Question

大家好，我有一个代码来从解析结果中获取句子的第一个名词。我写了以下代码。但似乎有些问题。如果语句没有中断for循环。任何人都可以帮我解决一下吗？

提前致谢。

public static String Find_Noun(Parse p)
{    
  label: 
    for(Parse k:p.getChildren())
    {   
        if((k.getType()).equals("NN")||(k.getType()).equals("NNP")||
           (k.getType()).equals("NNS")||(k.getType()).equals("NNPS"))
        {
            noun=k.toString();              
            System.out.println(noun);
            break label;            // I am aware that label is not needed,
                                    // but it doesn't work either way.
        }
        else
        {
            System.out.println("else  "+k);
            Find_Noun(k);
        }
    }
    return noun;
}

输入：

成为\ VBD a \ DT常规\ JJ客户\ NN of \ IN \ \ _ \ n \ n \ n郊区\ JJ花园\ NN

输出结果为：

else  became
else  became
else  a regular customer of a suburban garden
else  a regular customer
else  a
else  a
else  regular
else  regular
customer \\This is the string to be extracted
else  of a suburban garden
else  of
else  of
else  a suburban garden
else  a
else  a
else  suburban
else  suburban
garden
garden

Answer 1

问题在于您在每个非名词上递归调用Find_Noun。所以是的，它突破了你正在寻找的一次迭代的循环......但是它只是回到了堆栈的前一级。 / p>

我不清楚为什么你会再次递归，但如果你真的做需要递归，你需要一些方法来检测是否真的发现了递归调用是否是名词，如果是，则立即返回。所以可能是这样的：

// Renamed method to follow Java naming conventions
public static String findNoun(Parse p)
{    
    for(Parse k : p.getChildren())
    {   
        // Removed a bunch of extraneous brackets, and added whitespace
        // for readability. You should consider a set NOUN_TYPES
        // so you could use if (NOUN_TYPES.contains(k.getType()) instead.
        if (k.getType().equals("NN") || k.getType().equals("NNP") ||
            k.getType().equals("NNS")|| k.getType().equals("NNPS"))
        {
            return k.toString();
        }
        else
        {
            String possibleNoun = findNoun(k);
            // Return immediately if the recursion found a noun
            if (possibleNoun != null)
            {
                return possibleNoun;
            }
        }
    }
    // Nothing found in this node or children: tell the caller
    return null;
}

break命令不会停止循环

1 个答案: