Question

我要删除字符＆＃34; - ＆＃34;并替换为＆＃34;＆＃34;。不过这是一本字典，我不知道怎么做？

var arrays: [String:[String]] = ["2017-08-21":["red"], "2017-08-20":["red"], "2017-08-19":["red"], "2017-08-18":["red"], "2017-08-17":["red"]]
print(arrays)

Answer 1

最简单的解决方案是迭代字典中的键值对，并使用新键和旧值创建一个新字典。

var arrays: [String:[String]] = ["2017-08-21":["red"], "2017-08-20":["red"], "2017-08-19":["red"], "2017-08-18":["red"], "2017-08-17":["red"]]
var formattedDictionary = [String:[String]]()
for (key, value) in arrays {
    formattedDictionary[key.replacingOccurrences(of: "-", with: "")] = value
}
print(formattedDictionary)

Answer 2

Swift 4

随着Swift 4即将来临，我将添加一个Swift 4可行替代方案。

使用Dictionary的新init(uniqueKeysWithValues:)初始化程序，我们可以仅使用原始中的密钥修改键值对替换字典arrays的内容。字典。 E.g：

import Foundation

var arrays = ["2017-08-21": ["red"], 
              "2017-08-20": ["red"], 
              "2017-08-19": ["red"], 
              "2017-08-18": ["red"], 
              "2017-08-17": ["red"]]

arrays = Dictionary(uniqueKeysWithValues: arrays
    .map { ($0.replacingOccurrences(of: "-", with: ""), $1) })

print(arrays)
/* ["20170820": ["red"], 
    "20170817": ["red"], 
    "20170818": ["red"], 
    "20170821": ["red"], 
    "20170819": ["red"]] */

请注意导入Foundation以访问NSString String（通过桥接可用于Swift：s Dictionary。

或者，将上面的Dictionary密钥修改逻辑实现为extension Dictionary { mutating func updateKeys(_ transform: (Key) -> Key) { self = Dictionary(uniqueKeysWithValues: self.map { (transform($0), $1) }) } }的扩展名：

var arrays = ["2017-08-21": ["red"], 
              "2017-08-20": ["red"], 
              "2017-08-19": ["red"], 
              "2017-08-18": ["red"], 
              "2017-08-17": ["red"]]

arrays.updateKeys { $0.replacingOccurrences(of: "-", with: "") }

print(arrays)
/* ["20170820": ["red"], 
    "20170817": ["red"], 
    "20170818": ["red"], 
    "20170821": ["red"], 
    "20170819": ["red"]] */

允许＆＃34;就地＆＃34; （逻辑上，至少）通过一些提供的转换修改字典的键：

transform

请注意，将extension Dictionary { mutating func updateKeys(_ transform: (Key) -> Key, uniquingKeysWith combine: (Value, Value) -> Value) { self = Dictionary(map { (transform($0), $1) }, uniquingKeysWith: combine) } }应用于两个不同的唯一键自然会产生相同的关键结果，因此我们可能希望修改上面的扩展名以将其考虑在内：

(1)

应用例如在以下示例中，(2)和String个键是唯一的，但在应用上述示例中使用的transform后，解析为相同的var arrays = ["2017-08-21": ["red"], // (1) "20170821": ["blue"], // (2) "2017-08-20": ["red"], "2017-08-19": ["red"], "2017-08-18": ["red"], "2017-08-17": ["red"]] arrays.updateKeys({$0.replacingOccurrences(of: "-", with: "") }, uniquingKeysWith: { $0 + $1 }) print(arrays) /* ["20170820": ["red"], "20170817": ["red"], "20170818": ["red"], "20170821": ["red", "blue"], "20170819": ["red"]] */。

Value

在这里，我们选择连接transform数组，以防import datetime # Define hours that different times of day start start_of_afternoon = 12 start_of_evening = 16 start_of_night = 19 # Get current time now = datetime.datetime.now() if now.hour < start_of_afternoon: # before 12pm print ('morning') elif start_of_afternoon <= now.hour and now.hour < start_of_evening: # after 12pm and before 4pm print ('afternoon') elif start_of_evening <= now.hour and now.hour < start_of_night: # after 4pm and before 7pm print ('evening') elif start_of_night <= now.hour: # after 7pm print ('good night')将两个（或更多）键解析为同一个新键。

Answer 3

这是一个快速解决方案：

let dictionaryWithOldKeys: [String : [String]] = ["2017-08-21":["red"], "2017-08-20":["red"], "2017-08-19":["red"], "2017-08-18":["red"], "2017-08-17":["red"]]

var dictionaryWithNewKeys: [String : [String]] = [:]

for item in dictionaryWithOldKeys {    
    dictionaryWithNewKeys.updateValue(item.value, forKey: item.key.replacingOccurrences(of: "-", with: "."))
}

print(dictionaryWithNewKeys)

删除＆＃34; - ＆＃34;字典键中的字符

3 个答案:

Swift 4