-24", "Last_name": "Rose", "place": "India", "email": "spaeker1@example.com", "first_name": "Catherine", "salutation": "Mrs"]]), (key: "d", value: [["added": "2017-12-24", "Last_name": "Jose", "place": "India", "email": "spaeker1@example.com", "first_name": "Darwin", "salutation": "Mr"]])]

Answer 3

是否可以如上所述对字典数组进行分组？

let users = [
    ["added": "2017-12-24", "first_name": "Abdullah", "email": "spaeker1@example.com", "Last_name": "Jaleel", "place": "India", "salutation": "Mr"],
    ["added": "2017-12-24", "first_name": "Catherine", "email": "spaeker1@example.com", "Last_name": "Rose", "place": "India", "salutation": "Mrs"],
    ["added": "2017-12-24", "first_name": "Alok", "email": "spaeker1@example.com", "Last_name": "Raj", "place": "India", "salutation": "Mr"],
    ["added": "2017-12-24", "first_name": "Darwin", "email": "spaeker1@example.com", "Last_name": "Jose", "place": "India", "salutation": "Mr"]
]

func firstCharOfFirstName(_ aDict: [String:String]) -> Character {
    return aDict["first_name"]!.first!
}

let groupedUsers = Dictionary(grouping: users, by: firstCharOfFirstName)
print(groupedUsers)

--output:--
["C": 
[["added": "2017-12-24", "Last_name": "Rose", "place": "India", "email": "spaeker1@example.com", "first_name": "Catherine", "salutation": "Mrs"]], 
"D": 
[["added": "2017-12-24", "Last_name": "Jose", "place": "India", "email": "spaeker1@example.com", "first_name": "Darwin", "salutation": "Mr"]], 
"A": 
[["added": "2017-12-24", "Last_name": "Jaleel", "place": "India", "email": "spaeker1@example.com", "first_name": "Abdullah", "salutation": "Mr"],  
["added": "2017-12-24", "Last_name": "Raj", "place": "India", "email": "spaeker1@example.com", "first_name": "Alok", "salutation": "Mr"]]]

然后，您可以根据需要对每个组进行排序。