Word频率计数器问题与逻辑Java

Question

我正在建立一个基本的单词频率计数器。 下面列出了代码：

public static List<Frequency> computeWordFrequencies(List<String> words) 
{
    List<Frequency> list_of_frequency = new ArrayList<Frequency>();
    List<String> list_of_words = words;
    int j = 0;
    for(int i=0; i<list_of_words.size(); i++)
    {

        String current_word = list_of_words.get(i);
        boolean added = false;
        if(list_of_frequency.size() == 0)
        {
            list_of_frequency.add(new Frequency(current_word, 1));
            System.out.println("added " + current_word);
        }
        else
        {

            System.out.println("Current word: " + current_word);
            System.out.println("Current Frequency: " + list_of_frequency.get(j).getText());
            if(list_of_frequency.contains(current_word))
            {
                list_of_frequency.get(j).incrementFrequency();
                System.out.println("found... incremented " + list_of_frequency.get(j).getText() + " frequency");
                added = true;
            }
            else
            {
                list_of_frequency.add(new Frequency(current_word, 1));
                System.out.println("added " + current_word);
                added = true;
            }
        }
    }
}

我得到的输出是：

added I
Current word: am
Current Frequency: I
added am
Current word: very
Current Frequency: I
added very
Current word: good
Current Frequency: I
added good
Current word: at
Current Frequency: I
added at
Current word: being
Current Frequency: I
added being
Current word: good
Current Frequency: I
added good
Total item count: 7
Unique item count: 7
I:1
am:1
very:1
good:1
at:1
being:1
good:1

因此，我需要一个for循环来遍历“ list_of_frequency”，但是如果这样做，则会遇到其他问题，例如重复添加单词。 我的逻辑在这里吗，这个项目会有更好的方法吗？ 提前致谢！

Answer 1

您可以使用Collections类的频率方法执行此操作

这是一个示例：

public void wordFreq(){
String text = "hello bye hello a bb a bye hello";

        List<String> list = Arrays.asList(text.split(" "));

        Set<String> uniqueWords = new HashSet<String> (list);
        for (String word : uniqueWords) {
            System.out.println(word + ": " + Collections.frequency(list, word));
        }
}

Answer 2

您使事情变得过于复杂。

您只需要几行：

public static Map<String, Integer> getFrequencies(List<String> words) {
    Map<String, Integer> freq = new HashMap<String, Integer>();
    for (String word : words) {
        Integer i = freq.get(word);
        freq.put(word, i == null ? 1 : i + 1);
    }
    return freq;
}

Answer 3

将此代码添加到其他部分中。 你应该做的是

1. 循环检查单词是否已在列表中
2. 如果第1部分为真，则增加其频率

3. 否则将其放在频率为1的列表中

    for(j = 0; j < list_of_frequency.size; j++) if(list_of_frequency.get(i).getText().equals(current_word)) list_of_frequency.get(i).frequency++; // increment frequency //if word is already encountered before 

Answer 4

我认为要运行得更快，您应该使用从列表排序开始的另一种算法：

  1) sort your list of string (cf. java.util.Collections.sort())
  2) in pseudo code :
 iterate your sorted list
 current_word = word of current iteration
 if it's a new word (! current_word.equals( oldWord) )
 counter = 1
 if (current_word.equals( oldWord)) {
    counter++
     store current_word in variable oldWord 
 }
 when the word change create your Frequency(oldWord, counter) and add to the list of frequencies

因此，您不需要每次都检查频率列表，而一次插入一个单词就可以了。

由于list_of_frequency的所有条目都是唯一的单词，因此您也可以使用Set代替list_of_frequency的列表。

Answer 5

以此替换您的方法。 通过在分析数据时使用地图，您将获得更好的性能。

public static List<Frequency> computeWordFrequencies(List<String> words) {
    Map<String, Integer> counts = new HashMap<String, Integer>();
    for(String word : words) {
        Integer current = counts.get(word);
        if(current != null) {
            counts.put(word, current+1);
        }
        else counts.put(word, 1);
    }

    // Then, if you really need that list of Frequency
    List<Frequency> list_of_frequency = new ArrayList<Frequency>();

    for(String s : counts.keySet()) {
        list_of_frequency.add(new Frequency(s, counts.get(s)));
    }

    return list_of_frequency;
}

Answer 6

我将这样进行：

List<String> words = Arrays.asList("foo", "bar", "qux", "foo");

Map<String, AtomicInteger> frequencyMap = new HashMap<String, AtomicInteger>();
for (String word : words)
{
    AtomicInteger freq = frequencyMap.get(word);
    if (freq == null) {
        frequencyMap.put(word, new AtomicInteger(1));
    }
    else
    {
        freq.incrementAndGet();
    }
}

for (String word : frequencyMap.keySet())
{
    System.out.println(word + " :" + frequencyMap.get(word));
}

通过使用AtomicInteger，您可以轻松地增加频率计数器。