如何在Swift中计算给定文本中的句子数？

我想创建一个操场，计算给定文本的句子数

let input = "That would be the text . it hast 3. periods. "

func sentencecount() {
    let periods = CharacterSet.whitespacesAndNewlines.union(.punctuationCharacters)
    let periods = input.components(separatedBy: spaces)
    let periods2  = Int (words.count)
    print ("The Average Sentence length is \(periods2)") 
}
sentencecount()

这应该起作用：

   let input = "That would be the text . it hast 3. periods. "
   let occurrencies = input.characters.filter { $0 == "." || $0 == "?" }.count
   print(occurrencies)
   //result 3

---

只需在

charset

中添加字符，即可区分句子：

我假设

目前：

    let input = "That would be the text. it hast 3? periods."
    let charset = CharacterSet(charactersIn: ".?,")
    let arr = input.components(separatedBy: charset)
    let count = arr.count - 1

此处
arr
为：

["That would be the text", " it hast 3", " periods", ""]

将计数减少1，以获得实际的句子

 <强>注释：< /强>如果您不想考虑<代码>，“< /代码>，则将其从<代码>字符集> />代码> 
到目前为止，我可以看到您需要使用它们进行拆分。并按如下方式修剪空白：

func sentencecount () {

    let result = input.trimmingCharacters(in: .whitespaces).split(separator: ".")

   print ("The Average Sentence length is \(result.count)") // 3
}

祝你好运
 您可以使用
枚举子字符串（在：范围内）
并使用选项

let input = "Hello World !!! That would be the text. It hast 3 periods."
var sentences: [String] = []
input.enumerateSubstrings(in: input.startIndex..., options: .bySentences) { (string, range, enclosingRamge, stop) in
    sentences.append(string!)
}

另一种方法是使用子字符串数组而不是字符串：

var sentences: [Substring] = []
input.enumerateSubstrings(in: input.startIndex..., options: .bySentences) { (string, range, enclosingRamge, stop) in
    sentences.append(input[range])
}


亲爱的Matt我会编辑我的帖子谢谢我会检查；）它说：游乐场执行失败：错误：textcoach.playerd:76:30：错误：不明确地使用'filter'让发生率=input.characters.filter{$0='“|$$0==”？“}.count您有哪个版本的swift？非常感谢您，Ankit。这里我得到的最终解决方案要感谢您：func sentencecount（）->Double{let input=“这就是文本。它有3个句点。”let charset=CharacterSet（charactersIn:“.？，”），let arr=input.components（separatedBy:charset）let arrcount=Double（arr.count）let doubleone:Double=1 let count:Double=arrcount-doubleone打印（“平均句子长度为”，countofwords/count，“Words”）return（count）}@Valentin，welcome:）我一直在等待这个：）–请注意，您可以在：input.startIndex…
中将范围指定为“单边范围”
，进行比较
print(sentences)   // "["Hello World !!! ", "That would be the text. ", "It hast 3 periods."]\n"
print(sentences.count)  // "3\n"