java - 关于java数据结构的查询

Question

我想计算多个文件的词频。

此外，我在这些文件中有这些词

a1.txt = {aaa, aaa, aaa} 
a2.txt = {aaa} 
a3.txt = {aaa, bbb} 

所以，结果一定是aaa = 3，bbb = 1。

然后，我定义了上述数据结构，

LinkedHashMap<String, Integer> wordCount = new LinkedHashMap<String, Integer>();
Map<String, LinkedHashMap<String, Integer>>
fileToWordCount = new HashMap<String,LinkedHashMap<String, Integer>>();

然后，我从文件中读取单词并将它们放入 wordCount 和 fileToWordCount 中:

/*lineWords[i] is a word from a line in the file*/
if(wordCount.containsKey(lineWords[i])){
   System.out.println("1111111::"+lineWords[i]);
   wordCount.put(lineWords[i], wordCount.
   get(lineWords[i]).intValue()+1);
   }else{
   System.out.println("222222::"+lineWords[i]);
   wordCount.put(lineWords[i], 1);
}
fileToWordCount.put(filename, wordCount); //here we map filename
and occurences        of       words

最后，我使用上面的代码打印 fileToWordCount，

Collection a;
Set filenameset;

        filenameset = fileToWordCount.keySet();    
        a = fileToWordCount.values();          
        for(Object filenameFromMap: filenameset){
                   System.out.println("FILENAMEFROMAP::"+filenameFromMap);                                 
                System.out.println("VALUES::"+a);                                                
        }

和打印，

FILENAMEFROMAP::a3.txt
VALUES::[{aaa=5, bbb=1}, {aaa=5, bbb=1}, {aaa=5, bbb=1}]
FILENAMEFROMAP::a1.txt
VALUES::[{aaa=5, bbb=1}, {aaa=5, bbb=1}, {aaa=5, bbb=1}]
FILENAMEFROMAP::a2.txt
VALUES::[{aaa=5, bbb=1}, {aaa=5, bbb=1}, {aaa=5, bbb=1}]

那么，我如何使用映射 fileToWordCount 来查找文件中的词频？

Answer 1

你让事情变得比必要的更加困难。我会这样做:

Map<String, Counter> wordCounts = new HashMap<String, Counter>();
for (File file : files) {
    Set<String> wordsInFile = new HashSet<String>(); // to avoid counting the same word in the same file twice
    for (String word : readWordsFromFile(file)) {
        if (!wordsInFile.contains(word)) {
            wordsInFile.add(word);
            Counter counter = wordCounts.get(word);
            if (counter == null) {
                counter = new Counter();
                wordCounts.put(word, counter);
            }
            counter.increment();
        }
    }
}