ios - 将文本拆分为数组，同时保持 Swift 中的标点符号

Question

我想将文本拆分成一个数组，保留由其余单词分隔的标点符号，因此字符串如下:

Hello, I am Albert Einstein.

应该变成这样的数组:

["Hello", ",", "I", "am", "Albert", "Einstein", "."]

我已经尝试使用 sting.components(separatedBy: CharacterSet.init(charactersIn: ",;;:")) 但是这个方法会删除所有标点符号，并返回一个像这样的数组:

["Hello", "I", "am", "Albert", "Einstein"]

那么，我怎样才能像我的第一个例子那样得到一个数组呢？

Answer 1

它作为解决方案并不漂亮，但您可以尝试:

var str = "Hello, I am Albert Einstein."
var list = [String]()
var currentSubString = "";
//enumerate to get all characters including ".", ",", ";", " "
str.enumerateSubstrings(in: str.startIndex..<str.endIndex, options: String.EnumerationOptions.byComposedCharacterSequences) { (substring, substringRange, enclosingRange, value) in
    if let _subString = substring {
        if (!currentSubString.isEmpty &&
            (_subString.compare(" ") == .orderedSame
                || _subString.compare(",") == .orderedSame
                || _subString.compare(".") == .orderedSame
                || _subString.compare(";") == .orderedSame
            )
            ) {
            //create word if see any of those character and currentSubString is not empty
            list.append(currentSubString)
            currentSubString = _subString.trimmingCharacters(in: CharacterSet.whitespaces )
        } else {
            //add to current sub string if current character is not space.
            if (_subString.compare(" ") != .orderedSame) {
                currentSubString += _subString
            }
        }
    }
}


//last word
if (!currentSubString.isEmpty) {
    list.append(currentSubString)
}

在 Swift3 中:

var str = "Hello, I am Albert Einstein."
var list = [String]()
var currentSubString = "";
//enumerate to get all characters including ".", ",", ";", " "
str.enumerateSubstrings(in: str.startIndex..<str.endIndex, options: String.EnumerationOptions.byComposedCharacterSequences) { (substring, substringRange, enclosingRange, value) in
    if let _subString = substring {
        if (!currentSubString.isEmpty &&
            (_subString.compare(" ") == .orderedSame
                || _subString.compare(",") == .orderedSame
                || _subString.compare(".") == .orderedSame
                || _subString.compare(";") == .orderedSame
            )
            ) {
            //create word if see any of those character and currentSubString is not empty
            list.append(currentSubString)
            currentSubString = _subString.trimmingCharacters(in: CharacterSet.whitespaces )
        } else {
            //add to current sub string if current character is not space.
            if (_subString.compare(" ") != .orderedSame) {
                currentSubString += _subString
            }
        }
    }
} 


//last word
if (!currentSubString.isEmpty) {
    list.append(currentSubString)
}

想法是循环所有字符并同时创建单词。一个词是一组连续的字符，不是、、、.或;。所以，在循环创建单词的过程中，如果我们看到其中一个字符，我们就完成了当前单词，并且当前正在构造的单词不为空。 使用您的输入分解步骤:

1. get H(不是空格或其他终端字符) -> currentSubString = "H"
2. get e(不是空格或其他终端字符) -> currentSubString = "他"
3. get l(不是空格或其他终端字符) -> currentSubString = "Hel"
4. get l(不是空格或其他终端字符) -> currentSubString = " hell "
5. get o(不是空格或其他终端字符) -> currentSubString = "你好"
6. get . (是终端字符)
   • -> 因为 currentSubString 不为空，添加到 list 并重新开始构造下一个单词，然后 list = ["Hello"]
   • -> currentSubString = "."(我使用修剪的原因只是为了在我得到这个字符时删除 。但对于其他终端字符，我们必须保留下一个单词。
7. get (是空格字符)
   • -> currentSubString不为空，添加到list中，重新构建-> list = ["Hello", "."]
   • -> currentSubString = ""(修剪)。 ...等等。