Question

我正在开发一个项目来编写一个程序，找到文本中最常用的10个单词，但是我遇到了困难，不知道接下来应该做什么。有人能帮帮我吗？

我只走到这一步：

import java.io.File;
import java.io.FileNotFoundException;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;
import java.util.regex.Pattern;

public class Lab4 {
    public static void main(String[] args) throws FileNotFoundException {
        Scanner file = new Scanner(new File("text.txt")).useDelimiter("[^a-zA-Z]+");
        List<String> words = new ArrayList<String>();
        while (file.hasNext()){
            String tx = file.next();
            // String x = file.next().toLowerCase();
            words.add(tx);
        }
        Collections.sort(words);
        // System.out.println(words);
    }
}

Answer 1

您可以使用Guava Multiset，这是一个单词计数示例：http://code.google.com/p/guava-libraries/wiki/NewCollectionTypesExplained

以下是如何在Multiset中找到计数最高的单词：Simplest way to iterate through a Multiset in the order of element frequency?

更新我在2012年写了这个答案。从那以后我们有了Java 8，现在可以在没有外部库的情况下找到几行中最常用的10个单词：

List<String> words = ...

// map the words to their count
Map<String, Integer> frequencyMap = words.stream()
         .collect(toMap(
                s -> s, // key is the word
                s -> 1, // value is 1
                Integer::sum)); // merge function counts the identical words

// find the top 10
List<String> top10 = words.stream()
        .sorted(comparing(frequencyMap::get).reversed()) // sort by descending frequency
        .distinct() // take only unique values
        .limit(10)   // take only the first 10
        .collect(toList()); // put it in a returned list

System.out.println("top10 = " + top10);

静态导入是：

import static java.util.Comparator.comparing;
import static java.util.stream.Collectors.toList;
import static java.util.stream.Collectors.toMap;

Answer 2

创建一个地图以跟踪这样的事件：

   Scanner file = new Scanner(new File("text.txt")).useDelimiter("[^a-zA-Z]+");
   HashMap<String, Integer> map = new HashMap<>();

   while (file.hasNext()){
        String word = file.next().toLowerCase();
        if (map.containsKey(word)) {
            map.put(word, map.get(word) + 1);
        } else {
            map.put(word, 0);
        }
    }

    ArrayList<Map.Entry<String, Integer>> entries = new ArrayList<>(map.entrySet());
    Collections.sort(entries, new Comparator<Map.Entry<String, Integer>>() {

        @Override
        public int compare(Map.Entry<String, Integer> a, Map.Entry<String, Integer> b) {
            return a.getValue().compareTo(b.getValue());
        }
    });

    for(int i = 0; i < 10; i++){
        System.out.println(entries.get(entries.size() - i - 1).getKey());
    }

Answer 3

package src;

import java.io.File;
import java.io.FileNotFoundException;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
import java.util.Map.Entry;

public class ScannerTest
{
    public static void main(String[] args) throws FileNotFoundException
        {
        Scanner scanner = new Scanner(new File("G:/Script_nt.txt")).useDelimiter("[^a-zA-Z]+");
        Map<String, Integer> map = new HashMap<String, Integer>();
        while (scanner.hasNext())
            {
            String word = scanner.next();
            if (map.containsKey(word))
                {
                map.put(word, map.get(word)+1);
                }
            else
                {
                map.put(word, 1);
                }
            }

        List<Map.Entry<String, Integer>> entries = new ArrayList<Entry<String,Integer>>( map.entrySet());

        Collections.sort(entries, new Comparator<Map.Entry<String, Integer>>() {

            @Override
            public int compare(Map.Entry<String, Integer> a, Map.Entry<String, Integer> b) {
                return a.getValue().compareTo(b.getValue());
            }
        });

        for(int i = 0; i < map.size(); i++){
            System.out.println(entries.get(entries.size() - i - 1).getKey()+" "+entries.get(entries.size() - i - 1).getValue());
        }
        }
}

Answer 4

这是一个比lbalazscs更短的版本，它也使用Java 8的流API;

Arrays.stream(new String(Files.readAllBytes(PATH_TO_FILE), StandardCharsets.UTF_8).split("\\W+"))
            .collect(Collectors.groupingBy(Function.<String>identity(), HashMap::new, counting()))
            .entrySet()
            .stream()
            .sorted(((o1, o2) -> o2.getValue().compareTo(o1.getValue())))
            .limit(10)
            .forEach(System.out::println);

这将一气呵成：加载文件，按非单词字符拆分，按字词对所有内容进行分组，并为每个组分配单词计数，然后为前十个单词打印带有计数的单词。

有关非常类似设置的深入讨论，请参阅：https://stackoverflow.com/a/33946927/327301

Answer 5

在输入中创建一个来自文件或命令行的字符串，并将其传递给下面的方法，它将返回一个包含单词作为键的映射，并将值作为它们在该句子或段落中的出现或计数。

public Map<String,Integer> getWordsWithCount(String sentances)
{
    Map<String,Integer> wordsWithCount = new HashMap<String, Integer>();

    String[] words = sentances.split(" ");
    for (String word : words)
    {
        if(wordsWithCount.containsKey(word))
        {
            wordsWithCount.put(word, wordsWithCount.get(word)+1);
        }
        else
        {
            wordsWithCount.put(word, 1);
        }

    }

    return wordsWithCount;

}

如何计算文本中出现的单词数

5 个答案: