在该限制下，如何使用最短代码的java实现该词频统计

Question

#!/usr/bin/env python
import re, sys, collections

stops = open(’../stop_words.txt’).read().split(’,’)

words = re.findall(’[a-z]{2,}’, open(sys.argv[1]).read().lower())

counts = collections.Counter(w for w in words if w not in stops)

for (w, c) in counts.most_common(25):

print w, ’-’, c

上述示例代码为python版本的词频统计。功能如下：
导入stopwords和源文件，之后进行统计并输出。

问题：使用尽可能少的java代码来实现相同的功能，尽可能的调用函数

Answer 1

文件输入的功能你自己实现下，别的完全改写了。

import java.util.HashMap;
import java.util.regex.*;
import java.util.Map;
import java.util.*;

class HelloWorld {
    public static void main(String []args) {
        String s = "someone says: hello world! hello, this website hello the people this is the java language.";
        Pattern pattern = Pattern.compile("[a-z]{2,}");
        Matcher matcher = pattern.matcher(s);
        Map<String, Integer> map = new HashMap<>();
        while (matcher.find()) {
            String word = matcher.group(0).toLowerCase();
            if(!map.containsKey(word))
            {
                map.put(word, 1);
            }
            else
            {
                int times = map.get(word) + 1;
                map.remove(word);
                map.put(word, times);
            }
        }

        //System.out.println(map);
        List<Map.Entry<String, Integer>> list = new ArrayList<Map.Entry<String, Integer>>(map.entrySet());
        list.sort(new Comparator<Map.Entry<String, Integer>>() {
          @Override
          public int compare(Map.Entry<String, Integer> o1, Map.Entry<String, Integer> o2) {
              return o2.getValue().compareTo(o1.getValue());
          }
      });
        for (Map.Entry<String, Integer> mapping : list){
            System.out.println(mapping.getKey()+" - "+mapping.getValue());
        }    
    }
}

hello - 3
the - 2
this - 2
website - 1
world - 1
java - 1
someone - 1
is - 1
language - 1
says - 1
people - 1