辅音和元音 Swift

Question

我是 Swift 的新手，谁能向我解释为什么我一直遇到这个问题。我正在使用 Xcode 6.4，但这是我的问题，我希望我清除它，但我需要我的函数来接收大字符串，然后返回元组（numVowels，numConsonants） 计算以辅音/元音开头的单词数量 返回元组并打印函数调用的结果。 我不需要它来计算字符，只需要每个单词的第一个字符。我创建了一个 for 循环，它将所有内容都切换为小写。但现在我被困住了。

func count(string: String) -> (Vowels:Int, Consonants:Int) {
    var Vowels = 0, Consonants = 0
    for character in string {
        switch String(character).lowercaseString {
            case "a","e","i","o","u":
              ++Vowels
            case "b","c","d","e","f","g","h","j","k","l","m","n","p","q","r","s","t","v","w","x","y","z":
            ++Consonants
        default: break
        }

    }
    return (Vowels, Consonants)
}

Answer 1

如果您使用的是 swift 2：

只需改变这一行

for character in string {

到

for character in string.characters {

它会工作得很好

count("hello") // (.0 2, .1 3)

Answer 2

问题在于您为方法选择的名称。试试 count 以外的任何名称，应该没问题。

编辑/更新：Swift 4

func countCharacteres(_ string: String) -> (vowels: Int, consonants: Int) {
    var vowels = 0, consonants = 0
    for character in string {
        switch String(character).lowercased() {
        case "a","e","i","o","u":
            vowels += 1
        case "b","c","d","f","g","h","j","k","l","m","n","p","q","r","s","t","v","w","x","y","z":
            consonants += 1
        default:
            break
        }
    }
    return (vowels, consonants)
}

---

let name = "Dromel"
countCharacteres(name).vowels      // 2
countCharacteres(name).consonants  // 4

---

如果你只想计算每个单词的第一个字母，你需要将你的字符串分解成一个单词数组并提取每个第一个字母的小写，如下所示：

extension StringProtocol where Index == String.Index {
    var range: Range<Index> { return startIndex..<endIndex }
    var words: [String] {
        var words: [String] = []
        enumerateSubstrings(in: range, options: .byWords) { word, _, _, _ in words.append(word!) }
        return words
    }
}
extension StringProtocol  where Index == String.Index {
    var firstCharacterCount: (vowels: Int, consonants: Int) {
        var vowels = 0, consonants = 0
        for first in words.compactMap({$0.lowercased().first}) {
            switch first {
            case "a","e","i","o","u":
                vowels += 1
            case "b","c","d","f","g","h","j","k","l","m","n","p","q","r","s","t","v","w","x","y","z":
                consonants += 1
            default: break
            }
        }
        return (vowels, consonants)
    }
}

---

let sentence = "Hello World. Apple."
let counter =  sentence.firstCharacterCount
counter.vowels      // 1
counter.consonants  // 2

请注意，扩展名应放在项目内的新 Swift 文件中。

Answer 3

另一种选择是使用正则表达式来检索每个单词的第一个字母。 \b 选项用于查找单词边界，\w 是单个单词字符。因此，一旦你转义了反斜杠，\\b\\w 将捕获紧跟单词边界的第一个字母（也就是每个单词的第一个字母）：

func countLetters(string: String) -> (vowels: Int, consonants: Int) {
    var vowels = 0, consonants = 0
    let regex = NSRegularExpression(pattern: "\\b\\w", options: nil, error: nil)
    regex?.enumerateMatchesInString(string, options: nil, range: NSMakeRange(0, count(string))) { matches, flags, stop in
        switch (string as NSString).substringWithRange(matches.range).lowercaseString {
        case "a","e","i","o","u":
            ++vowels
        case "b","c","d","f","g","h","j","k","l","m","n","p","q","r","s","t","v","w","x","y","z":
            ++consonants
        default: break
        }
    }
    return (vowels, consonants)
}

--

如果您想处理重音字符（例如“étude”），您可能需要先删除这些变音符号：

func countLetters(string: String) -> (vowels: Int, consonants: Int) {
    let convertedString = NSMutableString(string: string)
    CFStringTransform(convertedString, nil, kCFStringTransformStripCombiningMarks, Boolean(0))

    var vowels = 0, consonants = 0
    let regex = NSRegularExpression(pattern: "\\b\\w", options: nil, error: nil)
    regex?.enumerateMatchesInString(convertedString as String, options: nil, range: NSMakeRange(0, convertedString.length)) { matches, flags, stop in
        switch convertedString.substringWithRange(matches.range).lowercaseString {
        case "a","e","i","o","u":
            ++vowels
        case "b","c","d","f","g","h","j","k","l","m","n","p","q","r","s","t","v","w","x","y","z":
            ++consonants
        default: break
        }
    }
    return (vowels, consonants)
}