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我正在尝试将 k-shingles 中的文本分开，遗憾的是我无法使用扫描仪。如果最后一个瓦太短，我想用“_”填充。我走了这么远：

public class Projektarbeit {

    public static void main(String[] args) {
        testKShingling(7, "ddssggeezzfff");
    }

    public static void testKShingling(int k, String source) {
        //first eliminate whitespace and then fill up with withespaces to match target.length%shingle.length() == 0
        String txt = source.replaceAll("\\s", "");

        //get shingles
        ArrayList<String> shingles = new ArrayList<String>();
        int i;
        int l = txt.length();
        String shingle = "";

        if (k == 1) {
            for(i = 0; i < l; i++){
                shingle = txt.substring(i, i + k);
                shingles.add(shingle);
            };
        }
        else {
            for(i = 0; i < l; i += k - 1){
                try {
                    shingle = txt.substring(i, i + k);
                    shingles.add(shingle);
                }
                catch(Exception e) {
                    txt = txt.concat("_");
                    i -= k - 1;
                };
            };
        }
        System.out.println(shingles);
    }
}

Output: [ddssgge, eezzfff, f______]

它几乎可以工作，但是在示例中使用给定参数的情况下，最后一个瓦不是必需的（它应该是 [ddssgge, eezzfff]

任何想法如何做到这一点更漂亮？

4

1 回答 1

1

要使发布的代码正常工作，您只需添加breakcatch 块的末尾：

catch(Exception e) {
     txt = txt.concat("_");
     i -= k - 1;
      break;
};

话虽如此，我不会使用异常来控制程序。例外就是：应该用于运行时错误。StringIndexOutOfBoundsException通过控制循环参数来避免：

public static void main(String[] args) {
    testKShingling(3, "ddssggeezzfff");
}

public static void testKShingling(int substringLength, String source) {

    //todo validate input
    String txt = source.replaceAll("\\s", "");
    //get shingles
    ArrayList<String> shingles = new ArrayList<>();
    int stringLength = txt.length();

    if (substringLength == 1) {
        for(int index = 0; index < stringLength; index++){
            String shingle = txt.substring(index, index + substringLength);
            shingles.add(shingle);
        };
    }
    else {
        for(int index = 0; index < stringLength -1 ; index += substringLength - 1){
            int endIndex = Math.min(index + substringLength, stringLength);
            String shingle = txt.substring(index, endIndex);
            if(shingle.length() < substringLength){
                shingle = extend(shingle, substringLength);
            }
            shingles.add(shingle);

        };
    }
    System.out.println(shingles);
}

private static String extend(String shingle, int toLength) {

    String s = shingle;
    for(int index = 0; index < toLength - shingle.length(); index ++){
        s = s.concat("_");
    }
    return s;
}

的替代实现testKShingling：

public static void testKShingling(int substringLength, String source) {

    //todo validate input
    String txt = source.replaceAll("\\s", "");
    ArrayList<String> shingles = new ArrayList<>();

    if (substringLength == 1) {
        for(char c : txt.toCharArray()){
            shingles.add(Character.toString(c));
        };
    }
    else {
        while(txt.length() > substringLength) {
            String shingle = txt.substring(0, substringLength); 
            shingles.add(shingle);
            txt = txt.substring(substringLength - 1); //remove first substringLength - 1 chars 
        }

        if(txt.length() < substringLength){  //check the length of what's left 
            txt = extend(txt, substringLength); 
        }
        shingles.add(txt); //add what's left 
    }
    System.out.println(shingles);
}
于 2020-05-15T05:38:29.923 回答