获取TreeMap中的三个最高值 [英] get the three highest values in a TreeMap

问题描述

我试图在TreeMap中找到三个最高值。我编写了一个代码，但是我想问一下你是否可以建议一个更有效的方法。
基本上，我在TreeMap中保存文本的每个单词以及它在文本中出现的次数。然后我使用比较器对值进行排序。然后我迭代新创建的Map，直到我到达最后三个值，这是排序后的最高值并打印出来。我将使用大文本，所以这不是一个很好的方法。
这是我的代码：

I am trying to find the three highest values in a TreeMap. I wrote a code that is kind of doing it, but I would like to ask whether you can suggest a more efficient way. Basically, I am saving each word of my text in a TreeMap along with the number of times it appears in the text. Then I am using a comparator to sort the values. Then I am iterating through the newly created Map until I reach the last three values, which are the highest values after the sorting and print them out. I am going to use large texts, so this is not a very good way. Here is my code:

class Text{
    public static void main(String args[]) throws FileNotFoundException, IOException{
        final File textFile = new File("C://FileIO//cinderella.txt"); 
        final BufferedReader in = new BufferedReader(new FileReader(textFile));                               
        final TreeMap<String, Integer> frequencyMap = new TreeMap<String, Integer>(); 

        String currentLine; 
        while ((currentLine = in.readLine()) != null) {  
            currentLine = currentLine.toLowerCase();  
            final StringTokenizer parser = new StringTokenizer(currentLine, " \t\n\r\f.,;:!?'"); 
            while (parser.hasMoreTokens()) { 
                final String currentWord = parser.nextToken(); 
                Integer frequency = frequencyMap.get(currentWord); 
                if (frequency == null) { 
                    frequency = 0; 
                } 
                frequencyMap.put(currentWord, frequency + 1);
            } 
        }  

        System.out.println("This the unsorted Map: "+frequencyMap);

        Map sortedMap = sortByComparator(frequencyMap);
        int i = 0;
        int max=sortedMap.size();
        StringBuilder query= new StringBuilder();

        for (Iterator it = sortedMap.entrySet().iterator(); it.hasNext();) {
            Map.Entry<String,Integer> entry = (Map.Entry<String,Integer>) it.next();
            i++;
            if(i<=max && i>=(max-2)){
                String key = entry.getKey();
                //System.out.println(key);
                query.append(key);
                query.append("+");
            }
        }
        System.out.println(query);
    }

    private static Map sortByComparator(TreeMap unsortMap) {
        List list = new LinkedList(unsortMap.entrySet());

        //sort list based on comparator
        Collections.sort(list, new Comparator() {
            public int compare(Object o1, Object o2) {
                return ((Comparable) ((Map.Entry) (o1)).getValue())
                       .compareTo(((Map.Entry) (o2)).getValue());
            }
        });

        //put sorted list into map again
        Map sortedMap = new LinkedHashMap();
        for (Iterator it = list.iterator(); it.hasNext();) {
            Map.Entry entry = (Map.Entry)it.next();
            sortedMap.put(entry.getKey(), entry.getValue());

        }
        return  sortedMap;
    }   
}

推荐答案

我将使用哈希映射计算频率，然后将它们全部循环，选择前3个。您可以通过这种方式最小化比较，而不必进行排序。使用选择算法

I would count the frequencies with a hash map, and then loop over them all, selecting the top 3. You minimize comparisons this way, and never have to sort. Use the Selection Algorithm

-edit ，维基百科页面详述了选择算法的许多不同实现。具体来说，只需使用有界优先级队列，并将大小设置为3.不要花哨并将队列实现为堆或任何东西。只需使用数组。

-edit, the wikipedia page details many different implementations of the selection algorithm. To be specific, just use a bounded priority queue, and set the size to 3. Dont get fancy and implement the queue as a heap or anything. just use an array.

这篇关于获取TreeMap中的三个最高值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！