将句子和问题迅速分开的函数

Question

我希望下面的函数将句子分隔成数组，将问题分隔到数组中，并插入"，“。还有"?“属于自己。目前，它正在同一数组中同时打印。有什么办法解决这个问题吗？

func separateAllSentences() {

    // needs to print just the sentences
    func separateDeclarations() { // AKA Separate sentences that end in "."
        if userInput.range(of: ".") != nil { // Notice how lowercased() wasn't used
            numSentencesBefore = userInput.components(separatedBy: ".") // Hasn't subtracted 1 yet
            numSentencesAfter = numSentencesBefore.count - 1
            separateSentencesArray = Array(numSentencesBefore)
            print("# Of Sentences = \(numSentencesAfter)")
            print(separateSentencesArray)
        } else {
            print("There are no declarations found.")
        }
    }

    // needs to print just the questions
    func separateQuestions() { // Pretty Self Explanitory
        if userInput.range(of: "?") != nil {
            numQuestionsBefore = userInput.components(separatedBy: "?")
            numQuestionsAfter = numQuestionsBefore.count - 1
            separateQuestionsArray = Array(numQuestionsBefore)
            print("# Of Questions = \(numQuestionsAfter)")
            print(separateQuestionsArray)
        } else {
            print("There are no questions found. I have nothing to solve. Please rephrase the work to solve as a question.")
        }
    }

    // TODO: - Separate Commas
    func separateCommas() {

    }

    separateDeclarations()
    separateQuestions()
}

控制台输出：

奈德骑着自行车去图书馆7英里。他在回家的路上抄了一条捷径，那条路只有5英里长。内德总共骑了多少英里？

句子#=2

“奈德骑自行车7英里到图书馆”，“他在回家的路上抄了一条只有5英里长的捷径”，“内德总共骑了多少英里？”

问题#=1

奈德骑自行车7英里到图书馆。\n他在回家的路上抄了一条捷径，只有5英里长。\n内德总共骑了多少英里？

奈德骑着自行车去图书馆7英里。他在回家的路上抄了一条捷径，那条路只有5英里长。内德总共骑了多少英里？

它应该打印出

句子#=2

问题#=1

句子：“奈德骑自行车去图书馆7英里。他在回家的路上抄了一条捷径，那里只有5英里长。”

问题：“内德总共骑了多少英里？”

Answer 1

我建议不要基于字符的存在而分开，而是使用.bySentences选项(它可以更优雅地处理不终止句子的标点符号)。然后遍历一次字符串，并附加到适当的数组，例如在Swift 3中：

var questions  = [String]()
var statements = [String]()
var unknown    = [String]()

let string = "Ned deployed version 1.0 of his app. He wrote very elegant code. How much money did Ned make?"

string.enumerateSubstrings(in: string.startIndex ..< string.endIndex, options: .bySentences) { string, _, _, _ in
    if let sentence = string?.trimmingCharacters(in: .whitespacesAndNewlines), let lastCharacter = sentence.characters.last {
        switch lastCharacter {
        case ".":
            statements.append(sentence)
        case "?":
            questions.append(sentence)
        default:
            unknown.append(sentence)
        }
    }
}

print("questions:  \(questions)")
print("statements: \(statements)")
print("unknown:    \(unknown)")

Answer 2

这个片段可以使用一些重构来替换通用代码，但它的工作原理是这样的。

let punctuation = CharacterSet(charactersIn: ".?")
let sentences = userInput.components(separatedBy: punctuation)

let questions = sentences.filter {
  guard let range = userInput.range(of: $0) else { return false }
  let start = range.upperBound
  let end   = userInput.index(after: start)
  let punctuation = userInput.substring(with: Range(uncheckedBounds: (start, end)))
  return punctuation == "?"
}
let statements = sentences.filter {
  guard let range = userInput.range(of: $0) else { return false }
  let start = range.upperBound
  let end   = userInput.index(after: start)
  let punctuation = userInput.substring(with: Range(uncheckedBounds: (start, end)))
  return punctuation == "."
}

首先查看闭包，range变量在用户输入中包含句子的索引。我们想让标点符号尾随这个特定的句子，所以我们从它的上界开始，并找到下一个索引超过它。使用substring，我们提取标点符号并将其与任一标点符号进行比较。或者？

现在，无论我们有疑问句还是语句句，我们都有返回真或假的代码，我们使用filter来迭代所有句子的数组，并且只返回一组问题或语句。

Answer 3

我做了一个最简单的解决方案

​

func separateDeclarations(userInput: String) { // AKA Separate sentences that end in "."
let array = userInput.components(separatedBy: ".")
let arrayQ = userInput.components(separatedBy: "?")

arrayQ.map { (string) -> String in
    if let question = string.components(separatedBy: ".").last {
        return question
    } else {
        return ""
    }
}

array.map { (string) -> String in
    if let question = string.components(separatedBy: "?").last {
        return question
    } else {
        return ""
    }
} 
}