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我正在尝试从 TreeMap 中的单个键检索多个值。这个想法是每个键都将链接到多个值并且应该是可搜索的。现在我遇到的麻烦是当我只能从 key 中取回一个值时。
键是一个字符串,值是一个自定义对象,称为 Song。歌曲包含多种元素。目标是从每首歌中逐字提取歌词,并将每个单词用作关键字。该键然后链接到包含该键的每首歌曲(值)。
我在 StackOverFlow 和网络上搜索了一般的技巧,我也看到了一些,但没有任何东西可以直接解决我的绊脚石。我看到的一个想法是将值更改为某种数组或列表。明天当我的大脑恢复活力时,我可能会尝试这样做。
无论如何,提前感谢您提供的任何提示和建议。是的,这是作业。不,我没有标记是因为我被告知作业标记不再常用。
代码:
public class SearchByLyricsWords {
private static Song[] songs;
private static TreeMap<String, Song> lyricsTreeMap = new TreeMap<String, Song>();
private static TreeSet<String> wordsToIgnoreTree = new TreeSet<String>();
private static File wordsToIgnoreInput = new File("ignore.txt");
private static String wordsToIgnoreString;
private static String[] wordsToIgnoreArray;
private Song[] searchResults; // holds the results of the search
private ArrayList<Song> searchList = new ArrayList<Song>();
public SearchByLyricsWords(SongCollection sc) throws FileNotFoundException {
// Create a string out of the ignore.txt file
Scanner scanInputFile = new Scanner(wordsToIgnoreInput);
String ignoreToken = scanInputFile.next();
ignoreToken.toLowerCase();
wordsToIgnoreString = ignoreToken + " ";
while (scanInputFile.hasNext()) {
ignoreToken = scanInputFile.next();
wordsToIgnoreString = wordsToIgnoreString + ignoreToken + " ";
}
// Split the string created from ignore.txt
wordsToIgnoreArray = wordsToIgnoreString.split("[^a-zA-Z]+");
// Fill a TreeSet from the wordsToIgnoreArray
for (int i = 0; i < wordsToIgnoreArray.length; i++) {
ignoreToken = wordsToIgnoreArray[i];
wordsToIgnoreTree.add(ignoreToken);
}
// Fill TreeMap with lyrics words as the key, Song objects as the value
songs = sc.getAllSongs();
for (int j = 0; j < songs.length; j++) {
Song currentSong = songs[j];
String lyrics = currentSong.getLyrics();
TreeSet<String> lyricsFound = new TreeSet<String>();
String lyricsToken;
String[] songLyricsArray;
songLyricsArray = lyrics.split("[^a-zA-Z]+");
for (int k = 0; k < songLyricsArray.length; k++) {
lyricsToken = songLyricsArray[k];
if (lyricsToken.length() <= 1) {
continue;
}
lyricsFound.add(lyricsToken);
}
lyricsFound.removeAll(wordsToIgnoreTree);
Iterator<String> iterator = lyricsFound.iterator();
while(iterator.hasNext()) {
String currentWord = (String)iterator.next();
lyricsTreeMap.put(currentWord, currentSong);
}
//System.out.println(lyricsTreeMap); // testing only
}
}
public Song[] search(String lyricsWords) {
lyricsWords = lyricsWords.toLowerCase();
TreeSet<String> searchTree = new TreeSet<String>();
String searchToken;
String[] lyricsWordsSearch = lyricsWords.split("[^a-zA-Z]+");
for (int l = 0; l < lyricsWordsSearch.length; l++) {
searchToken = lyricsWordsSearch[l];
if (searchToken.length() <= 1) {
continue;
}
searchTree.add(searchToken);
}
searchTree.removeAll(wordsToIgnoreTree);
Iterator<String> searchIterator = searchTree.iterator();
while(searchIterator.hasNext()) {
String currentSearchWord = (String)searchIterator.next();
Collection<Song> lyricsTreeCollection = lyricsTreeMap.values();
while (lyricsTreeMap.containsKey(currentSearchWord) == true) {
Iterator collectionIterator = lyricsTreeCollection.iterator();
while(collectionIterator.hasNext() && collectionIterator.next() == currentSearchWord) {
Song searchSong = lyricsTreeMap.get(currentSearchWord);
searchList.add(searchSong);
}
}
}
searchResults = searchList.toArray(new Song[searchList.size()]);
Arrays.sort(searchResults);
return searchResults;
}
最佳答案
TreeMap
每个键只保留一个值,as with all Map
implementations :
A map cannot contain duplicate keys; each key can map to at most one value.
您的备选方案包括
TreeMap<String, List<Song>>
相反,手动处理 List
值(value)观并保持更新TreeMultimap
来自 Guava,它(或多或少)像 TreeMap<K, TreeSet<V>>
一样运行,除了好多了。 (披露:我为 Guava 做出了贡献。)关于Java TreeMap : Retrieving multiple values from a single key,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13355721/
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