將字符串拆分爲具有特定長度的組

Question

如何將Swift中的給定String拆分爲給定長度的組，從右向左讀取？

例如，我有字符串123456789和組長度爲3.字符串應該分爲3組：123,456,789。字符串1234567將分爲1，234，567

所以，你可以寫在斯威夫特一些不錯的代碼：

func splitedString(string: String, length: Int) -> [String] { 

} 

BTW試過功能split()，但據我所知它找到了一些符號

只能

Answer 1

只是爲了我的條目添加到這個非常擁擠的比賽（SwiftStub）：

func splitedString(string: String, length: Int) -> [String] { 
    var result = [String]() 

    for var i = 0; i < string.characters.count; i += length { 
     let endIndex = string.endIndex.advancedBy(-i) 
     let startIndex = endIndex.advancedBy(-length, limit: string.startIndex) 
     result.append(string[startIndex..<endIndex]) 
    } 

    return result.reverse() 
} 

---

或者，如果你感覺功能-Y：

func splitedString2(string: String, length: Int) -> [String] { 
    return 0.stride(to: string.characters.count, by: length) 
     .reverse() 
     .map { 
      i -> String in 
      let endIndex = string.endIndex.advancedBy(-i) 
      let startIndex = endIndex.advancedBy(-length, limit: string.startIndex) 
      return string[startIndex..<endIndex] 
     } 
} 

Answer 2

這就是我從頭頂上想出來的。我敢打賭，有一個更好的方式來做這件事，所以我鼓勵你繼續嘗試。

func splitedString(string: String, length: Int) -> [String] { 
    var groups = [String]() 
    var currentGroup = "" 
    for index in string.startIndex..<string.endIndex { 
     currentGroup.append(string[index]) 
     if currentGroup.characters.count == 3 { 
      groups.append(currentGroup) 
      currentGroup = "" 
     } 
    } 

    if currentGroup.characters.count > 0 { 
     groups.append(currentGroup) 
    } 

    return groups 
} 

這裏是我的測試

let firstString = "123456789" 
let groups = splitedString(firstString, length: 3) 
// Returned ["123", "456", "789"] 

let secondString = "1234567" 
let moreGroups = splitedString(secondString, length: 3) 
// Returned ["123", "456", "7"] 

Answer 3

我做了這樣的事情，不能更好地創造任何東西看，但其結果的問題相匹配：

func splitedString(string: String, lenght: Int) -> [String] { 
    var result = [String](), count = 0, line = "" 
    for c in string.characters.reverse() { 
     count++; line.append(c) 
     if count == lenght {count = 0; result.append(String(line.characters.reverse())); line = ""} 
    } 
    if !line.isEmpty {result.append(String(line.characters.reverse()))} 
    return result.reverse() 
} 

Answer 4

這裏是一個版本使用NSRegularExpressions

func splitedString(string: String, length: Int) -> [String] { 
    var groups = [String]() 
    let regexString = "(\\d{1,\(length)})" 
    do { 
     let regex = try NSRegularExpression(pattern: regexString, options: .CaseInsensitive) 
     let matches = regex.matchesInString(string, options: .ReportCompletion, range: NSMakeRange(0, string.characters.count)) 
     let nsstring = string as NSString 
     matches.forEach { 
      let group = nsstring.substringWithRange($0.range) as String 
      groups.append(group) 
     } 
    } catch let error as NSError { 
     print("Bad Regex Format = \(error)") 
    } 

    return groups 
} 

Answer 5

這是另一個功能編程版本。

extension String{ 
    func splitedString(length: Int) -> [String]{ 
     guard length > 0 else { return [] } 
     let range = 0..<((characters.count+length-1)/length) 
     let indices = range.map{ length*$0..<min(length*($0+1),characters.count) } 
     return indices 
       .map{ characters.reverse()[$0.startIndex..<$0.endIndex] } 
       .map(String.init) 
    } 
} 

"1234567890".splitedString(3) 

Answer 6

有可能是一個更好的解決方案，但這個工程：

func splitedString(string: String, length: Int) -> [String] { 
    let string = Array(string.characters) 
    let firstGroupLength = string.count % length 
    var result: [String] = [] 
    var group = "" 

    if firstGroupLength > 0 { 
     for i in 0..<firstGroupLength { 
      group.append(string[i]) 
     } 
     result.append(String(group)) 
     group = "" 
    } 

    for i in firstGroupLength..<string.count { 
     group.append(string[i]) 
     if group.characters.count == length { 
      result.append(group) 
      group = "" 
     } 
    } 
    return result 
} 

splitedString("abcdefg", length: 2) // ["a", "bc", "de", "fg"] 
splitedString("1234567", length: 3) // ["1", "234", "567"] 

Answer 7

另一種解決方案使用字符串：

func splitStringByIntervals(str: String, interval: Int) -> [String] { 

    let st = String(str.characters.reverse()) 
    let length = st.characters.count 
    var groups = [String]() 

    for (var i = 0; i < length; i += interval) { 
     groups.append((st as NSString).substringWithRange(NSRange(location: i, length: min(interval, length - i)))) 
    } 

    return groups.map{ String($0.characters.reverse())}.reverse() 
} 

輸出爲：

for element in splitStringByIntervals("1234567", interval: 3) { 
    print(element) 
} 

是：

1 
234 
567 

Answer 8

斯威夫特4

我認爲擴展方法更有用。

extension String{ 

    public func splitedBy(length: Int) -> [String] { 

     var result = [String]() 

     for i in stride(from: 0, to: self.characters.count, by: length) { 
      let endIndex = self.index(self.endIndex, offsetBy: -i) 
      let startIndex = self.index(endIndex, offsetBy: -length, limitedBy: self.startIndex) ?? self.startIndex 
      result.append(String(self[startIndex..<endIndex])) 
     } 

     return result.reversed() 

    } 

} 

使用的例子：

Swift.debugPrint("123456789".splitedBy(length: 4)) 
// Returned ["1", "2345", "6789"]