Java：通过HashMap获得文本中的500个最常用的词
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我将单词计数存储到HashMap的value字段中，然后如何获取文本中的500个热门单词？
public ArrayList<String> topWords (int numberOfWordsToFind, ArrayList<String> theText) {
//ArrayList<String> frequentWords = new ArrayList<String>();
ArrayList<String> topWordsArray= new ArrayList<String>();
HashMap<String,Integer> frequentWords = new HashMap<String,Integer>();
int wordCounter=0;
for (int i=0; i<theText.size();i++){
if(frequentWords.containsKey(theText.get(i))){
//find value and increment
wordCounter=frequentWords.get(theText.get(i));
wordCounter++;
frequentWords.put(theText.get(i),wordCounter);
}
else {
//new word
frequentWords.put(theText.get(i),1);
}
}
for (int i=0; i<theText.size();i++){
if (frequentWords.containsKey(theText.get(i))){
// what to write here?
frequentWords.get(theText.get(i));
}
}
return topWordsArray;
}
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3 个回复
际恃啸称桅
。但是，更简单的是，您可以放弃循环：现在您有了一个Multiset，该怎么办？好吧，您可以调用，它为您提供了对象的集合。然后，您可以将它们放在中（它们放在中），然后使用对其进行分类。完整的代码可能看起来像（使用其他一些流行的Guava功能进行展示）：Multiset<String> words = HashMultiset.create(theWords); List<Multiset.Entry<String>> wordCounts = Lists.newArrayList(words.entrySet()); Collections.sort(wordCounts, new Comparator<Multiset.Entry<String>>() { public int compare(Multiset.Entry<String> left, Multiset.Entry<String> right) { // Note reversal of \'right\' and \'left\' to get descending order return right.getCount().compareTo(left.getCount()); } }); // wordCounts now contains all the words, sorted by count descending // Take the first 50 entries (alternative: use a loop; this is simple because // it copes easily with < 50 elements) Iterable<Multiset.Entry<String>> first50 = Iterables.limit(wordCounts, 50); // Guava-ey alternative: use a Function and Iterables.transform, but in this case // the \'manual\' way is probably simpler: for (Multiset.Entry<String> entry : first50) { wordArray.add(entry.getElement()); }荆怖赡
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