Question

我希望下面的函数将句子分成数组，将问题分成数组并插入＆＃34;，＆＃34; ＆＃34;。＆＃34;和＆＃34;？＆＃34;属于。目前它在同一阵列中打印。关于如何解决这个问题的任何想法？

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()
}

控制台打印出来：

奈德骑自行车7英里到图书馆。他在回家的路上走了一条捷径，只有5英里长。 Ned骑了多少英里？

[句子数量= 2]

[＆＃34; Ned骑自行车7英里到图书馆＆＃34;，＆＃34; \ n他在回家的路上走了一条捷径，只有5英里长＆＃34;，＆＃34; \ n如何Ned骑了多少英里？\ n＆＃34;]

[问题数量= 1]

<＆＃34; Ned骑着他的自行车7英里到了图书馆。\ n他在回家的路上走了一条长达5英里的捷径。\ n Ned骑了多少英里＆＃34;，＆＃34; \ n＆＃34;]

奈德骑自行车7英里到图书馆。他在回家的路上走了一条捷径，只有5英里长。 Ned骑了多少英里？

打印出来

[句子数量= 2]

[问题数量= 1]

句子：[＆＃34; Ned骑自行车7英里到图书馆。他在回家的路上走了一条捷径，只有5英里长。＆＃34;]

问题：[＆＃34; Ned骑了多少英里？＆＃34;]

Answer 1

此代码段可以使用一些重构来替换公共代码，但它可以按原样运行。

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变量包含用户输入中句子的索引。我们希望得到跟踪该特定句子的标点符号，因此我们从其上限开始并找到超过它的下一个索引。使用substring，我们提取标点符号并将其与之比较。还是？。

现在我们的代码将返回true或false，无论我们是否有问题或语句，我们使用filter迭代所有句子的数组并返回一系列问题或陈述。

Answer 2

我建议不要根据角色的存在进行分离，而是使用.bySentences选项进行分离（可以更优雅地处理不会终止句子的标点符号）。然后在您的字符串中迭代一次，附加到适当的数组，例如在Swift 3：

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 3

我制作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 ""
    }
} 
}

快速分离句子和问题的功能

3 个答案: