如何在字母和数字（或数字和字母）之间分割字符串？

Question

I'm trying to work out a way of splitting up a string in java that follows a pattern like so:

String a = "123abc345def";

The results from this should be the following:

x[0] = "123";
x[1] = "abc";
x[2] = "345";
x[3] = "def";

However I'm completely stumped as to how I can achieve this. Please can someone help me out? I have tried searching online for a similar problem, however it's very difficult to phrase it correctly in a search.

Please note: The number of letters & numbers may vary (eg There could be a string like so '1234a5bcdef')

Answer 1

You could try to split on (?<=\\D)(?=\\d)|(?<=\\d)(?=\\D) , like:

str.split("(?<=\\D)(?=\\d)|(?<=\\d)(?=\\D)");

It matches positions between a number and not-a-number (in any order).

• (?<=\\D)(?=\\d) - matches a position between a non-digit ( \\D ) and a digit ( \\d )
• (?<=\\d)(?=\\D) - matches a position between a digit and a non-digit.

Answer 2

How about:

private List<String> Parse(String str) {
    List<String> output = new ArrayList<String>();
    Matcher match = Pattern.compile("[0-9]+|[a-z]+|[A-Z]+").matcher(str);
    while (match.find()) {
        output.add(match.group());
    }
    return output;
}

Answer 3

You can try this:

Pattern p = Pattern.compile("[a-z]+|\\d+");
Matcher m = p.matcher("123abc345def");
ArrayList<String> allMatches = new ArrayList<>();
while (m.find()) {
    allMatches.add(m.group());
}

The result (allMatches) will be:

["123", "abc", "345", "def"]

Answer 4

使用两种不同的模式： [0-9]*和[a-zA-Z]*并分别将其分割两次。

Answer 5

If you are looking for solution without using Java String functionality (ie split , match , etc.) then the following should help:

List<String> splitString(String string) {
        List<String> list = new ArrayList<String>();
        String token = "";
        char curr;
        for (int e = 0; e < string.length() + 1; e++) {
            if (e == 0)
                curr = string.charAt(0);
            else {
                curr = string.charAt(--e);
            }

            if (isNumber(curr)) {
                while (e < string.length() && isNumber(string.charAt(e))) {
                    token += string.charAt(e++);
                }
                list.add(token);
                token = "";
            } else {
                while (e < string.length() && !isNumber(string.charAt(e))) {
                    token += string.charAt(e++);
                }
                list.add(token);
                token = "";
            }

        }

        return list;
    }

boolean isNumber(char c) {
        return c >= '0' && c <= '9';
    }

This solution will split numbers and 'words', where 'words' are strings that don't contain numbers. However, if you like to have only 'words' containing English letters then you can easily modify it by adding more conditions (like isNumber method call) depending on your requirements (for example you may wish to skip words that contain non English letters). Also note that the splitString method returns ArrayList which later can be converted to String array.

Answer 6

I was doing this sort of thing for mission critical code. Like every fraction of a second counts because I need to process 180k entries in an unnoticeable amount of time. So I skipped the regex and split altogether and allowed for inline processing of each element (though adding them to an ArrayList<String> would be fine). If you want to do this exact thing but need it to be something like 20x faster...

void parseGroups(String text) {
    int last = 0;
    int state = 0;
    for (int i = 0, s = text.length(); i < s; i++) {
        switch (text.charAt(i)) {
            case '0':
            case '1':
            case '2':
            case '3':
            case '4':
            case '5':
            case '6':
            case '7':
            case '8':
            case '9':
                if (state == 2) {
                    processElement(text.substring(last, i));
                    last = i;
                }
                state = 1;
                break;
            default:
                if (state == 1) {
                    processElement(text.substring(last, i));
                    last = i;
                }
                state = 2;
                break;
        }
    }
    processElement(text.substring(last));
}

Answer 7

Didn't use Java for ages, so just some pseudo code, that should help get you started (faster for me than looking up everything :) ).

 string a = "123abc345def";
 string[] result;
 while(a.Length > 0)
 {
      string part;
      if((part = a.Match(/\d+/)).Length) // match digits
           ;
      else if((part = a.Match(/\a+/)).Length) // match letters
           ;
      else
           break; // something invalid - neither digit nor letter
      result.append(part);
      a = a.SubStr(part.Length - 1); // remove the part we've found
 }

Answer 8

这个"d+|D+"不会代替繁琐的工作： "(?<=\\\\D)(?=\\\\d)|(?<=\\\\d)(?=\\\\D)" ？