Question

我目前正在使用Xcode 10.1，并试图计算给定句子中的元音和辅音数量。我全局声明了常量

let sentence = "Here is my sentence"

然后，我尝试使用名称为"sentenceInput"的参数调用函数，该函数在某些情况下将语句添加到函数中，这意味着对元音和辅音的数量进行计数并返回Int值。但是，在调用该函数时，我被告知只有1个辅音和0个元音。一般来说，对于编程和Xcode还是陌生的，我将非常感谢您的帮助。谢谢。功能代码：

func findVowelsConsonantsPunctuation(sentenceInput: String) -> (Int, Int, Int) {
    var vowels = 0; var consonants = 0; var punctuation = 0
    for character in sentenceInput.characters{
        switch String(character).lowercased
        {
    case "a", "e", "i", "o", "u":
        vowels += 1
    case ".","!",":",";","?", " ", "'", "":
        punctuation += 1
    default:
        consonants += 1
        }
    return (vowels, consonants, punctuation)
    }
}

Answer 1

请在下面尝试。您的for循环中有return条语句。因此，您将在第一次迭代后返回。我也将您的lowercased()移到了for循环上方。这样，每次迭代时小写字符的处理时间就会减少。

func findVowelsConsonantsPunctuation(sentenceInput: String) -> (Int, Int, Int) {
    var vowels = 0; var consonants = 0; var punctuation = 0
    sentenceInput = sentenceInput.lowercased()
    for character in sentenceInput.characters {
        switch character {
        case 'a', 'e', 'i', 'o', 'u':
            vowels += 1
        case '.','!',':',';','?', ' ', '\'', '':
            punctuation += 1
        default:
            consonants += 1
        }

    }
    return (vowels, consonants, punctuation)
}

Answer 2

您将return语句放入循环中，因此，第一次执行循环代码时，该函数将返回。这就是为什么只有一个辅音的原因，因为switch case语句仅执行一次。

您需要将return语句放在循环的外部之外，如下所示：

func findVowelsConsonantsPunctuation(sentenceInput: String) -> (Int, Int, Int) {
    var vowels = 0
    var consonants = 0
    var punctuation = 0
    for character in sentenceInput.characters {
        switch String(character).lowercase {
    case "a", "e", "i", "o", "u":
        vowels += 1
    case ".","!",":",";","?", " ", "'", "":
        punctuation += 1
    default:
        consonants += 1
        }
    }
    return (vowels, consonants, punctuation)
}

Answer 3

我建议您阅读Set。

请记住，您可以使用CharacterSet并创建3套。

// Make your vowels
let vowels = CharacterSet(charactersIn: "aeiouy")
let consonants = CharacterSet.letters.subtracting(vowels)
let punctuation = CharacterSet.punctuationCharacters

然后，您要跟踪元音，辅音和标点的计数：

// Make your vowels
let vowels = CharacterSet(charactersIn: "aeiouy")
let consonants = CharacterSet.letters.subtracting(vowels)
let punctuation = CharacterSet.punctuationCharacters

然后遍历它。

func sentenceComponents(string: String) -> (Int, Int, Int) {
    // Make your vowels
    let vowels = CharacterSet(charactersIn: "aeiouy")
    let consonants = CharacterSet.letters.subtracting(vowels)
    let punctuation = CharacterSet.punctuationCharacters

    // Set up vars to track our coutns
    var vowelCount = 0
    var consonantCount = 0
    var punctuationCount = 0

    string.forEach { char in
        // Make a set of one character
        let set = CharacterSet(charactersIn: String(char))
        // If the character is a member of a set, incremennt the relevant var
        if set.isSubset(of: vowels) { vowelCount += 1 }
        if set.isSubset(of: consonants) { consonantCount += 1 }
        if set.isSubset(of: punctuation) { punctuationCount += 1 }
    }

    return (vowelCount, consonantCount, punctuationCount)
}

let testString = "The quick brown fox jumped over the lazy dog."

sentenceComponents(string: testString)

更新：更整洁易读（也许？）

我不能忍受未标记的元组，所以这里有一个typealias的更新，它告诉您您所要获得的内容，而不必越过河流和穿过树林来弄清楚元组中的内容：

typealias SentenceComponents = (vowels: Int, consonants: Int, punctuation: Int)

func components(in string: String) -> SentenceComponents {
    // Make your vowels
    let vowels = CharacterSet(charactersIn: "aeiouy")
    let consonants = CharacterSet.letters.subtracting(vowels)
    let punctuation = CharacterSet.punctuationCharacters

    // Set up vars to track our coutns
    var vowelCount = 0
    var consonantCount = 0
    var punctuationCount = 0

    string.forEach { char in
        // Make a set of one character
        let singleCharSet = CharacterSet(charactersIn: String(char))
        // If the character is a member of a set, incremennt the relevant var
        if singleCharSet.isSubset(of: vowels) { vowelCount += 1 }
        if singleCharSet.isSubset(of: consonants) { consonantCount += 1 }
        if singleCharSet.isSubset(of: punctuation) { punctuationCount += 1 }
    }

    return (vowelCount, consonantCount, punctuationCount)
}

let testString = "Smokey, this is not 'Nam. This is bowling. There are rules."

// (vowels 17, consonants 27, punctuation 5)
components(in: testString)

Answer 4

func findVowelsConsonants (_ sentence: String)-> (Int, Int){
            let sentenceLowercase = sentence.lowercased()
            var tuple = (numberOfVowels: 0, numberOfConsonants: 0, numberOfPunctuation: 0)
            for char in sentenceLowercase{
                switch char{
                case "a", "e", "i", "o", "u":
                    tuple.0 += 1
                case "b", "c", "d", "f", "g", "h", "j", "k", "l", "m", "n", "p", "q", "r", "s", "t", "v", "w", "x", "y", "z":
                    tuple.1 += 1
                default:
                    tuple.2 += 1
                }
            }
            return (tuple.0, tuple.1)
        }

如何计算元音和辅音

4 个答案: