生成并输出所有可能的组合，每个唯一项只从三个List<String>生成一次

Question

我试着生成所有可能的句子组合。作为变量，我有两个字符串，一个字符串将作为主题，例如health和一个对象，比如fruit，但是我将有一个与一个"head“单词相关联的值的List<String>，因此为了与刚才提到的两个组件保持一致，它们将与list [improve, change, alter, modify]相关联。我希望生成这些句子的所有可能组合，并将每个句子添加到List<Sentences>中，如下所示：

Sentence example_sentence = new Sentence(verb, object, subject);
sentences.add(example_sentence);

现在，发生这种情况的更大的功能如下：

public Sentence dataPreprocessing(String raw_subject, String raw_object, String raw_verb, List<Sentence> sentences) throws IOException {
    WordNet wordnet = new WordNet();
    String verb = wordnet.getStem(raw_verb);
    String object = wordnet.getStem(raw_object);        
    String subject = wordnet.getStem(raw_subject);
    List<String> verb_hypernym_container = new ArrayList<>();       
    verb_hypernym_container = wordnet.getHypernyms(verb, POS.VERB);
    //wordnet.getHypernyms(this.object, POS.NOUN);  
    //wordnet.getHypernyms(this.subject, POS.NOUN); 
    Sentence return_sentence = new Sentence( verb, object, subject );
    return return_sentence;
}

我如何才能最有效地实现生成所有可能的句子的目标？

Answer 1

由于有固定数量的列表，最简单的方法是只使用嵌套循环：

List<Sentence> sentences = new ArrayList<>();

for(String verb_hypernym : wordnet.getHypernyms(verb, POS.VERB))
    for(String object_hypernym : wordnet.getHypernyms(object, POS.NOUN))
        for(String subject_hypernym : wordnet.getHypernyms(subject, POS.NOUN))
            sentences.add(new Sentence(verb_hypernym, object_hypernym, subject_hypernym));

return sentences;

或者，为了避免不必要的频繁调用getHypernyms：

List<String> verb_hypernyms = wordnet.getHypernyms(verb, POS.VERB);
List<String> object_hypernyms = wordnet.getHypernyms(object, POS.NOUN);
List<String> subject_hypernyms = wordnet.getHypernyms(subject, POS.NOUN);


for(String verb_hypernym : verb_hypernyms)
    for(String object_hypernym : object_hypernyms)
        for(String subject_hypernym : subject_hypernyms)
            sentences.add(new Sentence(verb_hypernym, object_hypernym, subject_hypernym));

return sentences;

Answer 2

一旦你有了一个名词和动词的列表，你可以使用流返回一个句子列表。这也为您提供了一个机会，以删除任何重复，排序或任何您需要做的流。

List<Sentence> sentences = subjectList.stream()
        .flatMap(object -> verbList.stream()
            .flatMap(verb -> objectList.stream()
                .map(subject -> new Sentence(object, verb, subject))))
        .distinct()
        .collect(Collectors.toList());