在JAVA中使用Regex拆分复杂的字符串

时间:2020-12-24 21:42:42

I want to Split following strings to an array with Regex in JAVA but I don't know how to do.

我想在JAVA中使用Regex将字符串拆分为数组,但我不知道该怎么做。

string1="advmod(likes-4, also-3)"  ==>    advmod, likes, also
string2="nsubj(likes-4, dog24-2)"  ==>    bsubj, likes, dog24
string3="num(dog24-3, 8-2)"        ==>    num, dog24, 8

Please help me to do this work? how split the string like "num(dog24-3, 8-2)" in three tokens num, dog24 and 8 and then putting they to an string array.

请帮我做这个工作?如何在三个标记num,dog24和8中将字符串分为“num(dog24-3,8-2)”,然后将它们放入字符串数组中。

Thanks a lot.

非常感谢。

4 个解决方案

#1


2  

This is generic:

这是通用的:

String string[] = {"advmod(likes-4, also-3)",// ==> advmod , likes , also
        "nsubj(likes-4, dog24-2)",// ==> bsubj , likes , dog24
        "num(dog24-3, 8-2)"};//==> num ,dog24 , 8
Pattern p = Pattern.compile("(\\w+)\\(([^-]+).*, ([^-]+)");
for (int i = 0; i < string.length; i++) {
    Matcher m = p.matcher(string[i]);
    while(m.find()) {
        System.out.print(i+": ");
        for(int j=1; j<= m.groupCount(); j++){
            System.out.print(m.group(j));
            if(j!=m.groupCount()) {
                System.out.print(", ");
            }
        }
        System.out.println("");
    }
}

Hope this helps, it works for me.

希望这有帮助,它对我有用。

This is the output:

这是输出:

0: advmod, likes, also
1: nsubj, likes, dog24
2: num, dog24, 8

#2


1  

For 3rd String

对于第三串

  String re1="(num)";   // Word 1
    String re2=".*?";   // Non-greedy match on filler
    String re3="(dog24)";   // Alphanum 1
    String re4=".*?";   // Non-greedy match on filler
    String re5="(8)";   // Integer Number 1

    Pattern p = Pattern.compile(re1+re2+re3+re4+re5,Pattern.CASE_INSENSITIVE | Pattern.DOTALL);
    Matcher m = p.matcher(txt);
    if (m.find())
    {
        String word1=m.group(1);
        String alphanum1=m.group(2);
        String int1=m.group(3);
        System.out.print("("+word1.toString()+")"+"("+alphanum1.toString()+")"+"("+int1.toString()+")"+"\n");
    }

#3


1  

You if you want to split, you could use this:

如果你想拆分,你可以使用这个:

str.split("\\(|-[0-9]+(?:,\\s+|\\))");

ideone demo.

#4


0  

You really haven't described your grammar, but assuming that it's something like looks like a Java method or a Prolog statement, try

你真的没有描述你的语法,但假设它看起来像Java方法或Prolog语句,试试吧

final static String TOKEN_CHARACTERS="[\w\d-]"
final Pattern p = Pattern.compile("^(" + TOKEN_CHARACTERS + "+)\((" + TOKEN_CHARACTERS + "+,\s*(" + TOKEN_CHARACTERS + ")\)$";

Then split on the -; I presume that it really is there for some reason, and it's not clear that it's always present (if so, you can change the pattern to hard-code the single - instead of considering it part of the token). If you allow additional space or such, adjust accordingly.

然后拆分 - ;我认为它确实存在是出于某种原因,并且不清楚它是否总是存在(如果是这样,你可以改变模式以硬编码单一 - 而不是将其视为令牌的一部分)。如果您允许额外的空间等,请相应调整。

#1


2  

This is generic:

这是通用的:

String string[] = {"advmod(likes-4, also-3)",// ==> advmod , likes , also
        "nsubj(likes-4, dog24-2)",// ==> bsubj , likes , dog24
        "num(dog24-3, 8-2)"};//==> num ,dog24 , 8
Pattern p = Pattern.compile("(\\w+)\\(([^-]+).*, ([^-]+)");
for (int i = 0; i < string.length; i++) {
    Matcher m = p.matcher(string[i]);
    while(m.find()) {
        System.out.print(i+": ");
        for(int j=1; j<= m.groupCount(); j++){
            System.out.print(m.group(j));
            if(j!=m.groupCount()) {
                System.out.print(", ");
            }
        }
        System.out.println("");
    }
}

Hope this helps, it works for me.

希望这有帮助,它对我有用。

This is the output:

这是输出:

0: advmod, likes, also
1: nsubj, likes, dog24
2: num, dog24, 8

#2


1  

For 3rd String

对于第三串

  String re1="(num)";   // Word 1
    String re2=".*?";   // Non-greedy match on filler
    String re3="(dog24)";   // Alphanum 1
    String re4=".*?";   // Non-greedy match on filler
    String re5="(8)";   // Integer Number 1

    Pattern p = Pattern.compile(re1+re2+re3+re4+re5,Pattern.CASE_INSENSITIVE | Pattern.DOTALL);
    Matcher m = p.matcher(txt);
    if (m.find())
    {
        String word1=m.group(1);
        String alphanum1=m.group(2);
        String int1=m.group(3);
        System.out.print("("+word1.toString()+")"+"("+alphanum1.toString()+")"+"("+int1.toString()+")"+"\n");
    }

#3


1  

You if you want to split, you could use this:

如果你想拆分,你可以使用这个:

str.split("\\(|-[0-9]+(?:,\\s+|\\))");

ideone demo.

#4


0  

You really haven't described your grammar, but assuming that it's something like looks like a Java method or a Prolog statement, try

你真的没有描述你的语法,但假设它看起来像Java方法或Prolog语句,试试吧

final static String TOKEN_CHARACTERS="[\w\d-]"
final Pattern p = Pattern.compile("^(" + TOKEN_CHARACTERS + "+)\((" + TOKEN_CHARACTERS + "+,\s*(" + TOKEN_CHARACTERS + ")\)$";

Then split on the -; I presume that it really is there for some reason, and it's not clear that it's always present (if so, you can change the pattern to hard-code the single - instead of considering it part of the token). If you allow additional space or such, adjust accordingly.

然后拆分 - ;我认为它确实存在是出于某种原因,并且不清楚它是否总是存在(如果是这样,你可以改变模式以硬编码单一 - 而不是将其视为令牌的一部分)。如果您允许额外的空间等,请相应调整。