Question

我正在尝试计算文本文件中单词的频率。但我必须采用不同的方法。例如，如果文件包含BRAIN-ISCHEMIA和ISCHEMIA-BRAIN，我需要计算两次BRAIN-ISCHEMIA（并离开ISCHEMIA-BRAIN），反之亦然。这是我的代码 -

// Mapping of String->Integer (word -> frequency) 
    HashMap<String, Integer> frequencyMap = new HashMap<String, Integer>(); 

    // Iterate through each line of the file 
    String[] temp;
    String currentLine; 
    String currentLine2;
    while ((currentLine = in.readLine()) != null) { 

    // Remove this line if you want words to be case sensitive 
    currentLine = currentLine.toLowerCase();
    temp=currentLine.split("-");
    currentLine2=temp[1]+"-"+temp[0];

    // Iterate through each word of the current line 
    // Delimit words based on whitespace, punctuation, and quotes 
    StringTokenizer parser = new StringTokenizer(currentLine);
    while (parser.hasMoreTokens()) { 
    String currentWord = parser.nextToken(); 

    Integer frequency = frequencyMap.get(currentWord); 

    // Add the word if it doesn't already exist, otherwise increment the 
    // frequency counter. 
    if (frequency == null) { 
    frequency = 0; 
    } 
    frequencyMap.put(currentWord, frequency + 1); 
    }
    StringTokenizer parser2 = new StringTokenizer(currentLine2);
    while (parser2.hasMoreTokens()) { 
        String currentWord2 = parser2.nextToken(); 

        Integer frequency = frequencyMap.get(currentWord2); 

        // Add the word if it doesn't already exist, otherwise increment the 
        // frequency counter. 
        if (frequency == null) { 
        frequency = 0; 
        } 
        frequencyMap.put(currentWord2, frequency + 1); 

        } 
    }

    // Display our nice little Map 
    System.out.println(frequencyMap);

但对于以下文件 -

缺血谷氨酸缺血性脑谷氨酸脑脑容忍脑容忍容忍脑谷氨酸缺血缺血谷氨酸

我得到以下输出 -

{谷氨酸脑= 1，缺血 - 谷氨酸= 3，缺血 - 脑= 1，谷氨酸缺血= 3，脑耐受= 3，脑缺血= 1，耐受脑= 3，脑 - 谷氨酸= 1}

我认为问题在于第二块。对此问题的任何启示都将受到高度赞赏。

Answer 1

从算法的角度来看，您可能需要考虑以下方法：

对于每个字符串，拆分，然后排序，然后重新组合（即采用DEF-ABC并转换为ABC-DEF.ABC-DEF将转换为ABC-DEF）。然后将其用作频率计数的关键。

如果您需要保留确切的原始项目，只需将其包含在您的密钥中 - 这样密钥就会：ordinal（重新组合的字符串）和原始项目。

Answer 2

免责声明：我偷了 Kevin Day 建议的甜蜜技巧来实施。

我仍想发帖只是为了让您知道使用正确的数据结构（Multiset/Bad）和正确的库（google-guava）不仅简化代码但也使它高效。

代码

public class BasicFrequencyCalculator
{
    public static void main(final String[] args) throws IOException
    {
        @SuppressWarnings("unchecked")
        Multiset<Word> frequency = Files.readLines(new File("c:/2.txt"), Charsets.ISO_8859_1, new LineProcessor() {

            private final Multiset<Word> result = HashMultiset.create();

            @Override
            public Object getResult()
            {
                return result;
            }

            @Override
            public boolean processLine(final String line) throws IOException
            {
                result.add(new Word(line));
                return true;
            }
        });
        for (Word w : frequency.elementSet())
        {
            System.out.println(w.getOriginal() + " = " + frequency.count(w));
        }
    }
}


public class Word
{
    private final String key;

    private final String original;

    public Word(final String orig)
    {
        this.original = orig.trim();
        String[] temp = original.toLowerCase().split("-");
        Arrays.sort(temp);
        key = temp[0] + "-"+temp[1];
    }

    @Override
    public int hashCode()
    {
        final int prime = 31;
        int result = 1;
        result = prime * result + ((getKey() == null) ? 0 : getKey().hashCode());
        return result;
    }

    @Override
    public boolean equals(final Object obj)
    {
        if (this == obj)
        {
            return true;
        }
        if (obj == null)
        {
            return false;
        }
        if (!(obj instanceof Word))
        {
            return false;
        }
        Word other = (Word) obj;
        if (getKey() == null)
        {
            if (other.getKey() != null)
            {
                return false;
            }
        }
        else if (!getKey().equals(other.getKey()))
        {
            return false;
        }
        return true;
    }

    @Override
    public String toString()
    {
        return getOriginal();
    }

    public String getKey()
    {
        return key;
    }

    public String getOriginal()
    {
        return original;
    }
}

输出

BRAIN-TOLERATE     = 3
ISCHEMIA-GLUTAMATE = 3
GLUTAMATE-BRAIN    = 1
ISCHEMIA-BRAIN     = 1

Answer 3

感谢大家的帮助。这就是我解决它的方法 -

// Mapping of String->Integer (word -> frequency) 
    TreeMap<String, Integer> frequencyMap = new TreeMap<String, Integer>(); 

    // Iterate through each line of the file 
    String[] temp;
    String currentLine; 
    String currentLine2;
    while ((currentLine = in.readLine()) != null) { 

    temp=currentLine.split("-");
    currentLine2=temp[1]+"-"+temp[0];

    // Iterate through each word of the current line  
    StringTokenizer parser = new StringTokenizer(currentLine);
    while (parser.hasMoreTokens()) { 
    String currentWord = parser.nextToken(); 

    Integer frequency = frequencyMap.get(currentWord);  
    Integer frequency2 = frequencyMap.get(currentLine2);

    // Add the word if it doesn't already exist, otherwise increment the 
    // frequency counter. 
    if (frequency == null) {
        if (frequency2 == null)
            frequency = 0;
        else {
            frequencyMap.put(currentLine2, frequency2 + 1);
            break;
        }//else
    } //if (frequency == null)

    frequencyMap.put(currentWord, frequency + 1);
    }//while (parser.hasMoreTokens())

    }//while ((currentLine = in.readLine()) != null)

    // Display our nice little Map 
    System.out.println(frequencyMap);

树图问题

3 个答案:

代码

输出