我使用 Java Comparator 按词频属性的降序对 Word 对象的 ArrayList 进行排序。创建 Word 对象时,首先使用 HashMap 从 .txt 文件中读取单词,然后将 HashMap 转换为 Word 对象的 ArrayList。然后我想按字母顺序对频率相同的单词进行排序。
while (reader.hasNext()) {
word = reader.next().toLowerCase();
word = word.substring(0, 1).toUpperCase() + word.substring(1);
word = word.replaceAll("[^a-zA-Z ]", "");
if (!word.contains("0") || !word.contains("1") || !word.contains("2") || !word.contains("3") || !word.contains("4") || !word.contains("5") || !word.contains("6") || !word.contains("7") || !word.contains("8") || !word.contains("9") || !word.contains("-") || !word.contains("_")) {
// This is equivalent to searching every word in the list via hashing (O(1))
if(!frequencyMap.containsKey(word)) {
frequencyMap.put(word, 1);
} else {
// We have already seen the word, increase frequency.
frequencyMap.put(word, frequencyMap.get(word) + 1);
}
}
counter++;
}
for(Map.Entry<String, Integer> entry : frequencyMap.entrySet()) {
Word word = new Word(entry.getKey());
word.frequency = entry.getValue();
wordList.add(word);
}
Collections.sort(wordList, Word.WordFrequency);
public class Word {
String value;
int frequency;
public Word(String v) {
value = v;
frequency = 1;
}
public String getValue() {
return value;
}
public int getFrequency() {
return frequency;
}
public static Comparator<Word> WordFrequency = new Comparator<Word>() {
public int compare(Word w1, Word w2) {
int w1Frequency = w1.getFrequency();
int w2Frequency = w2.getFrequency();
return w2Frequency-w1Frequency;
}
};
}
最佳答案
请参阅 thenComparing
方法,该方法允许您在存在平局时提供比较键:
// sort using 'd' will sort 1st alphabetically, then by length
// (this is a totally arbitrary example)
Comparator<String> c = String::compareTo;
Comparator<String> d = c.thenComparing(s -> s.length());
关于java - 如何使用 Java Comparator 使用两个参数对 ArrayList 进行排序?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53487307/