如果我有一些类似下面的句子，
"Hello everyone. This is a sentence."
如何使用 Swift 获取这样的数组
var words = ["Hello", "everyone", "This", "is", "a", "sentence"]
我还需要一种方法来记住每个单词在原始字符串中的位置，以及句号和逗号的位置。因此，如果我要将这个单词数组重新转换为相同的字符串，它将读取
"Hello everyone. This is a sentence."
谢谢！
【问题讨论】：
按单词分解的子字符串直接内置在 Cocoa 中：
let s = "Hello everyone. This is a sentence."
var arr = [String]()
s.enumerateSubstringsInRange(s.startIndex..<s.endIndex, options: .ByWords) {
ss, r, r2, stop in
arr.append(ss)
}
// now arr is ["Hello", "everyone", "This", "is", "a", "sentence"]
（但是你的规范的其余部分对我来说毫无意义，所以我省略了它。你可以看看如何计算出单词的来源，因为r 和r2 这两个范围告诉你所有关于它的信息.)
【讨论】：
有了这个，您可以以结构化的方式解析句子，这样您就可以修改单词并能够在任何时候重构完整的句子。
这样做的方法是：
按顺序在包含单词、标点符号和空格的标记列表中解析句子。
将每个标记包装在一个“标记”类中，这样可以更方便地访问，具体来说，可以让您快速查询标记是否是一个单词。
然后，您可以在需要仅包含单词的列表时使用isWord 进行过滤，并对其进行修改。
修改将反映在令牌数组中。所以当你想再次构造完整的句子时，你只需加入tokens数组即可。
实施：
let input = "Hello everyone. This is a sentence."
class Token {
var text: String
var isWord: Bool {
return !(text == " " || text == ".")
}
init(text: String) {
self.text = text
}
}
let options: NSLinguisticTaggerOptions = .OmitOther
let schemes = [NSLinguisticTagSchemeLexicalClass]
let tagger = NSLinguisticTagger(tagSchemes: schemes, options: Int(options.rawValue))
let range = NSMakeRange(0, (input as NSString).length)
tagger.string = input
var parts : [String] = [] // here we put all the parts including spaces and punctuation signs, such that we can reconstruct sentence at any time
tagger.enumerateTagsInRange(
range,
scheme: NSLinguisticTagSchemeLexicalClass,
options: options) {
(tag, tokenRange, _, _) in
let token = (input as NSString).substringWithRange(tokenRange)
parts.append(token)
}
println(parts) //"[Hello, , everyone, ., , This, , is, , a, , sentence, .]"
let tokens = parts.map{Token(text: $0)} // wrap the parts in a data structure to handle data more conveniently
let words = tokens.filter{$0.isWord} // get the tokens with only words, if you need them separatedly.
words[1].text = "world" // manipulate a word - this will reflect in the stored sentence
let text = join("", tokens.map{$0.text}) // "Hello world. This is a sentence."
【讨论】：
基于@matt 的想法，您可以使用enumerateSubstringsInRange 构建一个数组，还可以在数组中添加标点符号作为单独的项目。然后还创建一个函数，当您开始使用数组时，该函数可以使用字符集确定字符串是否为标点符号。 见下文：
func checkCharSet(part:String, cSet:NSCharacterSet) -> Bool{
let check = part.rangeOfCharacterFromSet(cSet)
return (check != nil) ? true : false
}
func isPunctuation(part:String) -> Bool{
let punctSet = NSCharacterSet.punctuationCharacterSet()
return checkCharSet(part, punctSet)
}
func isHexadecimal(part:String) -> Bool{
let hexadecimal = NSCharacterSet.alphanumericCharacterSet()
return checkCharSet(part, hexadecimal)
}
然后利用这两个函数，我们可以相应地构建我们的数组。
let s = "Hello everyone. This is a sentence."
var arr = [String]()
var part:String = ""
s.enumerateSubstringsInRange(s.startIndex..<s.endIndex, options: .ByComposedCharacterSequences) { (
ss, r, r2, stop) -> () in
if isHexadecimal(ss){
part += ss
}else{
arr.append(part)
arr.append(ss)
part = ""
}
}
打印和过滤结果：
let wordsOnly = arr.filter({isHexadecimal($0)})
let punctOnly = arr.filter({isPunctuation($0)})
println("\(arr)")
println("\(wordsOnly)")
println("\(punctOnly)")
【讨论】：
@Ixx，我不认为你的 Token 类很有用。以下是我要做的修改：
import Cocoa
let input = "Hello everyone. This is a sentence."
let options: NSLinguisticTaggerOptions = .OmitOther
let schemes = [NSLinguisticTagSchemeTokenType]
let tagger = NSLinguisticTagger(tagSchemes: schemes, options: Int(options.rawValue))
let range = NSMakeRange(0, (input as NSString).length)
tagger.string = input
var parts: [String] = [] // here we put all the parts including spaces and punctuation signs, such that we can reconstruct sentence at any time
var words: [String] = []
tagger.enumerateTagsInRange(
range,
scheme: NSLinguisticTagSchemeTokenType,
options: options) {
(tag, tokenRange, _, _) in
let part = (input as NSString).substringWithRange(tokenRange)
parts.append(part)
if tag == "Word" {
words.append(part)
}
}
println(parts)
println(words)
--output:--
[Hello, , everyone, ., , This, , is, , a, , sentence, .]
[Hello, everyone, This, is, a, sentence]
或者，为了获得最大的灵活性，您可以将 tokenType 添加到 Token 类并使用 tokenType 属性过滤令牌数组。对于 schemeNSLinguisticTagSchemeTokenType，标记类型是 Word、标点符号、空格或其他——这是 op 需要了解的唯一标记类型。
class Token {
var text: String
var tokenType: String
init(text: String, tokenType: String) {
self.text = text
self.tokenType = tokenType
}
}
let input = "Hello everyone. This is a sentence."
let options: NSLinguisticTaggerOptions = .OmitOther
let schemes = [NSLinguisticTagSchemeTokenType]
let tagger = NSLinguisticTagger(tagSchemes: schemes, options: Int(options.rawValue))
let range = NSMakeRange(0, (input as NSString).length)
tagger.string = input
var parts: [String] = [] // here we put all the parts including spaces and punctuation signs, such that we can reconstruct sentence at any time
//var words: [String] = []
var tokens: [Token] = []
tagger.enumerateTagsInRange(
range,
scheme: NSLinguisticTagSchemeTokenType,
options: options) {
(tag, tokenRange, _, _) in
let part = (input as NSString).substringWithRange(tokenRange)
parts.append(part)
tokens.append(
Token(text: part, tokenType: tag)
)
}
println(parts) //"[Hello, , everyone, ., , This, , is, , a, , sentence, .]"
let words = tokens.filter({$0.tokenType == "Word"})
println(
words.map({$0.text} //[Hello, everyone, This, is, a, sentence]
)
【讨论】：